help-26

[normal tapir]

how do we compare two elements, say (3,4) and (2, 5) ?

[true spindle]

the x and y values?

[normal tapir]

it wasn't a rhetorical question...

in order to talk about the sup and inf

[true spindle]

nah im just a little slow 😭

[normal tapir]

you need a way to compare elements of R^2

[true spindle]

parametric?

or whats it called

[normal tapir]

wat

[true spindle]

do you mean like compare if its less than

or greater than

[normal tapir]

i mean compare as partial order

[true spindle]

oh

i thought there wasnt any partial order

[normal tapir]

there is no normally defined partial order correct

which is why im asking

[true spindle]

yea i didnt know how to compare it either

thats why i thought they just all didnt exist

[normal tapir]

ah well you should go ask prof or check notes or w/e to see how it's ordered

otherwise the problem isn't doable

[true spindle]

lol got it

thanks for the help

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[cerulean quartz]

whats a faster more efficient way to diff this instead of quotient rule?

[cerulean quartz]

<@&286206848099549185>

[shell knoll]

yes

[remote agate]

my god what on earth

also

!15m

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[viral skiff]

real

[cerulean quartz]

got u

[cerulean quartz]

implicit diff is ugly bro😭😭😭

but i can just tell quotient rule is gonna be tedious

need a faster more effective way of doing it minimising error

[remote agate]

ah so this is the expression after differentiating once

[cerulean quartz]

yeah

[remote agate]

honestly i have no idea but it probably isnt intended to be done like that

theres no obvious simplification

[cerulean quartz]

ill show u how i got there with my working out

[remote agate]

that is an incredible amount of patience

[cerulean quartz]

i swear 😭😭😭

[gritty birch]

ok I thought this horrible fraction was the question that you need to diff

[remote agate]

i wouldve factored y^2 on the left before differentiating but idk

[cerulean quartz]

i need to diff that whole thing 😭

[remote agate]

and also you can differentiate it twice without making y' as the subject

[cerulean quartz]

hm

bro i lose track with quotient, like 80 variables you gotta account for

[remote agate]

id just stick to product rule

rearrange stuff so its easier to differentiate

[cerulean quartz]

ok

lemme try that

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[true turret]

Solving for X when a radical is the exponent

[true turret]

Found the answer but I don’t know how to actually work it out

3root(x) would also be x^1/3 right

[alpine mist]

probably start with logarithms and use some of those rules

[cursive mango]

Do you mean cube root of x?

[true turret]

Yes my bad

[cursive mango]

Then yeah

[true turret]

And then can’t you transfer it down?

I am having trouble isolating the x

[terse nest]

rewrite the bases

4 can be rewritten as 2² while 8 can be rewritten as 2³

[true turret]

And then the two exponents can be multiplied, right

[alpine mist]

oh yeah. i forgot about that, do that instead of logs.

[true turret]

Or am I canceling them out

[terse nest]

yep

[true turret]

Like this?

[terse nest]

yep

now, they have the same base

meaning, you can equate the exponents to each other

[true turret]

I get what you’re saying, but I’m having trouble writing it out

[terse nest]

writing which

[true turret]

Equating the exponents to each other

[terse nest]

you just do this

,,2\cbrt{x}=3

[thorny flame]

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

[true turret]

How?

How did you get that from the picture I posted

[terse nest]

,,2\sqrt[3]{x}=3

[thorny flame]

Renz

[true turret]

But what happened to the bases

[terse nest]

just take these 2 and equate them to each other

[true turret]

Oh my God I just got it

I am actually stupid

Thank you Renz

[terse nest]

no problem

time to sleep ig it 3am for me lol

[true turret]

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[haughty musk]

Quick question regarding this integral, why is it that when i separate the integrand that the +500 become a -500?

[haughty musk]

I first wrote it down as 20 | dx + 500 | 1/x+1 dx

because in the original function, it was 20 + 500 / x + 1

[neon iron]

that does seem like a mistake on their end

[haughty musk]

I was thinking of using symbolab and seeing how it would solve it

[neon iron]

,, \int \8{20 + \4{500}{x+1}} \dd x = 20\int \dd x + 500\int \41{x+1}\dd x

[haughty musk]

ok so i should solve it like that^

[neon iron]

yeah

[haughty musk]

which would be 20x

but 500 | 1 / x + 1

I can't turn that into (x+1)^-1 because it would be to the power of 0 when I add 1

would I then use U-sub in this case?

[neon iron]

yeah

u = x + 1 works

[haughty musk]

ok thank you!

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[neon iron]

So i showed that |T| = the right

like so

But now, how do i show that |T| = the left

i started it like this but I do not know where else to proceed

Now idk if I can make a combinatorial argument or If i can turn that into a choosing formula

dont really know what to do

[topaz sinew]

@neon iron Has your question been resolved?

[wary tulip]

think about how many pairs there are when z = 2, when z = 3 and so on

up to z = n + 1

counting each of these cases and summing them up will give you the sum on the left

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[neon iron]

can you apply greens theorum to this?

