help-13

[dire geode]

see #❓how-to-get-help

[buoyant sand]

mb vro 😭

i was giggling too much at ur response to not respond

[jagged tide]

he was right tho

[buoyant sand]

not what he asked tho

but ok

[jagged tide]

quazi's doubts are unclear

[quiet needle]

?

[jagged tide]

"why pi"

is that what you wanna know?

[split ice]

i would argue that its the other way around tbh, that one of the definitions for determining pi is "it's one of the zeroes of sin(x)."
if you are asking why sin(n*pi) = 0, that comes down to drawing a "degenerate triangle" that sits on top the x-axis, and noting that opp/hyp = 0/1 at, say, 1*pi

[jagged tide]

yeah

[quiet needle]

Do you know of any non geometric proof?

[split ice]

proof of what, specifically? that sin(n*pi) = 0?

[quiet needle]

Yeah, why specifically n*pi is the lowest and highest points in cos(x) and why it's 0 at sin(x)

[azure apex]

wym proof? the sine function is defined to be like that

[azure apex]

do u know the unit circle definition of the trig functions

[quiet needle]

Sin(x) = x-x³/3! + x⁵/5!... Why is x=nπ the point where it's 0?

[azure apex]

bc the sine function has zeros at x=npi

[quiet needle]

Yeah, but why?

Why exactly π

[jagged tide]

alright lets start it over

[azure apex]

Bc it is defined to be like that

[jagged tide]

begin from what sinx is

[azure apex]

do u know trig ratios on the unit circle

https://tenor.com/view/matematica-gif-7322246

[jagged tide]

thats where it originates

the y co ordinate of a point on an unit circle whose radius vector makes an angle theta with positive X axis is defined as sin theta

[buoyant sand]

u want an explanation on why subbing in x = npi into the taylor series expansion of sinx will produce 0?

[jagged tide]

now you can easily see why sin npi=0

[quiet needle]

Thank you 🙏

[buoyant sand]

specifically in the taylor expansion, right?

why do the terms cancel out

[quiet needle]

Yes 🙂‍↕️

[jagged tide]

Goodluck

[buoyant sand]

the taylor series is defined aas

$\sin (x) = \sum_{k=0}^{\infty} (-1)^k \frac{x^{2k+1}}{(2k+1)!}$

[wraith dagger]

Flatus

[quiet needle]

I know that

[azure apex]

once u see that sinx = x-x^3/3!+x^5/5!-... theres really no point in explaining why it should be 0 at npi 😭

you should know this if u know the taylor series

[jagged tide]

holy shit man forget the taylor series for now

[azure apex]

fr

[jagged tide]

begin with why sin npi=0

[quiet needle]

Ok, thanks for the help 😊🔫

[buoyant sand]

😭

[jagged tide]

hey that emoji is advertising self harm

we dont promote that here

[buoyant sand]

why r u guys such trolls LMAO 😭 🙏

[quiet needle]

Oh no 😱

[jagged tide]

hey im not "trolling"

💔

anyways

[quiet needle]

🤨

[jagged tide]

@quiet needle pls close the channel if you are done

[quiet needle]

What If I'm not?

[jagged tide]

well

[buoyant sand]

yeah idk
I have no clue how to prove that sin(npi) = 0 from the taylor expansion 🥀

$\sin (n\pi) = \sum_{k=0}^{\infty} (-1)^k \frac{(n\pi)^{2k+1}}{(2k+1)!}$

[wraith dagger]

Flatus

[jagged tide]

im crying

[buoyant sand]

😭

[quiet needle]

What's wrong lich?

[jagged tide]

its because $\sum_{k=0}^{\infty} (-1)^k \frac{x^{2k+1}}{(2k+1)!}$ CONVERGES to sinx for all x in R

[wraith dagger]

Lichtbach

[buoyant sand]

like here
you'd normally go from the right to the left and then know that sin(npi) = 0

[jagged tide]

and since sin npi=0

so yeah here we are

[quiet needle]

🙂👍

[jagged tide]

make a flow chart

define sinx by circular methods

[quiet needle]

Bro I know what sin(x) I just want to know why it only works with radians 💀

[jagged tide]

then notice that (sinx)'=cosx and (cosx)'=-sinx

[jagged tide]

what?

[quiet needle]

Why the Taylor expansion only works with radians

[jagged tide]

??

[buoyant sand]

uh

cuz degrees are made up

:D

they're kinda arbitrary

[jagged tide]

its because sinx/x goes to 1 only for radians

[buoyant sand]

like

basically

we made up degrees

cuz they're easier to look at

so like

they dont work

😭

[jagged tide]

thats why if we differentiate sinx, we get cosx only for radians

[buoyant sand]

does that make sense lwk @quiet needle

[jagged tide]

flatus thank you for your help ngl ts was frying my cells

[buoyant sand]

😭

[quiet needle]

I mean I guess, but the only real answer I've gotten is that sin(x) is defined like that.

[jagged tide]

Yeah

I mean thats how it exactly works

[buoyant sand]

is this about the radians?

[jagged tide]

you define the function, differentiate it

and find the mclaurin series

and boom

[split ice]

the fact that n*pi are the maxima/minima of cos(x) follows from the cofunction identity, and using the fact that sin(x) = 0 at n*pi.

to discuss sin(x) purely nongeometrically like this, i.e. in terms of its taylor series only, is not very helpful in my opinion. sin(x) is a very geometric function that is based on the unit circle, which is where pi comes from. there's not really a better explanation than this, because everything else (i.e. the construction of the taylor series) hinges on the fact that you have computed sin(pi) and sin(2pi), and understood that sin(n*pi) will always be equal to one of those 2 values.

