help-26

[wooden holly]

uh whats n in nC2

not exactly

[sweet shard]

<@&268886789983436800> spam

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@wooden holly Has your question been resolved?

[wooden holly]

<@&286206848099549185>

wrong ping

[warped thistle]

N is the number of circles

[wooden holly]

okay

so 4C2

so 6?

.close

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[topaz sinew]

@stable dawn Has your question been resolved?

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@stable dawn Has your question been resolved?

[vapid plover]

hey, let's do this systematically

using the properties of a Euler's totient function, and from how a factorial works

$\phi(p!) = \phi(p\cdot(p-1)\cdots2\cdot1).$
p is not divisible by any numbers less than it, thus phi is multiplicative over this whole product and $\phi(p\cdot(p-1)\cdots2\cdot1)=\phi(p)\cdot\phi(p-1)\cdots\phi(2)\cdot\phi(1).$

[thorny flame]

🇵🇸Mína🔆

[vapid plover]

Is this enough?

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@stable dawn Has your question been resolved?

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@stable dawn Has your question been resolved?

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[limpid flame]

Idk if it's A or B

Last time it was bullshit 😔

[cyan ivy]

I think it’s B

Nvm it’s a

[limpid flame]

Why

[cyan ivy]

I don’t think b is as good of an option cause it’s generalizing all customers from a set of 100

While the first is more accurate by saying the customers in the sample

And it’s also true you cannot conclude anything about the population

[limpid flame]

It was wrong

[cyan ivy]

my bad

[limpid flame]

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[lyric crest]

is it just me or none of these are correct?

[lyric crest]

by handshake the sum of degrees is twice the number of edges

or does the theorem change if we have a tree?

[strange whale]

well you have 50 edges ok

since we're talking about a tree, you can get the number of vertices from that

[lyric crest]

its 51?

[strange whale]

yea

[lyric crest]

ah I see

thanks

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[noble heart]

There are 12 boys and 10 girls in the class.
How many options you can choose 3 students of the same gender?

[long beacon]

like do you mean whats the probablity of choosing three students of the same gender?

[noble heart]

yes

and how many options are there to choose one gender?

???

[sweet shard]

can you not

!nosols

[topaz sinew]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[lyric crest]

I was going to explain afterwards but guess we do it the other way around

[lyric crest]

yk this formula? @noble heart

[noble heart]

Yes it's like this

[lyric crest]

exactly

So there are two genders

boys or girls

and we need three students of the same gender

that means we pick 3 students who are either boy or girl

lets consider cases now

case 1. We are picking girls

[noble heart]

but we can also choose 3 from both boys and girls at the same time

[lyric crest]

how do you choose 3 girls from 12 girls?

[lyric crest]

but then they wont all have the same gender will they?

[noble heart]

12!/(12-3)!3!?

[lyric crest]

yep,thats case 1

now case 2, the same for boys will be?

[noble heart]

yes

[lyric crest]

how will you do it for boys?

[noble heart]

only from a different quantity

[noble heart]

oh we made a mistake here, on the contrary boys 12

[lyric crest]

oh yeah my bad

[lyric crest]

this was for boys then

whats the number for girls then?

[noble heart]

so girls 10!/(10-3)!3!

[lyric crest]

exactly

now we have case 1 and case 2

[noble heart]

and do I need to add this

?

[lyric crest]

what do we do to them to get the final answer?

[lyric crest]

The question says 3 students of the same gender

[noble heart]

10!/(10-3)!3!+12!/(12-3)!3!

[lyric crest]

so its 3 boys OR 3 girls

then yeah we add our cases

OR means + basically

[lyric crest]

as for this... do you get why we dont need to worry about it?

[noble heart]

not really, because there is a chance to choose, for example, 3 boys from all of them at once

[lyric crest]

there is a chance, yes

but we are not calculating a chance

we are calculating the number of ways we can choose 3 boys or 3 girls

Here is a simple example

I show you a rock and a pen

in how many ways can you choose the pen from here?

no hard math

there is one pen and one rock

[noble heart]

1?

[lyric crest]

yes

you can choose the same pen in a different way can you

but what if I say close your eyes

and pick one (lets say you cant tell if its a pen by touching it haha)

what is the chance that you will pick the pen?

[noble heart]

1/2

[lyric crest]

yep

see the difference of these problems?

in the first one you know what you want to take and you take it, the question is in how many ways can you take it

in the second one its about you WANT to take it and the question is what is the chance that you will take it

[noble heart]

oh I understand

[lyric crest]

in this boys and girls problems in case 1 we calculated the number of ways we choose boys

cause lets say there are 12 boys

and we pick Jeremy, Josh and Jack

thats one way

then we can pick Josh, Jack and Nick

thats another way

and so on

[noble heart]

understood

[lyric crest]

Nice 😄

[noble heart]

thank you for the explanation!

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[ancient dagger]

Hey so I'm studying Generalized Likelihood Ratio Tests (GLRTs), and in examples whenever they try to reduce, it feels like they eliminate stuff that you really shouldn't eliminate, I'll give an example:

[ancient dagger]

the bottom picture is $L(\hat{\Theta}_0)$

[thorny flame]

Karl

[ancient dagger]

Like how can they just replace everything in the exp() with an n?