[loud oasis]

the curve doesn't appear to be closed

[neon iron]

Oh wait they didnt specify that ughsguh

so then I'd need to set up 3 line integrals?

yea gotta do it the hard way

pick a nice parameterisation

and compute them

sob

alr ty

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[supple solar]

is it possible to generalize the surface area of revolution with the double integral formula?
$\iint_D dA$ with domain D

[thorny flame]

5ail

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[limpid iris]

Hello, this answer explination appears wrong, can someone double check this?

[silent nest]

The answer is right (4 cups) but the explanation is wrong

The correct explanation is
2.5 cups serves 10 people, so 1 cup serves 4 people. To get to 16 people, we multiply both sides by 4, so 4 cups serves 16 people

[limpid iris]

Right thats what I thought.

but thanks for confirming

[silent nest]

👍

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[fossil minnow]

so I attempted this problem this is my work. It is incorrect, so I'm not sure how to go about this.

[smoky sparrow]

You need i j k in the top row of your determinant

But anyway it looks correct so far

[fossil minnow]

I know there should be i j k. I just left it alone cause this is just my raw work. and it isn't right

[smoky sparrow]

Ah, you need to multiply your middle component by -1

Cause remember the 2x2 determinants go +, -, +

[fossil minnow]

yea so i did that too,

and then m=-2

[smoky sparrow]

You didn't

[fossil minnow]

but that isn't right either

[smoky sparrow]

You just did m - 2

Should be -(m - 2)

[fossil minnow]

sorry, not in the work i showed

[smoky sparrow]

Oh wait you did

Sorry the minus sign was very small

Yeah isn't that correct then

[fossil minnow]

i tried with and without the sign

cause it still gave it to me as incorrect

yea so I need to get the entire question right to get a mark for it.

[smoky sparrow]

Ah it's probably your part a that is wrong

[fossil minnow]

i doubt it, cause I just proved that I can solve for the constants b,c,m

[smoky sparrow]

Oh interesting

[fossil minnow]

and checked off a anyways with the same work and it still says it's wrong

[smoky sparrow]

No wait did you prove it by using the working you showed me

[fossil minnow]

yea

[smoky sparrow]

That's not how you prove it

Like if you take x + cz = 0, then z = -x/c

So substitute this into ax + by + 2z = 0 and you have an equation in only x and y

[fossil minnow]

well it's not if x+cz=0 is if the flux is equal to 0

[smoky sparrow]

And ofc 9y + mx = 0

Shit not the divergence

Ah okay that gets messy

[fossil minnow]

yep idk

actually you were right

somewhat right

that isn't how you prove it

cause they were different questions

flux is found with the curl but the curl isn't necessarily zero

but the divergence is zero

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[hasty jewel]

surface integral for surface area of $z=11-x^2-y^2$ bounded below $z=2$

[thorny flame]

Jaxx

[smoky sparrow]

!status

[topaz sinew]

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

[hasty jewel]

2

[smoky sparrow]

Ah ok

Could you send your work then?

[hasty jewel]

i first tried doing it with the parametrization involving the cross product

$\int \int fds$ where $ds = |r_{u} x r_{v}|dudv$

[thorny flame]

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

Jaxx

[hasty jewel]

actually im not sure if it's fds

probably $\int \int ds$

[thorny flame]

Jaxx

[hasty jewel]

i did z=2 so that gives the circle at z=2 of $x^2 + y^2 = 9$

[thorny flame]

Jaxx

[smoky sparrow]

Nah should be f dS

[hasty jewel]

ok

so then $f = z=11-x^2-y^2$?

[thorny flame]

Jaxx

[smoky sparrow]

No wait, just ds

[hasty jewel]

ok

so oanyways

[smoky sparrow]

Yeah integrate 1 ds in the region, sorry

[hasty jewel]

it's a circle at z=2 so i switched to polar

$\int_{-3}^{3} \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} fdydx$ (cartesian not polar yet)

my issue here is finding f

because i know z=11-x^2-y^2

so is that f

or because im at z=2 is f =9-x^2-y^2

[thorny flame]

Jaxx

[hasty jewel]

when converting to polar i get $\int_{0}^{2\pi} \int_{0}^{3} frdrd\theta$ (polar)

[smoky sparrow]

Ah right I'm stuck here too

[thorny flame]

Jaxx

[hasty jewel]

also none of these are right btw

f = 11-x^2 -y^2 = 11-r^2 or f = 9-x^2-y^2 = 9-r^2

correct answer

i also tried parametrization with u,v in polar system so $u=r$ and $v=\theta$ but i got stuck when setting up that integral too

[thorny flame]

Jaxx

[hasty jewel]

because of the parametrization i need to find cross product of ru,rv for ds since ds = |ru x rv| dudv

so then my r becomes $r = <ucos(v),usin(v),11-u^2>$

[thorny flame]

Jaxx

[hasty jewel]

but then again i wonder if it's 11-u^2 or 9-u^2

anyways if i continue with 11-u^2 and compute the partials and then magnitude i get

$r_{u} = <cos(v),sin(v),-2u>$

[thorny flame]

Jaxx

[hasty jewel]

$r_{v} = <-usin(v),ucos(v),0>$

[thorny flame]

Jaxx

[hasty jewel]

cross then get magnitude, i get $\sqrt{4u^{4}+v^{2}}$

[thorny flame]