[jagged tide]

mhm

this

[azure apex]

wow we still on this

[split ice]

yeah well, yall kept yapping and i had smn to say so

[azure apex]

bruh

[split ice]

jokes :) you guys are here to help of course <3<3

[quiet needle]

Okay I guess, I haven't really gotten an answer but I don't think I will either 🫤

.close

[azure apex]

https://www.youtube.com/watch?v=uu2nJEbap80

[cedar kiln]

Channel closed

Closed by @quiet needle

Use .reopen if this was a mistake.

[buoyant sand]

is this about the radians or the taylor series

aight

[azure apex]

bruh

[azure apex]

watchi this vid to learn abt how the sine function works ig

well the trig functions in general

[buoyant sand]

idm explaining the radians but the taylor series is cooked

like 😭

[cedar kiln]

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[azure apex]

fr

[azure apex]

oh shit

[buoyant sand]

oop

.close quick before anyone sees

[azure apex]

what do i do 💀

[buoyant sand]

or give us a math question

[azure apex]

.close

[cedar kiln]

Channel closed

Closed by @azure apex

Use .reopen if this was a mistake.

[buoyant sand]

lmaoo

[cedar kiln]

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[final bronze]

basic arithemetic

i feel stupid

[final bronze]

this is my calculation

somehow the answer is 0.12

[indigo lagoon]

Do you just want to simplify the thing on the top?

[final bronze]

the correct answer somehow

[final bronze]

no i wanna know what i did wrong

HOW DID I GET 1080 AND HE GOT 0.12

[indigo lagoon]

This is absolutely correct

[final bronze]

so this is wrong?

@indigo lagoon

[indigo lagoon]

I don't know the orignal question, might as well post it here so we can review your entire work to find the mistake

[final bronze]

oh like the physics question?

[indigo lagoon]

Solely this, absolutely correct, flawless

[indigo lagoon]

yes

[final bronze]

Two point charges q1 and q2 are located at (0,0) and (4,3) cm. Q1 is 10 mew c and q2 is 30 mew c. Need to find magnitude and direction on q2 by q1. MY ANSWER IS 1080, THE CORRECT ANSWER IS 0.12 N.

BUT SEE HE STARTS OUT WITH RHE SAME VALUES

[indigo lagoon]

Ah freak, electronic-related physic

Let's wait for another helper

[final bronze]

OKAY

[sullen cipher]

Magnitude of force between these two?

[final bronze]

thats what we are supposed to find out

the formula goes 1/4pi epsilon q1q2/r^2

the value of 1/4pi epsilon is 9×10^9

[sullen cipher]

Do u have the original question

As stated

[final bronze]

this is it

[sullen cipher]

I got it

[sullen cipher]

No other formal format?

[final bronze]

nope, its written in my nb

this pictures all i really got @sullen cipher

[indigo lagoon]

,rccw

[sullen cipher]

They forgot to do the calculation of denominator

[final bronze]

who is they

[final bronze]

this is my nb

[sullen cipher]

Whoever wrote that

[final bronze]

as in these guys

[final bronze]

which picture

i have send multiple

[sullen cipher]

This

[final bronze]

oh and you are sure right? Cause in multiple notebooks its the same answer as it was copied from the board

[sullen cipher]

Yes cuz the( 5×10^(-2))² disappeared from the calculation

[final bronze]

EYYY alright THANK YOU SO MUCH MAN

[sullen cipher]

Still lemme calculate

[final bronze]

the physics server takes so long to answer 😭

[sullen cipher]

It is 1080

[final bronze]

THANK YOU

I WILL BE BACK

[sullen cipher]

It's not as much active as this

[final bronze]

.close

[cedar kiln]

Channel closed

Closed by @final bronze

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Please don't occupy multiple help channels.

[crimson delta]

<@&268886789983436800>

[cedar kiln]

Channel closed

Channel closed due to the original message being deleted.
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.

[spice pivot]

free money! wow!

[cedar kiln]

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[dull oxide]

,rcw

[tough finch]

In A26., the answer key says only A but I can't understand why it can't also be D

[royal zinc]

Which grade are you

[tough finch]

How does that matter

[spiral fog]

arccos is a function it doesnt have 2 values

[royal zinc]

trying to assess your thinking ability

no freaky stuff🙏

[tough finch]

But cosine of both 0.75 pi and 1.25 pi give -rt2/2

[royal zinc]

Its ok if you uncomfortable mentioning it tho

[spiral fog]

,w plot arccos x

[wraith dagger]

Results provided by WolframAlpha

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

Requested by casework

[spiral fog]

This is how the function looks

[tough finch]

I see

[royal zinc]

while arccos could have had two values but range of cos inverse is set from zero to pi as cos is not an invertible function

[tough finch]

Ohhhh

[royal zinc]

so there could be an interval where tan would have been negative

but it is not taken

as arccos has its range

cut short

[tough finch]

No but that is the interval taken

3pi/4 is 2nd quadrant

Tan is negative

[hexed badger]

It’s similar to x² having two x values (+ and -)
But sqrt(x) only has one value (only +)

[tough finch]

Alright I got it y'all thanks

[royal zinc]

didnt quite get you?

yeah exactly tan is negative

but if arccos hadnt have its range cut short

[tough finch]

I meant to say that in the interval [0, pi] for the value of x which is 3pi/4 tan is negative

[royal zinc]

there could have been

a value for which

[tough finch]

yes I got it

[royal zinc]

angle would have been in 3rd quadrant

and then

tangent of it would be negative

[spiral fog]

arccos x is defined to be an inverse of cos x , 0 <= x <= pi

[royal zinc]

to avoid confusion the range is made zero to pi

[last apex]

more like to make it a function

[tough finch]

If the range wasn't 0 to pi tho, it wouldn't be a function right?

