help-42

[fluid oxide]

yes

soo the answer would have a -

and we reduce to our cot(theta) instead of cot(boundary - theta)

boundary = pi here but can be 90, 270, 360, etc

[lusty sand]

okay i got that part just not the removal of 180/pi

wait

did you

say pi-(pi-1/x)

why did you count -1/x as a whole theta?

[fluid oxide]

No..

i counted 1/x as theta

[lusty sand]

why?

[fluid oxide]

because...

its convenient? lmao

If I wanted to know, say, $\sin135^\circ$

[potent lotus]

@fluid oxide

[lusty sand]

but theta=pi-1/x

not just -1/x

[fluid oxide]

I would rewrite it as $\sin(180^\circ-45^\circ)=\sin45^circ=\sqrt{2}/2$ by astc rule

[potent lotus]

@fluid oxide

[fluid oxide]

i mean

its just a variable

The thing we want to simplify is of the form $\cot(\pi-\alpha)$, where $\alpha=\frac{1}{x}$

[potent lotus]

@fluid oxide

[lusty sand]

i guess this makes sense

[fluid oxide]

i guess its hard to grasp if you havent studied trigonometry separately

astc rule in itself wouldnt make much sense

[lusty sand]

i guess youre right

think i'll practise more on those type of questions

thank you

[fluid oxide]

yeah

[lusty sand]

again let me write what you said then ill close the thread

.close

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[calm coral]

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[remote mural]

p*q = ?

[calm coral]

@remote mural Has your question been resolved?

[stray tundra]

compute both limits, and multiply?

[remote mural]

how do i compute them ?

<@&286206848099549185>

[stray tundra]

good fkin question

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[remote mural]

.reopen

[calm coral]

✅

[remote mural]

yes ig

help

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[calm coral]

@remote mural Has your question been resolved?

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<@&286206848099549185>

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[warped socket]

im lost

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[scarlet saddle]

help?

[scarlet saddle]

i dont know where to being

begin*

i know about P(A)*P(B)

but that doesnt work here

[drifting seal]

do you know the formula for P(A or B)

[scarlet saddle]

no

[drifting seal]

P(A or B) = P(A) + P(B) - P(A and B)

[scarlet saddle]

so add 1/7 and 1/3 and then subtract with 9/21?

,calc 1/7 + 1/3

[potent lotus]

Result:

0.47619047619048

[scarlet saddle]

,calc .47619047619048-9/21

[potent lotus]

Result:

0.047619047619051

[scarlet saddle]

,calc 1/21

[potent lotus]

Result:

0.047619047619048

[scarlet saddle]

answer choice 3?

[drifting seal]

7+3-9 = 1 so yea 1/21

[scarlet saddle]

@drifting seal any other tips?

for probability?

[drifting seal]

you literally just need to know a few formulas

[drifting seal]

this one

sum of probabilities equals 1

[scarlet saddle]

how about when knowing if the results are independent or not?

[drifting seal]

then you’ll get to conditional probability and have to learn another formula or two

and that’s it for intro probability

[drifting seal]

P(A|B) = P(A)

[scarlet saddle]

a*b?

[drifting seal]

or equivalently P(A and B) = P(A) * P(B)

[drifting seal]

no

the probability of A given B, if independent is just the probability of A since event B occurring will have no bearing on whether or not A happens

since they’re independent

[scarlet saddle]

oh ok

[drifting seal]

this formula comes from that

[scarlet saddle]

alr i see

[drifting seal]

P(A|B) = P(A and B)/P(B) and if independent P(A|B) = P(A) so we have P(A) = P(A and B)/P(B)

hence P(A and B) = P(A) * P(B)

[scarlet saddle]

do u know any videos?

[drifting seal]

look up mathmanbillg

i used him for ap stats

but his videos are also good for regular stats/probability

go to his probability unit

and watch a few

[scarlet saddle]

alr thanks

[drifting seal]

there’s also khan academy

[scarlet saddle]

true

[drifting seal]

you’re welcome

[scarlet saddle]

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[chilly fox]

im not sure how to do this

[median barn]

have you studied this ?

[chilly fox]

nope

[median barn]

then you should try studying it then

[chilly fox]

i dont have any resources to do so but its just compulsory homework

the video to help you doesnt necessarily account for everything and thats just how my homework site works unfortunately

not all of the help videos are relevant to the specific question

[median barn]

alright look

all you do is multiply and devide by the irrational denominator

so you get rid of the root by squaring it

try it

[chilly fox]

uhh

i dont quite follow

could you bring up an example -- obviously not of the same question i have

to explain how it works

or just be a bit more in depth

[chilly fox]

i dont understand

[kindred estuary]

https://www.khanacademy.org/math/algebra-home/alg-exp-and-log/miscellaneous-radicals/v/how-to-rationalize-a-denominator

[median barn]

what happens when you multiply smth by 1

[chilly fox]

you get the same thing

[median barn]

what's $\frac{\sqrt3}{\sqrt3}$

[potent lotus]

Hamdy Hisham

[chilly fox]

uhh

probably 1

i am very bad with roots and surds n stuff

[kindred estuary]

Try watching this

[median barn]

yup that's better

[chilly fox]

is that for an american curriculum

cuz im doing AQA GCSE higher tier

its quite a lot different to america's education so

[kindred estuary]

The concept still applies

The video will explain how to rationalize

[median barn]

bro tha'ts really elementary stuff you need to learn it anyways

[chilly fox]

mine is uhh

15 over root 3

i can do the bottom part cuz thats easy but

i know this is the wrong answer btw

idk how surds work

like the fundamentals

[calm coral]

@chilly fox Has your question been resolved?

