help-33

[desert dirge]

i would say the 10.424 is right as the first time

assuming it is, the second can be done by knowledge of sin

ie sin(x)=sin(pi-x)

[dense lion]

i did but it says it was wrong

[desert dirge]

ah i didnt see that the center of the wheel is the origin

the center is 103 above the ground
so y must be -23 if its at 80

[dense lion]

ohhhhh, that makes so much sense let me plug it in real quick

yup it was correct, thank you

[desert dirge]

np

[dense lion]

what would i have to do for the second time, add one more pi?

[desert dirge]

sin( arcsin(...) )= sin( pi - arcsin(...))

youd just replace your numerator with pi-arcsin(-23/100)+pi/2

i believe

[dense lion]

ahh, thank you once again, trig is not my strong suit

[marsh citrus]

@dense lion Has your question been resolved?

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[silent brook]

PL SHELP

[marsh citrus]

Please don't occupy multiple help channels.

[silent brook]

pls any1

[desert dirge]

.close

[marsh citrus]

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[desert dirge]

one channel

[silent brook]

srry im new to this server

can u pls help

plsss

[marsh citrus]

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[silent brook]

huh

[silent brook]

<@&286206848099549185>

PLEASE

[plucky pivot]

Keep in mind that angle DAB is equal to DCB

remember what the total amount of degrees makes a triangle as well

[silent brook]

i got angle dab and theone next to 41degrees, and using cosine rule i got length ab which equals to ac

😭

thats it

[plucky pivot]

angle DAB equals 120° which is also DCB

[silent brook]

so 120-X ?

[plucky pivot]

No, X1 ( the x visible in the drawing ) + X2 ( the x value that would be on the right side ) = 120°

they give you an angle inside that triangle as reference to solve for that angle, once you do you’ll be able to substract it from 120° and find X

Remember that DA equals CB

Let’s do this first

How many degrees equal a triangle

[silent brook]

180

[plucky pivot]

answering that should help you find the third angle for the obtuse triangle with the two given angles

[silent brook]

which one

[plucky pivot]

the one with 120° and 38°

[silent brook]

so 120-38

[plucky pivot]

find the third angle and then find what angle is identical to that one

no lmao

[silent brook]

[plucky pivot]

if a triangle equals 180 degrees

and it consists of three angles to equal that 180 and you have two of those angles already, how would you find the last one

[silent brook]

how do i have 2 of themm

ic anisualise

visualise

1 sec

bru wheres the 38

[silent brook]

supposed to b here

[plucky pivot]

find the third angle in that triangle

[silent brook]

22 for the missing angle top right

[plucky pivot]

I have to go sadly 😭

[silent brook]

i fifalready]

bri

i did find it

and the one next to 41

is 180-41

[plucky pivot]

yes that’s right, that 22° is equal to BDC

from there you should be able to find the rest

[silent brook]

WHAT

wth

nahh

fuk this i quit

but thx anyway

havea good day

[plucky pivot]

mb I’m not home rn

pm me, I’ll try to break it down for you when i get home

[marsh citrus]

@silent brook Has your question been resolved?

[marsh citrus]

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[marsh citrus]

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[willow nebula]

Ask

[void elm]

...do you have a question?

[willow nebula]

yeah, sorry, im a little slow

[void elm]

no problem

[willow nebula]

also the online textbook is being a pain

[void elm]

[void elm]

37?

[willow nebula]

Im just trying to figure out how to determine if each pair is an inverse function

im on 31

we doin odds

[void elm]

alright, what do you know about inverse functions?

[willow nebula]

the invers function part

[void elm]

there's a special property they satisfy, that you can use to check if two functions are inverses of each other

[willow nebula]

isnt that when I swap the X and the Y?

[void elm]

yes, but there's a way to state that more generally

the key is the word inverse

[willow nebula]

f^-1(x)?

[void elm]

that's the notation, yes

here's a hint

[willow nebula]

alr, thanks

[void elm]

if I have an input x, then apply f, then apply f^-1, I should get back x
if I have an input x, then apply f^-1, then apply f, I should get back x also

(that's what inverse functions do!)

they "undo" the original function, and the original function undoes them!

[willow nebula]

oh, neat

[void elm]

formally, we say that a function g(x) is the inverse of f(x) if
f(g(x)) = x, and
g(f(x)) = x

this is what we should check to see if we have inverse functions or not

...does that make sense?

[willow nebula]

not really to be honest

[void elm]

hmm, have you learnt about function composition yet?

[willow nebula]

you know what, imma just leave bc im stealing you away from someone else who will actually use your help

[void elm]

nono

please stay lol

[void elm]

I ask again:

[willow nebula]

no, not yet

[void elm]

okay, that's why you didn't understand haha

if not, we can do the good old "swap x and y"

[willow nebula]

alright

[void elm]

so you want to take f(x) and call it y, right?

then swap x with y, correct?

[willow nebula]

yeah, do I do that for both then compare them to see if they have the same outcome

?

[void elm]

you don't actually need to do it for both

[willow nebula]

oh

[void elm]

choose either f(x) or g(x)

it doesn't matter

take your pick

[willow nebula]

alright, im doin that rn then

[void elm]

[willow nebula]

well what do you know

thanks so much man

[void elm]

oh?

did you solve it by yourself?!