[vernal vale]

You might have more luck on the statistics server FYI

although i cant tell if this is just an algebra question

which step of equality are you looking at exactly?

like here?

@ancient dagger

[ancient dagger]

Sort of

[vernal vale]

how do you mean

which step are you confused about specifically

between which two equals

[ancient dagger]

So here you see in the exp?

[vernal vale]

this step?

[ancient dagger]

They make a ratio between $L(\hat{\Theta}_0)$ and $L(\hat{\Theta})$

[thorny flame]

Karl

[ancient dagger]

And they end up with this

How can they reduce everthing in the exp to an n?

[vernal vale]

you know it almost just looks like $\hat \sigma$ right

[thorny flame]

jan Niku

[ancient dagger]

Yeah sure

it's missing an n

OH

alright

sure

[vernal vale]

you can rewrite it as $- \frac{1}{2\hat \sigma ^2} \sum (X_i -\hat \mu)^2$

[thorny flame]

jan Niku

[vernal vale]

idk id have to think

thats what seems like is happening though

youre somehow eating the square

[ancient dagger]

No you multiply by $n / n$

[thorny flame]

Karl

[ancient dagger]

and then the $\hat{sigma}^2$ disapears and is left by $n$

this makes sense

[vernal vale]

oh, right

[ancient dagger]

But what about $L(\hat{\Theta}_0)$?

[thorny flame]

Karl

Karl

[vernal vale]

it looks similar

using that i think $\hat \sigma _0^2 = \sum \frac{(X_i -\mu _0 )^2}{n}$

[thorny flame]

jan Niku

[ancient dagger]

Is that true though?

[vernal vale]

i imagine this is the identity

[vernal vale]

its just a definition, idk how is it defined in what you have

[ancient dagger]

WHA

[vernal vale]

i mean im just guessing, it seems reasonable

to define variance like this

seems fine to me

using this will make the equality work out, i believe

[ancient dagger]

Okay here's why I feel weird about it:

Wait I gotta read more about this

It just doesn't sit well because the this is a restricted MLE

[vernal vale]

as far as the stats IDK its been a long time

for that stuff i'd say go to the stats server

[ancient dagger]

I mean what you say makes sense

[vernal vale]

https://discord.gg/2gXrMCgC

[ancient dagger]

DOPE

Thanks!

[vernal vale]

np good luck

[ancient dagger]

And thanks for your help!

[pastel salmon]

$\text{exp}\left[ -\sum_{i=1}^{n}\frac{\left( X_{i}-\mu \right)^{2}}{2\sigma^{2}} \right]=\\=\text{exp}\left[ -\frac{1}{2\sigma^{2}}\sum_{i=1}^{n}\left( X_{i}^{2}-2\mu
X_{i}+\mu^{2} \right) \right]=\\=\text{exp}\left[ -\frac{1}{2\sigma^{2}}\left( n\left( \sigma^{2}+\mu^{2} \right)-2n\mu^{2}+n\mu^{2} \right) \right]=e^{-\frac{n}{2}}$

[thorny flame]

Joanna Angel

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@ancient dagger Has your question been resolved?

[ancient dagger]

Yeah sure, but I don't get how it works with the restricted MLE

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[ionic parrot]

curl of f gives 2x^2, idk what to do next

is this too hard foryal 😭 i need help

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Helpers are just people volunteering their time to help you. Be polite and patient.

@ionic parrot Has your question been resolved?

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@ionic parrot Has your question been resolved?

[ionic parrot]

:)

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@ionic parrot Has your question been resolved?

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@ionic parrot Has your question been resolved?

[modern bloom]

Hello, what's the question ? 😵‍💫

[ionic parrot]

that is the question

using stokes i think

[modern bloom]

so it's asking for line integral of F on curve C

not surface integral

of surface enclosed by C

right?

[ionic parrot]

use stokes to evaulate integral F dr thats what the rest of it says

[modern bloom]

ok

well curl F is a vector, it's not scalar

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[dim marlin]

(x+y)^12 find the max coefficient

[dim marlin]

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[terse trellis]

So i just completed the square of the denominator to become this easier integral, but im unsure how to continue

[terse trellis]

apperantly my next step is:

but why and how

[restive inlet]

the completed the square form is useful for integrating fractions with a constant in the numerator
however here you also have an x-term present
but note that its quite close to the derivative of the denominator

$\int \underbrace{\frac{a \cdot g'(x)}{g(x)}}{\log} + \underbrace{\frac{b}{g(x)}}{\arctan} \dd{x}$

[thorny flame]

ℝαμΩℕωⅤ

[terse trellis]

🧐

okay

But how do they split 3x-8 into 2*3/2.....

[restive inlet]

firstly what's the derivative of the denominator

(don't worry about expanding the result)

[terse trellis]

2x+3

2(x+3)*

[restive inlet]

the x term here would have a coefficient of 2
however your initial numerator has a coefficient of 3

which is where the 3/2 comes from.
which will be your a in the above image

[terse trellis]

Ohh

Yes

okay

the rest i understand

i think

lets see

[restive inlet]

$\int \frac{\frac32 \cdot \blue{\overbrace{2(x+3)}^{g'(x)}}}{\red{\underbrace{(x+3)^2 + 1}{g(x)}}} + \frac{b}{\red{\underbrace{(x+3)^2 + 1}{g(x)}}} \dd{x}$

[thorny flame]

ℝαμΩℕωⅤ

[terse trellis]

yep

wow some impressive latex there

but i got the answer, its correct

thanks!!!!

