help-0

[viscid cove]

yeah that's a typo. Do you happen to know how that turns into $$(1+1)^{n-1}$$ ?

[ocean seal]

drunkzurg

[fathom violet]

binomial theorem

[viscid cove]

i'm learning it rn, could you elaborate?

[fathom violet]

well you know (a+b)^2 = a^2+2ab+b^2

[viscid cove]

uh huh

[fathom violet]

thats the same as (a+b)*(a+b)

a binomial expansion or the the binomial theorem

but you can put any number as the exponents

[viscid cove]

there are two 1's, i assume its some row in pascal's triangle

oh, one sec

[fathom violet]

ye

[viscid cove]

wait, where did we get the number of terms in the bracket as n though? i'm lost

[fathom violet]

okay first you just factor out n out of all the terms of the sum (in front of the brackets)

then you are left with the sum 1 + 2(n-1)/2! + 3(n-1)/3!+ ...

then you can reduce the factorial

[viscid cove]

so in the end term(n+1th) , there would be a n! at the top and maybe it was canceled by n! at the bottom?

[fathom violet]

ye

nah you right

[viscid cove]

wait, i saw that

why is it not (n-1)! ?

its starting from n(n-1)..... nevermind

well that's all, thanks for the help!

[fathom violet]

aite

[vague jolt]

If two different functions are bijective and have the same domain do their codomains have the same cardinality?

[alpine sable]

Help

[sleek elbow]

?

[wind bane]

yes. the codomain has the same cardinality as the domain in this case

[native temple]

2+2 =?

3+5=?

[alpine sable]

???

[sleek elbow]

you are very wrong on discord if you cant add 2+2 imo

[alpine sable]

LOL

They prob joking

U

H

[glass lichen]

yeah it's a troll

[ionic jewel]

if you can't do basic arithmetic you need to get an addition table and memorize it

or use your fingers

[sleek elbow]

if you can't do basic arithmetic you need to get an addition table and memorize it
or use your fingers

^^

[alpine sable]

Help??

<@&286206848099549185>

[tight locust]

QUESTION:

consider the infinite polynomial:

$f(x)=\prod_{n=-\infty}^\infty (x-n)$

[ocean seal]

EndTimes

[tight locust]

would this be equal to the sine curve?

sin(pi*x) specifically

[ionic jewel]

how did you get that?

[tight locust]

it's not nevermind

this polynomial just turns into a bunch of vertical lines

[ionic jewel]

i see

i thought it might expand out into the taylor series of sin or something lol

[tight locust]

it explodes between the zeros

[ionic jewel]

yeah wait it doesnt work for any integer x

because the entire expression just goes to 0

ah i see what you mean

[glass lichen]

It has the same solution set as sin(pi*x)

[tight locust]

x is the variable

it is supposed to have zeroes at all integers

[ionic jewel]

it does have zeroes at all the integers

just the rest of it doesnt work

[tight locust]

so how do you make an infinite polynomial behave nicely?

[glass lichen]

you... dont

[tight locust]

wdym?

plenty of functions are defined as infinite polynomials

but most of them are sums instead of products

[glass lichen]

well yeah, those are taylor/mclauren series

those are constructed to be elementary functions

[lapis sonnet]

Check out this video https://youtu.be/WL_Yzbo1ha4 Euler used this to prove a bunch of formulas about pi. You just need to include a coefficient

(He introduces it at around 4 minutes in the video)

[split slate]

I've already gotten that θ is 90, but what about α?

[cerulean comet]

33

[split slate]

According to my textbook, it's 40

But I just don't know the reason behind it

[tight locust]

yeah it's 40

not 33

[cerulean comet]

Oh

[split slate]

OH!

DANG IT

HOW DID I MISS THAT?!?!

Bruh

[cerulean comet]

Oh yeah I missed that when I said 33

[tight locust]

wait how did you get that theta is 90

it looks like a right angle but i can't recall the theorem or whatever you're supposed to use

[umbral igloo]

this is kind of a weird question but
are there any equations that look like breaking ocean waves (im not talking about sine / cosine waves)
kind of like these:
(Hopf equation)

(burgers equation)

please use the reply button, thanks! 🙂

[dark granite]

none that I know of

I chatted with my physics prof about this last year, and he pretty much said such a wave would only occur using functions like sin/cos

[umbral igloo]

is there a way to bend a sin/cos equation so that itll look like a breaking wave?

[dark granite]

I don't think so because cos/sin waves are pretty symmetric with respect to their peaks so the slight angle on the curves you attached (in the first screenshot) don't occur like that.

[umbral igloo]

oh alright, thanks!

[dark granite]

Try messing around with different values of $\alpha, \omega, \phi$ for $\alpha\sin(\omega t+\phi)$ and $\alpha\cos(\omega t+\phi)$.

[ocean seal]

logician_pdx

[dark granite]

That might give you some more intuition on what really happens @umbral igloo

[sudden crypt]

I wonder if there are cool f with sin(f(x)) because if f gives low values, its more streched out and if its high, it condenses more

maybe there are cool ones with polynomials?

[dark granite]

possibly yeah

[split slate]

Sorry for replying late, but:

Since the 90° angle in the other triangle is at point N and formed by lines coming from points J and L, θ is at point K which is also formed by lines form points L and J

[umbral igloo]

kk ill give it a try, thank you

[whole moat]

Anyone help me with this.. I did it but I'm kind of confused if I'm going in the right steps

[split slate]

So, any angle at the circumference subtended by the same two points will be the same, this θ is also equal to 90°

[manic glade]

which of these is the standard way to notate an equivalency class?

[primal furnace]

someone help

help me get an expression for the numerator pls

[tight locust]

have you found the roots

[primal furnace]

no

i got the ansewr

dw

1/a^2+1/b^2+1/c^2=(1/a+1/b+1/c)^2-2(1/a+1/b+1/c)

[sudden crypt]

? This isn't correct and I don't see how you would get the answer from that

[ionic jewel]

,w roots x^3-2x^2-4x+6

[ocean seal]

Results provided by WolframAlpha

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

Requested by 兔子#6562

[fringe yoke]

How do you solve that ?

[ionic jewel]

hmm, didnt think it would be that easy, not sure what the trick is though

[sudden crypt]

not even real roots lul

[ionic jewel]

dont see a way to convert in this case

[fringe yoke]

Usually I am able to use factoring by grouping for cubic stuff

but it doesn't seem to work here

wtf

[ionic jewel]

it doesnt usually work either

[fringe yoke]

[ionic jewel]

your teachers just give you the very small set of cubics that can be factored by grouping

[fringe yoke]

Hmm how could we solve this

well we know P(x) = (x-a)(x-b)(x-r)

[indigo jetty]

this problem is meant to be solved using vieta's formulas

[fringe yoke]

oh my

[fleet glacier]

(a²+b²+c²)²?