Jaxx

[hasty jewel]

so my integral would be $\int_{0}^{2\pi} \int_{0}^{3} f \sqrt{4u^{4}+v^{2}}dudv$

[thorny flame]

Jaxx

[hasty jewel]

$\int_{0}^{2\pi} \int_{0}^{3} (11-u^2) \sqrt{4u^{4}+v^{2}}dudv$

[thorny flame]

Jaxx

[hasty jewel]

f = 11-u^2

which this integral; is also wrong

any ideas

specifically where i went wrong

actually f isnt in this integral when i parametrize

should be $\int \int_{R} | r_{u} x r_{v} | dudv$

[thorny flame]

Jaxx

[hasty jewel]

still is wrong though $\int_{0}^{2\pi}\int_{0}^{3}\sqrt{4u^{4}+v^{2}}dudv$

[thorny flame]

Jaxx

[hasty jewel]

my bounds seem right so i think my issue is the parametrization of r then the subsequent crossproduct and magnitude

with u = r and v = theta my parametrization of r(u,v) = <x,y,z> = <rcostheta,rsintheta,11-x^2-y^2> = <ucos(v),usin(v),11-r^2cos^2(v)-r^2(sin^2(v)>

i cant find anything wrong with my parametrization

<@&286206848099549185>

[topaz sinew]

@hasty jewel Has your question been resolved?

[hasty jewel]

<@&286206848099549185>

[weak ridge]

Do you have a question?

[hasty jewel]

yes it's above

i cant figure out how to set up this integral with the parametrization u=r and v=theta

nevermind figured it out myself turns out i cant read the difference between u and v on a cross product calculator

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[golden patrol]

i am new in discord i have no idea to control

please help me

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[calm burrow]

can someone help me with this

[fickle pelican]

pythagoras theorem

[calm burrow]

i fogor

[fickle pelican]

A^2+B^2 = C^2

[calm burrow]

wait brv 2 min

[fickle pelican]

did u get it

[calm burrow]

im back

[ancient mica]

Don't you know pythagoras theorem

[calm burrow]

i was busy

r = x+y

all squared

[ancient mica]

no

yes

[calm burrow]

yuh i was seeing if the pants that i got tailored fit me well

now im back

[smoky sparrow]

Yeah, x^2 + y^2 = r^2

[ancient mica]

AC^2 = AB^2 + BC^2

[smoky sparrow]

Interesting you think of Pythagoras's theorem as being the equation of a circle with radius r

Centered at the origin

Good connection to make 👍🏻

[ancient mica]

yeah, do u understand

[calm burrow]

heres where i got stuck

[slate barn]

plug in the numbers

[calm burrow]

this the thing i fogor how to do tho ngl

[slate barn]

fair
🤡

[smoky sparrow]

AC is the length of the side connecting points A and C

So we don't know AC yet

We do know AB and BC tho

[calm burrow]

yuh im looking for the other numbers

[smoky sparrow]

okok

[calm burrow]

ab should be 90 right?

degrees

[smoky sparrow]

No, AB is not 90 units long

[smoky sparrow]

AB is a side

[calm burrow]

the 7?

[smoky sparrow]

angle B is angle ABC

[calm burrow]

ahhh

[smoky sparrow]

Yes

[remote agate]

but yes the angle ABC is 90 degress

[calm burrow]

ahh admn this what was confusinbg me

final answer should be AC = |------
|130

right?

[smoky sparrow]

No

What's side BC equal to?

Ah wait ok

So yeah, AC^2 = 130

[calm burrow]

i dont know the sign

[smoky sparrow]

So AC = sqrt(130)

[calm burrow]

yeah this

[smoky sparrow]

Well technically it could be negative and positive

But no side length is ever negative

So it has to be positive

[calm burrow]

yeah

[smoky sparrow]

Good on you for realising there are two solutions tho

Like it will help you when solving quadratics

Like x^2 = 4

[calm burrow]

thanks for yall help

preciate it

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[smoky sparrow]

No worries

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[crude thunder]

if im calculating the coefficients of a complex fourier series

$$
c_n = \frac{1}{T_0}\int_{(T_0)}s(t)e^{-jn\omega_0 t}dt
$$

and I want to use the sine and cosine symmetry, I have to multiply the integral by 2? That means

$$
c_n = \frac{2}{T_0}\int_{(T_0)}s(t)\cos(n\omega_0t)dt
$$

Also, does it affect the bounds of the integral?

[thorny flame]

konxmok

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@crude thunder Has your question been resolved?

[crude thunder]

<@&286206848099549185>

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@crude thunder Has your question been resolved?

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[wintry phoenix]

$\lim_{{x \to 0, , g \to \infty}} \ln(x) \cdot \left|x^{-1/g}\right|$

[thorny flame]

Alex

[wintry phoenix]

is this just -inf here or is there any trick Im missing?

[scenic flare]

Yes -inf., but I presume you meant to write x -> 0+

since the leftwise limit is undefined, meaning the entire limit is undefined

[wintry phoenix]

yes, sorry about that

[scenic flare]

np

[wintry phoenix]

i was thinking

why if the limit is lnx * x^(-1/a) being a any number u want is 0

but when u make it approach to infinity is -infinity

[scenic flare]

Why would it be 0

for a=1 the limit is -inf.