[royal zinc]

my bad, i was trying to make him understand

[royal zinc]

yup

[tough finch]

ok thanks 👍

bye bye

.close

[cedar kiln]

Channel closed

Closed by @tough finch

Use .reopen if this was a mistake.

[spiral fog]

I wonder if wolfram has branches

,w plot arccos_{-1} X

No nvm

[cedar kiln]

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[broken sentinel]

Idk where I went wrong

[livid hound]

you didn't copy down the question correctly

[inner pendant]

You’re supposed to multiply by $4^{-1}$ not add

[wraith dagger]

za

[broken sentinel]

So then would you multiply 0 to the x^3/2

Cuz aren't all derivatives from constants zero

[plush blade]

$4^{-1}$

[wraith dagger]

Arezu

[broken sentinel]

So then what would be f'(x) 4^-1

[inner pendant]

I think it’s more clear if you rewrite it as $-6x^{-1}-\frac{3}{4}x^{3/2}$

[wraith dagger]

za

[inner pendant]

There’s no reason to factor out the 4

[broken sentinel]

Aight thanks

Also this one

I lowkey don't even know where to start

Like I don't even know what it's asking

[dire geode]

slope of f at a point c is given by f'(c)

find f'(x) first

[broken sentinel]

Ok I did that

[dire geode]

solve 5 = f'(c) for c

[broken sentinel]

I don't get it

Where does c come from

.close

[cedar kiln]

Channel closed

Closed by @broken sentinel

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[lyric bough]

im trying to solve for equilibrium with the pivot point on 10lbs, but i keep getting 60’ as my answer which shouldnt be possible

[lyric bough]

thank you

i just wanna learn how to properly solve it so i can finish the other 3 problems after

how do i go about solving it correctly? these are the problems im going off of that we did in class

[cedar kiln]

@lyric bough Has your question been resolved?

[lyric bough]

did she use ai for that response?

[livid hound]

probably. sorry that you were subjected to that

hopefully someone else more qualified (not me) will be able to assist you

[lyric bough]

i hope so too

<@&286206848099549185>

[cedar kiln]

@lyric bough Has your question been resolved?

[sacred anchor]

,rotate

[sacred anchor]

do you have the original diagram?

[cedar kiln]

@lyric bough Has your question been resolved?

[sacred anchor]

the first diagram isnt in equilibrium

[lyric bough]

ik thats why im trying to solve it but i cant figure out where i went wrong

[cedar kiln]

@lyric bough Has your question been resolved?

[lyric bough]

could someone check my work?

[coral briar]

I need someone who can teach me trignometry in zoom

[fossil dawn]

!occupied, sorry. also a better question to ask in #study-discussion.

[cedar kiln]

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).

[cedar kiln]

@lyric bough Has your question been resolved?

[flint cape]

@lyric bough Wdym "❌"?

Two people have already checked your work here; do you have anything further regarding this question?

[lyric bough]

yeah 2 people were here but they didnt help me find if ny answers were right or not and hot to fix them if they were right

[digital bane]

they put a checkmark on your message though

[flint cape]

Like, what did you think this response meant?

[indigo lagoon]

They put checkmarks, meaning that your work has been reviewed and deemed correct by two people.

[lyric bough]

oh my bad i didnt see that

[indigo lagoon]

Do you have any other question aside from this one?

@lyric bough

[lyric bough]

that was all

[indigo lagoon]

Alright, don't forget to close the channel before leaving

[wicked mantle]

.solved

[cedar kiln]

Channel closed

Closed by @wicked mantle

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[steep steeple]

diff eq question, found both sides to be exact through partial derivitation

[steep steeple]

i got stuck on the next piece though

[cedar kiln]

@steep steeple Has your question been resolved?

[ancient lodge]

\textit{differentiation}, but \textit{deriviation}. \ \

However, for a differential equation of the form $M(x,y) \dd{x}+N(x,y) \dd{y}=0$, you need to check if $M_y=N_x$, not if $M_x=N_y$. \ \

Once you fix this, solving the differential equation should be a lot more straightforward.

[wraith dagger]

Civil Service Pigeon

[steep steeple]

gotcha

[cedar kiln]

@steep steeple Has your question been resolved?

[steep steeple]

i redid the equation, hopefully its more accurate

[ancient lodge]

,w (3x^2 y+2) dx+(x^3-1) dy=0

[wraith dagger]

Results provided by WolframAlpha

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

Requested by civil_service_pigeon

[ancient lodge]

Answer checks out

[steep steeple]

yay

thanks for the feedback!

.close

[cedar kiln]

Channel closed

Closed by @steep steeple

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[cedar kiln]

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[lethal lintel]

I thought it was just 0

because you would get 0, -3, and 4

from x, (x-4), and (x+3)

but it can't be -3 because there's a VA there

and it can't be 4 because that's another VA

[lethal lintel]

and anyways I tried giving all of these as an answer and it was wrong

[stoic crystal]

do you understand that the first case of the piecewise has a removable discontinuity, because (x-4)(x+3) = x^2 - x - 12

[lethal lintel]

yep

[stoic crystal]

what I mean is, if we have a horizontal asymptotes is with the last case of the piecewise

[lethal lintel]

hm idea

one sec

aw dangit it wasn't 0, 4

[stoic crystal]

for finding the horizontal asymptote you need to evaluate the limit when x is approaching -infty and infty

an horizontal asymptote exists if when you try finding both limits when x approaches-infty and infty you get to a k in R

[lethal lintel]

okay so like

I'm still a lil confused

I thought it was 4 four then

since when plugging in infity and all that

like

idk

I thought HAs were when the top part of the fraction equaled 0 y'know

i guess that's y intercepts tho

ah well

I wonder why it's that way for VAs then

it's just setting the bottom part of the simplified fraction equal to 0

[floral arrow]