[willow nebula]

well, not by myself

I had your help

[void elm]

actually, just to be sure

could you tell me what you did?

I just want to be 100% certain, haha

[willow nebula]

swapped x and y

[void elm]

and then...?

[willow nebula]

subtracted the 6 to put it next to the x

then divided everything by 4

looks exactly like g(x)

[void elm]

so you solved for y, after swapping x and y

that is indeed the correct approach!

if you don't get g(x) after solving for y, then you can safely say that f(x) and g(x) aren't inverses

if you do, then they are inverses

[willow nebula]

Ah, that makes sense!

[void elm]

[willow nebula]

So the answer would be yes

[void elm]

mhm

(I personally like using function composition over the "swap x and y", because doing the latter kind of hides what's going on)

but if you haven't learnt about function composition yet, then there's no helping it

in any case, does that solve your question?

[willow nebula]

ok, when I take math in college next year I will probably learn about that methood

nope, thats it

thanks for the help

[void elm]

okay, then you can close this channel if you're done

[void elm]

[willow nebula]

oki

[void elm]

happy to help!

[willow nebula]

how cloes lmao

[void elm]

type .close

[willow nebula]

ah, ok

thanks again

.close

[marsh citrus]

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[brave sapphire]

can someone help me with 19 and 21? I missed those notes at I have a test tmrrw

[gilded smelt]

cool art! (I really like the figure upper left of 23)

cscx = 25/7 (but tanx < 0)

so sinx = 7/25 (but tanx is < 0, so cos is negative)

we need to find sin2x, cos2x and tan2x

1 = sin^2x + cos^2x, cos^2x = 1- 7^2/25^2 -> cos^2x = 576/25^2 -> cosx = 24/25 -> cosx = -24/25, since tanx is <0 and sin is positive

you got the negative signs wrong for sin then, I think

so, then, sin2x = sinxcosx and cos2x = 2cos^2x-1, tanx = sinx/cosx

@brave sapphire is that enough for you to solve 19? or do you need more help?

[brave sapphire]

sorry back

let me see

[brave sapphire]

yup you’re right oops

[brave sapphire]

can you explain this part a little more?

[gilded smelt]

I mean you got the cosx part right lol

but sure

pythagorean identity,
1 = sin^2 + cos^2
cos^2 = 1 - sin^2,
cos^2 = 1 - 49/25^2
cos^2 = 576/25^2
cos^2 = (24/25)^2
cos = ±24/25 (sin is positive, tan is negative, so cos is negative)
cos = -24/25

[brave sapphire]

where are you getting the cos^2 from?

[gilded smelt]

pythagorean identity, in the unit circle, the sides are sinx and cosx, and the hypotenuse (radius of the unit circle) is 1. Ergo, side^2 + side^2 = hyp^2 -> sin^2 + cos^2 = 1

[brave sapphire]

mhm

[gilded smelt]

so, where are you getting stuck? any of the specific functions or in general?

[brave sapphire]

im still not sure how to get sin2x, cos2x, and tan2x

this is going just a tad over my head rn

[gilded smelt]

hmm ok ok

so we have sinx, we have cosx, you understand all that yes?

now, to get the rest we do trig identities

tan(2x) will be easy, because it's sin(2x)/cos(2x), and we need to find those two anyways

sin(2x) is up first then

sin(2x) = 2(sinx)(cosx) in a well known trig identity (which can be awful to remember)

ergo, what is sin(2x)?

[brave sapphire]

yes sorry one moment

336/625?

[gilded smelt]

yup (I forgot the 2 in front of the sin*cos above 😭 you caught the mistake though)

cos(2x) has a similar identity, 2(cosx)^2 - 1

what is cos(2x) then?

[brave sapphire]

ok one sec let me do that

I got 1777/625

[gilded smelt]

you added the 1 not subtracted

[brave sapphire]

oooh

so right

would that be 527/625

[gilded smelt]

I do believe so!

whoops we have made a mistake ;-; in sin(2x) the cos(x) is negative, so the whole thing is negative too

[brave sapphire]

woo!

[gilded smelt]

so,
sinx = 7/25
cosx = -24/25
sin(2x) = -336/625
cos(2x) = 527/625

[brave sapphire]

isn’t it squared though?

[gilded smelt]

what is tan(2x) then

[gilded smelt]

for the cos(2x) yes it's squared

in sin(2x) it's 2 sin(x) cos(x) which is 2 * positive * negative = negative

[brave sapphire]

ok

👍

So tan2x is -336/527?

[gilded smelt]

yes!

Do you want more help for 21 or do you think you can do it?

you can always come back after trying yourself

[brave sapphire]

can you stand by (aka I just keep the channel open you can leave if you want sksks) while I try it, so if I get stuck I can ask for help

[gilded smelt]

of course!

[brave sapphire]

I have a question

with half angle indentities how can you figure out if it’s positive or negative

or is it both

[gilded smelt]

what do you mean?

like, how do you know if sin2x or cos2x are positive or negative?

[brave sapphire]

yes

maybe I don’t actually know

[gilded smelt]

you don't need half angle identities for 21, and it's the same way they can give you sin = 7/25 and you don't know if cos is +24/25 or -24/25 without more information

ie they tell you tan < 0, so since sin is > 0 that means cos < 0

you cannot know without them giving more info

[brave sapphire]

No im confused

Because it tells you to find sin(x/2), etc.