[ionic jewel]

ah i dont happen to know vieta's formulas that would be quite the problem

[fringe yoke]

maybe I can manipulate the cubic formula to get 1/a^2 + 1/b^2 + 1/r^2

[sudden crypt]

actually I think it has something to do what I did earlier in #help-8

[fleet glacier]

Newton's sums could work

I can't do it rn

[sudden crypt]

I also think abc = -6 via Vieta or soenthing

[fleet glacier]

Yes

try (ab+ac+bc)²

Man I wish I had paper

[sudden crypt]

a+b+c = -2, ab+ac+bc=-4 and abc=6 should be right, can somebody check?

[fringe yoke]

,ask 1/(-1.866^2) + 1/(1.211^2) + 1/(2.655^2)

[sudden crypt]

we have e1, e2 and e3 and can calculate p1,p2 and p3
We can turn 1/a^2 + ... into ((ab)^2 + (bc)^2+ (ac)^2)/(abc)^2 and maybe something works

[sudden crypt]

yeah, that minus what we want is 2abc(a+b+c) which are all in the polynomial

so from the polynomial we have a+b+c=2, ab+ac+bc=-4, abc=-6
(ab+ac+bc)^2-2abc(a+b+c) = (ab)^2+(ac)^2+(bc)^2, which should be 40. Now 1/a^2 + 1/b^2 + 1/c^2 = ((ab)^2+(ac)^2+(bc)^2)/(abc)^2 = 40/36 = 10/9

Help me its 4:20AM and Im on my phone and cant sleep

[alpine sable]

(Gamma^2*Beta^2) + (Alpha^2 * Gamma^2) + (alpha^2+beta^2)
/(Alpha beta gamma)^2

@sudden crypt solve for this you will get the answer

[sudden crypt]

what are you talking about I already solved the problem?

[alpine sable]

Ohk

Sorry lmao

[alpine sable]

I just read this

[sudden crypt]

oh lmao

[alpine sable]

What does it mean by interchange the role of x and z

[tight locust]

i mean it literally gives you the inequality

[alpine sable]

[void swift]

Hello, how can I prove that a quartic polynomial can only have at most one line tangent to it at 2 distinct points?

I've determined already that the quartic will need to resemble an "M" or "W" shape on the graph for the line to be tangent at 2 points.
I also think that the Mean Value Theorem might have something to do with this, but I don't know.
I've racked my brain drawing graphs this afternoon trying to figure out how to prove this but have yet to get anywhere

This is the last question from this week's calc homework. The extra hint I received was to "consider doing interval analysis" but I'm just not getting it...

Aren't there other (empty) channels for you to use? @vague jolt

[wind bane]

You can't even do that composition unless B is a subset of A

[sage vale]

what is a gof

[wind bane]

And no. It would be g o f : A to C

[void swift]

Could one of the <@&286206848099549185> help me out with this question?

[wind bane]

have you considered what happens if you have two lines tangent to the graph, each tangent at two distinct points

[void swift]

Where on the graph would those points you mention be located?

[wind bane]

the points of tangency…

[void swift]

Well, yes, of course I know that, I'm asking are you asking me to think about those 2 points in any specific location

On the extremum? Close to the extremum?

I know that in the scenario I discussed the line can be tangent either to the 2 extremum or close to them, but I've no way of proving it

[wind bane]

if they are not tangent at the extremums, then you can use the intermediate value theorem to show that it is either tangent to one or more than two points i beleive

[primal furnace]

guys

i found out how to do it

[void swift]

The second scenario

[primal furnace]

bro do u know about sum/products of roots

i solved it by

ill send photo

[fringe yoke]

for quadratic

ye

not cubic

[primal furnace]

its kinda the same thing

u can get a general form

[wind bane]

consider cases where the points of tangency are to the left of one of the extremum or to the right of one of the extremum, or inbetween the two

[primal furnace]

ill send in a sec

[void swift]

In between the two? I don't think that's possible

[wind bane]

it’s not. you are working to show a contradiction.

[wind bane]

that’s what i said here

if they are not tangent at the extrmums

[void swift]

I think this would be the emphasis on proof of contradiction

To pretend that 2 lines is possible

[woeful pulsar]

You can always make that line horizontal by considering y'=y-mx

[void swift]

I can show that if the y-value of the two extremum are the same, then there can not be >1 line

According to rolle's

the problem part is asymmetric quartics

i will try to make 2 lines. maybe algebra?

[woeful pulsar]

easy to resolve, just change y -> y-mx

to make your line horizontal

essentially you are applying a shear transformation which preserves tangency

[paper bay]

is the channel free for me to ask my doubt?

[void swift]

I don't know if I'm making any sense if I'm just spewing nonsense

[void swift]

I'm interested in this, would I model a generic quartic function formula and then put a linear equation in

Like in a system of equations

[woeful pulsar]

you have a line

use that gradient of the line to modify your generic quartic

such that the tangent line becomes horizontal

[primal furnace]

i mistyped but here it is

[primal furnace]

this is da trick

also @fringe yoke if ur curious

[void swift]

Wait a minute, if it were horizontal... Wouldn't it just be y=constant then

Sorry for the silly questions, I've never heard of shearing before

[fringe yoke]

Yeah like im sure I could have done that

But holy shit

the cubic formula

is so fuckign tedious

I dont follow that well

but it's ok]

[primal furnace]

no cubic formula needed :)

[forest wasp]

I'm looking at this question right now but after viewing the solution I'm still a bit confused about one section of the solution. How do you know that the function must be recursive? (I'm confused about why f(n) must equal f(n-1) + f(n-2)). Thanks in advance

[woeful pulsar]

hmm maybe it's best if we have an example

if we have a tangent line y=mx+c, then we consider the transformation y'=y-mx. So we use this transformation for the graph and the tangent line

yeah it becomes y=constant

something like this

[woeful pulsar]

there are only two cases

[forest wasp]

Yeah, I got the part about two cases but I'm confused about why the function is recursive

[woeful pulsar]

the two cases are disjoint so you can add them together

[forest wasp]

ah, yeah I think I see it now, ty 🙂

[hexed bloom]

anybody care to help me out in this?

[ionic jewel]

https://www.geogebra.org/m/R98nZDWr

[solid lintel]

I want to draw a picture made up of a few ellipses, a semicircle and a few line segments as a single equation in a graphing calculator

I know how to draw a single line segment in them

But that method removes everything left or right of the segment

I came up with one theoretical method involving the sign function

But if you put something like x²+y² into geogebra or desmos they can't handle it at all

How do I resolve this?

[hexed bloom]

how'd it go from the step above to the one below?