[restive inlet]

pales in comparisson to what several here are capable of

np

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[toxic hill]

How would you identify a permutation vs a combination

[toxic hill]

Also in what situation would you divide a permutation by a certain value

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[alpine mist]

permutation: the order items are selected in matters
combination: the order items are selected in does not matter

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[thorny flame]

Quacker

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[coral fog]

Problem 9

[neon iron]

what have you tried?

[coral fog]

I get confused how to pick the variables

I know the outside integral should be constants

Idk how to start this exercise.

[neon iron]

you're given all the information you need for this iterated integral

$\dd{A}$ is just $\dd{x} \dd{y}$

[thorny flame]

maxim

[coral fog]

It can be dy dx aswell

[neon iron]

yes,

but that doesn't matter

[coral fog]

Yes

But how do I pick the bounds of the integral

in terms of dx dy , and dy dx

[neon iron]

look at the graph

you know y goes from 0 to 8

right?

[coral fog]

Yes

[neon iron]

oh actually it's

$\int_{x=0}^{x=2} \int_{y=x^3}^{y=8} \dd{y} \dd{x}$

thats what i think at least

haven't done multivariable calc in years

[coral fog]

It's incorrect 🙂

[neon iron]

what about this?

[coral fog]

yes

[neon iron]

ya

there you go

[coral fog]

Well I gotta undestand how

[thorny flame]

maxim

[neon iron]

ok so

y = x^3 is less than 8 for x < 2

do you agree?

[coral fog]

yes

OR put the two y's equal

right?

[neon iron]

cool

yes that's how we find x

we know y = x^3 = 8 when x = 2

and we know that this starts at x = 0

and goes to x = 2

hence the boundary for the integral is x = 0 to x = 2

as for the integral with respect to y

we know x^3 < 8 for x < 2

which means on the interval (0,2) x^3 < 8

so x^3 is the lower bound

does that make sense?

[coral fog]

Yeah

[neon iron]

cool

so this integral is simply $\iint_R \dd{A}$ where $\dd{A} = \dd{y}\dd{x}$ and $R = {0 \leq x \leq 2, x^3 \leq y \leq 8}$

[thorny flame]

maxim

[coral fog]

Oh

That makes even more sense

[coral fog]

Is it always possible to build this up ?

[neon iron]

wdym

[coral fog]

Like I can now do the integral

[neon iron]

yes you can

[coral fog]

find the bounds

from this

[neon iron]

finding the bounds

is ALWAYS

[coral fog]

it's kinda eaiser

[neon iron]

the hardest part

for multivariable calculus

it's the only hard thing about the subject imo

[neon iron]

if ur given the region R sure

[coral fog]

That's what the section is about

How can it be harder?

any idea

[neon iron]

it can get very hard

[coral fog]

reversing the order of integration?

[neon iron]

for example yes that's a pain in the ass

at least in my experience

[coral fog]

R.I.P

[neon iron]

but i haven't formally taken a class on multivariable calculus

[coral fog]

I have to do all this in 15 hours

[neon iron]

so i can't tell you much

[neon iron]

gl

[coral fog]

Alright thanks for your help !

[neon iron]

no worries

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[real finch]

can someone help me reduce z^-2 -2 + i = (1-i)^3

[real finch]

I just need to reduce to get the cis form

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[ruby mural]

Wdym

[sweet shard]

Write z = a + ib then group real and imaginary parts

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[mighty shell]

it should be Y' intersection X

but answerkey says

[reef fjord]

answer key is wrong :)

if X' means complement of X

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[merry citrus]

I tried that. Is correct? Thx!

[merry citrus]

This is the graphic:

[loud ingot]

Hello slajoj zizek

Looks good

[merry citrus]

I need a confirmation bc it's an avaluation exercice xd

[novel echo]

try plugging that value into both the derivatives to see if your answer is correct

[merry citrus]

So, what I did is equalize both derivates to get an x value. I don't have any idea if it make sense but I only know that the derivate is the way you can find thr slope of main function.

[merry citrus]

I'll try it

Yes, I get the same value so I think it's correct but I don't know why the solution is in equalizing the two derivates :/ @novel echo @loud ingot

[tribal latch]

if you plot the derivative functions in desmos you will see that is where they intercept

the first derivative of a function gives you the slope of the original function

that's why you equalize the 2 derivative functions

[merry citrus]

Ok I see, thanks for explaining to me! @tribal latch

Also thanks! @novel echo

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[sour egret]

Hello I was wondering how to do this. I don’t know about factor trees or anything ..