[wintry phoenix]

wait i must have look wrong

give me a second

[scenic flare]

kk

Any positive a does so

For negative a it should be 0

[wintry phoenix]

ok i didnt write correctly

i got confused and mixed 1/ and ^-

No wait im right

jeez my brain

is collapsing idk anymore

ok ill start with the easy

lnx/(1/x) is 0

because derivative of lnx is 1/x and derivative of 1/x is -1/x^2

now if i take 1/sqrtx instrad of 1/x is still 0

now 1/x^a being a whatever big number u hcoose is still 0

and finally if i sen the root of to infinity it becomes -inf

Im saying weird thing right?

[scenic flare]

was generally right I think

[wintry phoenix]

so the thing is that i made a mistake in the original problem right?

it should be

$lim (x,g) -> (0,inf) of ln(x) * |\frac{1}{x^{1/g}}|$

[thorny flame]

Alex

[wintry phoenix]

ok i still think this limit is -inf but dont understand why the others are 0

shouldnt this be like the moment where the speeds converge?

[topaz sinew]

@wintry phoenix Has your question been resolved?

[wintry phoenix]

$\lim_{{x \to 0, g \to \infty}} \frac{ln(x)}{\frac{1}{|x^{1/g}|}}$

[thorny flame]

Alex

[wintry phoenix]

Ok like this

is this 0 or -inf?

.close

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[heavy jewel]

is someone able to help me do this question

not too comfortable with surface integrals, but im pretty sure i know how they work

and i dont think i can use divergence thm, since the orientation is towards the origin (inward)

ok so i just put a -1 infront after using divergence thm and it was right, is that something that works for inward flux?

.close

[topaz sinew]

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[tough nest]

yes

[topaz sinew]

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[stable sinew]

what do i get from (1-sin^2x)^1/2 here? -cos(x) or cos(x)?

[stable sinew]

help?

[radiant marlin]

1-sin^2 is cos^2, but ideally you want to convert all the sin's into cos except one

it's +cos(x) from that

[topaz sinew]

@stable sinew Has your question been resolved?

[stable sinew]

i m just not sure about sqrt(cos^2x) cuz that is absolute value of cosx

[radiant marlin]

right its absolute value technically, but 0 to pi/2 is a nice interval

everything sin and cos is positive

[stable sinew]

ok, so how to i determine the sign of sin/cos if i have the interval?

[radiant marlin]

usually you just do it by knowing how they look on a graph

since changes at 0,pi,2pi, and cos changes at pi/2,3pi/2, 5pi/2

[stable sinew]

so this is graph of cosx, so if we had an interval from pi/2 to 3pi/2 then here i would had -cos(x) yes?

[topaz sinew]

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[cinder oxide]

I messed up with a sign somewhere

I believe the answer is B, if I make x3 = 4

[thin adder]

ugh reduced echelon form sucks

[cinder oxide]

for finding null space?

what would you recommend instead

just REF instead of RREF, or something else?

[thin adder]

nah nothing wrong with it. i just hate doing it

its like inverting matrices without a calculator

just painful xD

[cinder oxide]

i gotta get used to it tho

linear algebra final exam coming up in 2 weeks

[pseudo sonnet]

yo same

[cinder oxide]

i need B+ in this course and there is no curve

80% on midterm, 100% on assignments

but final exam is worth so much

[pseudo sonnet]

mine's worth a lot too

[cinder oxide]

so someone could skip all classes, skip all assignments, just come in for 3 hours and ace the final exam
they would be a legend

i feel really fucking stressed about this final exam tho, it's my last chance to get B+ in this course

university is not fucking around

to get into my program, they will kick you to the street if you don't make grade

https://tenor.com/view/gif-gif-19496342

studying my ass off, there is just so much proof work in this course

[worthy storm]

wow gavin newsom has aged

[wary tulip]

i’ve had classes where, if you get an A on every exam, you just automatically get an A in the class

including one class with only two exams

but that class is notoriously difficult

[pseudo sonnet]

my LA midterm + final are worth 80%

so an A on both gives an A in the course pretty much

[wary tulip]

no like mine had this policy explicitly written, separate from the “normal” grade scheme

[pseudo sonnet]

hmm interesting

@cinder oxide has your question been resolved?

[topaz sinew]

@cinder oxide Has your question been resolved?

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[topaz sinew]

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

[craggy copper]

2

[topaz sinew]

Show your work, and if possible, explain where you are stuck.

[craggy copper]

uh

I wrote the functions in terms of their Maclaurin series

acutally let me take a picture

i think my series is correct? but im not sure how im meant to find the value of n

sorry im confused

the maclarin series for e^x is x^n/n!, so then for e^-x, we have (-x)^n/n!

and in this case it would be (-u)^n/n! i think

[mint crescent]

I highlighted the wrong one

$\cos x=\sum^{\infty}_{n=0} \frac{(-1)^n x^{2n}}{(2n)!}$

[thorny flame]

Civil Service Pigeon

[mint crescent]

So $\cos(x^{10})=\sum^{\infty}_{n=0} \frac{(-1)^n x^{20n}}{(2n)!}$

[thorny flame]

Civil Service Pigeon

[mint crescent]

,w maclaurin series cos(x^10)

[thorny flame]

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Requested by civil_service_pigeon

[craggy copper]

right

i fixed it

but i guess my bigger issue is how to find n

[mint crescent]

fix up your derivative first

then we'll work from there

[craggy copper]

ok give me a moment

[mint crescent]

<@&268886789983436800>

[craggy copper]

would this be correct?