[floral rampart]

15/sqrt(3) is being multiplied by one

(sqrt(3)/sqrt(3) is equivalent to one; it's the same number divided by itself just like 2/2)

to do the multiplication you multiply the top stuff and leave it on top and multiply the bottom stuff and leave them on the bottom

you are correct with the bottom because square root means what number times itself will get three. Sqrt(3) is a number that would be three if multiplied by itself, which it is

hence just three

the top is 15 times sqrt(3)

which is not the same as 3 times sqrt(15)

hence why your answer is wrong

think about it like this:

sqrt(4) = 2

sqrt(16) = 4

sqrt(4) times 16 = 2 times 16 = 32

which isn't the same as:

sqrt(16) times 4 = 4 * 4 = 16

however if it was sqrt(15) times sqrt(3) you'd get sqrt(3)

see:

sqrt(4) * sqrt(16) = sqrt(64) = 8
or
sqrt(4) * sqrt(16) = 2 * 4 = 8

notice how it works either way

anyway in your case

you would leave 15 times sqrt(3) all over three

then it's time to simplify

you essentially have 15 of some number [sqrt(3)], with that product divded by three

that would leave you with five of that number

hence

5 times sqrt(3)

[austere olive]

Good shi

[calm coral]

@chilly fox Has your question been resolved?

[calm coral]

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[hasty creek]

I'm trying to this integral with partial fractions, and for some reason my A and B are switched

[hasty creek]

This is the books integral, for some reason the numerators are switched and I cant figure out why

[hollow totem]

because you should have $\f{5x-1}{x^2+1}$, not $\f{5x+1}{x^2+1}$

[potent lotus]

Dreyuk

[hasty creek]

oh my goodness

thank you so much

i spent 30 mins trying to figure out where i went wrong hahaha

thank you again ‼️

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[kindred estuary]

Also isn't it x^2 - 1 in the denominator?

[empty flicker]

Yall

How can I ask for help

[calm coral]

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[hollow totem]

oh that too yes

[hasty creek]

woopsy

ig i got it all mixed up 😵‍💫

thank you all again

[calm coral]

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[zinc falcon]

I understand if you can't be bothered to read all this. the question is asking for the derivative of the composed function using the chain rule. I wrote the general form of the derivative, but going through the first one I did not get the same answer as the book. Any help?

[zinc falcon]

this is the book answer

i didn't bother to compute D_2 and D_3 becasue D_1 is wrong

[pale blade]

D1 being the x partial?

[zinc falcon]

yeah

idk why we do it like that

its goofy

[pale blade]

wait a sec

where exactly did you go wrong

[zinc falcon]

i don't know

[pale blade]

looks like you got the x partial using the composition right, but youre off on the y partial?

no i mean which part of the question did you asnwer incorrectly

[zinc falcon]

im not sure either

[pale blade]

like of the 6: x, y, z partials with composition and x, y, z partials with chain rule, which one of them did you make a mistake/stop in?

cause your x partial using composition is right

[zinc falcon]

when calculating D1 or partial/ partialx i did not get the same result as the answer in the book

[pale blade]

when calculating D1 with the composition or chain rule?

cause when you calculated it with the composition (first pic), its correct

[zinc falcon]

oh right i didn't even notice i got that correct lmao

yeah with the chain rule

thats what my prof gave

[pale blade]

you might have applied the chain rule incorrectly

Yeah you’re supposed to differentiate g wrt u and v, not x.

The point of using the chain rule is that g is a function of u and v to begin with, it has no (initial) connection to x y or z

[zinc falcon]

ahh yeah okay thats my issue then

[pale blade]

So whenever you interact with (differentiate) g(u,v), it obviously has to be with respect to its constituent variables: u and v

[zinc falcon]

so then i would differentiate wrt to u and v and then plug in the values

okk yeah i see

what you did

[pale blade]

yup

[zinc falcon]

damn alright thank you very much

made it very clear

[pale blade]

a good rule of thumb when applying the chain rule is to only differentiate functions wrt its constituent variables. So you should exclusively differentiate f wrt x,y, or z and g wrt u or v.

[zinc falcon]

i think i got confused becasue my teacher gave us Djg(f(a))

like this

so i thought we would do it for the variables of f

[pale blade]

your prof has very odd notation

[zinc falcon]

yes

i don't like it

[pale blade]

yeah, that Djg(f(a)) means differentiate first and then plug in u and v as functions of x,y,z, but i can see how that might be cconfusing

[zinc falcon]

idk if you know but do you know if james stewart calculus teaches this well?