Horizontal and vertical asymptotes are very different

Horizontal (and oblique) asymptotes are only about what the graph does toward either infinities

Vertical asymptotes are about what the graph does at one specific point

[stoic crystal]

^

[lethal lintel]

okay, gotcha

still not 100% sure how to solve for HAs, my calc teacher is kind of slow moving and so we spent a ton of time on HAs then very very little on VAs

wait no

other way around

a lot of time on VAs, not much on HAs

[floral arrow]

So a function can only have at most 2 horizontal or oblique asymptotes, but arbitrarily many vertical ones

[lethal lintel]

like 7 VA examples vs 1 HA

[lethal lintel]

makes sense

[floral arrow]

Because HAs are basically just limits at infinity

[lethal lintel]

mhm

[floral arrow]

If f has a limit L at +inf, then it has the horizontal asymptote y=L, that's all

Same for -inf

[flat mica]

a common mistake is to define HAs as curves that the function approaches but never touches as x->infinity. the definition using limits is the correct one

you didn't say the wrong definition, I just hear the mistake a lot

[floral arrow]

(and L needs to be a real number, not +-inf)

[flat mica]

so im pointing it out

[lethal lintel]

I'm so confused like what do I actually do for this problem 🙏

[flat mica]

which problem

oh I see

[floral arrow]

What's $\lim_{x \to \infty} f(x)$ ?

[stoic crystal]

evaluate the limit as x approaches infinity from both sides for f(x)

[wraith dagger]

Nel

[lethal lintel]

alr one sec

[flat mica]

for the bottom one

that's the easier of the two

bottom meaning x > 4

[lethal lintel]

what

I understand VAs just not HAs

[flat mica]

yes i read that

I mean to say

[lethal lintel]

I mean

[lethal lintel]

1 and (1/0)

[flat mica]

do you agree that for the limit as x->infinity , you only need to consider the piece of the function defined for x > 4?

or is that not obvious

[lethal lintel]

p

answer was 1

alr I get it now

awesome

so just like

[floral arrow]

Right, 1 is the only valid L

[lethal lintel]

alr alr the next question is very similar lemme try it out

wait okay no this one is also confusing

[stoic crystal]

I mean if you consider the first case, you will get lim f(x) approaches infinity so there is not a horizontal asymptote there

[lethal lintel]

alr so the answer is 1/infinity

but why is it not also 1/-infinity

[flat mica]

that's not the answer wtf

answer should be a number

[flat mica]

that's wrong

Don't care that it has a green checkmark, any calc professor worth their salt would mark it wrong

[lethal lintel]

so i should email my professor about it

[floral arrow]

Yeah that software is evaluating it which is just wrong

[stoic crystal]

the horizontal asymptote is what those limit evaluates to

recall the definition of horizontal asymptote

[lethal lintel]

right right

plug in infinity into the equation

and -infinity

see what happens

[stoic crystal]

what do you get?

[lethal lintel]

so you'd get 1/infinity

and 1/-infinity

[stoic crystal]

when you evaluate those limits

[flat mica]

but you don't do math with infinite like that in calculus

because it's not a number

[lethal lintel]

we do for his class

[flat mica]

wth

[floral arrow]

"plugging infinity" is just wrong

[lethal lintel]

🤷‍♀️

[stoic crystal]

what is this limit as x approaches infinity? 1/x

[lethal lintel]

1/infinity

[flat mica]

if your prof really taught it that way then there's something very wrong

[lethal lintel]

lemme send his notes rq

maybe I misunderstood in class

[flat mica]

even if you treat infinity as a number, the limit of a function as it approaches a number isn't the function evaluated at the number

[lethal lintel]

idk he's been teaching for like 40 years

[flat mica]

unfortunately that doesnt mean he's a good teacher

but it could be you misunderstood

[lethal lintel]

https://osu.instructure.com/courses/191216/files/80537259/download?download_frd=1

h

*oh

that wont work

[flat mica]

i need a username and pword for it, can you take a screenshot?

[lethal lintel]

ye

I'm kind of annoyed with him for only covering HAs for 1/9 of the pages and 10 mins in class in a class that's an hour and a half

where the lesson was meant to be VAs and HAs

he's my worst prof this year

kay so he definitely didn't have us plug in infinity

but like

idk how that ended up happening in my brain ngl

[floral arrow]

You made a shortcut

It's a classic mistake, I'm not surprised

[lethal lintel]

aight

so like what should the answer be instead of 1/infinity

and why

[floral arrow]

0, because that's the real number it tends to

1/inf is not a number

[flat mica]

and even if it were, you cant just evaluate limits by evaluating the function at the point

[floral arrow]

If you want an actual reason, take the definition of a limit at infinity. For example:

(where L=0)

[cedar kiln]

@lethal lintel Has your question been resolved?

[lethal lintel]

sorry I had a guy who's here to work on my car

one sec

[lethal lintel]

ohh right

yeah our prof would have told us that

whoops

.close

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[junior zenith]

Hello I am having a bit of trouble with the inductive hypothesis for this problem

[flat mica]

what's the part that's difficult?

do you know what strong induction means?