[mint crescent]

,w \frac{4}{8!}

[thorny flame]

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[mint crescent]

yup

now, note that we have f as a polynomial (except with infinitely many terms, but pardon my abuse of language)

and we need to find when f'(0) isn't equal to zero

[mint crescent]

try out some simpler polynomials and figure out how many times you need to differentiate until this happens

[craggy copper]

sorry, im a bit confused, isnt this series for f, not f'?

[mint crescent]

oh yeah mb

I forgot to say we should find f' first lol

brain skipped a step there

[craggy copper]

right, so by looking at this series, the first term (apart from n = 0) will have a power of x^40

and i suppose that i need to differentiate 40 times to get rid of the x

is that correct?

[mint crescent]

yup, the minimum value of n is 40

[craggy copper]

okay, nice

then how would i find the value...?

or rather

should i be comparing this to the definition of maclaurin series?

the coefficients that is

[mint crescent]

honestly just start differentiating

there's a pattern

[craggy copper]

alright

ok, i tried both methods, i get (40!)/6

i think thats correct?

[mint crescent]

✅

[craggy copper]

alright

thank you so much

[mint crescent]

anything else?

[craggy copper]

nope

have a nice day

.close

[mint crescent]

you too 👋

[topaz sinew]

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[stable wren]

can someone help me simiplify this to 1 trig function?

[gritty birch]

try writing tan in terms of sin and cos

[marsh lava]

Try taking common and use the formulae

[marsh lava]

yes, exactly

$\tan x =\frac {\sin x}{\cos x}$

[thorny flame]

মাতুসপুতুসকুতুলিয়া

[marsh lava]

$\tan (-x)=-\tan x$

[thorny flame]

মাতুসপুতুসকুতুলিয়া

[stable wren]

should I mutiple the denominator together and get -(sin^3x)/cos?

then do the recipcal

[marsh lava]

maybe you should write sin squared x in terms of cos

so it will shorten things

[stable wren]

how do you write that in terms of cos?

[gritty birch]

do you know what is the formula relating sin^2 and cos^2?

[stable wren]

oh yeah so like 1-sin^2x

oh wait let me think

nevermind I got nothing, but yes I do know that sin^2x + cos^2x = 1

or is it 1-cos^2x

[gritty birch]

ok good, try doing the reciprocal as you mentioned, see if you can simplify it and then see how you can use $\sin^2 x + \cos^2 x =1$

[thorny flame]

lgkoo

[gritty birch]

that is sin^2 yeah, basically rearranged from sin^2 + cos^2 = 1, same thing

[stable wren]

I have this so far

should I use the reciprocal now?

[gritty birch]

try not using sin^ = 1 - cos^2 yet, the strategy is to turn all tan into sin/cos first

there's still tan in the numerator

and yeah use the reciprocal too

[stable wren]

So this?

[gritty birch]

yes

now take the reciprocal in the denominator (or equivalently multiply top and bottom by cos(x) if that makes sense to you)

[stable wren]

so I do that to both of the terms in the numerator right?

[gritty birch]

yes

[stable wren]

Than since they are the same denominator I can have them over one give fraction now right?

or could you use the sin^3x and cancel out somethings?

[gritty birch]

yup you cancel a factor of sinx

[gritty birch]

yes that is also correct

[stable wren]

Now I have this

now I am stuck again

[gritty birch]

first of all can you send the picture not rotated pls, and second of all I think your mistake was

you forgot your cos in the denominator when you moved to the next line

[gritty birch]

,rotate

[stable wren]

Does the negative get multiplied in now?

[gritty birch]

yes

[stable wren]

don't you get (sin^2x)/(sin^2x) = 1?

because -cos^2x + 1 = sin^2x, and that is all over sin^2x

[gritty birch]

yes

that's the answer

[stable wren]

cool

thanks that helped a lot

[gritty birch]

np

[stable wren]

.close

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[rocky pumice]

how did they pick (0,1)

[terse nest]

its a negative angle

-3/2 pi is -270 degrees right

[rocky pumice]

oh ts u again

lol

[terse nest]

since youre more.familiar with degrees lol

ye

when we have negative angle, we go clockwise

[rocky pumice]

do i have to do any calculation if i didnt bother with the angles

[terse nest]

wdym

[rocky pumice]

figuring out the location of both csc and tan

i got pie/2

[terse nest]

you still gotta solve for the location

[terse nest]

yea, pi/2 is coterminal with -3pi/2

meaning, -3pi/2 has the coords (0,1)

where x=0 and y=1

[rocky pumice]

okay i get it

so look for coterminal?

