help-36

[copper roost]

but

-pi/2 to pi/2 isnt a full period

like -pi/2 period should end at 1pi

oh my bad

my mistake sorry

[final saddle]

@bleak edge Has your question been resolved?

[faint edge]

Gonna test close this

.close

[final saddle]

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[solar summit]

which triangles are similar

.open

?

hello

[errant estuary]

can someone help me

[torn pond]

anyone know how to do this? i can do everything besides the intervals and x int

[ivory vessel]

please don't multipost

[ivory vessel]

what have you tried?

[ivory vessel]

spit it out

[solar summit]

I’m just like

Overwhelmed by everything on it 😭

Idk how to approach

[ivory vessel]

tell me, how do you check whether 2 triangles are similar?

[final saddle]

@solar summit Has your question been resolved?

[river oasis]

help

[solar summit]

many ways

wdym

[ivory vessel]

such as?

[solar summit]

angles are the same

[bronze vector]

I need help with 6-108

[plain kestrel]

The equation should be x(2x+5)

Which should be multiplied out to be 2x^2 + 5x

Set that equal to 50 and solve as a quadratic

[vital kettle]

For this Q, are the boys and girls themselves distinguishable? I presume so: I get n! * n! for the favorable outcomes, and 2^(2n) for the total outcomes. But the math doesn't simplify out correctly. Where did i go wrong?

[final saddle]

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[crimson ermine]

Alexa wrote the first $16$ numbers of a sequence:
[1, 2, 2, 3, 4, 4, 5, 6, 6, 7, 8, 8, 9, 10, 10, 11, …]Then she continued following the same pattern, until she had $2015$ numbers in total.
What was the last number she wrote?

[soft zealot]

Human

[crimson ermine]

for this problem i first noticed that every 5th term has a difference of 3

[tired walrus]

should we understand that these are the natural numbers in order except that every even number is repeated

[crimson ermine]

yea

[tired walrus]

what do you mean by "every 5th term has a difference of 3"?

im afraid that if you look at the 3rd term (2) and the 8th term (6) they do not differ by 3

[crimson ermine]

4 is a fifth term

and 1 is the first term

so 4-1 3

7 is the 10th term

7-4 3

[tired walrus]

did you miss two equals signs

[crimson ermine]

is this correct?

[tired walrus]

anyway you're saying that $a_5 - a_1 = 3$ and $a_{10} - a_5 = 3$

[soft zealot]

Ann

[crimson ermine]

yes

yes

[tired walrus]

these two statements are correct but somewhat incoherent

[crimson ermine]

is that wrong?

[tired walrus]

less wrong, more unhelpful

[crimson ermine]

but like

[tired walrus]

positions 1 and 5 are not the same distance apart as 5 and 10 btw

[crimson ermine]

doesnt this mean that every (multiple of 5)th term can be written as 1 + 3(a), when the number which multiplies 5 is a

[tired walrus]

no it doesn't

it fails for the 20th term. the 20th (5*4) term is not equal to 1+3*4

you are off-track

[crimson ermine]

ohh

[tired walrus]

like you are right that the sequence has this sort of steadily increasing behavior but you are wrong about its rate

[crimson ermine]

how should i approach this problem

[tired walrus]

it makes much much more sense to look at terms by threes

because you've got groups that look like 2n-1, 2n, 2n

in fact you can see that $a_{3n} = 2n$ always

[soft zealot]

Ann

[crimson ermine]

so there are 671.5 groups of 2 even and 1 odd number?

[tired walrus]

671.5?

[crimson ermine]

2015/3

[tired walrus]

2015/3 is not equal to 671.5 that much i can tell you for sure

[crimson ermine]

oh 671.67

[tired walrus]

there are 671 groups of three, and then two numbers left over at the end.

[final saddle]

@crimson ermine Has your question been resolved?

[crimson ermine]

is that the answe?

672?

[tired walrus]

no it's not the answer, and no the answer isn't 672

[crimson ermine]

why is the answer 1344

[tired walrus]

you tell me. how did you get 1344?

[crimson ermine]

answerkey

[tired walrus]

k

[crimson ermine]

why

[tired walrus]

do you understand why $a_{3n} = 2n$

[soft zealot]

Ann

[tired walrus]

maybe to be a bit more informative, $a_{3n-1} = a_{3n} = 2n$

[soft zealot]

Ann

[crimson ermine]

yea

[tired walrus]

yeah ok so

the closest-from-below multiple of 3 to 2015 is 2013 (=3*671)

thus the 2013th term is 2*671

[crimson ermine]

ohh so the next term is 1343 and then 1344

[tired walrus]

yes

[crimson ermine]

.close

[final saddle]

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[timber plume]

hey I’m unsure why I got everything correct except the answer was in negative

[timber plume]

,rotate

[severe verge]

mistake at 3rd step in left column

[timber plume]

I redid the question and got it incorrect again

[timber plume]

@severe verge spot the mistake in here

[terse crypt]

x - (x+h)

[terse crypt]

Distribute carefully

[timber plume]

So I did this part correct?

@terse crypt

[terse crypt]

U must

Combine the fractions

Before cancelling

[timber plume]

I don’t understand

I did combine the fractions

[timber plume]

Should I combine it separately on the side of the paper, not in the limit formula it self

.close

[final saddle]

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[glossy zephyr]

if i correctly understood you, thats basically a vector field composition

also, its usually easier to understand functions, is to check from where they map from and where they map to

[empty nest]

can you kindly elaborate on that in this case

[empty nest]

yes

[glossy zephyr]

lets per say that you have a field
F(x,y), where x and y are vector components that both depend on some functions x(u,v) and y(u,v)
this whole thing, in reality just traces from (u,v) to a transformation in (x,y)

R^2 -> R^2 basically.