[warm wren]

divide all of them by ABC

[hexed bloom]

ahh gotcha thx <3

[burnt zodiac]

any tips on this would be greatly appreciated thanks just need to know i'm doing the steps correctly

[tough hatch]

what activity is this?

[urban pollen]

cot (-16pi/3) = 1/tan(-16pi/3)
-16pi/3 = -5⅓pi, which means you would go clockwise direction on the cartesian plane to end up in the 2nd quadrant
thus tan (1/3pi) = tan (60°) = sq root 3
and since the angle lies in the second quadrant, tangent is negative so
cot (-16pi/3) = - 1/tan(-16pi/3) = -1/sq rt 3

[tepid jungle]

X is a function of Y, and X is a random walk

why is Xn independent of Xm- Xn when Xm is just a later step of Xn?

[burnt zodiac]

@tough hatchanova activity

? it's one way anova acitvity huh

[spare spade]

I need help

Is anyone online

[shadow furnace]

Hello

@spare spade

[noble sinew]

You said you reject null but right after not statistical significant, so which one is it?

[spare spade]

2x (2x>2 - 3x -36) Find the coordinates

Help please

[woeful pulsar]

are you trying to simplify the inequality? try making sure you do the same thing to both sides that does not affect the inequality sign

[rough acorn]

simple q but idk what rules get to that answer.

[tough hatch]

4=2^2
1/8 = (1/2)^3 = 2^(-3)

[rough acorn]

Ahhhh

thx!

[rough acorn]

how dumb am I? sin(45) * 6 = 3 root 2,

that should give the hieght?

[gray isle]

where's sin(45) coming from?

[shadow furnace]

Yeah, where do we get sin(45) from?

[proven flare]

Guys can someone help me

I need to understand what this is saying

[dull onyx]

the channel is obviously occupied

[proven flare]

Oh okay

Where do I ask my question?

[dull onyx]

in a different channel

[proven flare]

Okay

[rough acorn]

NVM

[shadow furnace]

What does AC angles mean?

[rough acorn]

dw, i meant the angles for A and C on the LHS but i messed it up

[inner raven]

(ln(x))^i = a + b*i

Could someone solve for a and b please?

[shadow furnace]

We can't solve things

But we could help you solve it yourself

[inner raven]

Basically

I was trying to solve for x in cos(x) + i*sin(x) = -ln(x)^i

[ionic jewel]

that's quite a different problem

[inner raven]

And I thought that, assuming that -ln(x)^i was a complex number, maybe finding the real and the immaginary part would have helped me solving the equation

[ionic jewel]

the problem is that ln(x)^i is a function and a+bi is not

[inner raven]

Mhhh yeah you're right

How could I try solving for x in the starting equation then?

[inner raven]

I mean this one

[ionic jewel]

hmm well i can go through my thought process, not sure it'll get us anywhere but i don't know the solution off the top of my head

I'm going to use exponential form for both sides for ease

[inner raven]

Okok, I'd be glad if u try : )

[ionic jewel]

what i was trying didn't end up working

I'll keep thinking on it

[inner raven]

Ok thank you

I'm trying some other ways too but I don't think I'll ever figure it out

[ionic jewel]

$e^{iz} = -\qty(\ln{z})^i$

[ocean seal]

bunny

[inner raven]

Ok so

[ionic jewel]

taking the natural log of both sides might not be terrible actually

bring down the exponent and get rid of the e

[inner raven]

You basically reduced cos(x) + isin(x) to e^(xi)?

[ionic jewel]

isn't that right

or am I super confused rn

I'm pretty sure that's right

[inner raven]

Yes I get it

That should be right

[ionic jewel]

$iz = i\ln\qty(-ln z)$

[ocean seal]

bunny

[ionic jewel]

divide both sides by i

$z = ln(-ln z))$

[inner raven]

Ok so

[ocean seal]

bunny

[inner raven]

We took i away

And that's so great

[ionic jewel]

raise both to e again while we are at it

$e^z = -ln(z)$

[ocean seal]

bunny

[ionic jewel]

this sure looks a lot nicer now

[inner raven]

Maybe you should use Lambert function someway

[icy sun]

z < 1?

[inner raven]

Yeah it does

[ionic jewel]

where

did i miss some restriction?

[icy sun]

Cause ln(z) < 0

[inner raven]

Mhh

[ionic jewel]

ah

[icy sun]

Or are we not solving in reals

In which case I'll skeddadle cause idk complex logs

[inner raven]

Actually

[ionic jewel]

this has exactly one solution, which is real

[inner raven]

I think we should consider complex solutions as well

[inner raven]

Oh ok nevermind

[ionic jewel]

according to Wolfram at least

idk how to solve it

[icy sun]

gasp wolfram?

[ionic jewel]

$e^z + ln(z) = 0$

[ocean seal]

bunny

[icy sun]

Let's see if bounding gets me anywhere

[inner raven]

Thank you all

[ionic jewel]

better check my algebra i speedran this problem

[inner raven]

Yeah we'll check everything

[icy sun]

Yeah, 1/e < z < 1

The algebra is right, 99% sure

[ionic jewel]

,calc 1/e

[ocean seal]

Result:

0.36787944117144

[ionic jewel]

its less than that

,w e^z = -ln(z)

[icy sun]

Doesn't look it

[ionic jewel]

of course it can't just give the number

[ocean seal]

Results provided by WolframAlpha

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

Requested by 兔子#6562

[ionic jewel]

0.27

[icy sun]

Hmm

What did I miss

[ionic jewel]

idk how did you do that

[icy sun]

Oh wait, I'm dumb

0 < z < 1/e

hey!

[inner raven]

Shouldn't it be 0 < z < 1/e

[icy sun]

[ionic jewel]

[inner raven]

Yeah

[icy sun]

I'm not convinced there is an easy solution for this

Maybe not even an algebraic one

[ionic jewel]

$e^{e^{i\theta}} + ln(r) + i\theta = 0$

[inner raven]

Maybe we should use calculus someway

[ocean seal]

bunny

[inner raven]

But I have no clue

[ionic jewel]

yeah that's not gonna go anywhere

my solution, idk about calculus

[icy sun]

z = 1/z for removing the - if you want

[inner raven]

WELL

[thorn kindle]

it really isn't. try using the taylor series

[icy sun]

I was going to do that after z = 1/z

[ionic jewel]

that's what

[icy sun]

Thanks for stealing my thunder

[ionic jewel]

x^n/n!