[novel echo]

look up prime factorisation

then try to attempt this problem or come back here for further help

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[neon iron]

One thing im currently trying is cubing both sides

is this a valid step or no

[vernal matrix]

It won’t get rid of the cube root inside of the log, at least, so you shouldn’t

[neon iron]

hm

I also thought about

logging the 2/3

but that wouldn't give me 2/3

atleast log base 10

so I wouldn't be able to cancel and so on

[vernal matrix]

Maybe think about an alternative way to write a cube root?

[keen venture]

Do you have your log identities with you?

[neon iron]

I write it as 1/3rd

let me continue that way then

[neon iron]

yes

$log(x^2+48x)^{1/3}=2/3$

[thorny flame]

mj

[neon iron]

the 2/3 looks really tempting

I want to play with it

not sure if that'll get me anywhere though

[vernal matrix]

You have “the log of a power”

[neon iron]

yeah but you don't have two bases

so that wouldn't work

[vernal matrix]

Which identities do you happen to have to hand? Do you have a list of them?

[neon iron]

well, I have some memorized

trying not to rely on them too much though

I have loga+logb=log ab

loga-logb= log a/b

and the exponential one of course

[vernal matrix]

that one would make your life much notably easier

May as well state it too

[neon iron]

well

loga(m)^c

=clogam

$\frac{1}{3}log(x^2+48x)=2/3$

[thorny flame]

mj

[neon iron]

not sure if i did it right

but if I did..

[reef fjord]

ye that's right

just had to look at your original problem cause the way you texed it was unclar

[neon iron]

$log(x^2+48x)=2$

[thorny flame]

mj

[neon iron]

Im thinking now

what could I log to get 2

log base 10 to get 2

ah it's 100, but how am I supposed to know?

is it a memorization thing?

[limpid frigate]

10^2 = 100

[neon iron]

right

$log(x^2+48x)=log(100)$

[thorny flame]

mj

[neon iron]

$(x^2+48x)=(100)$

[thorny flame]

mj

[neon iron]

$x^2+48x-100=0$

[thorny flame]

mj

[vernal matrix]

Alternatively, notice that your equation here is then equivalent to $x^2 + 48x = 10^2$ by definition of log (base 10)

[thorny flame]

@vernal matrix

[limpid frigate]

For the future it is probably going to be a memorization thing like your multiplication facts (e.g. knowing 6 * 7 without needing to compute it)

[neon iron]

$(x+50)(x-2)$
$x=-50$
$x=2$

[thorny flame]

mj

[neon iron]

guess it's something you guys can just spot

need a bit more practice with these

[vernal matrix]

You may also want to check whether the solutions you found are actually valid too

[neon iron]

there a rent any restrictions..

[vernal matrix]

(i.e. can you put both into the original equation and have no issues?)

[neon iron]

actually

you can't have a -ve log

and cube root doesn't remove the 've

-ve

[vernal matrix]

Yep- can’t take logs of nonpositive* numbers

[neon iron]

hm.. finding the domain is a problem within itself

[vernal matrix]

It is a bit of a problem - but you only really need to check that for the solutions you found, you don’t get “the log of a negative number”

[neon iron]

oh wait a minute

x^2 is

non negative

but then you have +48x

[vernal matrix]

Yep, that could potentially cause you issues though

[neon iron]

how would you find

the domain

without guessing and checking

it has to be pure coincedence that it is 48

😭

[keen venture]

How would you find it WITH guess and check?

[fair socket]

It's range should be $(infinity,0)U(-48,-infinity)$

[thorny flame]

Blunder69

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[reef fjord]

which d

o saw you said 53 above also

because you're assuming u and v are functions

7v(x)-2u(x)

yes

do you know how to find tangent line?

the equation of the tangent line to a function

like y-f(a)=f'(a)(x-a)

well, if you know how to find tangent line, all you need to do is take the negative reciprocal for slope and that's your normal line

if y=f'(a)(x-a)+f(a) is the equation of your tangent line, then y=-(f'(a))^-1(x-a)+f(a) will be the equation of the normal line

ye

:)

also the -15 is wrong

ye

!showwork

[topaz sinew]

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

[reef fjord]

point slope form is useful for doing this sorts of problems

are you using point slope form for your lines?

[reef fjord]

but if it means largest negative, then yeah, -infinity

it means magnitude, then 0 is the smallest slope

ah wait

they found it by taking the derivative

used power rule

they didn't evaluate it everywhere

basically they're looking for where f'(x) is smallest

i.e. minimizing f'(x)

if you've learned about minimum and maximum you know they occur at critical points, and boundaries of interval (since boundaries are at infinity, don't consider them)

critical points are where the derivative is undefined at the value, or it is 0

since we're trying to find the critical points of f'(x)

we take the derivative again

and set it equal to 0

f''(x)=0

[topaz sinew]

@neon iron Has your question been resolved?