If u and v are in reality vectors themselves, lets say they depend on some functions u(i,j) and v(i,j)
they still map from R^2 -> R^2

since for each ordered pair (i,j) there is a single F(x,y), just with intermediate steps, and more over, you could replace the x's and y's from F(x,y) into F(u,v) and then F(i,j)

[royal tangle]

I think you are you confusing maps and their compositions, you should write each map with their domain to clear up this confusion, and explicitly write implied compositions

[empty nest]

oh i understand it my bad

i was a bit confused

thank you

do i just .close

.close

[final saddle]

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[plain marsh]

so... hmm... i have a past exam file containing 6 questions that the teacher gave to us yesterday

And the look i gave after i read it is like
"how tf do i pass this..."

[plain marsh]

i need to know how to do them... and whooooo boy do i not know how to do them </3

[gritty drift]

I mean make an attempt first then someone will correct you, like I'm sure you can produce an educated guess for 1a

[plain marsh]

sure thing!

i think this is it? Idk how to present it 😭

[terse crypt]

f’(5) IS the instantaneous rate of change of the bacterial population with respect to time after 5 hours

[terse crypt]

The idea is correct yes

[plain marsh]

NICEEEE

[gritty drift]

You couldve just typed it out but yeah that's the gist

positive f'(5) means bacteria population is increasing at time = 5hr and negative f'(5) means the opposite

with the magnitude represemting how fast it's increasing or decreasing

[plain marsh]

Ohhhhh... huh... never realized how that works

[final saddle]

@plain marsh Has your question been resolved?

[vital harness]

the wire one is interesting, I'm going to try it

@plain marsh i've just had a look, and I can do those

So if you need help, feel free to ping

[plain marsh]

Ohhh thanks!

I am currently stuck in the rain and can't get to my laptop </3

[vital harness]

oh

well, I'll let you have a think about these problems

[plain marsh]

It really rain at the worst possible time

[vital harness]

and then come back to me when you have an idea, alr ?

at least for the first question

[plain marsh]

Yep!

[final saddle]

@plain marsh Has your question been resolved?

[plain marsh]

1b)
For unlimited space and nutrient, f'(10) will be larger than f'(5) since the rate of change will exponentially be faster
But for limited, i think f'(10) rate of change will be less than f'(5) since f'(10)'s rate of change will slow down, depend on how much space... and nutrient... there is...

oh crap... derivative

i forgot-

[vital harness]

yep, I believe that is correct

[vital harness]

mmh... you must know it though

(f o u)' = u' (f' o u)

[plain marsh]

AUGHHHHH...
Cries
im cheating a bit with this by using the definition of derivative

[vital harness]

thta's the general fomula

[vital harness]

well thta's like redemonstrating it

[vital harness]

here

f o u = 1/u

and u(x) = x^2

so (f o u)'= (x^2)' * (1/u^2)

[plain marsh]

i am... hesitant to ask why it is like this-

[vital harness]

it's ok

[vital harness]

it comes from this

$d(f o u(x))/dx=du/dx*df/du$

not formatting well

um

do you know what o means ?

@plain marsh

[plain marsh]

um.... noooo...

[vital harness]

rght

it means composition

like

[soft zealot]

Silloky

[plain marsh]

huh...

[vital harness]

is composition of the sqrt function

and the x^2 function

right ?

[plain marsh]

yep yep it is

[vital harness]

👍

thta's what o means

if you prefer

let f and u be functions

the derivative of this composed function

is the derivative of the inner function

times the derivative of the outer function applied to the inner function

so

[plain marsh]

so like... (f∘u)(x) = f(u(x)) ?

[vital harness]

yeah ! that is correct

so

[soft zealot]

Silloky

[vital harness]

and

[soft zealot]

Silloky

[vital harness]

right ?

[plain marsh]

ohhh i remember thisss

[vital harness]

g : x |-> x^2 |-> 1/x^2

say g is your overall function

the first arrow is the u function

and the second is the f

u good ?

[vital harness]

this is overly complicated for the question

you won't finish your exam if you do this

hello ?

you there ?

[plain marsh]

helllo-

[vital harness]

oh

[plain marsh]

i am indeed here 😭

[vital harness]

ok 👍

[vital harness]

knowing this

what do you think g'(x) is ?

(sry i mixed up the g and f from the original photo ; their f is my g)

[plain marsh]

damn- 😭

[vital harness]

so

[soft zealot]

Silloky

[vital harness]

ok ?

[plain marsh]

yuh huh...

[vital harness]

what's the inner functon ?

?

[plain marsh]

uhhhhhhhhhhh

[vital harness]

i have literally written it above

[plain marsh]

I am kinda... very stupid-

[vital harness]

perfect

that is exactly right !

well done$

though you mixed up f and g

[plain marsh]

aw damn

[vital harness]

but the result is correct

[plain marsh]

niceeeeeeee

[vital harness]

your first line

g= f o u

and g(x)= 1/x^2 = f(u(x)) where f(u)=1/u and u(x)=x^2

other than that, it's correct

the second ^hotograph is easier if you want to try

[plain marsh]

ok thanks!

[plain marsh]

do you know how to do the... whatever question 2 is-

[vital harness]

yeah !

just for info, in what grade are you ?

i have no idea, I'm from France

[plain marsh]

uhhh... first year uni!

i am so far behind <3

[vital harness]

oh

ok

well, umm... good luck catching up 😄

[vital harness]

so

a derivative is only defined where the function is defined

you can't derivate a function where it doesn't exist

nor where it isn't continuous

can you find those points on the graphs ?