[gilded smelt]

whoops you are correct

I misread the question, assumed it was the same for both

[brave sapphire]

[gilded smelt]

actually in this case they give you exactly what you need to know

they say the angle is 90 ≤ x ≤ 180

[brave sapphire]

mhm

[gilded smelt]

so x/2 is therefore 45 ≤ x/2 ≤ 90

so x/2 is in quadrant 1, the top half of it, where sin and cos are both positive

[brave sapphire]

Ok

is this correct so far then?

[gilded smelt]

yup

sin = 15/17, tan = -8/15, looks right to me

[brave sapphire]

Ok so for cos I got this

I don’t know why I made it negative ignore that

alright it’s getting pretty late for me and I understand how to do the problems now, so thank you!!!!!!!!!!

you were a very big help

[gilded smelt]

no problem, glad I could help :)

also, I said it before, but I really like that art style! I don't see much like it anymore

[brave sapphire]

oh thank you so much!!!here are the full silly images <3 (/p)

good night! thank you again for your help!

.close

[marsh citrus]

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[marsh citrus]

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[meager badger]

How do i integrate this function>

i understand that its 5 + ∫(0 -> 3) of a(t)

But im having trouble integrating this

[still temple]

Seems like a calc problem

[meager badger]

yeah

[still temple]

Then put the integral

In ur calculator

[meager badger]

this is a no calc problem

[still temple]

some u sub

[meager badger]

but even with u sub

[still temple]

Im unsure what the u sub would be

[meager badger]

i end up with u^3 and ill still have a t variable

[still temple]

Yea

Wait

[twilit grove]

the antiderivative is not trivial; there's probably a more clever way to do it

[meager badger]

What other way is there?

[serene spoke]

No other way

Heu

Hey

I ve found a solution kind of can work

See first break the cube

I mean a³ + b³ formula

[meager badger]

so then t^3/2 + 1?

oh

[twilit grove]

well acceleration is positive, so it must certainly be > 5. rules out (A) and (B)

[serene spoke]

Wait few min imma figure out

[twilit grove]

acceleration at t=0 is 3. so velocity at t=1 will be 8 (approx)

and it keeps increasing, so that should leave you with one possible answer

[meager badger]

howd u get that?

[twilit grove]

if I'm understanding the problem correctly

just evaluate the acceleration at t=0

[meager badger]

and then howd u get that v = 8?

did u add what u got from a to v(1)

[twilit grove]

it's just an approximation. but if v=0 at the beginning, and the acceleration at t=0 is 3, you add to 5

[serene spoke]

11.710 ?

[meager badger]

yea thats what the ans sheet says

[serene spoke]

For actual solution you should figure out integral else

Approx works

V = u + at

U = 5

[meager badger]

Did u figure out the actual integral?

[serene spoke]

And considering a at 3 multiplying that gives around 8

Nah im trying to

[twilit grove]

it's not trivial

[serene spoke]

Yeah

[twilit grove]

,w integral (t + 3) / (sqrt(t^3 + 1)) dt

[elfin berry]

Results provided by WolframAlpha

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Requested by cwatson

[meager badger]

what does that mean?

[twilit grove]

it means you have to estimate, like I suggested

[serene spoke]

Hmm

Wait

See

√t³ + 1 > √t² +1

Does this holds ?

[twilit grove]

no

[serene spoke]

positive values only

=

[twilit grove]

still no

[serene spoke]

Why ?

[twilit grove]

plug in t=0.5

[serene spoke]

Oh yes

0-1

Then √t³ + 1 < t³ +1 ?

If the above holds ^ then we will have I >= (t+3)/(t³+1)

Which can be solved

[twilit grove]

still not easily

[serene spoke]

Bit solvable

Its easy by partial fractions

[twilit grove]

easy?

[serene spoke]

Yeah

[twilit grove]

,w integral (t + 3) / (t^3 + 1) dt

[elfin berry]

Results provided by WolframAlpha

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

Requested by cwatson

[serene spoke]

t³ + 1 = (t + 1 )(t² -t + 1)

Now you just need to seperate using partial fractions

So its the estimated f(n) and should give the result

Constant will be 5 +( 4√3/3)* pi/6

[marsh citrus]

@meager badger Has your question been resolved?

[marsh citrus]

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[still temple]

Uh oh its sideways

[glass silo]

,rccw

[still temple]

4 tan x - 3 cot x = 4?

[glass silo]

Will rotate it counterclockwise for you

[still temple]

Yes

[glass silo]

Also you may find it easier to get everything in terms of, say, tan

[still temple]

Hmmm

ye

Omg it worked

Thank you guys

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[harsh vine]

can someone please help me with this

[harsh vine]

not sure what substitution to use

is it tan?

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[lyric kelp]

how do i know if this converges

if it goes positive and negative

[mighty gyro]

it is 2 infinite gps

[whole compass]

We can use one

No need for two

[mighty gyro]

or yea

nvm

r is -3/4

[whole compass]

Do you know sum of infinite GP?

