help-38

[sharp heart]

So, you said 3 was in A n B.

That's the part where the circles intersect.

Now look at every part except that.

Then add them together.

[severe vigil]

yes the complement would be {12, 15, 20}, so wouldn't solving for n be the total number of numbers within that set? making it 3? does the n mean to add them?

[sharp heart]

You didn't answer the first two questions with that reasoning.

You didn't say that the sets you wrote have 1 element and then answered 1.

[severe vigil]

no i know because the set has {3} which is 1 element. since 1 didnt work i put 3, the actual set number and it worked, so i did the same for b

when i answered 1 it was obviously wrong

[sharp heart]

Right. The numbers there are a cardinality of that subset.

So, the intersection of the circles has 3 elements in it. Its cardinality is 3.

[severe vigil]

this is why i did that

[sharp heart]

If you look at A, one part of A has 12 elements in it. The other part has 3 elements.

So, in total, there are 15 elements in A.

[severe vigil]

wait you lost me there

[sharp heart]

The numbers on the diagram aren't set elements.

They're set cardinalities.

They tell you the number of elements in that part.

[severe vigil]

how do you distinguish that?

[sharp heart]

They're not written {3} and {12}.

[severe vigil]

interesting, the professor doesn't write the elements as {x} when writing them in the venn diagram

that's probably where im confused

[sharp heart]

Right, but here we have one number per part.

[severe vigil]

okay i think i understand, ill lyk in a couple mins if i get the other quesitons

[sharp heart]

What happens if you click the circle i next to the diagram?

[sharp heart]

Ahh, OK.

[severe vigil]

adding the numbers i got from the sets worked out anyway to complete the questions, i appreciate you lol cause chatgpt was getting the same thing i was getting

[sharp heart]

Glad you got it 🙂

[severe vigil]

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[novel dagger]

Why do you take the averge value as opposed to just summing up all the probabilities for each value of theta?

[viscid flower]

theyre being lazy

its just a way of describing

really theres a limit and an integral

[viscid flower]

but this is what theyre actually doing

you could try it yourself, and build up to that, if you want

lemme scribble

[novel dagger]

like i get the integral

but dont undestand why you divide by pi

[viscid flower]

divide by pi?

[novel dagger]

yeah my integral gave 3pi/4

integrating over 0 to pi

[viscid flower]

ah, well, its an average

[novel dagger]

1-theta/2 wrt theta

[viscid flower]

$\overline f = \frac{1}{b-a} \int _a ^b f (x) \dd x$

[solid kiln]

jan Niku

[viscid flower]

so i think youre getting b=pi and a=0

[novel dagger]

yeah but like why are you taking the avg value in the first place

[viscid flower]

its just how it works out

i wanna scribble to show

but we can write out like

well i wanna scribble since creating an evenly spaced sum can be a pain in the ass and error prone

okay

[ivory rapids]

F(x)dx

I don’t have the variables Srry

[empty orchid]

Do you need help, or what?

[viscid flower]

theyre posting defn of something? idk

you know sage

i dont think deriving this is gonna be all that helpful

[novel dagger]

like what i dont get is conceptually why are we not just adding up all the probabilities for each value of theta (just integrating)

cuz wouldnt that just give up the total probability

[viscid flower]

well, i guess, what we want is total probability

as in law of total prob

im just trying to be sure

[novel dagger]

I think where i might be going wrong is its conditional or smth

[viscid flower]

yea

i mean

you are doing a bunch of conditionals right

because its like

lets write something easy

[novel dagger]

like for each value of theta it should be pr(that value of theta occurs)*pr(3rd point lies wherever)

[viscid flower]

lets say g for "point good" make sense

[novel dagger]

then add that up for all thetas

[viscid flower]

so youre looking at something like this

A here is "the point is on the same semi circle"

and B_n is each possible configuration of the two other points

you can maybe see where P(B_n) comes in?

[novel dagger]

isnt b_n continuous in this case?

[viscid flower]

since the probability of the other two points is uniform, then $P(B_1) = \dots = P(B_n)$

[solid kiln]

jan Niku

[viscid flower]

yea, were abusing notation

were abusing the sum to mean integral, either

some kind of totalling operator

can you work out what $P(B_n)$ is?

[solid kiln]

jan Niku

[viscid flower]

lets say in the discrete case

lets say you only consider n=2

so, 3 points

r = 1/2, r = 3/4, and r=1

whats the probability of each configuration here

(we know that P is a probability, so $\sum _i P(B_i) = 1$ must be true)

[solid kiln]

jan Niku

[viscid flower]

funny typo lol

[viscid flower]

combine this

[viscid flower]

with this

@novel dagger

hopefully i havent totally lost you

[novel dagger]

im lost tbh😅

[viscid flower]

okay lets start more easy

[viscid flower]

you know what uniform means?