[pale blade]

stewart is solid. it's not too theoretical/confusing. I used it for my cal 1/2. some people find it too light on theory but I dont think thats a problem at your level

[zinc falcon]

alright i'll have a look

it would be in chain rule i guess

[pale blade]

would also reccomend pauls online notes
https://tutorial.math.lamar.edu/Classes/CalcIII/ChainRule.aspx

[zinc falcon]

that's it right?

[zinc falcon]

i'll check this out as well thank you

[pale blade]

yeah

[zinc falcon]

dope thank you very much

have a good day/night

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[hasty creek]

How would I integrate this? U sub doesnt seem to be working well

[eternal stirrup]

That is ideally the u-sub you’d want

you’ve got left the integral of sin u cos u, how is that not “working well”?

[hasty creek]

because where is that e^x supposed to go

[eternal stirrup]

U wrote du = e^x dx

So replace it

Or alternatively

U wrote u = e^x

So use it

[hasty creek]

Wdym replace it? Using u sub it should be $\int(usin(u)cos(u))\frac{du}{e^x}$

[potent lotus]

dingypine

[hasty creek]

my issue is that i dont know where that e^x at the end should go

[eternal stirrup]

What did u say e^x was again?

[hasty creek]

u

[eternal stirrup]

Okay use it

[hasty creek]

and the derivative of e^x is u though

so du = e^x dx

[eternal stirrup]

Yup

[hasty creek]

solving for dx results in

$\frac{du}{e^x}=dx$

[potent lotus]

dingypine

[hasty creek]

can you replace it with u again after doing that?

[eternal stirrup]

Okay sure

[eternal stirrup]

Yeah

I mean even if you’re unsure about that

You can avoid all this by staying with e^x dx = du

And noticing that you have this factor in your integral from before

I.e sin e^x cos e^x e^x dx = sin u cos u du

[hasty creek]

So just to make sure, this works then?

[eternal stirrup]

Yeah

[hasty creek]

insane

ok well thank you

that makes everything a lot easier

[eternal stirrup]

It’s not insane?

[hasty creek]

just an expression, ive never encountered it before

so its just surprising this works

ok well thank you thats all

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[eternal stirrup]

Substitution is surprising?

Welp, ur welcome

[hasty creek]

Substituting the denominator under du, yeah

[eternal stirrup]

e^x is still u

And in this informal sense it’s not really treated as a differential form anyways so it’s just the same as 1/u du

So u can do formal algebra on this

[hasty creek]

good to know, thank you

[calm coral]

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[languid wave]

hi! i understand how to use the integral test for this, but i dont understand what they mean of "have limit 0"

[pallid halo]

do the terms converge to zero

in other words, is it true that $$\lim_{n \to \infty} \frac{n}{n^2+1} = 0$$

[potent lotus]

Bungo

[languid wave]

ohh i see

then yes

the limit does go to zero

[pallid halo]

right

[languid wave]

just redid the integral test and got that it diverges

[pallid halo]

yep

[languid wave]

so it's

[pallid halo]

can also use the comparison test with the harmonic series

[languid wave]

B?

[pallid halo]

correct, B

[languid wave]

this question is specifically using integral test

so im just gonna use that

[pallid halo]

ok cool

[languid wave]

alright

i got more, do i ask here?

[pallid halo]

sure

[languid wave]

or ask in a new help channel

alr

[pallid halo]

ok

which one do you think it is?

[languid wave]

i have no clue

[pallid halo]

do you know what the n'th partial sum is?

in terms of the a_n's

[languid wave]

no

[pallid halo]

it's the sum of the first n terms

in other words, S_n = a_1 + a_2 + ... + a_n

[languid wave]

ah alr

so how would i approach this problem?

[pallid halo]

maybe try thinking about how you could get just a_n

if you have all of the partial sums available to you

and if that doesn't work, you could try just computing the right hand sides for the four options and see which one gives you a_n

[languid wave]

im still stuck

[pallid halo]

what did you try so far?

[languid wave]

i was thinking maybe if i can rewrite the a_1 + a_2 + ... thing into a sum such that i can isolate a_n?

[pallid halo]

how would you do that?

[languid wave]

i dont know how i can rewrite a_1 + a_2 + ...

that's what stumped me

[pallid halo]

you mean write it using a summation form?

like:

$$S_n = \sum_{i=1}^n a_i$$

[potent lotus]

Bungo

[languid wave]

yeah im stuck idk where to go now 💀

[pallid halo]

S_n is the sum of the first n terms

you want only the n'th term

so is there something you could subtract from S_n to get a_n?

[languid wave]

S_n-1?

[pallid halo]

yep

[languid wave]

ohh

[pallid halo]

simple as that

[languid wave]

so it's B

[pallid halo]

correct

[languid wave]

alr

got another one

i am not great with alternative series error bound 😭

[pallid halo]

do you know a general formula for the alternating series error bound?

[languid wave]

uhh

not on top of my head but i can get it from my notes

$\left|\sum_{k=1}^\infty a_k-\sum_{k=1}^n a_k\right|\le|a_{n+1}|$

[potent lotus]

gekka

[pallid halo]

ok good

so if you approximate using 3 terms, the error is no larger than the 4th term (in abs value)

in your case your series starts at n=3

so if you use 3 terms you are using n=3,4,5

so your error should be bounded by what?