[tropic oxide]

the inductive hypothesis is that the formula c_k = 2^k + 3^k holds for all k up to and including n-1; your goal is to prove it also holds for k=n

[junior zenith]

oh I thought I had to prove it holds for all k up to n, and that I have to prove it holds for n=k+1

or rather prove it holds for all n up to k

[tropic oxide]

rather big misunderstanding

of what the word "hypothesis" means in this context

[junior zenith]

Like for the basis case I was proving that it holds for a range of numbers, I chose 2 up to 5

[tropic oxide]

you don't need more than 2 base cases

or in general, as many as the order of your recurrence

[junior zenith]

so I first need to prove that the equality holds for n = 2 and n = 3

then I need to assume that the equality holds true for n=k and all the values below k

assume it holds true for 2 <= n <= k

so basically I have to prove algebraically that

$5C_{n-1}-6C_n-2=2^n+3^n$

$5C_{n-1}-6C_{n-2}=2^n+3^n$

[wraith dagger]

Lasagna

[junior zenith]

I'm confused because this is what the question asks in the first place

Ok so for the base cases:

when n=2:

$C_2=5C_{2-1}-6C_{2-2}$

[wraith dagger]

Lasagna

[junior zenith]

$=5C_1-6C_0$

[wraith dagger]

Lasagna

[junior zenith]

$=(5)(5)-(6)(2)$

[wraith dagger]

Lasagna

[junior zenith]

$=25-12$

[wraith dagger]

Lasagna

[junior zenith]

$=13$

[wraith dagger]

Lasagna

[junior zenith]

and

$for n=2$

for n=2

$2^2+3^2=4+9=13$

[wraith dagger]

Lasagna

[junior zenith]

$13=13$

[wraith dagger]

Lasagna

[junior zenith]

therefore the equality holds for n=2

now for n=3

$C_3=5C_{3-1}-6C_{3-2}$

[wraith dagger]

Lasagna

[junior zenith]

$=5C_2-6C1$

$=5C_2-6C_1$

[wraith dagger]

Lasagna

[junior zenith]

$=(5)(13)-(6)(5)$

[wraith dagger]

Lasagna

[junior zenith]

$=65-30=35$

[wraith dagger]

Lasagna

[junior zenith]

and when n=3:

$2^3+3^3=8+27=35$

[wraith dagger]

Lasagna

[junior zenith]

$35=35$

[wraith dagger]

Lasagna

[junior zenith]

therefore the equality holds true for n=3

Now that I've proven the base cases, I must make an assumption

I'm not sure what that assumption should be...

according to codecademy.com: "the inductive step involves showing that if all elements up to and including n have some property, then n + 1 has that property as well."

[cedar kiln]

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[plush blade]

What did i do wrong?

a*(b/2a) = ab/2a = b/a

[hollow trail]

how did the last step work

[plush blade]

well if there is ab/2a ,why cant i just reduct the a from the top and bottom?

[hollow trail]

you can't subtract a from top and bottom, only divide a from top and bottom

[plush blade]

i dont get it

[hollow trail]

the way that cancelling fractions work is that dividing something by itself = 1

[plush blade]

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[worldly flower]

can anyone explain odd/even functions to me? Specifically, I need an explanation on f(x)= kx if k≠0 and HOW it's odd

[celest badger]

A function f is even if for all x in f's domain, f(x) = f(-x)

A nice geometrical interpretation would be symmetry about the y-axis

In other words, if I plug in 3 into the function

I get the same output as if I had plugged into -3

the function makes no distinction between negative numbers

On the other hand

A function f is odd if for all x in f's domain, f(x) = -f(-x)

The geometrical interpretation here is symmetry about the origin ig

for a line y = kx for k not equal to 0

plugging in, say, 3

would give the output 3k

pluggin in -3 would give the output -3k

this works for more than just 3 though

plug in any arbitrary x value

and you'll see that plugging in the negation of the value

gives you the negation of the output you would've received from plugging in the positive version of the input

[worldly flower]

I just can't seem to wrap my head around this I'm so sorry

I think it's all the letters that throw me off

[worldly flower]

so it hypothetically -3k because it's not equal to 0?

[cedar kiln]

@worldly flower Has your question been resolved?

[worldly flower]

no but I'm just gonna take my loss on this

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[obtuse coral]

For #70, what does that symbol mean? What’s it called?

[dire geode]

it's just a greek letter, treat it like any other variable

$\phi$

[wraith dagger]

riemann

[obtuse coral]

It’s called phi?

[dire geode]

$\phi$ and $\theta$ are common uses of angles

[wraith dagger]

riemann

[obtuse coral]

Got it, thanks

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[random swift]

!status

[cedar kiln]

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

[peak ibex]

step 4

[stoic crystal]

show your work or explain what you did

[peak ibex]

idk im lost

[stoic crystal]

a line is of the form y = mx + b

[random swift]

you're on step 4, so you have an answer. what's your answer?

[stoic crystal]

when x = 0, what is b?

[random swift]

we're not going to do your homework for you.

we can help.

[peak ibex]

y=1/2x

= 7

mb

uhhh y=1/2+7

[random swift]

nope. you've got c right, but m is wrong.

you know the formula for a line, right?

[peak ibex]

yeh

[random swift]

what is it?

[peak ibex]

wait nahh i cdont

why do i have tot ake algebra 2 in eight

[random swift]

https://www.mathsisfun.com/equation_of_line.html

$$y = mx + c$$

[wraith dagger]

Shuba

[peak ibex]

im in the us

[random swift]

• m tells us how steep the slope is
• c tells us the y intercept.

[peak ibex]

yeah i know that

[random swift]

how do you work out c?

[peak ibex]

finding the y intercept

[random swift]

yep. nice

how do you work out m?

[peak ibex]

finding the slope

[random swift]

how do you find the slope?

[peak ibex]

finding y and x

[random swift]

what do you mean by x and y?

[peak ibex]

y is 0,7

but i cant see x'

[random swift]

but y depends on x, right?

$$m = \frac{\mbox{change in $y$}}{\mbox{change in $x$}}$$

[wraith dagger]

Shuba

[peak ibex]

i dont know this

[random swift]

Take this for example.

When y changes by 6 units, x changes by 4 units.