[plain marsh]

I helping a friend with something rn, will be back in around 5-10m!

[vital harness]

alr

i'm doing some physics anyway

[plain marsh]

Oh shit you're doing physics too!?

[vital harness]

yep

got an oral presentation to do for my Baccalauréat (= A level = end of high school test)

i'm doing it on solar sails

[plain marsh]

Yooo that's sick

[vital harness]

thx

yeah i got to prepare 2

one in physics

and one in maths

maths is using spatial geometry for rank aggregation

[plain marsh]

Zamn you are.. smart-

[vital harness]

hello ?

[plain marsh]

hello..? 😭

[vital harness]

standard nightingale behaviour

@plain marsh would rather stay on the server ; that ok with you ?

[plain marsh]

that is very ok

[final saddle]

@plain marsh Has your question been resolved?

Channel closed

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• After 15 minutes, feel free to ping <@&286206848099549185>.
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[atomic tinsel]

the dynamic frictional constant between two surfaces is 0.4 if the object slides freely what is the maximum vertical displacment it can achive in second inclined plain

[final saddle]

Please don't occupy multiple help channels.

[vital crag]

!1c

[final saddle]

Please stick to your channel.

[atomic tinsel]

but i closed it

[atomic tinsel]

can you help me btw

[vital crag]

Yes after I told you

[atomic tinsel]

okay then?

[magic coyote]

Just admit you were wrong, why do you want to engage

[old quarry]

i think work energy theorem

[wary coral]

What about getting it's acceleration and getting the velocity and the bottom point

[atomic tinsel]

how to apply it

[wary coral]

Get the acceleration of the block

[old quarry]

first state the theorem

[atomic tinsel]

work done= work spent for friction?

[old quarry]

no state the actual theorem not for this question

[atomic tinsel]

idk

[old quarry]

total work done = change in kinetic energy

[wary coral]

Is the answer 1m

[atomic tinsel]

no

it cannot be

[old quarry]

no..friction is acting how will it still reach the same height

[wary coral]

Wait

[old quarry]

bro either u help him or let me help him

[wary coral]

What about 3/7

[atomic tinsel]

yes

[atomic tinsel]

How did you get that answer?

[wary coral]

Ok so it is simple find the net acceleration of the block when it is in the first triangle (initial one)

[atomic tinsel]

then?

[wary coral]

Find the velocity of the block when it is in the lowest point

[atomic tinsel]

how to find that velocity

[wary coral]

Use v^2 = u^2 + 2ax

[atomic tinsel]

[wary coral]

Where u and v are initial and final velocity and x is the displacement and a is the net acc

[atomic tinsel]

how did you get a answer for x and u

[wary coral]

Since the block starts from rest the initial velocity is 0

[atomic tinsel]

ok

then how did you get that displacement

[wary coral]

Wait is the answer I gave correct?

[atomic tinsel]

yeah i think it is approximate to the answer in answer sheet it is 0.43

[wary coral]

Ok so using trigonometry get the length of base

Which is the x displacement

[atomic tinsel]

idk

it not given

[wary coral]

Ok we know the height is 1m

[atomic tinsel]

but in our sceniro it is the distance from starting point to lowest point isn't it?

[wary coral]

Yes

[atomic tinsel]

how did you obtain that

[wary coral]

use trigonometry and get the displacement

we know sin x = length of opposite side / length of hypotheuse

[atomic tinsel]

okay you mean x*sin 45=1?

[wary coral]

Yes

[atomic tinsel]

okay then?

[wary coral]

Substitute the values in the eqn

And get the velocity of block at the bottom point

[atomic tinsel]

after that?

[wary coral]

Do the same for the second one

Find the acceleration of the block on the second triangle

[atomic tinsel]

but isn't we need to consider the impulse when it hit bottom?

[wary coral]

Hmmm

I didn't think of it

[atomic tinsel]

what did you assumed in second inclined plain

[wary coral]

The final velocity of the block at first plane is the initial velocity of block at second plane

[atomic tinsel]

okay then we apply the motion equation again ,then?

[wary coral]

You fill get the block's displacement in the second plane

[atomic tinsel]

okay got it

[wary coral]

Using trigonometry like before find the vertical height

[atomic tinsel]

okay thanks for the help

.close

[final saddle]

Channel closed

Closed by @atomic tinsel

Use .reopen if this was a mistake.

[vital harness]

@plain marsh whagt about the other questions ? you good with them ?!

[plain marsh]

AHHHH NO I AM NOT BUT ITS A BIT TOO LATE

[vital harness]

oh ok

[final saddle]

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[spare moth]

how to do this?

[spare moth]

i didnt get the part u crossed out, how did u get π - sin inverse √3/2

[loud sundial]

$\sin(x)=\sin(\pi-x)$

[soft zealot]

Civil Service Pigeon

[loud sundial]

$0<\alpha+\beta<\pi$, and since the first quadrant case didn’t work, we consider the second quadrant case

[soft zealot]

Civil Service Pigeon

[loud sundial]

Which comes from this

[steep spruce]

Alright then

[spare moth]

am i supposed to memorise all these? theres soo many

[loud sundial]

A decent number of them follow from each other so eh

I think all I have learnt off is the periods of sin, cos, tan and whether they’re odd or even

And the cofunction identities

The rest is quick to derive once you know that

[spare moth]

that gives me ans c but the ans is a ?