?

don't know the Taylor series for log(z)

[icy sun]

same. Was going to google it

[ionic jewel]

$\sum_{n=0}^\infty \frac{x^n}{n!} + \ln(z) = 0$

[ocean seal]

bunny

[ionic jewel]

,w Taylor series ln(z) at z = 0

[inner raven]

This got much more difficult than I thought it was first

[icy sun]

the series depends on where it is centered

[icy sun]

It's difficult to compute

[ionic jewel]

maclaurin superiority

[ocean seal]

Results provided by WolframAlpha

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

Requested by 兔子#6562

[icy sun]

It's easy to see where it would roughly intersect, what it looks like etc.

[ionic jewel]

that's still not helpful Mr Wolfram

[icy sun]

But exact values is rip

[thorn kindle]

for re(z)>0, z!= 0

[icy sun]

thanks

[ionic jewel]

uhhh

[thorn kindle]

uhhh

[ionic jewel]

good luck canceling that out with the e^x series

[icy sun]

ahem

yeah

[thorn kindle]

you can't have a closed form but you can represent it with an arbitrary amount of precision

[ionic jewel]

so the moral of the story is that we should have just put the original problem into wolfram

[inner raven]

Yeah you're right

I'm gonna mail blackpenredpen

[ionic jewel]

who

[woeful pulsar]

a youtuber, but why

[inner raven]

I remember he solved something quite similar

[woeful pulsar]

oh, you mean with the W

Lambert W

[ionic jewel]

really now i actually have to look it up

[inner raven]

Basically in e^(x - a) = ln(x)

[icy sun]

I did

[ionic jewel]

I've seen that function so much and i don't know what it isb

[woeful pulsar]

it's just the inverse of ze^z

[icy sun]

oh this would be every useful I think

[inner raven]

That's actually quite similar to our starting equation, just put pi instead of a and use some algebra

[inner raven]

Yeah

That's useful if you have to figure something like 2^x = x^2

[ionic jewel]

just take the log of both sides works every time

if i see any exponent you already know I'm sending in the ln's

[thorn kindle]

that's cheating. you have to equate infinite series of polynomials

[ionic jewel]

don't worry I'll use the series representation of log to do it

[woeful pulsar]

start by calculating the number of tourists?

[thorn kindle]

10 restaurants, n theatres, 4 tourists/restaurant, 6 tourists/theater, 5 restaurants/tourist, 3 theatres/tourist

[alpine sable]

is there any sample paper

so that i can pratice

i am in 8th grade

[thorn kindle]

practice what

[alpine sable]

questions 🙂

[thorn kindle]

what questions?

[alpine sable]

im literally having the hardest time of my life figuring out why the section i highlighted is true

can anyone help?

[steel horizon]

😮 yesssssss thankssssssssss

[woeful pulsar]

can you try that out?

to do that start by calculating the number of pairs (restaurant, tourist) such that the tourist visits the restaurant

[steel horizon]

🤔

i seee

[steel horizon]

i mean now im covering up my backlogs so im sorryy 😦

@woeful pulsar any special tip to cover my backlogs ?

[woeful pulsar]

what backlogs?

[woeful pulsar]

@steel horizon can you perhaps try this step?

[woeful pulsar]

try implicit differentiation to find the gradient of the line and you get (y-k)=m(x-h)

[alpine sable]

of rational numbers

[steel horizon]

im actually covering my backlogs, im reallyy sorryy mate
BUt i will do it i promise

[alpine sable]

i did, and the best thing i got was this (x0 and y0 was replaced with h,k cause its easier to write)

do the equations look similar? is there a way to convert this to the one i sent in that pic?

idk ughhh mine looks messy

[woeful pulsar]

hold on how did you get that

did you do something like (y-k)=m(x-h)

oh wait

i see it has to pass by h, k

and (h, k) is not necessarily a point on the graph

[woeful pulsar]

yeah you can definitely multiply both sides out by the square root in the denominator

[alpine sable]

uhhh do not ask LOL i usually make math problems longer than they should be

ignore this one, its useless lol

[quaint pond]

hello, could someone help me understand the solution to this problem:

the answer apparently is

[alpine sable]

also what is the (y-k)=m(x-h) method you're talking about? aaa i dont think ive seen anything like it tbh

[woeful pulsar]

oh yeah sorry i think it was somewhat off

because I thought the point of tangency was (h, k)

[woeful pulsar]

use stars and bars, 5 stars 5 bars.
1|2|3|4|5|6
where the star is gives the number

[alpine sable]

ok so last question, i got this

which is apparently the same equation as the one in the image

how do i reduce it to a second power equation? it doesnt look clean at all

[quaint pond]

what would that represent

[languid moat]

I'm a beginner what topic should I start first?

[woeful pulsar]

depends, what were the last topics you did

[languid moat]

idk I want to rebuild my math foundation so where should I start

[woeful pulsar]

khanacademy is a good resource

it's free, easy and you can do it anytime

[languid moat]

i have this book called "basic mathematics by serge lang" should i complete it

it starts with addition up to functions should i read it, it seems simple but it might help rebuild my foundations

[woeful pulsar]

i think khanacademy is decent too since it gives practice problems

yeah you can try reading it and see how it goes for you

[languid moat]

yep I think I should read it, I really want to have a strong foundation so it would be easier to learn harder topics

I hope I can participate here in the future

[full igloo]

how do i find x here?

[woeful pulsar]

this channel is really crowded lol

check timestamps

[woeful pulsar]

are you sure you didn't leave out any information

[tight locust]

all you need is arc AB

[woeful pulsar]

hmm lemme check

yeah all the information is there

consider triangle ABE

[tight locust]

x = 40 degrees

[tight locust]

no need.

[woeful pulsar]

don't just give answers

ensure you engage the question asker

[urban pollen]

angle abc is 90 degrees cause of right angle in semicircle

then angle bad is 50° cause angle in centre = 2 angles at the circumference

[full igloo]

ah

[urban pollen]

then looking at triangle abe angle aeb is 40°

[full igloo]

bad=50 already found that

but i stuck after that..

[urban pollen]

ah

[urban pollen]

^^

[full igloo]

ah

[urban pollen]

then focus on only triangle ABE

180° - 90° - 50° = 40°

[full igloo]

ah so its 40°?

thank you :D

[urban pollen]

yepp

welc :))

[full igloo]

might eventually ask more

[urban pollen]

i'll try my best HAHA

[tight locust]

(130-50)/2

[full igloo]

noted

stuck again :(

[woeful pulsar]

what are you trying to find here?

[full igloo]

BDC and CAD

[woeful pulsar]

angles subtend same arc

[full igloo]

what do you mean?

i cant understand that well..

[woeful pulsar]

these angles are the same size

[full igloo]

BDC is 60°?

how about CAD? is it any different?