[languid wave]

a_6?

[pallid halo]

yea, or more precisely, |a_6|

[languid wave]

so we plug in n=6 to a_n?

[pallid halo]

yea

[languid wave]

alr i got A

yeah im not great with alternating series error bound

[pallid halo]

it's the same deal, the error bound would be |a_9| if you use S_8 to approximate the series

and you have enough info to get a_9 from what they gave you

(remember the previous problem we did)

[languid wave]

hmm

im still stuck aaaa

[pallid halo]

think about this earlier problem:

[languid wave]

yeah im stuck :/

[pallid halo]

if i give you S_9 and S_8, you know how to get a_9, right?

we just did a problem like this

[languid wave]

that's the thing idk how to do a_9

since it didnt give me what a_k is

:/

[pallid halo]

no but they gave you S_8 and S_9, they're in the table

how do you get a_9 from those?

[languid wave]

oh is it just S_9-S_8

[pallid halo]

yep

so what is a_9 in this case?

[languid wave]

0.302

C

[pallid halo]

yep correct

[languid wave]

ik that this uses alternative series test

and ik that this series must converges

because of that test

{a_n} is decreasing

and that this series is limit 0

but im not sure how to determine if this series converges absolutely or conditionally

[pallid halo]

if you take absolute values of the terms you just get a_n

and a_n > 1/n

does $\sum_{n=1}^\infty \frac{1}{n}$ converge or diverge?

[potent lotus]

Bungo

[languid wave]

diverges

so it converges conditionally?

[calm coral]

@languid wave Has your question been resolved?

[pallid halo]

yea, can you explain why it doesn't converge absolutely?

[languid wave]

if it converges absolutely, then the function will converge if all of the terms are positive

so |a_n|

[pallid halo]

right

so why doesn't it converge absolutely in this case?

[calm coral]

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[golden gulch]

What type of error is made if for instance my null hypothesis is that the mean is x and the alternate hypothesis is that the mean is less than x but it turns out that the true mean was greater than x?

[golden gulch]

type 1 would be when we reject the null when its true but in this case the null isnt necessarily true

and it definitely cant be type 2

but it doesnt feel like it should be type 1 either

[calm coral]

@golden gulch Has your question been resolved?

[calm coral]

@golden gulch Has your question been resolved?

[calm coral]

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[ionic igloo]

FUTURE VALUE?

[calm coral]

@ionic igloo Has your question been resolved?

[ionic igloo]

<@&286206848099549185>

[calm coral]

@ionic igloo Has your question been resolved?

[calm coral]

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[shell notch]

I don't understand how to do the factorization of (3α-2) (α+3)+(α+6) (α+3)? I know I'm supposed to do =(a+3) but I don't know how to continue

[shell notch]

It's just "Factorize the expressions"

[rocky flower]

Notice both terms have the (a+3)

[shell notch]

Yes

[rocky flower]

Then that’s a factor.

Do you get how to factor ab + ac btw?

[shell notch]

No

[rocky flower]

Basically a(b+c) = ab + ac

The left hand side is how you would factor

[shell notch]

Okay

[rocky flower]

Now apply the same idea for this problem

[shell notch]

A I see

Tyy

[rocky flower]

No problem :))

[shell notch]

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[olive matrix]

where did I make a mistake here?

[leaden thunder]

Your handwriting is hard to read. Can you type your work

[unkempt drift]

careful

$\log(x^2) \ne (\log x)^2$

[potent lotus]

south, just south

[unkempt drift]

you should just sub $u = \log x$ and hence get a quadratic in terms of $u$

[potent lotus]

south, just south

[olive matrix]

oh, that would be much easier

thank you

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[unkempt drift]

yeah the reason why riemann said that is that you probably confused yourself

your working goes backwards in places like log(x/x^2)

before you undo it

I don't think it's literally hard to read your handwriting

it's just very confusing to follow

[olive matrix]

it's probably a mix of both

[unkempt drift]

yeah

[olive matrix]

my handwriting is absolutely terrible

[unkempt drift]

I've seen worse trust me

[olive matrix]

haha, I have yet to see that

thank you for the kind words though

[calm coral]

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[nova hound]

may someone please help me do this

[calm coral]

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[nova hound]

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[proud cave]

Question about graph theory: if you have a directed multigraph, does that include loops? Or do they have another terminology if you add loops (like with pseudograph)?

[proud cave]

<@&286206848099549185>

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[marsh yacht]

heyo could anyone help me

[modern glen]

In case of directed multigraph, there is no loops but it allows multiple edges between the nodes...

[calm coral]

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[old epoch]

for the defintion of this $\bigcup$, i saw that this is performed on a set of sets

[potent lotus]

LXDL

[old epoch]

but then here they have some weird notation

with $\bigcup_{x \in X}$

[potent lotus]

LXDL

[old epoch]

on the equivalnce class [x]

but [x] is a set, not a set of sets

so whats exactly going on here?

[potent igloo]

what are you trying to prove?