[terse nest]

ya

between 0 and 2pi

[rocky pumice]

i look for the positive coterminal

[terse nest]

-3pi/2 aint between 0 and 2pi

so we add 2pi, we get pi/2

and looking at pi/2 on the unit circle, we get (0,1)

[rocky pumice]

so add 2pi always?

or do i have to worry about choosing pi or 2pi

like the supplementary and complementary

[terse nest]

just add or subtract any multiples of 2pi

[terse nest]

no you can only add or subtract 2pi

[rocky pumice]

ok i understand

i have another problem here i wanna try but dont assist yey

sin (0) times cos (pi)

[rocky pumice]

ok im stumped

sin (0)

so far i have sin = y / r cos x/r and i know that r = 1 since r = hypotenuse

[terse nest]

then?

[rocky pumice]

i got it

i got a question tho

when i did pi + 2pi for cos (pi)

i got 3pi

[terse nest]

no you dont need to add 2pi there

since pi is between 0 and 2pi

no need to find coterminal angles for it

[rocky pumice]

ok but even if i did

would it have gotten the same

[terse nest]

since its between 0 and 2pi, meaning it will be in the unit circle

[terse nest]

ye you will get the same

cos(3pi) and cos(pi are equal

but thats unnecessary

[rocky pumice]

yeah ic all right

[terse nest]

since youve already given an angle between 0 and 2pi

[rocky pumice]

i got 0 since 0 times -1 is 0

[terse nest]

correct

[rocky pumice]

im having trouble with reference angles

[terse nest]

like

[rocky pumice]

i find myself converting to angles

[terse nest]

finding the reference angles?

[rocky pumice]

yeah

[terse nest]

1sec

leme find my precalculus reviewer 😭

the formula you are going to use will depend on where the angle is located

for example, in quadrant 1, the reference angle will be the angle itself

in quadrant 2, the reference angle will be 180 minus the angle

etc.

[rocky pumice]

i totally understand that

but what about when its radian

[terse nest]

its the same

but you do

quadrant 2:
pi - angle
quadrant 3:
angle - pi
quadrant 4:
2pi - angle

[rocky pumice]

ik its the same but i end up converting to degrees anyway

[terse nest]

like just use the radian version of 180 and 360 deg

[rocky pumice]

so if i have 6pi/5 i subtract it to pie/2?

or pie

[terse nest]

6pi/5 is quadrant 3

correct?

meaning we use the formula for quadrant 3

,,\theta' = \theta - \pi

[thorny flame]

Renz

[terse nest]

,,\theta'= \frac{6\pi}{5}- \pi

[rocky pumice]

so find the location and whichever its closest to in the unit circle subtract it by that?

[thorny flame]

Renz

[terse nest]

location of what

[rocky pumice]

location of the given

6pi/5

[terse nest]

didnt i gave you a summary of that

like quadrant 1:
between 0 and pi/2 etc.

basically
(only works for angles between 0 and 360 or between 0pi and 2pi)
Quadrant 1:
between 0 and 90
between 0 and pi/2
between 0 and 0.5pi

Quadrant 2
between 90 and 180
between pi/2 and pi
between 0.5pi and pi

Quadrant 3
between 180 and 270
between pi and 3pi/2
between 1pi and 1.5pi

Quadrant 4
between 270 and 360
between 3pi/2 and 2pi
between 1.5pi and 2pi

found it

6pi/5 is 1.2pi

1.2pi is between pi and 1.5pi

so quadrant 3

[terse nest]

since its quadrant 3, we use this formula

[rocky pumice]

.close

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[rocky pumice]

.reopen

@terse nest

[topaz sinew]

✅

[rocky pumice]

i was doing unit circle questions can u check if im right

1.) cot (-5pi)

=undefined

2.) csc (-5pi/6) =2

3. ) sin (15pi/2)

=1

4.) tan (pi/6) =

sqrt 3/ 3

[terse nest]

correct

[rocky pumice]

hek yeah

[topaz sinew]

@rocky pumice Has your question been resolved?

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[long dome]

I have no idea how to get from 1/(1-x) which is the x^n series

To that pile of complete garbage in the problem

you can multiply by x/3 but then you have x/(3-3x)

and 3x completely destroys the problem

because you can't get to 4x^2

and I don't know how to substitute something in

or what you would subtitute

[worthy storm]

try writing it like this:
$$\frac{x}{3}\left(\frac{1}{1 + (4/3)x^2}\right)$$

[thorny flame]

Bungo

[worthy storm]

the thing inside the brackets is a geometric series

[long dome]

how

what

[worthy storm]

how

what

[long dome]

there's an x^2 on the bottom and thats it

[worthy storm]

that's not true

[long dome]

how is that geometric

[worthy storm]

there's a 1 + (4/3)x^2

pattern match, compare it to 1/(1-y)

y = ?

[long dome]

-4/3x^2

[worthy storm]

yep

so now you can write a series for it

[long dome]

(-4x^2)^(n+1)/ 3

are you supposed to go forward or backward from the given complicated function?

My professor taught us to start at 1/(1-x) and then add things in

but other places show starting with the complicated function and moving to 1/(1-x)

[topaz sinew]

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[neon iron]

hello im confused pls help

[topaz sinew]

@neon iron Has your question been resolved?

[normal tapir]

what are you confused on

[neon iron]

first of all, for its apparently a joint pmf not pdf as its discrete

and for b, i have to get E(X, R) - E(X)E(R)

i dont get how to sum them up, as the supports also go to infinity

its made up of negative binomila left and binomial right, and i dont know which one if f(x) and which is f(r)

[topaz sinew]

@neon iron Has your question been resolved?