So m = 6/4 = 3/2

Similarly, when x changes by 2 units, y changes by 3 units. So m = 3/2

Does that make sense?

[peak ibex]

yeah

[random swift]

Here's another example. When x changes by 10 units, y changes by 6 units. So m = 6/10 = 3/5.

Have a look at your original graph.

[cedar kiln]

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[cedar kiln]

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[radiant furnace]

how do i turn sin and cos functions to complex exponential? for example how can i simplify $cos(3x)$ or $sin(11x)$? i'm aware of euler's formula, but that says whem sin and cos are multiplied together, not when they are individual terms

[wraith dagger]

dis_da_mør

[radiant furnace]

i'm allowed to use Re and Im

[hot cipher]

oh, then just do Re(e^ix) for cos x or Im(e^ix) for sin x

[novel veldt]

trig identities to reduce them down to cos and sin then use eulers

[radiant furnace]

what do you mean by this? how can i use it on the functions i gave?

[hot cipher]

to be more specific and iirc:
cos [ f(x) ] = Re(e^if(x))

[radiant furnace]

ohh ok

does the same apply to sin?

[hot cipher]

sum like this

[hot cipher]

but with Im

[radiant furnace]

so sin(x) = Im(e^(ix))

[hot cipher]

Id swear theres another explicit formula for this

[radiant furnace]

r = sqrt(x^2 + y^2)?
theta = atan2(y/x)?

[hot cipher]

here it is

[radiant furnace]

alright

[hot cipher]

theta is more like atan2 (b/a) where z = a+bi

but yeah

[radiant furnace]

yeah by x i mean a and by y i mean b

cool thanks

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[cyan nacelle]

question 95

[cyan nacelle]

question 95

this is my working

the answer according to textbook is 13.44% and i dont know where i did something wrong

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[gray stirrup]

how do i do this question

[rain pine]

I think answer is "point 1 and point 4", because the accelerations are in the "same direction" as the velocity, so they becomes bigger then.

[hot cipher]

Ill take it as you already know how acceleration works and what "speed" is
🥀

[gray stirrup]

ty

[gray stirrup]

ye

[hot cipher]

usually try to explain the problem even its trivial for you

[hot cipher]

Well, in general, if the acceleration vector is perpendicular to the velocity vector, the particle isnt speeding up, its just rotating

Obviously, if its the same direction as the velocity, it speeds up, and if its opposite, it slows down

[gray stirrup]

ohhhhhh

[hot cipher]

In a more general way, you can know how this operates by doing dot product

if its 0, its rotating (or doesnt change if a = (0,0))

if its positive, the mod of the velocity (speed) is increasing, and if its negative, it decreases

The value also is important, since the higher it is, the more it speeds / slows

[rain pine]

visually, if "the angle between acceleration and speed" less than 90 degrees, the speed increases

the acceleration causes by "force". So imaging you are pulling something ahead, then it move faster in this direction

[gray stirrup]

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[royal gazelle]

Help

[hot cipher]

What is that you need help with?

[royal gazelle]

like the concept of the question itself

[hot cipher]

well, lets start with the basics, do you know what a diff equation is?

[royal gazelle]

yessir

i can do the part a

but how do i sketch it

[hot cipher]

Well, the points that have dy/dx = 0 are essentially those where you have a flat slope

Here you two ways, the hard which is searching pairs (x,y) where the function = 0, or looking at the graph to the slopes which are horizontal and eye ball it to connect them

knowing how to do both is useful btw

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[royal gazelle]

.reopen

[proper imp]

do you gusy know what concept this is

[hot cipher]

wrong place to ask this, but looks an awful lot like relativity and geodesics

[proper imp]

ok

where to ask for future refernce

[hot cipher]

The math help is mostly reserved to asking help for math problems you dont understand, for general purposes, just do the discussion channels.

[proper imp]

ok

thank you

how to close this help

.close

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[frank prairie]

I am confused with this whole problem and I have a test tomorrow and just need help learning the basics like how to know what number I’m supposed to divide, multiply etc. Or how I know how to divide multiply etc.🥺

[pastel vault]

you can do 8g - 5g right

[buoyant sand]

is that a 9

or a g

[buoyant sand]

i feeeel like it isn't a 9 tho

[pastel vault]

oh wait

yeah must be 8g - 5g then

[frank prairie]

G🥺

[frank prairie]

Isn’t it literally just 3g I need to pay for attention when taking notes I can’t do this

[pastel vault]

yep!

so $3g - 4 \le -3 + 3g$

[wraith dagger]

south

[pastel vault]

and then what term is the same on both sides?

[buoyant sand]

uh

@pastel vault

do u notice smth about this question 😭

it's a bit funky

[pastel vault]

I know, so I bet the purpose is to determine if the inequality is always true, always false, or conditionally true

I've seen that type of question before

[buoyant sand]

i see

[frank prairie]

3g🥺

[pastel vault]

yep, and then?

[frank prairie]

Does that mean you gotta do something with 3g to both sides

[pastel vault]

indeed!

what can we do to cancel out 3g on both sides?

[frank prairie]

Subtract both 3g’s by 3g

[pastel vault]

yep, and so you get what?

[frank prairie]

0

[pastel vault]

really?

3g - 4 <= -3 + 3g
-3g -3g

[frank prairie]

4 < -3 ?