[calm coral]

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[old epoch]

im just asking in general about a notation

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[brittle basin]

what do u multiply to get d2y/dx2 from dy/dx

[brittle basin]

do you just:

dy/dx • dy/dx

[old plume]

d2y/dx2 is the second derivative of y wrt x.
It's different from (dy/dx)(dy/dx)

[brittle basin]

ah ok

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[ripe musk]

Can you help me study

[ripe musk]

Iam doing 9th grade math

[leaden thunder]

Help channels are for specific math questions.

If you're working on a math problem and don't know how to do it just post it

[ripe musk]

Oh ok

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[worn anvil]

Can someone check my method for 2iv

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[zealous stream]

Linear algebra:

We are given a real matrix $A$ such that all its eigenvalues are $|\lambda_i|<1$. Let $P > 0$ be the solution to the following discrete-time Lyapunov equation:
$$A^TPA - P = -A^TA$$

Show that $P$ and $A^{-T}PA^{-1}$ share the same eigenvectors.

[potent lotus]

ego

[zealous stream]

I don't see how to make use of the |lambda| < 1 condition

[calm coral]

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[tired locust]

Hi, I am not sure what do if this qestion I don't know the properties of a parallel lines to side lenghts in a trapozid

[tall moon]

AFCD and ABED are paralellorograms

[tired locust]

yeah

are the sides lengthes eqaul then

but that make no sense

[tall moon]

what doesent make sense?

[tired locust]

if the sides lenght are parrell are they equal in a parrallogram

cause i a mnot usre

sure

[tall moon]

yea

[tired locust]

oh

i dumb

ok

i get back to y ou id i need mroe help

more

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[grim snow]

<@&286206848099549185>

.close

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[worthy matrix]

ive been doing this exponential function (#9) and both sides are going up, which is not supposed to happen. did i do something wrong?

[leaden thunder]

show how you got your numbers

[worthy matrix]

i did 1/6 to the power of whatever x is in the chart @leaden thunder

and then multiplied it by 2

so for x=2 i did 2/6=1/3 and multiplied that by 2 which gave me 2/3

[leaden thunder]

,calc (1/6)^(-2)

[potent lotus]

Result:

36

[leaden thunder]

Doesn't match what you're describing

[worthy matrix]

how is it 36?

idek what im doing wrong maybe i multiplid wrong? but then my y intercept is still right

hm let me try smt

[leaden thunder]

,tex .exp rules

[potent lotus]

riemann

[leaden thunder]

6^2 = 36 so 1/(1/36) = 36

[worthy matrix]

ok it was my multipyling i got it now

thx

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[hollow sphinx]

not sure how to do this, I split up the fraction into -3s/denominator + 1/denominator then im not sure what to do since I cant factor the bottom

[serene geode]

complete the square in the denominator

i think then you can use known transforms from whatever table you have

[hollow sphinx]

I got this

[serene geode]

i guess take the transform to check it lol

[hollow sphinx]

The lhs one is right but the rhs one is wrong, where did I mess up?

[serene geode]

the rhs one looks right to me?

[quaint sphinx]

-3s + 1 = -3(s + 3) + 10

the right has to be multiplied by 10

[hollow sphinx]

why am I not allowed to just split up the -3s and the 1 ?

[serene geode]

you are allowed to do that lol

[hollow sphinx]

weird, when I put it into a calc they have the right one multiplied by 10, or are there several answers?

yeah it was wrong

[hollow sphinx]

this makes sense but I dont understand why im not allowed to do what I did and just split them?

[calm coral]

@hollow sphinx Has your question been resolved?

[hollow sphinx]

<@&286206848099549185>

[calm coral]

@hollow sphinx Has your question been resolved?

[hollow sphinx]

<@&286206848099549185>

[hollow sphinx]

<@&286206848099549185>

<@&286206848099549185>

[calm coral]

@hollow sphinx Has your question been resolved?

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[dreamy shale]

phi(ab) is abX + I wherea phi(a)phi(b) is abX^2 + I, am i meant to be claiming phi(a)phi(b) - phi(ab) is in the ideal and go from there and somehow prove something? i dont get how they could be equal aside from the restrictions put on the ideal since X^2 - X +abX =/= X^2 - X + abX^2 as a simple example

[mortal orbit]

abX^2 = abX since X^2 = X inside the quotient R[X]/I

[dreamy shale]

i think i just dont get ideals maybe

[mortal orbit]

do we agree that X^2-X is in the ideal I?