[woeful pulsar]

you can use a similar argument

[full igloo]

ah but i cant seem to find it

is there any way to find it?

i got BDC now so thats not a problem

[woeful pulsar]

how did you get BDC

[full igloo]

using this??

both are on Arc BC apparently

BAC=60 so i suppose BDC=60

[woeful pulsar]

yeah now use that with a different arc to find CAD

[full igloo]

wait i found it :O

[stable orbit]

Q11 quadratic equations

Assignment

[gray isle]

what have you tried?

[hallow flax]

I would suggest, to start by writing down the conditions you have in mathematical terms, so to extract all information from the problem, then, the problem might be easier.

[woeful pulsar]

model the question with an equation

[urban pollen]

let the number of students be x and the number of apples distributed to them be a, 300/x = a and 300/x+10 = a - 1

x = 50

[woeful pulsar]

remember your brackets

also don't just give answers

engage the person who asked the question

[woven plaza]

n= 300/x
n-1=300/x+10
Where x is number of students and n is number of apple each student gets in first case

Solve for x

[urban pollen]

oh wait yaaa thats a shorter method ah sorry

[woven plaza]

[solid lintel]

Anyone?

[woven plaza]

What kind of picture

[solid lintel]

Wait I'll send it

[woven plaza]

If you have a fixed goal then you can start with basic equations and transform them into whatever curve you want

[solid lintel]

https://media.discordapp.net/attachments/808571144767668236/867688477369434122/unknown.png

[woven plaza]

Yeah that's a semi circle

[jagged imp]

are you sure? x^2+y^2=1 works just fine on geogebra if thats the kinda thing you were going for

[solid lintel]

Just straight lines and semi circles and ellipses

[woeful pulsar]

well it's possible as a piecewise defined function that is noncontinuous

[solid lintel]

I want everything in one equation of that's possible

[woven plaza]

The eyes will be circle of radius same as the "b" of elliptical equation

[woeful pulsar]

of course you'll need to use a parametric equation probably

[solid lintel]

See I had this idea

[woeful pulsar]

oh wait batman curve

[woven plaza]

Have you drawn the rectangle and hands?

[solid lintel]

I could add (1-|sign(f(x)|)(1-sign(y)) to the entire equation

[woeful pulsar]

yeah we can also use a f(x, y)=0 equation

that too

[woven plaza]

It seems hands are at 45° so hands are simply shifted formof y=x and y=-x lines

[woeful pulsar]

but maybe find equations for each of the parts first

[woven plaza]

Don't need to go that complex

[solid lintel]

And that would give me a semicircle

[solid lintel]

I already have those

[solid lintel]

Yes something like that

[solid lintel]

This works theoretically

But geogebra hates it

[solid lintel]

Do you have an idea?

[proven flare]

Hey guys I have a few questions about summation

Can you guys help me understand these or give me resources to help me

[split slate]

How might I find x?

I don't even know where to begin with this

[alpine sable]

You could use Sine rule

If you haven't learnt that yet you could just use your trig ratios for right angular triangles

[split slate]

What's the sine rule?

But considering I don't know what that is...

The other option seems more viable

[split slate]

Could you still explain further, please?

Like?

Find the length of the dotted line

[alpine sable]

yeah and then use that to find x

[split slate]

Oh ok

Thanks!

[ocean seal]

Πολλά άτομα είναι

[split slate]

Yeah...

I'll look into it later

very late at night

[alpine sable]

and to solve for side lengths just get the reciprocate of that

[pallid sail]

there isn't much to look into

you got the sides and the angles

you just need to utilize the formula that was given to you

[alpine sable]

https://gramoly.org/integral_championship/

hey, check out this competition

A car traveling at 53 km/h hits a bridge abutment. A passenger in the car moves forward a distance of 65 cm (with respect to the road) while being brought to rest by an inflated air bag. What force (assumed constant) acts on the passenger’s upper torso, which has a mass of 39 kg?

Why do we care about the average velocity here? I meant (53 + 0)/2 km/h, and we find the acceleration using the average velocity..?

We could just use v_f^2 - v_i^ = 2ad, where v_f = 0 and v_i = 53 km/h (but converted to meters/sec)

To find the acceleration?

[vale wigeon]

we care about velocity to find the change in kinetic energy, which gives us the work done by the stopping force

we don't know how long the passenger took to be brought to a stop nor do we need to

@alpine sable

[alpine sable]

I never said we did

We use v_f^2 - vi^2 = 2ad to find the acceleration when the car was stopping

[vale wigeon]

hm wait

i think you will have the same result either way

ah yes

yes you will

[alpine sable]

No I got 6500 N

[glass lichen]

as opposed to...?

[alpine sable]

6600 N

[glass lichen]

Yeah... both kinematics and energy give ~-6502N

so likely a typo

im assuming you got the 6600N from the answer key

[alpine sable]

I got like 6474 or something

Not 6502

But I rounded up to 2 sigfiggs

53 km/hr is 14.7m/s. The average speed while decelerating is vav = 7.4m/s. The time of deceleration is t = x/vav = (0.65m)/(7.4m/s) = 8.8×10−2 s. The deceleration is a = ∆v/t = (−14.7 m/s)/(8.8×10−2 s) = −17×102m/s2. The force is F = ma = (39kg)(1.7×102m/s2) = 6600 N.

This is the solution's manual solution

[glass lichen]

$a=\frac{v_f^2-v_i^2}{2d}=\frac{-(\frac{53}{3.6})^2}{2(\frac{65}{100})}\approx -166.73$

[ocean seal]

Mosh

[alpine sable]

,w 0^2 - (53 * 10/36)^2 = 2 * a * (0.65)

[ocean seal]

Results provided by WolframAlpha

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

Requested by n/c#1137

[alpine sable]

,w 167 * 39

[ocean seal]

Results provided by WolframAlpha

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

Requested by n/c#1137

[alpine sable]

That's not 6502 though

,w 166 * 39

[glass lichen]

yeah cause you rounded in an intermediate step

[ocean seal]

Results provided by WolframAlpha

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

Requested by n/c#1137

[alpine sable]

What intermediate step?

[glass lichen]

You used a=167 and a=166

that's rounded

[alpine sable]

What else should it be?

We only care about two sig figs?

[glass lichen]

use the actual value....

you round at the end

[alpine sable]

,w 166.726 * 39

[ocean seal]

Results provided by WolframAlpha

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

Requested by n/c#1137

[alpine sable]

Okay thanks I didn't realize I shouldn't round like that

[glass lichen]

,w (-(53/3.6)^2/(2*0.65))*39

[ocean seal]

Results provided by WolframAlpha

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

Requested by moshill1#9771

[alpine sable]

But still look at the solution above by the solution's manual

It's way different

,w 39 * (1.7 * 10^(2))

[glass lichen]

yeah they used average stuff

[ocean seal]

Results provided by WolframAlpha

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

Requested by n/c#1137

[alpine sable]

Okay but only one of these can be correct?