[lyric kelp]

idk what gp is

[lyric kelp]

ah

yeah i do know this ytes

[whole compass]

Here a is first term

First find a value

[lyric kelp]

9/16-27/64+81/256

these are 3 firts terms

but idk how i can find a and r if theyre switching betrween positive and negative

[whole compass]

r is negative

r.r is positive

[lyric kelp]

ahhhhhhhh

okay tysm!

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[shell hill]

how is this equal to 1/abs(x)

[frail orbit]

compute $\f{d}{dx}\left(\f{x^2-1}{2x}\right)$ and maybe it will be more clear

[elfin berry]

🫎 Moosey Mycobacteriophage 🫎 🧬

[shell hill]

$\frac{4x^{2}-2\left(x^{2}-1\right)}{4x^{2}}$

[elfin berry]

RulzerFly

[frail orbit]

mhm

[shell hill]

it's not getting any clearer

[frail orbit]

well

note that you get 1+() for that

can you write it as 1+()

[shell hill]

$1-\frac{2\left(x^{2}-1\right)}{4x^{2}}$

[elfin berry]

RulzerFly

[frail orbit]

mhm

now distribute the negative

so its a +

@shell hill

[shell hill]

$-\frac{2\left(x^{2}-1\right)}{4x^{2}\sqrt{1+\left(\frac{x^{2}-1}{2x}\right)^{2}}}$

[elfin berry]

RulzerFly

[frail orbit]

hmm

huh

[shell hill]

why ?

[frail orbit]

okay let's just back up to this step

and note we have $\frac{1-\frac{2\left(x^{2}-1\right)}{4x^{2}}}{\sqrt{1+\left(\frac{x^{2}-1}{2x}\right)^{2}}}$

then note we can write this as

[elfin berry]

🫎 Moosey Mycobacteriophage 🫎 🧬

[frail orbit]

$\frac{1}{\frac{1}{1-\frac{2\left(x^{2}-1\right)}{4x^{2}}}\sqrt{1+\left(\frac{x^{2}-1}{2x}\right)^{2}}}$

[elfin berry]

🫎 Moosey Mycobacteriophage 🫎 🧬

[frail orbit]

then you can bring the thing in the denominator into the square root by squaring it

[shell hill]

if you plug that into that sqrt it will be under abs and you can't simplify

because $\sqrt{x^{2}}=\left|x\right|$

[elfin berry]

RulzerFly

[frail orbit]

mhm

[shell hill]

any other idea

[frail orbit]

ok i've figured out a much simpler way of dealing with all the variable nonsense

[shell hill]

how ??

[frail orbit]

let u=(x^2-1)/2x

[frail orbit]

then try to write this in terms of u

[frail orbit]

you can get to this point, then realize this is $1-\frac{2\left(x^{2}-1\right)}{(2x)^{2}}$

[elfin berry]

🫎 Moosey Mycobacteriophage 🫎 🧬

[frail orbit]

so you can multiply numerator and denominator by (x^2-1) to get (x^2-1)^2/(2x)^2=u^2

then you can solve for x^2-1 in terms of u to get it even simpler

until you get 1-u/x

[shell hill]

yeah yeah thanks for help

[frail orbit]

eventually you get to this step $\frac{1}{\sqrt{\frac{\left(1+u^{2}\right)}{\left(1-\frac{u}{x}\right)^{2}}}}$ and you can substitute back for u from there

[elfin berry]

🫎 Moosey Mycobacteriophage 🫎 🧬

[frail orbit]

it actually simplifies very nicely :)

simplifies into 1/sqrt(x^2) which is precisely 1/|x|

[shell hill]

yeah i used a math app it gives me abs(x)/x^2

[frail orbit]

which is 1/|x|...

[shell hill]

how ??

[frail orbit]

sqrt(x^2)/x^2

sqrt(a)/a=1/sqrt(a)

so sqrt(x^2)/x^2=1/sqrt(x^2)=1/|x|

[frail orbit]

Let $u=\f{x^2-1}{2x}$. $1-\f{2(x^2-1)}{(2x)^2}=1-\f{2(x^2-1)^2}{(2x)^2(x^2-1)}=1-\f{2u^2}{x^2-1}=1-\f{2u^2}{2ux}=1-\f{u}{x}$

[elfin berry]

🫎 Moosey Mycobacteriophage 🫎 🧬

[marsh citrus]

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[spring stirrup]

how do i prove?

what are they doing here?

@sharp wing

<@&286206848099549185> s

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[dusk moth]

I don't know what to do

My main plan was this

Pi/8 (area of a half circle with radius 1/2) minus integral from 0 to 0.45 of 0.5(2theta)^2

Ik it's wrong bc of how polar integration works, BUT I don't know what's right

[marsh citrus]

@dusk moth Has your question been resolved?

[marsh citrus]

@dusk moth Has your question been resolved?

[marsh citrus]

@dusk moth Has your question been resolved?

[marsh citrus]

@dusk moth Has your question been resolved?

[outer vapor]

the upper part to the x-axis is
(0.5^2-(x-0.5)^2)^0.5 from 0 to cos2(0.45) = 0.3410
is that right?
but no clue how to determine the lower part r=2theta

[marsh citrus]

@dusk moth Has your question been resolved?