[loud sundial]

,w find sin(x)+cos(y) if sin(x+y)=(sqrt 3)/2, cos(y)=1/2, 0<x<pi/2, 0<y<pi/2

[soft zealot]

Results provided by WolframAlpha

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

Requested by civil_service_pigeon

[loud sundial]

Congrats you have a bad answer key

[spare moth]

damn 💔😭 okay thank u though also can u tell me what u mean by this

[steep spruce]

Try to understand why they’re correct instead of memorizing

[loud sundial]

Ex. tan(3pi/2 - theta) = tan((3pi/2 - theta) - pi) = tan(pi/2 - theta) = cot(theta)

[spare moth]

okay now that went over my head 💔

[paper kite]

memorizing that is a bad idea

[spare moth]

then how do u suggest

[paper kite]

for example,

this is the sign of sin

between 0 and 180 it is positive

sin(pi + theta) = -sin(theta) because

sin(180 + any number) gets you to the negative section

not any number necessarily, sin(180 + theta) will get you to the negative section

[spare moth]

ok thanks :- ) i understand this method now but i kinda dont get how to use it like over here in the question 60 fails bc alpha > 0 so what thought process am i supposed to use after like how will i know to get 120

if that makes sense ?

[paper kite]

what's the question

[spare moth]

here

[paper kite]

and where are you stuck at?

[spare moth]

we're saying that a + B cant be equal to 60 right so the person said to find the next value bc a cant be 0 but how do i know which identity to use to get the next angle

😢

[paper kite]

how'd you even come up with the
a + b = 120

did u not expand (sin a + b) = sqrt(3)/2?

[spare moth]

a + b = sin^-1 (√3/2)

i put 60 bc thats the only value i remember but it doesnt work bc a cant be 0 bc b is already 60

[spare moth]

no its a mcq exam that's too long of a method

[paper kite]

🤷‍♂️ not familiar with that method

[spare moth]

<@&286206848099549185> if anyone else can help. ...

.close

[final saddle]

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[naive shoal]

In number 3

[naive shoal]

What causes the sqaure root to make a vertical syrech

Stretch*

[tardy void]

use property of logs

[naive shoal]

Oh yea

Shot

[tardy void]

You can write square root as 1/2

and bring it outside the log

and compare

[naive shoal]

Okay thanks

[final saddle]

@naive shoal Has your question been resolved?

[final saddle]

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[spare moth]

how do i do this

[spare moth]

i only know this is going like (cosx)(cos2x)(cos3x)(cos4x)

[pale niche]

id use the fact that $\cos(x) = -\cos(\pi - x)$

[spare moth]

i thought of that but idk what to do with it

[pale niche]

do you know how to evaluate $\cos\left(\f{\pi}{5}\right)$?

[spare moth]

i only know how to do cos for like 0,45, 60,30 so on and can also break like cos 15 -> Cos (45-30) Et c

[pale niche]

okay cool, so you are aware of the cosine doubling identity

and cosine addition

so you can repeat $\cos(2x) = 2\cos^2(x) - 1$ with $x = 36^{\circ}$

[pale niche]

until you get to a known value

in this case it's prob 180

[spare moth]

wont that make it more complicated .. cos (2 x 36 ) = 2cos^2(36)-1

[marsh whale]

you can multiply and divide by sin(pi/5)

do the algebra and can you see what to do with the cos(3pi/5) ?

[spare moth]

but whats the reason for doing that

[marsh whale]

okay so basically 2 reasons :
3pi/5 can be changed into 8pi/5, so you have the cos of the angles of 1,2,4,8 multiples of pi/5 so that suggests some sort of reduction with sin(2x) formula
the options are related to powers of 2

ig best reason I can give you is to just try it out

[spare moth]

oh lord what is this

i hate maths 💔

[fossil ruin]

See just multiply and divide by 2sin(pi/5), as 2sinA.cosA = sin(2A), then keep multiplying and dividing by 2, to get this formula

[spare moth]

ohh wait i get what ure saying let me try it

[haughty prairie]

tanθ+sinθ=m tanθ-sinθ=n
Then prove secθ=√mn cosec^2 θ

[spare moth]

wait so if i multiply/divide cospi/5 with 2sinpi/5 do i do cos2pi/5 with 2sin2pi/5

[marsh whale]

!occupied

[final saddle]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[marsh whale]

yes cospi/5 and sinpi/5 will combine to get a sin(2pi/5) and you continue combining that with cos(2pi/5) , cos(4pi/5) etc

[fossil ruin]

See when u multiply and divide the first term, which is cos(pi/5) with 2sin(pi/5), u will already get sin(2pi/5), which can be further combined with cos(2pi/5) by multiplying and dividing by 2, this will continue, and at last you will get cos(8pi/5) on numerator and denominator, then after cancelling them, the answer will be 1/16..... make sure to convert cos(3pi/5) into cos(8pi/5)

[spare moth]

okay got it thanks!

[spare moth]

thanks gojo satoru lol ...

.close

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[left trail]

@onyx peak hi I'm so sorry for the late reply I was working on it and didn't check discord. I think I have made some progress but not done yet. Also it wasn't letting me use the old help channel

[left trail]

I just figured it out. Thanks again for all the help

.close

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[wind flame]

can someone please help me out on this question?

[primal sierra]

!status

[final saddle]

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

[wind flame]

3

[primal sierra]

what did you get for your answer

if you have any work done then plz send a pic aswell

[wind flame]

1 sec im trying to find it

[primal sierra]

it should be -6x - 8y = -204, not -202

[wind flame]

oh ok

thank you

[gilded trench]

X=25

[wind flame]

could you also help me with this? i got stuck on b and c. this is the work ive done.