[glass lichen]

ask your teacher

[alpine sable]

It's from a book

https://www.askiitians.com/forums/Mechanics/an-electron-is-projected-horizontally-at-a-speed-o_121717.htm

How was the a = F/m calculated?

,w (4.5 * 10^(-16))/(9.11 * 10^(-31))

[ocean seal]

Results provided by WolframAlpha

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

Requested by n/c#1137

[alpine sable]

It should be 4.9 * 10^14??? Or am I missing something?

But the solution's manual also says 4.9 * 10^15, so I am confused?

[glass lichen]

it's called a typo...

[alpine sable]

Two typos (same!!!) in two different sources I guess

I'm shooketh

[mint rock]

Doing some crypto analytics and have arrived at something of the form A^x -Bx = C. Is this algebraically solvable?

Quick search says that it is not, but then I do not know how to approximate it either.

[quaint trout]

What is the exact question given to you?

There is no such thing as cross product in R^5

[sudden crypt]

well not with 2 input vectors, it is defined for 4 input vectors

[quaint trout]

No, if the question is "Find the cross product of a and b" and this is a and b

There is no answer

That's not a thing

You can't do anything

[sudden crypt]

the dot product however is defined, maybe you swapped the terms?

[quaint trout]

So is the question you were given to find the cross product or to do something else and you wanted to try to do it using a cross product?

Can you send a picture?

Oh, it's probably a trick question

You are supposed to say it's not defined

Or something

👍

[alpine sable]

q 21

can someone have a look

im strugglin so bad TT

[gusty ravine]

yea the middle part is right

after u arrange the middle part you're gonna be left with 1 vowel and 5 consonants, so u just need to permute these 6 in the other spots @alpine sable

[alpine sable]

the ans is 4320

i don't get why

[gusty ravine]

i dont see why its not 4! * 6!

[opaque zodiac]

: Construct a matrix that triples the height, width, and length of an objective in 3D. What exactly this question wants ?

[torn lion]

A matrix $M$ such that $M\begin{pmatrix}1\0\0\end{pmatrix}=\begin{pmatrix}3\0\0\end{pmatrix}$, $M\begin{pmatrix}0\1\0\end{pmatrix}=\begin{pmatrix}0\3\0\end{pmatrix}$ and $M\begin{pmatrix}0\0\1\end{pmatrix}=\begin{pmatrix}0\0\3\end{pmatrix}$.

[ocean seal]

Nonassociative Semigroupoid ✓

[opaque zodiac]

im doing this on programming maths so this is actually maths right ?

[torn lion]

I don't know what programming maths is, but it looks like a fairly standard linear algebra problem to me.

[opaque zodiac]

is this like theory thing ?

it seems simple thats why i ask

[torn lion]

You don't need to prove anything, if that's what you mean by "theory thing". You just need to find the matrix.

[opaque zodiac]

i thought of it, but thought was too simple asnwer

ah

[torn lion]

It is a rather simple problem.

[alpine sable]

Any idea on this sum?

[torn lion]

So basically $S=\sum_{1}^{n}(3i^{2}-i)$, right?

[ocean seal]

Nonassociative Semigroupoid ✓

[alpine sable]

yes

[torn lion]

Do you know $\sum_{1}^{n}i$ and $\sum_{1}^{n}i^{2}$?

[ocean seal]

Nonassociative Semigroupoid ✓

[torn lion]

They're pretty standard identities

[alpine sable]

n(n+1)/2 and [n(n+1)(2n+1)]/6

[torn lion]

Combine that with the fact that $\sum(f(i)+g(i))=\sum f(i)+\sum g(i)$ and $\sum af(i)=a\sum f(i)$, where $f,g$ are functions of $i$ and $a$ is some constant, and you should be good.

[ocean seal]

Nonassociative Semigroupoid ✓

[alpine sable]

Alright

Thank you

[torn lion]

That is, you can say $S=\sum_{1}^{n}3i^{2}-\sum_{1}^{n} i$, etc.

[ocean seal]

Nonassociative Semigroupoid ✓

[opaque zodiac]

just to make sure, the answer was as the bot constructed ?

[alpine sable]

I understood

[glass lichen]

M=3I

[alpine sable]

I do not do math on a purpose now ,I just like it but I had no idea on this one ..

[opaque zodiac]

3*i?

[torn lion]

I don't know what your bot constructed.

[glass lichen]

I, yes

[torn lion]

But it's basically the answer that Mosh gave.

[glass lichen]

$T[v]=3v\implies [T]_B^B=3I$ for canonical basis B

[ocean seal]

Mosh

[torn lion]

Yeah, no worries, everyone starts learning from somewhere. Good luck with the rest of your math!

[alpine sable]

What’s 17 + 204871773874

I know this can be solved with the quadratic formula but because the formula is beyond the curriculum of the present chapter is there any other way to solve it?

[glass lichen]

discriminant has to be what sign?

[torn lion]

The quadratic formula is basically just completing the square, if that's in the curriculum.

[alpine sable]

Oh I got it nvm

Like 6 or something thanks anyway

[alpine sable]

This is polynomials chapter quadratic equations is the next

Wha

Sorry

[glass lichen]

@alpine sable you need the quadratic to not have roots.. what does that make the discriminant

[torn lion]

Do you know how to calculate the discriminant of a quadratic?

[opaque zodiac]

is that basically proving what we just said ? im kinda confused haven't done anything like this

[glass lichen]

It's find matrix representation based on the mappings that were pointed out

[alpine sable]

Yes I do know

D = b^2 - 4ac

If D is bigger than 0 then the thing has real roots

[torn lion]

So what you want is for there to be no real roots, which means you want D to be negative.

[glass lichen]

We love people that ignore what you say

[alpine sable]

Negative

[glass lichen]

yes, so solve b^2-4ac<0

[opaque zodiac]

mmm i think i haven't done it this way, thanks either way i will try understand it tho.

[torn lion]

It sounds like you're doing a math-for-compsci-nerds class, right?

[alpine sable]

Yeah I mean

I know how can I do that

But the thing is

That thing is beyond the current chapter

[tawdry saffron]

how to calculate the maximum value of x+y of a circle equation

[torn lion]

In that case it will probably just be the computation that I had mentioned in the beginning.

[opaque zodiac]

im doing maths for game programming, a lot of 3D stuff that i have never done in my life feels harder than i thought

[alpine sable]

There must be alternative

Fivht

Right*

[torn lion]

I mean, what techniques did you learn in the chapter?

[tawdry saffron]

if (equation) what is the maximum value of x+y

how to start with such a question?

[torn lion]

This channel is already in use twiceover. Can you please wait your turn or go to another channel?