[outer vapor]

the r=2θ part go as follow

x = rcosθ = 2θcosθ
y = rsinθ = 2θsinθ

lower part Area
= ∫y dx
from here we need to find dx/dθ
= 2cosθ+2θ(sinθ)
and dx
= [2cosθ+2θ(sinθ)]dθ
so ∫y dx
= ∫2θsinθ [2cosθ+2θ(sinθ)] dθ from 0 to 0.45
and the upper part - lower part is the answer

the process should be right cause the answer addup

[marsh citrus]

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[dusk moth]

Ty

[dusk moth]

I didn't use the y =rsin(theta) identity for some reason 😭

But now ik

.close

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[wraith sphinx]

how to solve (vi)

It’s a problem of advanced algebra. And it’s too difficult for me

Can anyone please help me 🥺, deadline is today

[marsh citrus]

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[wraith sphinx]

not yet

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[jovial bear]

sorry dumb question vietas formulas and grouping in pairs aren't the same right? you can use either to find roots but the methods are different

[jovial bear]

like with vieta its through simultaneous to solve for roots but with grouping in pairs its through factoring right

.close

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[cunning fiber]

@fiery field #help-32 message

[cunning fiber]

I can’t rlly tell what you did

But you should’ve gotten this

Your final answer is wrong tho

But if you got the blue highlighted part then you just messed up your arithmetic

[marsh citrus]

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[leaden monolith]

#1229263782770114581 message

[marsh citrus]

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[dull yew]

can someone help me

[dull yew]

i don’t understand this step

[dull yew]

i did this

but i don’t get why he does (root3 +3)(3+root3)

@whole compass

[whole compass]

Hey

[whole compass]

.

See this

[whole compass]

You know (a-b)(a+b)?

[dull yew]

yea

[whole compass]

They do that in denominator

To get rid of square root

[dull yew]

oh ok

[whole compass]

That's y I sent this

[dull yew]

okay

then do i just foil it

?

@whole compass

[whole compass]

Yes

That's correct

What are u getting after simplification

[marsh citrus]

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[marsh citrus]

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[still temple]

Hi , i need help with solving this system

[novel juniper]

honestly I'd just graph 'em

if not find y in terms of x

square both sides and solve

but that will be a huge headache IMO

[still temple]

i just need x in terms of y

[desert socket]

why does the first equation look like cos(3x)

I smell trig

[novel juniper]

oo, good catch

yeah, that will be way better

[still temple]

that's the original equation

[desert socket]

defo trig sub

[novel juniper]

$x=sin(u)$ then

[elfin berry]

ƒ(Why am. I here)=I don't Know

[desert socket]

cos(u) I'd say though

[still temple]

what's the difference?

[desert socket]

tbh both will work the same

cos(u) is just a bit easier to see (IMO)

[stoic slate]

It is easier with cos(u)

[still temple]

i'll try it

can we solve it without using this substitution?

by system or algebraic manipulation

[marsh citrus]

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[marsh citrus]

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[obsidian kayak]

question 12

[vernal forge]

@novel juniper

[obsidian kayak]

<@&286206848099549185>

ahh leave it

.close

[marsh citrus]

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[marsh citrus]

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[cerulean oxide]

how am i supposed to find out if this is divergent or convergent

[serene spoke]

Intuition

[broken dome]

Did you try to calculate the first couple of terms?

Maybe there's a pattern

[cerulean oxide]

how can i deduce a pattern from 2 pieces of information

[broken dome]

Try calculating c3, c4, c5, c6

[cerulean oxide]

c_3 = 3/2*

[broken dome]

How did you get that?

[serene spoke]

C1,c2 givem

1+2/2

[broken dome]

That's also wrong

[cerulean oxide]

how

c_2 + c_1 = 3

/2

[broken dome]

Those brackets, are those supposed to be the floor function or did you crop the image weirdly?

[serene spoke]

Weirdly one

[cerulean oxide]

oh yeah i forgot about those

[broken dome]

ye but why put brackets at all

[cerulean oxide]

3/2 becomes 2 i guess

[serene spoke]

Floor fn ?

[broken dome]

if not for floor

[cerulean oxide]

i think it is floor

[broken dome]

It becomes 1, no?

[cerulean oxide]

1*

yeah

idk why i thought of it as 2.5

but yes 1.5 -> 1

[broken dome]

ok now calculate c4

What did you get?

[cerulean oxide]

also 1

[broken dome]

ok, now c5?

[cerulean oxide]

well

that will aslo be 1

1+1/2

[broken dome]

ok now can you tell me what c55555 will be?

[cerulean oxide]

1

[broken dome]

eyup

[cerulean oxide]

now you will just have 1

and the one before will be 1

1+1

/2

[broken dome]

eyup

[shut needle]

Hello

[broken dome]

aw lwitlwe spwidwer

[cerulean oxide]

that was surpringsingly easy

[broken dome]

Whenever you have problems like these, always try to check some values and see if it will get hung up in a pattern

[shut needle]

I'm looking for help

[cerulean oxide]

#help-14

@shut needle

[broken dome]

You need your own channel

use an available one

[shut needle]

I'm being sent around like crazy

[broken dome]

this one is occupied

[cerulean oxide]

is there a smart way to find cluster points?

[leaden flint]

check #❓how-to-get-help

[broken dome]

cluster points? what are those?