[gilded trench]

Dawg wtf is that💔🥀

[wind flame]

idk bro 💔

[gilded trench]

English is like my 69th language so lemme read

[faint fern]

Just equate them?

Solve for x

[wind flame]

wait what this what i get ifi do that

[faint fern]

You evaluated 3x + 2x/3 wrong

Try again

(Referring to the bit after 3x + 1 = ...)

[wind flame]

im sorry im a bit slow

where is that on the paper.

?

wait but wouldnt that still equal to 5x/3

bc 3x/3+2x/3=5x/3

[gilded trench]

I deadass hate linear sh

Ts pmo

[wind flame]

fr bro

i still dont understand

how am i wrong

[gilded trench]

I know one sec

[digital basin]

Hello, where do I ask for math questions?

[gilded trench]

This is how ts looks (not 100% accurate)

[faint fern]

$$ 3x + \frac{2x}{3} $$

[soft zealot]

StrangeQuarkAL

[gilded trench]

Lemme do the other

[wind flame]

ohhhhhhhhh

[faint fern]

Not $$ \frac{3x}{3} + \frac}2x}{3} $$

[soft zealot]

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

[wind flame]

that makes complete sense

[faint fern]

Yeah you get the idea

[wind flame]

ty bro

[faint fern]

Np

[wind flame]

im sorry im slow i thought it said over 3

[faint fern]

Just a mistake

It's all good

[gilded trench]

Imma js go to sleep it’s 23:00 in CET good luck I was once in ur shoes bro you got this

[wind flame]

wait

[wind flame]

ty bro

now i got 5/11=x

is that wrong

nvrmind i re did it

i got x=-1

ty for ur help guys im relying on u to get above 90 tom

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[naive shoal]

How come for 4, all of my lines are liners

Can't find a max and min with this

[rapid sky]

You need to zoom out.

For instructions: https://mathclasscalculator.com/calculator-tutorials/how-to-adjust-the-graph-window-on-ti-84-plus/.

[naive shoal]

How far man

I alr tried that

It says the max is 3, 14

I just zoomed out to 200 by 200 and cannot see a max

3, 14 should fit 😭

[rapid sky]

Ahh, cos(2pi/5) is a constant.

You need cos(2pi/5 (x + 3)) with that stuff all inside the cos parentheses.

[naive shoal]

How come it ain't written like that on my paper

[rapid sky]

Well, in computer systems, we generally have cos(all the stuff to take the cosine of).

In writing, we can have cos all the stuff to take the cosine of.

Like cos(x) vs cos x.

[naive shoal]

For my previous questions I didn't put it all in brackets 😭

For example, for 1 I just put the x in brackets

And the +1 was outside of the sin

[rapid sky]

Oh, that's correct.

[naive shoal]

So how come we just do it now

[rapid sky]

With writing, it's a bit inexact.

Like that means 2 sin(x) + 1.

[naive shoal]

Yea

[naive shoal]

So how do I write 4a in my calc

[rapid sky]

If you had 2 sin x + pi/2, you'd have 2 sin(x + pi/2).

The difference is that pi/2 is usually an angle or part of it, so it goes inside the trig function.

1 isn't usually an angle, so it goes outside.

[naive shoal]

So only if it has a pi under it 💔

Stupid asf

[rapid sky]

(y = 6 \cos\qty(\frac{2\pi}5 (x + 3)) + 8)

[soft zealot]

Chai T. Rex

[rapid sky]

x is usually used as an angle in these.

And 2pi/5 is an angle.

But 8 isn't.

[naive shoal]

That is Messi g up.my brain

It makes no sense e

[rapid sky]

It takes some getting used to.

[naive shoal]

That sucks for my final

Was tryna ace it

But I can't just remember dumb shit like this

At least it's multiple choice so I'll kinda know if something's off

[rapid sky]

If it shows up as a line when you graph it, you probably forgot to put x or theta inside the cos's or sin's or whatever's parentheses.

[naive shoal]

Oh yea

The x has to be inside

That's a good way to remember

[rapid sky]

Like (6 \cos \frac{2\pi}5 (x + 3) + 8)

[soft zealot]

Chai T. Rex

[rapid sky]

The 2pi part is an angle.

So, it goes inside the cos.

The x is usually used as an angle.

So, it goes inside the cos.

And since it's in parentheses (with (x + 3)), the whole parentheses come inside.

But 8 isn't really an angle when you use radians.

[naive shoal]

This look good?

[rapid sky]

Yes.

It's good to learn this stuff now.

One because it'll help you on your upcoming test.

Another because later math courses will write it without parentheses, and you'll have to know how to do it.

[naive shoal]

Good to know

This wasted like an hour of my time bro 😭

Tysm

[rapid sky]

It just takes some getting used to, so don't worry too much about not being able to learn it.

[naive shoal]

Lemme just make sure I got it tho

[rapid sky]

No problem.

[naive shoal]

Alright I got it

.close

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[timber plume]

Could someone explain me the basic rules of differentiation (rule 3 and rule 4), please dumb it doown becuase ive been really struggling with this concept since yesterday

[faint plover]

any constant multiplied will always stay and wont be affected by differentiation

lets say f(x)= 4x-2

and g(x)= 2x-1

f(x)=2g(x)

deriving both does the same look

f'(x)=4
g'(x)=2

f'(x)=2*2 wich is 4.

its correct

as for the second rule

(f(x)+g(x))'= f'(x)+g'(x)

you can distribute the derivation if there are many things

[robust mulch]

note the derivation/deriving is different from differentiation/differentiating

when talking about derivatives, you should use differentiating, not deriving

kinda annoying rule, but it avoids confusion

[faint plover]

lets say

f(x)= 4x -x^2 + x

f(x)= 4 -2x +1

[faint plover]

thanks

i dont study maths in english so i kinda mess up names all the time

[final saddle]

@timber plume Has your question been resolved?