[tawdry saffron]

oh sorry

[torn lion]

#help-1 through to #help-8 are all open right now.

Looks like 9 opened recently too

[alpine sable]

None

I mean

None which seem helpful

Just some stuff about polynomials

[torn lion]

What stuff about polynomials did you learn?

[alpine sable]

Basic stuff

A summary which is supposed to carry outside the syllabus stuff

[torn lion]

Honestly I'd just use the discriminant.

[alpine sable]

Vieta's formula too

If that has anything to do with this

[alpine sable]

I would too but like I expect there to be a explanation inside the scope of this thing

[torn lion]

Lots of things named after Vieta. Which formula did you learn?

[alpine sable]

Alpha + Beta = -b/a
Alpha * Beta = c/a

And for cubic polynomials too

[torn lion]

Okay, so that provides something.

[alpine sable]

I tried writing it that way

But cut it down

Because roots are supposedly unknown, and it says any x so no clue

[torn lion]

If we have no real roots, i.e. $\alpha$ and $\beta$ are complex, what does that tell us about $-b/a$ and $c/a$?

[ocean seal]

Nonassociative Semigroupoid ✓

[alpine sable]

Complex numbers are outside the scope of my entire curriculum

[torn lion]

[alpine sable]

But this got nothing to do with them ig

[torn lion]

Use the discriminant. Write a page explaining why the discriminant works if need be.

[alpine sable]

Sorry don't know

[alpine sable]

Alright ig

[torn lion]

Honestly fuck it, your teacher should be happy that you're taking initiative and reading ahead to the next chapter anyway, and a right answer is a right answer regardless of how you got there.

[alpine sable]

I don't have a teacher lmao I mean our school book has questions like

Find the roots of
x^2 + 2x + 1

Thanks

[torn lion]

Who grades the answers to the questions?

And gives lectures, etc.?

[alpine sable]

These questions are from a coaching institute or smth and I don't take coaching + school holidays are going on and our teacher doesn't like doubts with questions sadly

[fair crater]

That's a really weird teacher...

[torn lion]

Honestly a right answer is a right answer is just what I have to say to that.

[fair crater]

Generally teachers like engaged students

[torn lion]

As long as how you got there is valid, of course.

[glass lichen]

Im confused how, if you havent learned quadratic theory in any capacity, why are you doing quadratics/being told to do them

[alpine sable]

I haven't been told by anyone to do these questions I have already completed the school textbook and am using this booklet to do questions

[torn lion]

I'm confused why you're learning about general polynomials before learning about quadratics. I learned about polynomials as a generalization of quadratics when I was in highschool.

[glass lichen]

So you're opting to do questions you have no fucking clue how any of it works?

[alpine sable]

I know quadratics

[glass lichen]

you said you didnt

[alpine sable]

I already mentioned it

That I do

[alpine sable]

"I know.."

[torn lion]

Then you have the answer.

[glass lichen]

then what was the problem of "not being at the quadratics chapter" yet?

[alpine sable]

I expected an explanation without using that equation

[glass lichen]

You asked a question about a quadratic...

[alpine sable]

Isn't a quadratic polynomial still a polynomial

[glass lichen]

yes

[alpine sable]

Then?

[glass lichen]

then what?

it'a a quadratic, so all the quadratic theory tricks come into play

[fair crater]

@alpine sable Without you listing the 'acceptable methods' it will be difficult for us to comment on the appropriateness on use of methods

[alpine sable]

I already listed them tho

[fair crater]

You would essentially be asking strangers to deduce your syllabus which is like, magic on top of magic

[alpine sable]

There's just one

[fair crater]

I think you just said 'no quadratic formula'?

[alpine sable]

Vieta's thing

[glass lichen]

something like factor theorem / remainder theorem, or generalized Vieta's arent usually used in quadratics

[fair crater]

😐 listing no goes makes it hard to know the 'ok-to-go'?

[alpine sable]

.

Why are you guys all on me

[glass lichen]

cause you're really not making much sense

[alpine sable]

Idk

[glass lichen]

it's a quadratic, so use stuff for quadratics

if it were a cubic, then use stuff for cubics

[alpine sable]

Bro

Forgive me

[torn lion]

father for I have sinned

[alpine sable]

[fathom matrix]

dam

[alpine sable]

Ok so I solved that, and like I get

(K-4)(K-2) but the answer key says 3

No clue why

[indigo jetty]

i think this question is not worded properly

if it asked for K as an integer then the answer is 3

[alpine sable]

Is it talking about the net force $\sum F_x = F_{m2} - F_{m2m1}$ having acceleration $2.6 m/s^2$ or just $F_{m2}$?

[ocean seal]

n/c

[lunar relic]

My friend and I are conflicted on whether or not it's possible to trivially integrate this integral. He says to treat y like a constant and integrate it, but then differentiating it, I get something different than what we start with.

[alpine sable]

You're not treating y as a constant when you differentiate it

[lunar relic]

right, I'm not taking the partial derivative

[alpine sable]

okay?

I'm not sure what you're expecting here

You treat it as a constant when you integrate, but not when you differentiate

Why would you expect the answers to be the same?

@lunar relic

[lunar relic]

yeah, my question is can you even treat y as a constant when integrating? If not, how would you integrate this?

I don't know if my friend's assumption is correct, which is just integrate without touching y. What I'm trying to show is that if we say y is constant, integrate it, then take the derivative, since y is a function of x, it doesn't give the same thing back because of the dy/dx. If I take the partial derivative, then y can be assumed to be a constant since we're simplifying the problem.

[alpine sable]

It does give back the exact same thing

y is constant, so dy/dx = 0

[lunar relic]

Oh huh

[alpine sable]

Either y is or isn't constant

it can't be both

You can integrate it once more if you need to

Where you treat x as constant

Anyway, in your example, assuming y is a function that does not depend on x, it would still be the correct answer

[lunar relic]

Which it indeed is a function of x

[alpine sable]

Okay so then you can't treat it as a constant

[lunar relic]

so in that case guess my friend and I are back to step 0 on integrating this lol

[alpine sable]

You also can't integrate arbitrarily abstract things

Like you can't say ∫ x f(x) dx = x^2/2 f(x) + C

And pretend f(x) is constant

That doesn't make sense

You need to know what y is to integrate it

That being said

You can always use integration by parts on problems like this, usually.

[lunar relic]

yeah the issue is that we're both doing differential equations and we don't know what y is so we can't substitute in. This is from our exam and I wasn't sure if we could even integrate this since we don't know what y is other than the fact that it's some function of x

actually wait let me think about this

[alpine sable]

Perhaps the correct answer is in the form of integration by parts.

[safe kite]

need some one to walk me through how to work this voice explaination would help alot

[full cypress]

anyone know any websites where i can practice algebra?