[cerulean oxide]

Let (an) be a sequence. The number a is said to be a cluster point of (an) if there exists a subsequence of (an) that converges to a. Formally, a is said to be a cluster point of (an) if for every ε > 0 there exists a natural number N such that |aφ(n) - a| < ε for all n ≥ N. Here, (aφ(n)) is a subsequence of (an), where φ(n) : ℕ → ℕ is a strictly increasing mapping.

[broken dome]

Ah those

[cerulean oxide]

yeah, not sure what the name is

[broken dome]

yeah english isn't my first language sorry

[cerulean oxide]

$$b_n = e^{i\frac{2\pi n}{5}} $$

if i had this

[elfin berry]

// 1800

[cerulean oxide]

i could probably use intuition to find the number, but what is the fool proof way

if there is any

[broken dome]

define i in this context?

[cerulean oxide]

how masny lcuster points

is there for b_n as n goes to infinity

[broken dome]

yeah but what's that i?

[cerulean oxide]

sqrt(-1)

i would assume 5

[broken dome]

ah so imaginary constant ,thought it might be something else

[cerulean oxide]

one trip around a circle is 2pi

so i assume it will be 5

[broken dome]

this is a complex number in euler form, no?

[cerulean oxide]

it is

you could rewrite it with sin/cos

might be smart actually

[broken dome]

I'd argue that for every n that's divisible by 5 you'd get the same value for this number

right?

Since it would land you on the same angle due to the period of sin and cos

best to rewrite in terms of angles for better visual

[cerulean oxide]

yeah, i think that is the idea

$$ b_n = \cos\left(\frac{2\pi n}{5}\right) + i\sin\left(\frac{2\pi n}{5}\right) $$

[elfin berry]

// 1800

[broken dome]

Right, so if n = {0, 5, 10, 15, 20, 25....} would this have the same value for all those numbers?

[cerulean oxide]

mhh

yeah

actually

0 and 5 would be equal

10 would only be pi, no?

[broken dome]

so would that value be a clustering point since there's infinitely many N's that have it?

10 would give you 2 * 10 * pi/5 = 4pi = 2 * 2pi

ahhh formating

[cerulean oxide]

yeah nvm

but i still think that there 5 cluster points

[broken dome]

aight, prove it

[cerulean oxide]

ok

1sec

import numpy as np
import matplotlib.pyplot as plt

n_values = np.arange(5)
b_n = np.exp(1j * 2 * np.pi * n_values / 5)

theta = np.linspace(0, 2 * np.pi, 100)
x = np.cos(theta)
y = np.sin(theta)
plt.plot(x, y, color='blue')

plt.scatter(b_n.real, b_n.imag, color='red')
plt.axis('equal')
plt.title('Cluster Points on Unit Circle')
plt.xlabel('Real')
plt.ylabel('Imaginary')

plt.show()

try this

[broken dome]

I meant a mathematical proof

like, try to see what happens when n = {1,6,11,16,21,26..}

[cerulean oxide]

oh

xd

idk if i can do that

[broken dome]

why not?

we already checked all numbers divisible by 5

so now let's check the other numbers

hint : there's 5 sets of em and they're all gonna end up being a cluster point

hint 2 : you can rewrite that set as being n = {0+1, 5+1, 10+1, ...}

hint 3 : are you even here anymore?

[cerulean oxide]

no

i am doing somethign

i will be back shortly

[broken dome]

[marsh citrus]

@cerulean oxide Has your question been resolved?

[marsh citrus]

@cerulean oxide Has your question been resolved?

[marsh citrus]

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[wind swallow]

cant get it the whole question

<@&286206848099549185>

[void lintel]

draw a line through the point c parelleling AD

[wind swallow]

yes

[void lintel]

then

you can point dot E on AB intersect with that line

[lime light]

DC=4 => AE=4 BCE= equilateral 9+4=13

[void lintel]

then

[lime light]

thought process

[void lintel]

because CD // AB => /_ DCE = /_ CEB

[wind swallow]

I only didnt get where did 9 + 4 come from

[lime light]

BCE is equilateral

all sides equal 9, so BE should equal 9

[void lintel]

sorry

I retry

[wind swallow]

yes then ?

[void lintel]

ce is half line

[lime light]

AE+EB = AB

[void lintel]

of BCD

[wind swallow]

Yes Im so dumb

I got it now

ty guys

[lime light]

ur not, if not for @void lintel i wouldnt have figured it myself

[wind swallow]

ty anyways

.close

[marsh citrus]

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[void lintel]

u r welcome

[lime light]

np

[marsh citrus]

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[mint sonnet]

help?

[normal vessel]

what does period of the function mean?

[mint sonnet]

i dont know how to define it in a cohesive way but its how long the graph will get to its original place on diff units?

[normal vessel]

oh

simple enough then

[mint sonnet]

so now what?

?

[normal vessel]

if it's trigonometric it always moves by the same distance

[mint sonnet]

but how do i find the period?

[normal vessel]

wait i made a mistake lmao

[mint sonnet]

?

[normal vessel]

16.9 for x

y goes back to -5.5

[mint sonnet]

x?