[reef fjord]

do you know how to find f''(x)?

you just take the derivative of f'(x)

@neon iron then set f''(x)=0

and solve for x

this gets you the x-value where f'(x) is smallest, then you just plug that into f'(x)

the second derivative is just the deriative of the first deriative

if it helps you get let g(x)=f'(x)

and g'(x)=f''(x)

https://tutorial.math.lamar.edu/classes/calci/higherorderderivatives.aspx

when you look at the sign f''(x) it tells you the general 'shape' of the curve as well

f''(x)>0 means the curve in that region is generally 'u' shaped, concave up

f''(x)<0 means the curve in that region is generally 'n' shaped, concave down

https://tutorial.math.lamar.edu/Classes/CalcI/ShapeofGraphPtI.aspx

https://tutorial.math.lamar.edu/Classes/CalcI/ShapeofGraphPtII.aspx

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[mellow venture]

I think I finally got it
$$ - u^{''}_n (x) - \lambda u_n (x) =0,$$
with $u_n (0) = u_n (L)$ and $u' n (0) = u' (L).$
where $\lambda = \kappa ^2 \left( N + 1 \right) ^2$.

[thorny flame]

Fractalogist

[mellow venture]

solutions are
$u_n (x) = c_1 cos((N+1) \kappa x)) + c_2 sin((N+1) \kappa x))$

[thorny flame]

Fractalogist

[mellow venture]

for first boundary we get
$$c_1 = c_1 cos((N+1)\kappa L)) + c_2 sin((N+1) \kappa L))$$

[thorny flame]

Fractalogist

[mellow venture]

second
$$c_2 = -c_1 sin((N+1) \kappa L)) + c_2 cos((N+1)\kappa L))$$

[thorny flame]

Fractalogist

[mellow venture]

clearly c1=c2=0, what if we set one constant to be zero, and other one arbitrary, for example for first boundary we set c1=0 then we get
$0=c_2 sin((N+1) \kappa L)$
since $c_2 \neq 0$ this implies $(N+1) \kappa L = n\pi$

[thorny flame]

Fractalogist

[mellow venture]

what would our eigenfunction be then?

would it be $u_{\kappa, n} = c_2 sin(n\pi x)$, but the eigenfunctions should depend on $\kappa$ and n at same time i think because the notation that they gave for the eigenvalue is $\lambda_{\kappa, n}$

[thorny flame]

Fractalogist

[mellow venture]

<@&286206848099549185>

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@mellow venture Has your question been resolved?

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@mellow venture Has your question been resolved?

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@mellow venture Has your question been resolved?

[topaz sinew]

@mellow venture Has your question been resolved?

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@mellow venture Has your question been resolved?

[ivory wave]

What kind of math are you studying

Looks hard

[sweet shard]

#odes-and-pdes or #advanced-pdes stuff

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[vast stirrup]

how do they get it to x-2/4

[gray storm]

On the LHS of the first step?

[vernal matrix]

The $\frac{(x - 2)^{n+1}}{(x - 2)^n}$ simplifies to $(x -2)$, as does $\frac{2^{2n}}{2^{2n + 2}} = \frac1{2^2}$

[vast stirrup]

oh wait i got it i just had a mistake in the algebra

im sorry guys

[thorny flame]

@vernal matrix

[vast stirrup]

thanks so much for your fast responce

[gray storm]

You are welcome

[vast stirrup]

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[summer meadow]

This is a crane and the arm between C and B rotates around C. CB and AC is always constant. and AB changes its length . I want to make a graph showing the how AB changes as you change angle v. (the text isn't important) and i use radians for angle v

[topaz sinew]

@summer meadow Has your question been resolved?

[soft lily]

I'm working on it

I'm close to an answer

[summer meadow]

take all the time you need

[soft lily]

okay I think I got it

simplest thing to do would be to consider everything to be on a cartesian plane

where c is at the origin

so the location of A would be -44,33

since we're subtracting 5 from 38 because A is 38 off the ground, and C is 5 off the ground

then the location of B would be 67COSv,67SINv

then we make a triangle out of those coordinates

67COSv+44 and 67SINv-33

so c, which would be the distance between A and B would be the square root of ( (67COSv+44)^2 + (67SINv-33)^2 )

sorry, I never figured out how to use latex on this server xd

does that seem accurate?

[summer meadow]

im a bit confused why is A = -44,33

[soft lily]

if C is the origin, A is 44 to the left of C, that would be X = -44

and Y = 33, because A is 33 above C

[summer meadow]

oh 44,33 is the cords

[soft lily]

ye

sorry

my b

[marble skiff]

i can confirm the solution

if you want i can offer a non-coordinate way but the solution @soft lily gave is perfect (it's sort of the same except without coordinates)

[summer meadow]

i would love to hear more solutions

[summer meadow]

also i dont fully understand this step

[marble skiff]

so first using pythagorean theorem we get AC = 55

[summer meadow]

yeah

[marble skiff]

so now from sin (ACD) = 33/55 we can determine the angle acd

then since the three angles v, bca and acd sum to pi (since they lie on a line), the angle BCA is equal to pi - v - ACD

and now you have to use the law of cosines in triangle ACB

and the sqrt would be the exact same as swarm's answer

actually i'm sure that expanding that answer would give you the answer i have

[summer meadow]

okay but how do i fx make it into a graph where xAxis = angle v and yAxis = length of AB

[marble skiff]

that is your function that you're looking for

when you plot either swarm's or my equation, you get the length of |AB| for every v

the function doesn't look very nice sure but it gives you an |AB| for each v

you can try plotting it and see how it looks

[summer meadow]

it looks like this

but is this in radians

cause that might be why

2 sec

[marble skiff]

i think that graph is correct actually

but you might need to change some settings, especially scale the x axis accordingly

i mean |AB| can get quite big while |v| ranges from somewhere between 0 to pi (approx. 3.1415)

so the scaling isn't that right

[summer meadow]

This is how it looks now

I limited the x value

[marble skiff]

nice

yeah looks fine

[summer meadow]

This is exactly what i was looking for thx @marble skiff and @soft lily (sry if you didn't want me to ping you)

[marble skiff]

no problem :)

[wary zephyr]

Sorry to be a bother but I've been at this single question for an hour, what would at thr very least be the correct formula?