[final saddle]

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[lethal drum]

hi

[lethal drum]

how do i calculate this

the effective capacitance

[robust mulch]

this isnt in situ in a circuit

it depends on where your have your current flowing in or out

[final saddle]

@lethal drum Has your question been resolved?

[pale niche]

i heard the ash guy was an expert on circuits

[lethal drum]

this is all it says

[robust mulch]

ok, i guess you just consider all 4 in series

[lethal drum]

arent the top and bottom ones considered paralell?

theyre right ontop of each other

[robust mulch]

the way im interpreting it is calculate the effective capacitance of the loop

[final saddle]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[weary sun]

For this

a.) Show that the magnitude of the electric field 10.0cm from an isolated charge $q_1=-10.0μC$ is $9.00\times10^{6} NC^{-1}$.

Easy

b.) State with reason, the distance from the charge at which the electric field would have a magnitude of $3.60\times10^{7}NC^{-1}$.

How do we use proportionality to show b.)

[tired walrus]

electric field strength follows the inverse-square law does it not?

[soft zealot]

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

[weary sun]

yeah i give up

[weary sun]

i don't know what that is

Like I know E is proportional to 1/r^2 if thats what you mean

But how can I apply that

[tired walrus]

correct

that's what i meant with inverse-square law

[weary sun]

What I know is that a magnitude of $3.60\times10^{7}NC^{-1}$ is 4 times the amount of the original 9.00 quantity thing

[soft zealot]

giordan

[tired walrus]

$E \propto \frac{1}{r^2}$

[soft zealot]

Ann

[weary sun]

ye

[tired walrus]

so if E increases by a factor of 4

what must happen to r

[weary sun]

Yeah thats the problem

I was thinking about it and then I got confused

Becuase I'm usually good with proportionality but I just have no idea what would happen

Hmmm

[tired walrus]

r would have to decrease by a factor of two.

[weary sun]

Why is that exactly

[tired walrus]

$E_1 = \frac{k}{r_1^2}$ and $E_2 = \frac{k}{r_2^2}$

[soft zealot]

Ann

[tired walrus]

and we also know $E_2 = 4E_1$ as you mentioned

[soft zealot]

Ann

[tired walrus]

$\frac{E_2}{E_1} = \frac{kr_2^{-2}}{kr_1^{-2}} = \frac{r_1^2}{r_2^2} = 4$

[soft zealot]

Ann

[weary sun]

Why is there no q here

do they cancel out

[tired walrus]

it's folded into the proportionality constant k

[weary sun]

oh ok

Because we usually write $\frac{1}{4\pi\epsilon_{0}}$

[soft zealot]

giordan

[weary sun]

multiplied by q

But yeah that makes sense

Ok so

[tired walrus]

doesnt matter

[weary sun]

Yeah

Sorry I'm just trying to comprehend your work

[tired walrus]

like literally all the physics absolutely doesnt matter, this is just working with the inverse square proportionality

[weary sun]

What do we do next from here

[tired walrus]

$\paren{\frac{r_1}{r_2}}^2 = 4$

[soft zealot]

Ann

[tired walrus]

we've found the ratio between the old and the new distance

r1/r2 = 2

[weary sun]

so r_2 is half r_1

Ahhhh

Thats crazy

ok thank you so much

i understand

but is there an easier way to do this

because the worked solutions literally state it, so i feel like there may be an easier way to do this at a quicker pace

ok

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• Show your work, and if possible, explain where you are stuck.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[tender osprey]

Pardon but this is for a friend I know nothing about this

If you could just explain everything in one message it would be convenient

[vital crag]

The fuck

Tell your friend to join to ask.

[tender osprey]

Ok sir

Can I @ him?

[frosty kiln]

Sup

I’m in

[tender osprey]

This is him

[frosty kiln]

Wait can I curse?

[tender osprey]

Idk

Bye

[frosty kiln]

Alr

[tender osprey]

Do I have to close this and make him open one?

[frosty kiln]

Idk if it matters

If it gets solved it gets solved

[final saddle]

@tender osprey Has your question been resolved?

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

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[storm haven]

Help why are the underlined ones true??

[lime crest]

and i thought my question was bad

[severe verge]

oh my god

lemme take the time to study this

[lime crest]

i can kinda understand why f is surjective but why injective

[ashen pond]

The bijectivity comes from the fact it is a permutation ig?
But it’s kind of a circular reasoning

[lime crest]

'observe' and you gotta have a third eye to see it

[severe verge]

I got it

[ashen pond]

Same

[storm haven]

Lol yes

[ashen pond]

Do you need a proof? Like mathematically correct? Or is an explanation fine?

[storm haven]

Explanation

[severe verge]

if f(x1) and f(x2) are the same value then f f(x1) (x1) and likewise has the same value when applied the first time and you apply the function the same amount of times

[storm haven]

What on earth is "blah blah are the same then the first time then blah blah"

No offense I don't understand your english

"Then the first time then"

[severe verge]

my brain f up

[storm haven]

"Then then"

Ok wait let me see

It's more readable now

[severe verge]

I got too excited sometimes lol

[storm haven]

What is ur notation bro could u like write it instead?