[normal vessel]

next x is 28.3

uhhh

[mint sonnet]

im confused

[normal vessel]

ohhh wait i get it now

the period is 33.8

[mint sonnet]

yea it is

but how

[normal vessel]

so the period is the distance that the function takes to reach back the next position

the function always moves the same distance in each position

[mint sonnet]

so minus max from minimum

and multiply by 2?

[normal vessel]

since it went from -5.5 to 11.4

the distance between that is 16.9

[mint sonnet]

what about this?

[normal vessel]

to go back to it's first spot will take 2 turns

which is 33.8

[mint sonnet]

wait

,calc 5.8-(-3.4)

[elfin berry]

Result:

9.2

[mint sonnet]

,calc 9.2*2

[elfin berry]

Result:

18.4

[mint sonnet]

is it 18.4?

[normal vessel]

i don't understand this

is 2/3 pi supposed to be X?

[mint sonnet]

yea but i dont think that matters

[normal vessel]

ok i found it

since the first point of appears to be 4pi/3 -2(2pi/3)

which is 0

then the period is 2(4pi/3)

which is 8pi/3

[mint sonnet]

yea but im on another problem now

ur smart

like this, since this is max do i subtract?

[normal vessel]

8.38

[mint sonnet]

i would do -2pi - (-1/2pi)?

and now multiply by 4

and get -6pi?

well 6pi

since it cant be neg?

[normal vessel]

yep

6pi

[mint sonnet]

r u sure?

[normal vessel]

yes

you can make sure with a calculator

[mint sonnet]

yea it was lol

can we go again?

[normal vessel]

-2pi + 6pi is 12.5

[mint sonnet]

like with this

well he amplitude is 5, right?

[normal vessel]

i think you need to get the middle of the -10.7 and -5.5 line

[mint sonnet]

wdym?

[normal vessel]

(-8.1/2) is the max of the function

[mint sonnet]

how do u know?

[normal vessel]

and you can basically do -8.1 - 4(-5.5) which is 13.6

therefore the period of the function is 13.6-(-8.1)

which is 21.7

[mint sonnet]

,calc 13.6-(-8.1)

[elfin berry]

Result:

21.7

[mint sonnet]

so period is 21.7?

[normal vessel]

the middle of that line is where the max of the function is

the half of the line between -10.7 and -5.5

[mint sonnet]

its not 21.7

its 10.4

[normal vessel]

wait tiny mistake

[mint sonnet]

im on another problem

lets see

[normal vessel]

the distance between -8.1 and -5.5 is actually 2.6

[mint sonnet]

ok so -7-(-7)

[normal vessel]

so the period is 4(2.6)

[mint sonnet]

,calc -7-(-7)

[elfin berry]

Result:

0

[normal vessel]

which 10.4

sorry

[mint sonnet]

0*2=0

so there is no period?

[normal vessel]

how did you even reach that conclusion

[mint sonnet]

so i used y axis i need to use x

lets do it again

[normal vessel]

it's the same way as the last one

[mint sonnet]

,calc pi-(2/3pi)

[elfin berry]

Result:

1.0471975511966

[mint sonnet]

so its 10/3 pi?

[normal vessel]

middle of the line between first midline intersection and second intersection is where the max of the function is

[mint sonnet]

i got 10/3 pi

[normal vessel]

the period is the distance between the max and the intersection x4

i didn't count it yet

gimme a sec

so the max of the function is -2pi/3 + 5pi/6

which is pi/6

[mint sonnet]

pi/6?

[normal vessel]

the distance between pi/6 and pi x4 is the period

[mint sonnet]

dont we multiply by 2 since its the midline?

[normal vessel]

which is 4x(5pi/6)

[mint sonnet]

huh?

for the period i got 10/3 pi

[normal vessel]

ye that's it

5pi x 4/3

[mint sonnet]

lets try this

[normal vessel]

where are you getting all these anyways

[mint sonnet]

how do we do this?

[mint sonnet]

hw

[normal vessel]

this is litteraly the most basic one

[mint sonnet]

do we find midline?

is it pi?

period?

[normal vessel]

pi/4 + 2(2pi/4)

which is 5pi/4

the distance between pi/4 and 5pi/4 is the period

which is pi

[mint sonnet]

its pi period?

[normal vessel]

wait a sec

i didn't see it right

pi/4 + 2(3pi/4) is the second max

or you can just count the distance between max and min then multiply it by two

[mint sonnet]

its noy pi?

[normal vessel]

it is pi

[mint sonnet]

period?

[normal vessel]

distance between pi/4 and 3pi/4 is 2pi/4

multiply that by 2 and you get pi

[mint sonnet]

yep

lets try this one together

i think i know how to

so

so

[normal vessel]

this one is easy

i found a more efficient way

[mint sonnet]

,calc 1.2-5.2

[elfin berry]

Result:

-4

[normal vessel]

distance between min and half

[mint sonnet]

,calc -4*4

[elfin berry]

Result:

-16

[normal vessel]

multiply that by 4

[mint sonnet]

its 16?

[normal vessel]

ye

[mint sonnet]

sure?

[normal vessel]

ye

[mint sonnet]

ok last one

ok i think i know

so

,calc -5/4pi-(-7/4pi)

[elfin berry]

Result:

1.5707963267949

[mint sonnet]

,calc 1.5707963267949*2

[elfin berry]

Result:

3.1415926535898

[normal vessel]

most efficient way is to count the distance between them and multiply it by 2

[normal vessel]

thats pi

[mint sonnet]

so its pi?

thats the period?