[neon iron]

get ur own help channel

[wary zephyr]

I did

[neon iron]

no this is someone elses

look at the name of the channel

[lime perch]

go to #❓how-to-get-help

[summer meadow]

do i need to close this?

[lime perch]

if its done, then yes

[summer meadow]

how?

[lime perch]

use .close

[summer meadow]

.close

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[vernal matrix]

That does form the upper bound for the x values, for sure (and note that log_3 is increasing)

You can similarly find the lower bound on the values of x

More simple than that

You said that log_3(3) is 1

Can you make a similar statement, but with -1 instead?

No, not quite

Remember your log rules: if $\log_3(x) = -1$, then what is $x$?

[thorny flame]

@vernal matrix

[vernal matrix]

but we have x/3 instead of x here

So you had $\log_3\pqty{\frac{x}3} = -1$, and found that $\frac{x}3 = \frac13$

[thorny flame]

@vernal matrix

[vernal matrix]

how do you get that?

if you mean x=1, then yep

That's pretty much it, $x$ has to be between 1 and 9 inclusive, you get that $1 \leq x \leq 9$

[thorny flame]

@vernal matrix

[vernal matrix]

Yep, and the fact you know that $\log_3$ is increasing allows you to then say that
[
1 \leq x \leq 9 \implies \frac13 \leq \frac{x}3 \leq 3 \implies -1 \leq \log_3(x) \leq 1
]

[thorny flame]

@vernal matrix

[vernal matrix]

No problem

interesting

Assumedly it's an identity type question or something

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[proven narwhal]

a bit confused, im trying to find the projection of the vector Pp0 onto the direction vector (-3,2,-3) but im getting a 0 when trying to compute their dot product

what have I done wrong?

[tidal wing]

projection can be 0 tho

[proven narwhal]

does that mean the point Q is (0,0,0) ?

[tidal wing]

no

it means that point Q is the same as point P

Pp0 is perpendicular to the line

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[tough anchor]

hello

[tough anchor]

why did they choose a semicircle

i am confused on the residue theorem and how they can change the limits to match the contour of ^^

ok i read more of the textbook and going from -R --> R makes sense and to enclose a pole

but does it matter which pole? and why dont we need to do both

[crude anchor]

look at it a different way

the contour integral over C

is a more complicated problem and distinct

but it can be solved easily using residue theorem

so since there is a "simple" method to solve the contour integral thats great

now they find the original integral I in the solution of the controur integral

and then using that and showing that the part of contour integral over the arc is zero when p approaches infinity

you can find the answer to the original problem

why is it a semicircle? why doesnt it contain a second pole?

those questions are simply answered by why should it be?

they setup a separate problem, the contour integral over the semi-circle, with the goal of finding the answer to the real integral I in it

and they use the residue theorem to help get to that solution

you could use a different contour that encolses a different pole

but i cant think of any contour that would make this easier

the both part of the semicircle represents the real integral I that your trying to solve

so thats why the semicircle is nice

you could in principle have you used a square though

that would just be way more annoying since then polar form cant be used

[tough anchor]

okay i think i understand it more

tyvm

[crude anchor]

glad to hear

[tough anchor]

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[junior drum]

how do you approach part 2

[craggy haven]

combine the two matrices, only the middle column

[junior drum]

that's what kept tripping me up because wouldn't you need to add the entire matrices

so b would become 2b and e would become 2e, etc

[craggy haven]

oh uh

i don't really remember how that works tbh sorry :/

[junior drum]

gotcha

oh yeah one other question I was stumped on was this one

what would the general steps be to solve smth like that

we've only ever done it in class when the examples at the top were columns of an identity matrix

[neon iron]

uh well like

if I'm not mistaken [
\m\det{EA} = \det E \det A
]
where $E$ is an elementary matrix

[neon iron]

what you have is a bunch of transformations of A that transform A by multiplying by some elementary matrix

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[junior drum]

gotcha

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[fickle kite]

suppose G is a group with a binary operation * , if I know that a*b in G can I say that a and b are in G ?

[worthy storm]

not automatically, suppose G is a subgroup of some larger group, then ab can be in G without a or b being in G

[warm crest]

this statement is not always true, if you arbitrarily pick a group that only contains even numbers and * is the usual product of two even numbers, then the ab will always be part of the group, however not a and b have to be even numbers, only a or b, its not exclusive

[fickle kite]

oh ok I see

thanks a lot 🙂

👍

.close

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[proven narwhal]

not sure how to start

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<@&286206848099549185>

[paper dragon]

hello

[neon iron]

for that I remember, you must find a line through P. In this line you must find the orthogonal (in the line) through Q and impose that distance

Remember that if Q is a solution of that line, then Q will have a 0 distance. Q is in the span of a line

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[robust flame]

for the x polynomial the answer is (x-2)(x-4)(x+1)

i know x=y^2/3

i just dont know how to solve question c

[pastel oracle]

just make that substitution

[lime perch]

what is the relationship between x and y

[robust flame]

x=y^2/3

[lime perch]

well, you know for what values of x, the equation is 0 right?

so then what is y

[robust flame]

would i do 4^2/3=y?

and 2^2/3?