F(x1) (x1)

[severe verge]

idk how to latex lol

[storm haven]

Oh nooooooo

[storm haven]

This is not logical

[severe verge]

lemme write it on paper

[lime crest]

$f^{f(x_1)}(x_1)$

[soft zealot]

Copter

[storm haven]

Ohh

[lime crest]

oh shit that worked

[storm haven]

He wrote it as f(x1) (x1) just now 💀

Wait I missed the f before it

Yeah there was double f

[lime crest]

i think this is what he meant lol

[severe verge]

ye

[storm haven]

Yup well why is the function injectove tho?

I still don't see it

[ashen pond]

If f were not injective, the cycle structure would break as some differen elements might map the same element
If f were no surjective, some elements wouldn’t be in any cycle, so the equality could not hold for those
So overall it has to be bihectirve to make the orbit length conditions make sense ig

[severe verge]

same case for x2

[storm haven]

Wait wait

Let me reas

Oh wait that makes sense

Okok

Ok, but there's more underlined ones

Why must d divide f(n)

I understand it has to divide f^f(n) (n)

[ashen pond]

The mapping is cyclic, so every element is a shift of its former one

Then, because the function is a cycle of length d and the mapping just shifts the position within the cycle, this divisibility is kinda pushed back across the cyclic

Got it? Idk another way to say that

[storm haven]

"Shift"? What do you mean about that

[ashen pond]

Like pushed one position further

[storm haven]

Huh what

I don't get it

[ashen pond]

Ok

F cycles through the elements in a loop with length d

If f(n) is divisible by d, by repeatedly applying f, the divisibility properties stay for all cycle elements

Because f just moves you around the cycle

[severe verge]

ooh I get it now

it's a bit abstract but once you think about it all makes sense

[final saddle]

@storm haven Has your question been resolved?

[lime crest]

gotta love the help channels where we're all confused on the same qn

[ashen pond]

Idk if they is not understanding my explanation can someone else try it I have no other way to put it

[severe verge]

I can draw a diagram

[storm haven]

so, you are saying : if f(n) is divisible by d, so is f(f...f(n))...)?

i dont see how

[storm haven]

pls

[severe verge]

having another problem in progress lol

I'm also reading other papers

[storm haven]

oh ok

<@&286206848099549185>

[lime crest]

guhh i cant help but i dont think thats how it works😭

[storm haven]

what really???

[lime crest]

iirc its like one helper ping per channel?

[storm haven]

ohh shoot

[lime crest]

yeahh i was confused too at first

[storm haven]

😭 my ques hasnt been answered

[lime crest]

sadge

maybe you'll have better luck in the olym maths discord network?

[storm haven]

just the last few underlined statements

[storm haven]

that kind of server exists??

[lime crest]

yea yea its specifically for olyms

ill send you

[lime crest]

uhh how many have you done now

do you have d | f(n) => d | n?

[storm haven]

nah i still domnmt get that

[lime crest]

awhh

[storm haven]

do u know how??

[lime crest]

uhh ill try my best

[lime crest]

check dms btw

[severe verge]

I got that but uhhh one thing I don't get

[storm haven]

alr

[storm haven]

??

[storm haven]

just realized i also dont undersatnd this

[severe verge]

I don't think from that we can get f(n) divisible by d

[severe verge]

ah

[storm haven]

uhm...

i dont understand hehe

im so stupid i dont understand why d divides any thing

[storm haven]

okay i get it so far

[lime crest]

i think i can do if d | f(n) then d | f^{f(n)}(n) but not the opposite lmao

[severe verge]

I know that if f(n) is divisible by d then n has to be divisible by d

[storm haven]

eh show how

[severe verge]

but how do I get f(n) divisible by d

[storm haven]

prob we can reverse engineer somehow?

[storm haven]

like we try to reverse engineer copter's method

i cant believe the solution they gave is so shady

they just give the facts as if they are obvious

[severe verge]

,rccw

[storm haven]

what??

[severe verge]

if f(n) is divisible by d

Then f^f(n) (n)=n

then n also has to be divisible by d

[storm haven]

wait, that makes sense

[storm haven]

this, i dont understand.

wait oh yeahhh

that makes sense

[severe verge]

well n after d steps will come back to n right?

[storm haven]

yup i get it

[lime crest]

since $f^{d}(n) = n$ for all $n \in C$ the original condition in the question implies$\$ $f^{f(n)}(n)$ must be a multiple of d bc n turns to itself after $f^{f(n)}(n)$ applications of f, and d is the minimum such exponent, so then $d|f^{f(n)}(n)$

[soft zealot]

Copter

[lime crest]

dunno if this makes sense

[storm haven]

oh yeah but why does d divide f(n)

[severe verge]

ye that's the question

[lime crest]

i supposed d | f(n) and then that implies d | f^f(n)(n)😭

[storm haven]

i know tthat d | f^f(n) (n) and d | f(n) => d|n

[storm haven]

oh yeah

wait we need to see how tthat could be reversed

[lime crest]

ehh similar logic i think

[storm haven]

wait i dont really get it 😭

[storm haven]

you didnt use d|f(n) here? or did i just now see it?

[severe verge]

they didnt show it

[lime crest]

yea

[storm haven]

wdym?

wait i feel so stupid

where is the d|f(n) statement?? 😭

[lime crest]

since C is a cycle of length d then f^d(n) = n for all n in C

or no?

[severe verge]

that's true

[storm haven]

but that is different from d|f(n)?