[normal vessel]

ye

[mint sonnet]

yep

thx

[normal vessel]

np

[mint sonnet]

.close

[marsh citrus]

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[static pendant]

I have a math problem, and I don't know how to begin.

The function f(x) =- (x - 3)^2 +9 can be used to represent the area of a rectangle with a perimeter of 12 units, as a function of the length of the rectangle, x. What is the maximum area of the rectangle?

The answer choices are
A. 3 square units
B. 6 square units
C. 9 square units
D. 12 square units.

[still temple]

calc question?

[still temple]

draw the shape of $-x^2$ what is the highest point?

[elfin berry]

Book Reader

[still temple]

where does that highest point go when you transform it into $-(x-3)^2 + 9$

[elfin berry]

Book Reader

[static pendant]

Algebra

[marsh citrus]

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[peak arrow]

hii, i cant figure out why my answer is wrong. the tasks asks you to make an algebraic expression that shows how many pieces you’ll need to make figure number n. here’s my work:

[peak arrow]

please help 🥲

I thought the answer was 10n^2 + 14n + 4

[sharp sequoia]

How many pieces are there in the first few diagrams?

[peak arrow]

f1: 32 pieces
f2: 67 pieces
f3: 115

I THINK, im going to count again just in case 🥲

I think that’s right ^

[sharp sequoia]

Well just looking by inspection this looks incorrect for the n=1 case.

[peak arrow]

oh

could you explain why

<@&286206848099549185> haven’t received a response in the last 15 minutes so

please help 🥲

[sharp sequoia]

Sorry, I was busy looking at other channels.

[peak arrow]

no worries sorry

[twilit grove]

I'm not too sure, but it might be easiest to have a different equation for each section of the insect

[peak arrow]

ah yes I did do that

[twilit grove]

so the antennae will be 2(n + 1)

[sharp sequoia]

So basically try plugging in n=1 into your equation of 10n^2+14n+4

[peak arrow]

wdym

[sharp sequoia]

You would see that your answer would be incorrect

[peak arrow]

yes I did that

[sharp sequoia]

Since you counted it as 32

Ok.

[peak arrow]

that’s how I knew it was incorrect

[sharp sequoia]

Now let's work on the approach.

I would agree with @twilit grove 's approach of dissecting such insect into three sections.

[peak arrow]

I did try to do that

let me show

so the green parts: 2n+2
grey parts: n(n+1) / 2 * 6 + 1
yellow/black parts: (n+1)(n+2) / 2 * 6 + 1 + (n+1)(n+2) / 2 - 1

@sharp sequoia there sorry

[sharp sequoia]

Did you try combining the like terms?

[peak arrow]

like simplifying it?

[sharp sequoia]

Supposing your equations above are correct, yes.

[peak arrow]

yes

in the last picture you can see my attempt at doing it

[sharp sequoia]

Let me just check if your equations are correct

One second.

Sorry do you mind sending a vertically inverted image?

I kept twisting my head hahaha

Just to see the work.

[peak arrow]

oh yea of course

ignore this part

[marsh citrus]

@peak arrow Has your question been resolved?

[marsh citrus]

@peak arrow Has your question been resolved?

[marsh citrus]

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[marsh citrus]

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[karmic ore]

<@&286206848099549185>

[proud lance]

and then when ur done u can turn x = a + bi, into cis form and ur done and then take the ref angle

[karmic ore]

Is this for number 16?

[proud lance]

yep

[karmic ore]

Is this to solve a only? Or a and b?

[proud lance]

u cna solve for all the equations

[karmic ore]

My teacher said we only have to solve 16 a

[karmic ore]

Ooo

Alr alr

[proud lance]

its just a little bit creative too think of wut equations to use

k im leave if u have nay question jsut ping helpers again

[karmic ore]

Ok

Thanks

[marsh citrus]

@karmic ore Has your question been resolved?

[marsh citrus]

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[still temple]

it is known that if polynomial p(x) is divided by x^2+2x+3, the remaining would be 3x-14. As for (x-1), the remaining would be 1. What remains of p(x) if its divided by (x^2+2x+3)(x-1)?

[still temple]

I suggest you to try making equations of the conditions given in the problem, so you will have:
P(x) = Q(x)(x^2+2x+3) + (3x-14)
P(x) = Q'(x)(x-1) + 1
P(x) = Q''(x)(x^2+2x+3)(x-1) + Y(X)
and we are looking for Y(X)
(not that by Q'(x) and just mean other polynomial , it is not a derivative or something )

[still temple]

already did, after that what?

I still dont solve it , but it looks like you may need a good factorization idk
But it seems that you must to begin with
P(x) = Q(x)(x^2+2x+3) + (3x-14)
P(x) = Q'(x)(x-1) + 1
and then do the "mathematical magic" and then get something like this P(x) = Q''(x)(x^2+2x+3)(x-1) + Y(X)
that's what I am looking for

[marsh citrus]

@still temple Has your question been resolved?

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[still temple]

ill skip the question for now i think, gonna try it again later or ask my teacher about it

[marsh citrus]

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[buoyant pawn]

this is the solution