[lime perch]

almost

a=b^(1/2) <=> a^2=b

$a=b^{\frac{1}{2}}\Leftrightarrow a^2=b$

[thorny flame]

Flappie

[lime perch]

does this help?

[robust flame]

recpricol power?

[lime perch]

yes

[robust flame]

ah i see

[lime perch]

so y = x^?

[robust flame]

so 4^3/2

3/2?

[lime perch]

yes

[robust flame]

thanks

.close

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[ivory wave]

!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

[ivory wave]

1

[marble gate]

so you know these two triangles are similar, right?

[ivory wave]

I need to find the value of b

[ivory wave]

yes

[marble gate]

which means the little triangle is a scaled down version of the big triangle

[reef fjord]

similar triangles my beloved

[ivory wave]

do I need to find the angle first?

[reef fjord]

nah, its enough to know they all have the same angles

[ivory wave]

interesting I see

so b is 9?

[reef fjord]

no...

[ivory wave]

[marble gate]

similar, not the same

[ivory wave]

12/4=9b?

@reef fjord

,w 12/4=9b

[thorny flame]

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Requested by kensukeken

[reef fjord]

oh i see

[ivory wave]

the answer says its 3

[reef fjord]

it should be 9/b

not 9b

[ivory wave]

oh

[reef fjord]

,w 12/4=9/b

[thorny flame]

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Requested by imyourmoosey

[reef fjord]

:)

[ivory wave]

got it thank you

.close

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[robust flame]

can someone assist me on the last question

[robust flame]

e

2x^3-7x^2+3x+2=0 factorises to (2x^2-5x-2)(x-1)

[hushed ore]

Can you use a calculator?

You must, since it says give answers to two decimal places, so just find the roots of 2x^2-5x-2

[robust flame]

how do i find that

it doesnt solve

[hushed ore]

Use the quadratic formula to find an expression for the roots

and then input those into your calculator

[robust flame]

now what do i do?

[hushed ore]

Do you have an expression for the two roots? Then you should just be able to type them into your calculator to get a decimal version

then you're done

[robust flame]

ah ok

whats the y value?

[hushed ore]

0

roots mean y is 0, so you're finding the roots of 2x^2-5x-2

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[slender sonnet]

Need help with 5

[topaz sinew]

@slender sonnet Has your question been resolved?

[slender sonnet]

<@&286206848099549185>

[polar torrent]

@slender sonnet is tg = tangent? and ctg = cotangent?

Also what is the statement of the original question?

[neon iron]

Is it to simplify?

[topaz sinew]

@slender sonnet Has your question been resolved?

[modest birch]

please help??

<@&286206848099549185>

[outer delta]

x^3

Increasing means it only goes up

In x^2 it's decreasing on negative x values

@modest birch

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[modest birch]

something like this but how come its increasing in negative here 2??

[modest birch]

there is something im not understanging here... left-right they r both increasing

so can @outer delta please explain this plss

[outer delta]

Yes

You read graphs left to right and if you're numbers are increasing (y-axis) the graph is described as increasing. If they're decreasing the graph is described as such

This graph increases on all such points x

@modest birch

2 - x decreases on all
Cos x and sin x increase and decrease
X^2 first decreases till it's parabola than increases for infinity

[outer delta]

Also with power to third any negative you plug in is also going to be negative cause (-) times (-) = (+) times (-) = (-)

So it's larger in the negative direction

Powers are just multiply said number by itself 'the power' amount of times. You can simply plug in negative or positive and see what happens doesn't have to be an actual number

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[floral crest]

How do I solve this?

[modest birch]

icicicicicc @outer delta thankyouuu

[outer delta]

Why is 3 squared but not actually squared (if that makes sense?

[pastel salmon]

I show you an integral that it will be very useful when you transformed your integral:

$\int_{}^{}\frac{dx}{\left( x^{2}+1 \right)^{2}}=\int_{}^{}\frac{\left( x^{2}+1 \right)-x^{2}}{\left( x^{2}+1 \right)^{2}}dx=\\\int_{}^{}\frac{dx}{x^{2}+1}-\frac{1}{2}\int_{}^{}x\cdot \frac{2x}{\left( x^{2}+1 \right)^{2}}dx=\\=\arctan\text{}x-\frac{1}{2}\left[ x\left( -\frac{1}{x^{2}+1} \right) +\int_{}^{}\frac{dx}{x^{2}+1}\right]=\\\frac{1}{2}\arctan\text{}x+\frac{x}{2\left( x^{2}+1 \right)}+C$

[thorny flame]

Joanna Angel

[topaz sinew]

@floral crest Has your question been resolved?