[topaz scaffold]

Wolfram Alpha?

[topaz scaffold]

Linear algebra or the other one?

Regular algebra

Idk what's the name

[full cypress]

regular

[full cypress]

highschool algebra

[topaz scaffold]

Oh

Khan academy ig

Or Wolfram

Depending on what kind of questions you want

[full cypress]

imma check out wolfram

it has hard questions right?

[topaz scaffold]

I only know that it has questions

I've never tried the algebra ones tho

[full cypress]

ight thanks

[gentle turtle]

Please correct my solution..
Why my solution different from calculator -4x+1/3x-2

[ionic jewel]

,w inverse (2x-1)/(3x+4)

[ocean seal]

Results provided by WolframAlpha

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

Requested by 兔子#6562

[ionic jewel]

your solution is right

oh you forgot to change the y on the bottom to an x but idk if that's just your handwriting

[scenic crystal]

I want to get sum of the squares of the first ten natural numbers like = 1^2 + 2^2 + 3 ^ 2 .....10^2 = 385, Is there any formula to do this?

[ionic jewel]

summation notation, see:

$\sum_{n=1}^{10} n^2$

[ocean seal]

bunny

[ionic jewel]

oh there's also a closed form let me find it

,w sum from n = a to b of n^2

[ocean seal]

Results provided by WolframAlpha

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

Requested by 兔子#6562

[ionic jewel]

-1/6 (-1 + a - b) (-a + 2 a^2 + b + 2 a b + 2 b^2)

with a as the lower bound, and b as the upper bound

in your case a = 1, b = 10

[scenic crystal]

Thanks man 😄

[ionic jewel]

you can of course simplify that if you always want to start at a = 1 or something

yw

[gentle turtle]

Thank you very much

[ionic jewel]

yw

[dawn galleon]

<@&286206848099549185>

[viscid cove]

have you plotted them in the graph?

[alpine sable]

For this problem, why can you not use the power rule for y^2? Why did the instructor use the chain rule?

[dawn galleon]

bruh go to another channel

channel in use

#❓how-to-get-help

[lapis valley]

Because u are differentiating y not x

d(y^2)/dx = d(y^2)/dy * dy/dx = 2y* dy/dx

[alpine sable]

Ty

[alpine sable]

You literally didnt type for 18min

[viscid cove]

come on guys, rules..

[alpine sable]

Tried a lot of ressources, but yet don't understand integrals as much, can somebody help me get a clean understanding?

[viscid cove]

next channel pls

[ionic jewel]

what do you not understand about them

area under a curve between the bounds

[alpine sable]

Pretty much everything

[ionic jewel]

well unfortunately I'm not available to teach the entirely of integral calculus

[viscid cove]

@dawn galleon i might have the answer, you need to cross check it though

[alpine sable]

I just need the main concept of integrating, not all of it lol

An explanation of what it is

Basically

[viscid cove]

jesus there's literally a general channel

so, the period is 365, cause seasons repeat in a year

[dawn galleon]

after you plot it, we would need it to create a function to model that data. Because its temperature over the year it would be periodic, so we might start with something like y = Asin(b(x + c) + d, where A a and b are all variables that you'll need to change around to make it fit the data

[viscid cove]

its a little rough due to the given data but its follows a sine curve

[ionic jewel]

geometrically $\int_a^b f(x) , dx$ finds the (signed) area between the x axis and the curve f(x) between x = a and x = b

[ocean seal]

bunny

[viscid cove]

yeah yeah

[ionic jewel]

I found that starting by learning Reimann sums helped nicely so maybe those those up too

[viscid cove]

we have the period=365, ratio= sine, amplitude=....

[alpine sable]

Alrighr

[viscid cove]

(highest temp-lowest temp)=19--8=27

and yeah the midline and shift, one sec

[dawn galleon]

ye

[viscid cove]

there's a stat aspect to it so find the mean value and the median

[dawn galleon]

Period is 365?

[viscid cove]

uh huh

well the midline passes though one of them, if the given data is symmetric both median and mean will be the same( i think)

[primal ether]

the data doesn't seem symmetrical about the middle

[viscid cove]

it must be symmetric about some line

[primal ether]

if the range is 27 the coefficient of sin should be 13.5 right?

[dawn galleon]

what should the amplitude be?

[primal ether]

and adjust the constant to fit

[viscid cove]

13.5

[primal ether]

this is the best i've gotten

[dawn galleon]

how did u get that?

[primal ether]

as the amplitude is 13.5, and we know that there is a peak at temp 19

we can use the fact that a sin x + b has a peak of a+b

[dawn galleon]

did u use the equation Asin(b(x + c) + d,

[primal ether]

woops

13.5sin(x-100)+5 is what i got lol

[dawn galleon]

there was no period the b value?

[primal ether]

The sine graph has a period of 360 degrees

We can disregard the period as 360 is close enough to 365

[dawn galleon]

Ohh can u send the link of ur desmos graph

[primal ether]

https://www.desmos.com/calculator/ozukmdelxz

The last 2 points don't align

[dawn galleon]

thanks

[viscid cove]

But if he was supposed to do stat operations for the means and such, wouldnt the approximation be too much?

yours is still better though, i suppose its fine

[primal ether]

the data isn't

the best

[viscid cove]

the last two points are a bitch

$$y=13.5\sin\left(\frac{\pi}{182.5}\left(X-106.5\right)\right)+5.833$$

[ocean seal]

drunkzurg

[viscid cove]

100 is still too much imo, you cant explain where it came from

[primal ether]

good approx

no sine graph can satisfy the last 5 points though

[viscid cove]

amplitude: (max-min)/2, period: 365 because seasons repeat, x translation 106.5 because the minimum point is in 15 and each quarter of the period is 365/4=91.25 and the y translation is comes from (max+min/2)

true

[feral iron]

Does any1 see a formula to make column A match column B from 0-359°?

[viscid cove]

match?

why would you want that?

[feral iron]

Not match, but match the relationship in the spreadsheet, so when the ship angle is 90°, the camera angle should be 0°

[viscid cove]

no no i dont think its anything like that

[feral iron]

It is something like (Ship angle - 45)x2

[viscid cove]

i might be wrong, could you send the actual question or language associated with it?

[feral iron]

Its my own question - ship always needs to face up in camera view (top down). I have tested all of the above values and they work, but I dont know how to make it work for increments bettween the values

[viscid topaz]

Bruh, (ship angle + 360 - camera angle ) % 360 = 90

[feral iron]

The camera angle needs to be set via a formula, depending on ship angle rather than manually entered

[viscid topaz]

Same deal, just rearrange the formula to solve for camera angle, right?

[feral iron]

Correct