[pastel vault]

nearly

$-4 \le -3$

[wraith dagger]

south

[frank prairie]

How’d the 4 become negative

[pastel vault]

we started off with this

so the 4 was already negative

[frank prairie]

ok ok

[frank prairie]

Doesn’t this mean this is the answer because it’s a true statement and the variable is gone

[pastel vault]

the variable is gone, nice observation

[pastel vault]

so it doesn't matter what g is, g could be anything and the inequality is ...?

true

you said it yourself, just realised

true for all g

[frank prairie]

Do you know any questions that could be inequality’s but like with division or changing the inequality because you do something with a negative

That’s kind of what I’m struggling on

[pastel vault]

yeah, https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:solve-equations-inequalities/x2f8bb11595b61c86:linear-equations-variables-both-sides/e/linear_equations_3

so it helps to do equations, the ones with = first

then it's the exact same with inequalities, you use the same inequality sign

except if you multiply or divide both sides by a negative number, in which the inequality sign flips

[pastel vault]

oh yeah so you want these questions, one sec

[frank prairie]

How do you know which number your supposed to do something with and how do you know what numbers your supposed to subtract multiply etc with

[pastel vault]

okay, so addition and subtraction undo each other
multiplication and division undo each other

is that part clear?

[frank prairie]

Yeah

[pastel vault]

so then we first used the concept of like terms

8g and 5g are like terms, cause they're both some number multiplied by g

[pastel vault]

and then notice we had +3g on both sides here

+3g is the same as 3g; we don't need to write the positive sign when 3g comes first

so look for addition, it's in "addition and subtraction"

"addition and subtraction"

cross addition out, then the operation you need on both sides is subtraction

subtract 3g on both sides

or let's say I give you b/10 = 20

look for division, it's in "multiplication and division"
cross out division: "multiplication and division"

so we need to multiply both sides

multiply both sides by 10

b/10 * 10 = 20 * 10
b = 200

[frank prairie]

I don’t get this one how come we do multiplication

[pastel vault]

because multiplication and division undo each other

if you want to undo the division, you must multiply

[frank prairie]

Oh the / is for division

[pastel vault]

yeah

[frank prairie]

I’m so done

[pastel vault]

well that's how we write it over text

and * is multiplication

of course don't write it like that on paper

so on paper the fraction symbol is always the clearest

$\frac{b}{10} \cdot 10 = 20 \cdot 10$

$\cdot$ is the same as $\times$, but we use $\cdot$ cause $\times$ can be confused with $x$, the variable $x$

[wraith dagger]

south

[frank prairie]

Oh yeah I forgot

So a problem with multiple numbers on each side how do I know which one to add, multiply etc on

[pastel vault]

do you have an example problem?

or should I just use this one

[frank prairie]

That one’s fine

[pastel vault]

yeah, so we can combine like terms

2d and 5d are like terms; 4 and 10 are like terms (cause they're both numbers)

so given that we have +2d, what I said above tells you that you need to subtract 2d on both sides

and given that we have +4, you need to subtract 4 on both sides

[frank prairie]

So do we subtract 4 from 2d and 5d

[pastel vault]

4 and 2d aren't like terms

same with 4 and 5d

[frank prairie]

Oh

So you subtract like terms from each other

So subtract 4 from 4 and 10

So 0 and 6

[pastel vault]

yep!

so remember to update the equation

2d + 0 = 6 + 5d

alright, keep going

[frank prairie]

Wouldn’t you get rid of the 0

[pastel vault]

sure, 2d = 6 + 5d

[frank prairie]

So 2d= 11d?

[pastel vault]

no, 6 and 5d aren't like terms

6 is a number
5d is a number times a variable

[frank prairie]

So you never do something with 2 numbers on one side except their both like terms or 5 x variable

[pastel vault]

yes

so notice how 2d = 6 + 5d is simpler

cause we only have one number term (the 6)

before we had two number terms, the 4 and the 10

[frank prairie]

Yeah

[pastel vault]

now if we can make there be only one variable term, like only 4d or 7d....

[frank prairie]

Does that mean we would do something with 2d and 5d because there’s no other like terms

[pastel vault]

yeah

also +2d and +5d are both addition

[frank prairie]

So 7d

[pastel vault]

not the right thing to do here, even though 2d + 5d = 7d yes

[frank prairie]

Wait

Since there both positive wouldn’t you need to cancel it out with subtraction

[pastel vault]

yeah!

[frank prairie]

But 5 - 7 would be a negative

So - 2d?

[pastel vault]

no, there is no 7 anywhere

[frank prairie]

Wait

Wrong way

2 - 5 so -3

-3d?

[pastel vault]

yep!

[frank prairie]

I was looking at this while I was typing and got confused

[pastel vault]

2d = 6 + 5d
-5d -5d

-3d = 6

[pastel vault]

that's ok (we all get confused right)

[frank prairie]

Negative x negative is a positive?

[pastel vault]

yes

[frank prairie]

So d= 2?

[pastel vault]

well actually, it helps to write this as (-3 times d) = 6

[pastel vault]

no

[frank prairie]

-2?

[pastel vault]

yes

cause -3 is negative, -2 is negative

6 is positive

[frank prairie]

So it cancels out eachother and makes it 3x2 and makes it 6

[pastel vault]

negative * what? = positive

what? = negative

yes, the negatives would cancel when you multiply

[frank prairie]

So when you do a negative minus a negative what happens

[pastel vault]

good question!

it becomes a positive

[frank prairie]

Oh ok

[pastel vault]

$-5 - (-7) = -5 + 7$

[wraith dagger]

south

[frank prairie]

So anything negative ? Negative is always a positive

[pastel vault]

yeah, it just happens to be that way

okay, so when you undo addition, you get subtraction

when you undo subtraction, you get back addition

[frank prairie]

And same way for multiplication and division

[pastel vault]

yeah

so actually let me show you something cool

$\frac{1}{\frac{1}{5}} = 5$

[wraith dagger]

south

[frank prairie]

Oh what the

[pastel vault]

1 over 1/5 is 5

[frank prairie]

How

[pastel vault]

if I divide a pie and each piece is 1/5th of the pie

I have 5 pieces

like, if I have 6 pies and each piece is 2 pies

I have 3 pieces
so that shows you 6 divided by 2 = 3

[frank prairie]