[dreamy shale]

yes

[mortal orbit]

so X^2-X+I = I in R[X]/I right?

by definition of R[X]/I

[dreamy shale]

the statements i have on R/I i just that if r + I = I then r is in I

[mortal orbit]

ok

it's actually an if and only if

so

(abX^2-abX)+I = ab(X^2-X)+I

but ab(X^2-X) is in I

so ab(X^2-X)+I = I

so abX^2 + I = abX + I

You have 5 seconds to stop the troll before I call mods

well check yourself

<@&268886789983436800>

you know mods can see messages you delete right? :)

[dreamy shale]

i think i have to look at quotient rings and groups and stuff again

X^2 = X in R/I made more sense to me when i thought of it as being 3 = -1 in Z/4Z

[mortal orbit]

well think about it as you're "adding multiples of 4" on each side

[dreamy shale]

thank you 13 year old who thinks theyre really funny

[mortal orbit]

3+ 4k = -1 + 4k

<@&268886789983436800> pinging again

[mortal orbit]

so it's always been like 3 and -1 differ from an element in the ideal

[dreamy shale]

do i close this or should i leave it open so this little comedian here is the most recent thing? then its clear what went on with mods and whatnot

[brazen elbow]

if you are not helping, then please dont talk

[torpid brook]

said comedian has been killed

[dreamy shale]

ah ggs

.close

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[mortal orbit]

thank u

[mortal orbit]

so your original question is solved?

[calm coral]

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[latent gulch]

Guys, i’m trying to get the domain of f^-1 by finding the range of f. Can i get the range of f by making x the subject?

[drifting seal]

yea i mean you found f^-1 so now state the domain of that

[latent gulch]

I haven’t found f^-1 tho

[drifting seal]

you did though

you found x in terms of y

[latent gulch]

Wait thats f^-1?

[drifting seal]

well switch the x and y but yea

[latent gulch]

Ohh

[drifting seal]

notice that y ≠ 1

graphically there is a horizontal asymptote there

[latent gulch]

1/2?

Oops nvm

[drifting seal]

no i mean y ≠ 1

[latent gulch]

Yes

[drifting seal]

you can demonstrate this by

$1 = \frac{2x+1}{2x-1}$

[potent lotus]

knief

[drifting seal]

this would imply that 2x-1 = 2x+1 or -1 = 1 which is clearly false

[latent gulch]

Yes

[drifting seal]

but yea you could’ve done it your way

[latent gulch]

Wait so is that the range

[drifting seal]

nono thats the only point excluded

it would be all reals except 1

[latent gulch]

Ohhh

So the range is y not equals to 1

[drifting seal]

yea

[latent gulch]

And that is the domain of f^-1

[drifting seal]

yea R \ {1}

[latent gulch]

Whats R{1}

[drifting seal]

i had a backslash

[latent gulch]

R{1}

[drifting seal]

it just means the set of real numbers minus 1

[latent gulch]

Oh

Okaayyy

[calm coral]

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@latent gulch Has your question been resolved?

[latent gulch]

btww thank you so much

.close

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[drifting seal]

you’re welcome

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[languid iris]

Help

[calm coral]

Please don't occupy multiple help channels.

[naive kindle]

what do you want help with

[languid iris]

How to line it

With the graph

[tawdry flame]

Um

[tepid terrace]

I recommend using Geogebra

[tawdry flame]

😭

[tepid terrace]

Search up Geogebra then input your slope, then make a line or whatever you need there.

[tawdry flame]

Just draw one point at 0.0

Ten another at 1,a

[naive kindle]

this is not really helping

[tawdry flame]

And connect them

[tepid terrace]

I don't know what he needs 😭

He's inconsistently replying.

[languid iris]

Bruh where is that at

[tepid terrace]

https://www.geogebra.org/calculator

[languid iris]

Holy confusing

That makes no sense

[tepid terrace]

My professor recommended this back then.

I've been using it ever since.

[languid iris]

Can you tell me how to do it tho rq ima tryna learn quick I got a test

[tawdry flame]

😭

I dont think thats what they meant

[dusky flax]

i dont think geogebra can draw graphs on paper though ):

[dusky flax]

in general, start from the origin

if you have a slope like 3

go right once

and go up thrice

in general, if the slope is something like a/b

you start from the origin

go right b units

go up a units

and mark that point

and finally connect it with the origin with a straight line

try it with a simple positive integer first

[tawdry flame]

Connect (0,0) with (b,a)

When you have a/b

[dusky flax]

yea

[calm coral]

@languid iris Has your question been resolved?

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[red smelt]

Y'all can someone help me in integration

[red smelt]

Aight

Ignore the (25)

It's just the points for the question

[tacit brook]

4(cos(2x))^2

= 2* 2(cos(2x))^2

= 2* ( cos(4x) +1 )

And for 1/(cos x )^2
= (secx)^2

This should be enough to give you a start

[red smelt]

Aight thx mate

I think I did something wrong

@tacit brook

[calm coral]

@red smelt Has your question been resolved?

[red smelt]

Yes

[calm coral]

@red smelt Has your question been resolved?

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[night tapir]

im doing the same thing lol

[night tapir]

I need help

[red smelt]

Good luck my guy

[night tapir]

Thanks:)

[mossy abyss]

What you don’t understand?

[night tapir]

part (f)

[mossy abyss]

last just add - in front of integral

And swap the lower bound to upper bound

Solved

[night tapir]

I dont understand why it would be negative?

Anything under the x-axis would mean the area is negative

[mossy abyss]

Because 4 is on lower bound

It’s area

[night tapir]

so basically it would be - part(a)?