[neon iron]

also my combimed sum always ends up to be 0

cause of r = 0

[neon iron]

i simplified the sum of f(XR) to this

but because of the extra r on the right, i cant simplify to the E(X) of a binomial distribution

[topaz sinew]

@neon iron Has your question been resolved?

[topaz sinew]

@neon iron Has your question been resolved?

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[topaz sinew]

@ionic valley Has your question been resolved?

[ionic valley]

<@&286206848099549185>

[neon iron]

Physics?

[ionic valley]

nah

matrices stuff

[topaz sinew]

@ionic valley Has your question been resolved?

[topaz sinew]

@ionic valley Has your question been resolved?

[urban vessel]

You will only be able to solve this if the matrix is invertible. This means the determinant cannot be zero. So if you reduce to row-echelon form, you will be able to find the determinant quite quickly and hence find the value that R cannot take (the value that makes the determinant 0)

If this helps?

[topaz sinew]

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[stone turtle]

sorry how could use us g(2) using the graph

[stone turtle]

mhmm.

.close

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[stone turtle]

.reopen

[topaz sinew]

✅

[stone turtle]

sorry just need help on this one

[snow nexus]

if you take the derivative of both sides youll get g'(x)=f(x)
and f(x) is the graph shown
then for the concavity youll do the 2nd derivative which would be g''(x)=f'(x)

[vocal carbon]

Increasing is when it's going up, concave up is when the curve is upward
A local maximum is the maximum value on some intervals
An inflection point is where the concavity changes

[stone turtle]

I mean how do you find g(2)

given the f'(t) = g(x)

they say to use area under curve

but like how do u estimate the area under the curve here

wait

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[somber venture]

hi there i’m confused about 4

i did a similar question in 2 but

this is reversed

so i want to verify

would it just be like

for example a horizontal translation 3 units right?

[brisk blaze]

My advice is: make a list of transformations and then reverse it

[somber venture]

ooo 😮

[brisk blaze]

So moving three units right becomes 3 units left

[somber venture]

so if i have a reflection across the x axis

that would be?

reflection across y?

when reversing?

[brisk blaze]

A reflection across the x-axis, your just reflecting it back

[somber venture]

ohhhhhh

OHHH

i see

so i would still write it as “reflection across the x axis”

[brisk blaze]

Indeed

[somber venture]

if i identity a horizontal expansion by a factor of 2 and i want to reverse that, how would i do so?

where i’m at rn

(the transformations i wrote i already reversed.)

<@&286206848099549185>

[brisk blaze]

Divide by 2

[somber venture]

sorry how would i write that

so it turns into a horizontal compression..?

by a factor of?

1/2?

[brisk blaze]

Horizontal shrink by factor of 2

I think by a factor of 1/2 would be a stretch

[somber venture]

oh i see

i thought we can’t have a HC by 2

i thought it has to be below 1 so like HC by a factor of 1/2

[brisk blaze]

Probably, diff schools use diff notation

[somber venture]

does this seem right..??

(21, 6) as my final answer,

[somber venture]

ah

sorry i think this might be correct.

[brisk blaze]

Would the 2f(x) be a V Streach or am I trippin

[somber venture]

yeah yeah that’s why i changed it to 1/2 here

noticed that

if we were going to y=f(x) -> other equation we would have a VE by 2 but since we are reversing it, it turns to VC by 1/2 right?

[brisk blaze]

Yes

[somber venture]

okay good stuff

do my coordinates look okay

[brisk blaze]

I got (-15,-2)

[somber venture]

oh

[brisk blaze]

Wait no Nevermind

I did scales first

Your starting cords are wrong, it’s 9,-2

[somber venture]

oh shoot

it’s actually -9, 2

my bad

i got 0, 1

[brisk blaze]

Looks good

[somber venture]

awesome!!!!!!

tysm 🙏🏻🙏🏻🙏🏻

also can anyone verify if this method is correct?

solving for y and seeing if they equal one another

got it

.close

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[severe vector]

need help

[severe vector]

whats wrong with this?

f(0)=0

[topaz sinew]

@severe vector Has your question been resolved?

[severe vector]

<@&286206848099549185>

[paper walrus]

is it possible that the area becomes negative

at some point

[meager summit]

it is asking about f not F

[paper walrus]

no it is asking about F for the minimum value

[meager summit]

the graph shows f(0)=1

[paper walrus]

so I assume the area becomes negative at some point if you were to calculate it, and it looks like maybe x=5 is the minimum

because that is when the derivative of F or in other words f is 0

so its a critical point

[meager summit]

this question is kinda weird if it's mixing up f and F

[severe vector]

I'm just as confused

[paper walrus]

ok so the actual lines on the graph are the function f

and then the area under the curve of the graph is the function F

meaning f is the derivative of F due to the fundamental theorem of calculus

[meager summit]

does not the text before the blank say f(0)?

so it should be asking for the y value shown on the graph at x=0

[severe vector]

f(0)=1 then

[paper walrus]

yeah that is true i didnt notice the online hw lol

[paper walrus]

yea

[meager summit]

I dont understand how it is using this for a maximum value test though

[paper walrus]

yeah this is really bad calculus the hw is doing lol

[meager summit]

shouldn't it be asking for roots of the derivative instead of whatever those are???