[floral crest]

That's similar but how do I transform it?

[pastel salmon]

first you have to use substituion in yoru case: t = x+1

and next , you have to take out 3^2 = 9

to get 1

adn next apply the other substituion

and finally you get my integral

[floral crest]

Thanks, I'll try to do it tomorrow

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[steady escarp]

what have u tried

where and why are u stuck

[nimble stream]

Yeah, in the future, please don't just post a picture of your homework assignment with no description or explanation.

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[tight rivet]

Graph x^2(x^2-25)

[tight rivet]

Do I just have to keep plugging random points for x or is there an easier way to know how graph looks?

[round gust]

differentiation

[shut obsidian]

Can you determine the zeros?

For that, notice something about x^2 - 25

[tight rivet]

X=5 and -5

[shut obsidian]

Yes, since this can be rewritten into x^2(x-5)(x+5)

There is a third zero too

[tight rivet]

0?

[shut obsidian]

Yes

[shut obsidian]

What's the behaviour for x -> infinity?

[tight rivet]

Y goes infinity in x^2

[shut obsidian]

What do you mean "in x^2"?

[tight rivet]

The graph

[shut obsidian]

It goes to infinity, yes

[tight rivet]

For x^2

[shut obsidian]

Well but we're looking at f(x) = x^2(x-5)(x+5)

Does that go to infinity too for x to infinity?

[tight rivet]

Yeah?

[shut obsidian]

Yes

[shut obsidian]

What about x to -infinity?

[tight rivet]

Y goes -inf

[shut obsidian]

Are you sure?

The leading term in x^2(x-5)(x+5) is x^4.

You should look at the leading term when determining what it approaches for x -> +/- infinity.

[shut obsidian]

Does x^4 go to -infinity for x -> -infinity?

[tight rivet]

Idk how x^4 looks

[shut obsidian]

It looks just like a parabola, basically

,w plot x^4 for -10 < x < 10

But more "pushed down" near the center

,w plot f(x) = x^2, g(x) = x^4, h(x) = x^6 for -5 < x <5

[tight rivet]

Ok so basically looks like x^2

[shut obsidian]

Yeah, almost

[tight rivet]

So do I just plug in random points for x then?

[shut obsidian]

We're looking at how it looks for x -> -infinity currently.

What is x^4 approaching for x -> -infinity

If it behaves just like x^2

[tight rivet]

Y is going positive inf

[shut obsidian]

Yes

[shut obsidian]

Ok, another remark: we said earlier that the zeros are -5, 5 and 0.

Note that -5 and 5 are of odd degree, 0 is of even degree

Because we have f(x) = x^2(x-5)(x+5)

Technically, we'd have two terms that are 0 at x = 0 (x * x)

That's why we call it of even degree

Zeros of even degree always look u or n-shaped

For example, x^2, a standard parabola, also has a zero of even degree at x = 0

or look at this example,

,w plot (x-2)^2 * (x+2) for -5 < x < 5

x = 2 is a zero of even degree.

So it looks u-shaped there

(it could also be n-shaped depending on the factor of f)

[tight rivet]

Ok so I just plot -5 5 and 0

[shut obsidian]

So with this, we know f looks something like this

We don't know yet if it should be the green of purple option but we know it looks u or n shaped

So next, we should determine the extrema of f

Do you know how to do that?

[tight rivet]

No

[shut obsidian]

Have you had calculus?

Derivatives

[tight rivet]

No

But I know basic derivatives

But I probably can’t use that on test cus I’m in precalc

[shut obsidian]

Is this something from your class? Since in that case, we shouldn't use derivatives, I guess

If you haven't had them

[tight rivet]

Yeah don’t use it

[shut obsidian]

Ok, well then we can confirm if it is the green or purple option by plugging in just one point, actually. One x between 0 and 5 or -5 and 0.

If it is positive, then it's the green option, if it is negative, then purple

[tight rivet]

I drew this

[shut obsidian]

That can't be because we have zeros of odd degree at -5 and 5

That means they go through the x-axis

Only at x = 0, we have a zero of even degree

Also what you drew doesn't include the zero at x = 0 at all

[shut obsidian]

So, we have this. From here, we need to plug in some sample x.
Usually, the highest/lowest points lie somewhere in the middle of two zeros

So plug in x = -2.5 and x = 2.5

[tight rivet]

I used 2 and got -84

For both

@shut obsidian

[shut obsidian]

Yep

[tight rivet]

Ok so what do I do with that

[shut obsidian]

So combining all the information, it'll look about like this.

But note that the minimums are not accurate

since we just said "the minimum will be about in the middle of the two zeros"

[tight rivet]

Ok do u know how this works? Why does 5pi/6 not work even tho it is in -90 degree quadrant?

@shut obsidian

[shut obsidian]

Do you mean why sin^-1(1/2) gives you pi/6 on your calculator and not 5pi/6?

[tight rivet]

Yeah

Video says range is 90 to -90

And 5pi/6 is in -90

[shut obsidian]

It's not

[midnight fox]

-90 degrees to +90 degrees means -pi/2 to +pi/2

[tight rivet]

Oh wait it’s in quadrant 2

Nvm

[shut obsidian]

Yes