[mossy abyss]

So basically you have from 0 to 4 S area

Than if you 4->0

Ofc it’s -S

Geometric interpretation

Integral swapping bound get minus in front of integral which is algebraic interpretation

Yes

You can think defined integral as a function

So basically

[night tapir]

so the answer would be - sign of part (a) basically

[mossy abyss]

Yes

[night tapir]

OKay i got it:)

Thanks

[mossy abyss]

$I(x)=\int^x_0 f(t)dt$

[potent lotus]

meow

[night tapir]

:)👍

[mossy abyss]

So understand it as a function

[night tapir]

Yes

[mossy abyss]

Physical interpretation

[night tapir]

[mossy abyss]

Work done

Overcome gravity to do work

[night tapir]

Thank you:_DD)

[mossy abyss]

So negative

You’re welcome

😊

[night tapir]

.close

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[exotic cosmos]

A list of n integers has an arithmetic mean of 56, where n > 100. The arithmetic mean of 10 of the integers in the list is 60. Which of the following is closest to the arithmetic mean of the other n - 10 integers in the list?

A) 50
B) 52
C) 56
D) 60
E) 66

[exotic cosmos]

right off the bat we can eliminate C D and E right?

or well maybe only D and E because n > 100

idk if i'm being dumb but i can't do the algebra lol

,, \frac{600 + \sum_{n-10}}{n} = 56

[potent lotus]

mmmm7

[marble pendant]

Yes correct

And you need $\frac{\sum_{n-10}}{n-10}$

[potent lotus]

George (Wumpus Man)

[marble pendant]

Can you find this in terms of n?

@exotic cosmos

[exotic cosmos]

not really sure i see anything

[exotic cosmos]

but yes

i have to express $\sum_{n - 10}$ in terms of n?

[potent lotus]

mmmm7

[exotic cosmos]

i even cheated and used the answer choices but even that didn't work

$\frac{\sum_{n-10}}{n-10} = k \implies k(n - 10) = \sum_{n-10}$

[potent lotus]

mmmm7

[exotic cosmos]

where k is one of the multiple choice answers

and plugged that into the main thing we had

,, \frac{600 + k(n - 10)}{n} = 56 \implies 600 - 10k = n(56-k) \implies n = \frac{600 - 10k}{56 - k}

[marble pendant]

Solve for the Sigma here

[potent lotus]

mmmm7

[exotic cosmos]

so why does this not work?

[marble pendant]

It does

You just need to see it

[exotic cosmos]

how?

[marble pendant]

Gimme a min

[exotic cosmos]

btw k is supposed to be an answer in the multiple choice given

i cheated a bit but i still never got the answer lol

the multiple choice gives us the answer to sum_{n - 10}/ 10

so i called that k

https://cdn.discordapp.com/attachments/948770026914213909/1308835958824767508/1266602198020919368.png?ex=673f63f9&is=673e1279&hm=b445abe68361e6bd48557f80791d0f9866d59f977d9f0afcaded60380dd8c347&

basically this

this is good, no?

then algebra leads you here

n is an integer and n > 100

[marble pendant]

$\frac{2+2x}{3+x} = 2 + \frac{-4}{3+x}$

[exotic cosmos]

u can plug the k's given in the question but ... none of them give both n as an integer and also n > 100

$\frac{2+2x}{3+x} = 2 + \frac{-4}{3+x}$

[potent lotus]

mmmm7

George (Wumpus Man)

[marble pendant]

Lol

See what I'm tryna do?

[exotic cosmos]

hmm wait before we proced, could you check my "cheat"?

why does it not work

[marble pendant]

Let's do my method first

You will see on your own

[exotic cosmos]

okay

[marble pendant]

Solve for Sigma and then express in the form I was trying to say

[marble pendant]

This form

[exotic cosmos]

https://cdn.discordapp.com/attachments/948770026914213909/1308834534686589100/1266602198020919368.png?ex=673f62a6&is=673e1126&hm=750576915ae31a33fee7b4ceab7b2f25e1e7da7ff01e7880de155ab63b81a63b&

u mean solve for sigma from here?

or where

[marble pendant]

Yup

Don't introduce the k tho

[exotic cosmos]

sigma = 56n - 600

[marble pendant]

Correct

Now the mean?

[exotic cosmos]

what mean

[marble pendant]

Mean of n-10 values

In terms of n

[exotic cosmos]

sigma/(n-10) = (56n - 600)/(n-10)

[marble pendant]

Correct

[marble pendant]

Now in a form like this

[exotic cosmos]

,, 60 - \frac{4n}{n - 10}

[potent lotus]

mmmm7

[exotic cosmos]

i guess?

[marble pendant]

Good

But try to eliminate the n from the numerator

[exotic cosmos]

wdym?

,, 6 - \frac{4}{1 - \frac{10}{n}}?

Lol

[potent lotus]

mmmm7

[marble pendant]

Numerator lol

[exotic cosmos]

i'm confused

can u tell me the end goal

[marble pendant]

See how I eliminated x from the numerator?

[exotic cosmos]

maybe the sheds some perspective

[marble pendant]

Have you heard of approximations?

[exotic cosmos]

like?

[marble pendant]

Like how sometimes small values like 1/1000 can be ignored

1 + 1/1000 is approximately 1

[exotic cosmos]

not really

do we have to use approximations for this?