[severe vector]

?

[paper walrus]

oh i kinda see what its doing but shouldnt it be F'

wait this is so wrong

[meager summit]

I think whoever made this should be exiled

[paper walrus]

agreed

why are they even evaluating the endpoints of the derivative graph?

that doesn't even really matter in this case

[meager summit]

ask your teacher because this problem is messed up big time

they are mixing up lowercase and capital f

[severe vector]

figured, thank you

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[raven robin]

really bad with questions such as these

[toxic zinc]

hi

[raven robin]

yo

[frigid wyvern]

i believe shape ABCD is a kite

[brave acorn]

Do you know what angle a tangent makes with the radius of circle

[raven robin]

ur porlly right

dk how i didnt think of that

[raven robin]

90?

[worthy storm]

@noble laurel

[raven robin]

I'm really bad with these type of questions

[brave acorn]

Yes, so ABC and ADC are both 90,which means they are equal

[frigid wyvern]

still need help with c?

[brave acorn]

Since the angle are equal, the quadrilateral is a kite

[pearl fog]

what do you think? we know that BAD=50, BCD=130, ABC=90, ADC=90, BC=DC

wait shit sorry for spoiling number 3

[raven robin]

lol

[pearl fog]

but if you want to know how, BAD+BCD=180

[raven robin]

opposite angles of a kite add up to 180?

[pearl fog]

uhh idk, i dont think so tho

but a quick google search told me that

[raven robin]

90+90+130 = 310
360 - 310 / 2 =25?

[raven robin]

is that a rule or something?

[pearl fog]

idk

[brave acorn]

ABC and ADC are both equal to 90
Sum of all angles of a quadrilateral is 360, so angle bac + abc + bcd + adc = 360
So bac+bcd = 180

[pearl fog]

thats bac no?

[raven robin]

oh its bad not bac

[pearl fog]

its easy to prove tho

[raven robin]

do u mean bad+bcd=180?

[brave acorn]

Yes, I was just proving that

[raven robin]

meant for this

[pearl fog]

do you have more questions

[raven robin]

not really think I got it

seems lile i was just overcomplicating it

thx though that was quick

[pearl fog]

im bored af so i tried checking channels to see if i can see if i can help someone, turns out i could

[brave acorn]

Yeah mb

[raven robin]

i got more questions im stuck on if ud like lol

[pearl fog]

send em

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[full kiln]

I swear theres no way to get T2

[loud oasis]

assuming an ideal everything, tension (magnitude) should be uniform in the string

[full kiln]

What does that mean

[loud oasis]

the magnitude (not direction) of two tension forces from the same string should be equal

[full kiln]

Why can u assume that?

[loud oasis]

because it is very light, so we treat it as effectively massless

and doesn't stretch

we are also assuming the pulley doesn't exert any friction

[full kiln]

So T1 = T2 such that they cancel out to 0 because it doesnt stretch?

[loud oasis]

so we can imagine a little chunk of rope. it is weightless and frictionless, so the only force on the bit rope is the tension from the other rope. since mass is ~ 0 the tension coming in = tension coming out (newton's second law), and by newton's third law it exerts the same tension on the next little bit of rope

so basically as long as we have a "light, inextensible string" and "smooth pulley" we can assume the tension is the same (magnitude, not direction)

[full kiln]

Okay thanks

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[neon iron]

Please an answear before it pushes itself down.

[wooden osprey]

!xy

[topaz sinew]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[neon iron]

Find length of u and t. The square root of 55 isnt even there marked.

[topaz sinew]

@neon iron Has your question been resolved?

[neon iron]

@wooden osprey

[wooden osprey]

whar

[wooden osprey]

these formulas might be helpful for you

[neon iron]

OH

[neon iron]

.close

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[glad sierra]

why does saying $|f(z_1)- f(z_2)| \geq |\frac{f’(z0)}{2}||z_{2} - z_1|$ means f is one one is D?

[thorny flame]

Dubs

[glad sierra]

i suppose this statement says, how ever close I make z_1 to z_2, the distance between their corresponding images is always more than some constant

does this mean z_1 != z2 implies f(z_1) != f(z_2)?

or say $0<|x_1 - x_2|< \delta \implies |f(x_1)-f(x_2)| >0$ for all $x_1, x_2$ in D

is it a real analysis way of saying f is one one in D?

[thorny flame]

Dubs

[winter egret]

A function is one-to-one if z1 != z2 implies f(z1) != f(z2)

If z1 != z2, then the right hand side of the inequality is nonzero. Thus |f(z_1) - f(z_2)| is certainly nonzero. But since |x|=0 iff x=0, we find that f(z_1) - f(z_2) must also be nonzero, which implies that f(z_1) != f(z_2), which was to be shown

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[last yacht]

Wondering if anyone can give me some advice on a project I'm doing on galois connections and closure operators

[last yacht]

its attached below (very very roughly, I only started about a week ago)

I'm just wondering (before i ask my prof) whether or not I'm going in the right direction

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[median egret]

Has anyone prooved this formula

[median egret]

Frustrum of pyramid

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@median egret Has your question been resolved?