[severe verge]

yea

[storm haven]

so like, i mean, @lime crest said he used d|f(n) to get d|f^f(n) (n) but i dont see the d|f(n) statement (or anything equivalent)

[lime crest]

i think im stupid hold on

hmpf

[storm haven]

nah i think its just me who is too blind to see the d|f(n) statement 😭

[severe verge]

at least we know that at least one element in C has to be divisible by d

[storm haven]

what?

is that f^f(n) (n)?

[severe verge]

because C after d steps repeats right?

[storm haven]

yeah?

[severe verge]

so that must appear in C

[storm haven]

yup exactly

[severe verge]

but having that doesn't help so far

[storm haven]

yeah

but they used it later on in the proof

everything about this problem is so stressing

WAIT I SEE SOMETHING

[severe verge]

yes?

[storm haven]

we dont have to proof that d|f^f(n) (n) => d|f(n)

cause all they said is that IF d|f(n) then f|n

nothing said that d|f(n) must be true

its just inprovable

umprovable

nonprovable?

idk

does that make sense tho?

[severe verge]

doesnt make any sense to me

[storm haven]

uhm

[severe verge]

ik

[storm haven]

tthey didnt write blah blah => d|f(n)

they just wrote d|f(n) => d|n

so, all i need to know now is : Why does d | some element of C => d | all elements of C?

[severe verge]

this is so confusing

[storm haven]

lol just try to comprehend first

doesnt it make sense?? Its an IF d | f(n) not a d | f(n) MUST BE TRUE

[severe verge]

you see that n is just some random element in the loop C

[storm haven]

yea?

oh wait i think i see it

yeas

[severe verge]

if n leads to f(n) divisible by d, it must be true for every other point in the loop

so every element in the loop is divisible by d

[storm haven]

yup i get it now

i just need to review the solution as a whole

[severe verge]

dw inequalities will make you feel hopeless

in terms of approach

[storm haven]

"dw"? whats that

[severe verge]

don't worry

[storm haven]

oh lol

im so living under a rock

[severe verge]

I skipped nearly every hard inequalities in exams

[storm haven]

well thanks @severe verge and @lime crest and @ashen pond

.close

[final saddle]

Channel closed

Closed by @storm haven

Use .reopen if this was a mistake.

[severe verge]

and @ashen pond

[storm haven]

oh yeah

[final saddle]

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[soft zealot]

∲ビジョン∲
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[final saddle]

Channel closed

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If you did not intend to do this, please open a new help channel,
as this action is irreversible, and this channel may abruptly lock.

[vocal gale]

wouldn’t these results still hold if the region where the Hessian isn’t zero is connected?

[magic coyote]

.

[final saddle]

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Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[forest pilot]

Using the definite integral, calculate the area of ​​the bounded figure
curves y = x2 + 2x; y = 3x 4 and the y-axis (draw a picture!)
https://www.desmos.com/calculator/3jjrkgp3j0

[forest pilot]

now do idk what i have to caculate in this

[tired walrus]

this one i think

the area of this region

your copypaste was borked but at least you had the desmos thing

[forest pilot]

yeah thats what i was thinking but then when i ploped the question into ai it was also caulate the x[-4 ,-1]

[tired walrus]

!nogpt

[final saddle]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[forest pilot]

yeah i know i js use it to check my answers

[tired walrus]

no

don't

[forest pilot]

so i form he question i js have to get that area right since it aligns with all the paramotors

[storm haven]

@forest pilot u know integrals right

[forest pilot]

English not my first bro '='

yes

[storm haven]

Ok so what have u tried?

!status

[final saddle]

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

[forest pilot]

no i did the area js wasnt sure if i understood the question right cause ai bs was also caculating the area form x[-4,-1]

[tired walrus]

please do not call me "bro"

[forest pilot]

ok sry wouldnt do it again

[tired walrus]

you don't need it from -4 to -1. that region does not border the y axis

[forest pilot]

yeah thats what i was thinking js wasnt sure

is there any way of fact checking if u did the question right?without ai

[tired walrus]

no

do the graphing and stare at the graph

no way around it

no ai

forget ai

[storm haven]

Yeah, don't trust AI

X is between [?, 0]

Right?

What is ?

[forest pilot]

-1 i understand

[storm haven]

Ok, then take integral

[forest pilot]

i know how to read grafs its js i didnt understand if the question wanted me to also get the area of -4 and -1

[storm haven]

Wdym?

Between -4 and -1?

[storm haven]

Oh what are u struggling on?

It's a pretty simple integral

[forest pilot]

idk how to hightlight in desmos

[storm haven]

It's fine

What are you struggling on??

[forest pilot]

nothing my prob is solved i been trying to tell u :/

[storm haven]

Oops

Okok

[forest pilot]

but how do u hightlight in desmos

[storm haven]

What r ur equations?

Tell me first

I'll tell u how to highlight

[storm haven]

(The quadratic and line)

Ok, u type : -3x-4 < y < x^2 +2x

Try to type it and look

And maybe add another restriction

-3x-4 < y < x^2 +2x {x<0}

Get how to highlight areas now? Just use inequalities

If you want, use \leq

[forest pilot]

\leq?

[tired walrus]

$\leq$

[soft zealot]

Ann

[tired walrus]

<=

[storm haven]

Yeah sorry

[storm haven]

Noo

Type {}

Use the brackets

{x<0}

[storm haven]

Yep

That's how basically

You should close the help channel if u have no more ques

[forest pilot]

but i didnt wanna hightlight that part tho

[storm haven]

Which part?

The one Ann highlighted?

[forest pilot]

A2