[marble pendant]

Yes

That is why it is said n>100

,, 56 - \frac{40}{n - 10}

[potent lotus]

George (Wumpus Man)

[marble pendant]

This is what in saying it will become

Now see that if n = 100, this will become 55.555

And as n becomes larger

This will get closer and closer to 56

That is why we can ignore the fractional part

And say that it will approximately be 56

[exotic cosmos]

not sure how u got this

[marble pendant]

One moment

$\frac{56n - 600}{n - 10} = \frac{56n - 560 + 560 - 600}{n - 10} = \frac{56(n - 10) + 560 - 600}{n - 10} = 56 - \frac{40}{n - 10}$

[potent lotus]

George (Wumpus Man)

[marble pendant]

FIRST TRY

$\frac{56n - 600}{n - 10} = \ \frac{56n - 560 + 560 - 600}{n - 10} = \ \frac{56(n - 10) + 560 - 600}{n - 10} = 56 - \ \frac{40}{n - 10}$

[exotic cosmos]

😭 am i supposed to spot all this

[potent lotus]

George (Wumpus Man)

[marble pendant]

Why is latex like this 💀

[exotic cosmos]

it's a 1 minute question 😔

[marble pendant]

You get better at such stuff with time lol

[exotic cosmos]

\begin{align*}
\frac{56n - 600}{n - 10} &= \frac{56n - 560 + 560 - 600}{n - 10} \\\\ &= \frac{56(n - 10) + 560 - 600}{n - 10} \\\\ &= 56 -  \frac{40}{n - 10}
\end{align*}

Lol wait that's not what u wanted is it

😭

[marble pendant]

It is

[marble pendant]

Damnn

It's ok

[potent lotus]

mmmm7

[exotic cosmos]

okay there

anyway that's that

[marble pendant]

I think you get how I got this right?

Yeah that's what I wanted thankss

[exotic cosmos]

okay wait i did it a bit differently

and it doesn't eliminate the n

this all seems like guesswork 😭

wait

[marble pendant]

You eliminated the constant part

[exotic cosmos]

so if i add and subtract by some factor of n then i'll have the numerator with an n

interesting

[marble pendant]

Yup

[exotic cosmos]

and if i add and subtract constants then i'll have a constant up in the numerator

never had to do something like this lol

[marble pendant]

The reason I added 560 was cause we had 56n

And I wanted 56(n-10)

[exotic cosmos]

that part i get

because i did the same thing but with 60n - 4n + 4n

[marble pendant]

Yeah

Ok so what step are you on?

Like till where do you understand?

[exotic cosmos]

no okay i get it now

and limit as n goes to infinity of 40/(n-10) is 0

uhhhh okay

[marble pendant]

Lol

[marble pendant]

Now let's see this

The cheat method works too

[exotic cosmos]

so adding and subtracting some factor of n

is never good

i'm just comparing your form and my form

[marble pendant]

But as we saw n isn't Integer, it is approximated to one

[marble pendant]

I don't know about never, but in our case eliminating it is better

[exotic cosmos]

i mean if it's just about finding the limit

then both works

i just don't get why we consider the limit tbh

100 and "infinity" are pretty far apart

uh okay i guess my way works too

as in getting till here

[marble pendant]

0.44 is pretty small for 56

[exotic cosmos]

the limit of the second term is indeed 4

so 60 - 56

[marble pendant]

That's why we can ignore it

[exotic cosmos]

and for n = 100 you have like 4.444ish

so 60 - 4.4444ish is around 55-56 anyway

[exotic cosmos]

makes sense 😭

[marble pendant]

I think it's like if < 5% then ignore

[exotic cosmos]

i just never thought

i'd be looking at n as some dependent variable

and thought this was uniquely solvable or something

which is why i guess my algebra never worked

[marble pendant]

Ohh lol

The only reason I thought of this was cause we do this a lot in physics

Ignore most things cause they are "too small"

[exotic cosmos]

LOL

okay fair enough i guess

hmm so let's check my cheat too

which seems valid

i think

maybe my interpretation is off

https://cdn.discordapp.com/attachments/948770026914213909/1308835958824767508/1266602198020919368.png?ex=673f63f9&is=673e1279&hm=b445abe68361e6bd48557f80791d0f9866d59f977d9f0afcaded60380dd8c347&

k is one of the answer choices given

plug this back into our main thing

[marble pendant]

We end up with the same thing lol

[exotic cosmos]

but i can't plug in 56

😭

cuz division by 0

[undone harness]

GUY WHO CAN FIND X

[exotic cosmos]

[undone harness]

WHAT

[exotic cosmos]

no context so it's impossible

also could you open your own channel

[undone harness]

YEA

[exotic cosmos]

#help-40

[undone harness]

IN THE CALC GRAPHIC

THERE IS SIN(X) AND COS(X)

[exotic cosmos]

post your question there with all the context

[undone harness]

AND A CIRCOL R=PE

K

[marble pendant]

!occupied

[calm coral]

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).

[marble pendant]

Hmm lemme see

No it is correct