So for the 1 over 1/5 would you basically just be flipping the 1/5 to 5/1

[pastel vault]

yeah, so that's where the flip rule comes from

I think I should honestly just show you

$\frac{1}{1/5} \cdot 1$

$=\frac{1}{1/5} \cdot \frac{5}{5}$

$=\frac{1 \cdot 5}{1/5 \cdot 5} = \frac{5}{1}$

[wraith dagger]

south

[pastel vault]

$\cdot$ is multiply

[wraith dagger]

south

[frank prairie]

Ok I see

[frank prairie]

Can you find any problems like this but with inequalities

[pastel vault]

https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:solve-equations-inequalities/x2f8bb11595b61c86:multistep-inequalities/e/linear_inequalities

then it's the exact same with inequalities, you use the same inequality sign

except if you multiply or divide both sides by a negative number, in which the inequality sign flips

[frank prairie]

So if it was

<

You divide it multiply a negative it becomes >

Well

Wait

$-42 + 31 \ge 17z+23$

[wraith dagger]

Bakudo

[frank prairie]

So would you get 31 and add it to 31 and 23 because there’s both like terms because there positive numbers?

[pastel vault]

no, you need to subtract

[frank prairie]

Oh yeah

Because their both positive?

[pastel vault]

also you probably mean $-42z + 31 \ge 17z+23$

[wraith dagger]

south

[pastel vault]

yeah

[frank prairie]

Omg

$-4z + 31 \ge 17z+23$

[wraith dagger]

Bakudo

[frank prairie]

My thing copied the Z’s as 2

[pastel vault]

ohhhh

[pallid veldt]

The population of a village increases every year by 2% due to births and 3% due to immigration. At the end of the two years, the population of the village becomes 44100. (a).find the total increase in population due to births.

[frank prairie]

So subtract 31 and 23 by 31 and get 0 and -10

[pallid veldt]

help me out with this

pls

tomorrow is my exam

[frank prairie]

$-4z \ge 17z + -10$

[wraith dagger]

Bakudo

[frank prairie]

Is that it?

And then -4z - or + 17z

[pastel vault]

don't use multiple channels pls

stick to your own

[pastel vault]

23 - 31 = -8

[frank prairie]

Oh

Wait what did I write wrong

Oh

I honestly don’t know where I got -10 from

$-4z \ge 17z + -8$

[wraith dagger]

Bakudo

[frank prairie]

So would I add or subtract 4z and 17z

[pastel vault]

subtracting 17z on both sides would be easiest

[frank prairie]

Why is that

[pastel vault]

you should try it and see for yourself!

[frank prairie]

So 21z and 0

[pastel vault]

no, not 21z

[frank prairie]

Why not?

Wait

[pastel vault]

-4z >= 17z - 8
-17z -17z

[frank prairie]

It’s not a negative

It’s just subtraction

So -13z

[pastel vault]

yes the order is really important, so -4 - 17

[pastel vault]

no

[frank prairie]

Oh

Why?

17 - 4 is 13

So wouldn’t it be the same the other way around just negative instead

Wait

-21z

$-21z \ge -8)$

[pastel vault]

indeed, so -4 - 17 = -(4 + 17)

[wraith dagger]

Bakudo

[frank prairie]

But we never did multiplication or division so the sign never flipped

[pastel vault]

yes

[frank prairie]

So wouldn’t this be false

[pastel vault]

why do you think it would be false

[frank prairie]

Wait

Z

😔

so Z is a negative

[pastel vault]

no, z could be any number

[pastel vault]

you mean that -21 is negative

[frank prairie]

No

Because -21 and if Z was a positive number it would still be a negative would it not?

[pastel vault]

let's not go there

[frank prairie]

I’m confused

[pastel vault]

what should you do on both sides now?

[frank prairie]

I don’t know because they’re not like terms

[pastel vault]

we're past the like terms stage, yes

-21 multiplied by z

what can you do to undo the multiplication?

[frank prairie]

Division

[pastel vault]

yeah!

so divide both sides by what?

[frank prairie]

-21?

At least I think

[pastel vault]

yeah

[frank prairie]

-21/-21 is 1?

-21/-8 is 2.625

[pastel vault]

wrong order

[frank prairie]

-8/-21?

[pastel vault]

yes

[frank prairie]

-8/-21 is 0.381 and -21/-21 is 1 but since we divided it switched it around so it would be 1<0.381

Because we divided by a negative number?

[pastel vault]

yes, so the sign flips

you should just leave your answer in fraction form

so $z \le \frac{8}{21}$

[wraith dagger]

south

[frank prairie]

So wouldn’t it be wrong because 1 is more than 8 over 21

[pastel vault]

so this was the original inequality right?

[frank prairie]

Yea

[pastel vault]

now go and substitute z = 1 into both sides

see what happens

if you're claiming that z = 1 satisfies the inequality, you should end up with something that is always true

[frank prairie]

Oh

I forgot we didn’t do anything with the Z yet

I don’t have any idea what Z is

[pastel vault]

so z is just any number

any real number to be more precise

[frank prairie]

So infinite solutions

[pastel vault]

we're saying that for the original inequality to be true, that z must be less or equal to 8/21

[frank prairie]

So Z is 8 over 21

Or anything below that

[pastel vault]

yes

[frank prairie]

But 8 over 21 x 1 is 0.381 which is -8 divided by -21

[pastel vault]

so?

[frank prairie]

And 0.386 divided by 8 over 21 is 0.3166

So 0.381 < 0.3166 is still false

And if you round it it’s 4 < 3

0.4 < 0.3

[pastel vault]

you can't expect to do random operations and have it make any sense