[novel dagger]

this is where i got up to

[viscid flower]

yea, actually, i did this too

i dont think its very helpful though, since you only get to P(A)

but, its alright

its just ... its not the most elucidating thing

I guess I'm only going to explain it in a discrete case

@novel dagger we know discrete probabilities over some interval

right

if theere are 5 points, and they all have the same chance

they are 1/5 each

[novel dagger]

yeah

[viscid flower]

that is P(B_n) here

P(B_n) in your case is the chance that some configuration of the first 2 points happen

each one is equally likely

so if there are n of them

[novel dagger]

yeah that makes sense

[viscid flower]

each one is 1/n

we can rewrite this

$P(A) = \sum _i P( A \mid B_i ) \frac 1n$

we are so much abusing notation.

[solid kiln]

jan Niku

[viscid flower]

this makes it more clear

n is just a number, yea? not an index

n is how many points we are considering, in this discrete example

it can come out

$P(A) = \frac 1 n \sum _i P(A \mid B_i)$

[solid kiln]

jan Niku

[viscid flower]

this looks a lot like an average

and it is. It's the exact average theyre talking about in the problem

make sense?

[novel dagger]

yeah

[viscid flower]

okay

so now

you either believe the transition to average in integral form lol

or, you work out P(A | B_i) and do the limit

im not sure

but im tired

[novel dagger]

wait what is the limit tho?

is it i tending to inf?

[viscid flower]

$P(A) = \lim _{n \to \infty} \frac 1 n \sum _{i=0}^n P(A \mid B_i)$

[solid kiln]

jan Niku

[novel dagger]

ohhh right

[viscid flower]

this is just an integral

nothing fancy

except the function is a probability and its conditional

[novel dagger]

yeah right that makes sense

thanks sm

[viscid flower]

[novel dagger]

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[umbral river]

I am stuck on b) for this Differential Equations problem. I can't figure out how to show that y2 is asymptotically stable. I tried plugging it back into the dy/dt equation, but I (of course) just get 0, as that's where it came from. I don't know how to prove it for a non-trivial number such as y2.

[edgy willow]

stability has to do with the 2nd derivative

so you want to plug into the derivative of r(1-y/K)y-Ey and see if it's negative or positive

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[umbral river]

I thought stability was if the rate of change is toward or away from it?

SO the 1st derivativer

[edgy willow]

that's true too, you can check points a little above and below y2 and see if it's towards y2

but the 2nd derivative is the formal way to show it, otherwise you'd need a graph or something

[umbral river]

OH ok that helps a lot

Thank you

[umbral river]

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[leaden falcon]

Hello my doctor says at x=2 it’s differentiable but it’s not continuous

[leaden falcon]

How is this possible

[fair beacon]

it's not differentiable at x = 2 (for the reason you stated)

[limpid dawn]

A differentiable function implies it's continuous

[leaden falcon]

Yes exactly

I’m in class en

Rn*

I’m talking to him

And he’s telling me it’s differentiable

I keep on explaining it to him and he’s not giving me a reason

He tells me

Since the function from the left and right are the same so it’s differentiable

Doesn’t matter what f(2) equals

[limpid dawn]

But it's not continuous just because left and right

[leaden falcon]

Exactly

I hate this doctor

He’s under qualified and we are all complaining

He just “understood” my point

[limpid dawn]

Ask him to prove it then

[leaden falcon]

And corrected his mistake

I did and he realized he’s wrong

I’ve been trying to convince him for 30 minutes

[limpid dawn]

Good job Bobo

[leaden falcon]

Thanks

Thanks for ur help

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[midnight vessel]

$\int{\frac{\sec^2(x)}{(\sec(x)+\tan(x))^{\frac92}}~\mathrm{d}x}\\$

I tried substituting $u=\tan(x)$ but that didn't work out. I tried to get the numerator in the form of the derivative of the denominator but that required a different substitution of $\sec(x)$ later. After substitutions, the integrals looked even uglier and more difficult.

[solid kiln]

GoldBarley

[past widget]

Time for me to payback

1 + sin x = (sin x/2 + cos x/2)²

Send (cos x)^{9/2} above, times (sec x)² gives (cos x/2 - sin x/2)^{5/2} (cos x/2 + sin x/2)^{5/2}

Cancel terms from denom and you have

(cos x/2 - sin x/2)^{5/2} / (sin x/2 + cos x/2)^{13/2}

Correct me if there's some mistake but it's a dv/v type if you know what I mean

[midnight vessel]

I got here.

[past widget]

cos x = (cos x/2 - sin x/2) (cos x/2 + sin x/2)

[midnight vessel]

Ah, got it, that part confused me a bit.

Ok got it, thanks for the help. It's easy after that.(I realised it's not that easy 😭 )

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[orchid wagon]

partial fraction decomposition

wait

[wanton rune]

let u=x+3

[orchid wagon]

ok this is simpler

[ember ivy]

make (x+3)^-2 ????

[wanton rune]

no, just let u=x+3

and note this means x=u-3

[ember ivy]

t=x+3, notu

*not u

[wanton rune]

sure, you can use whatever variable you want

[lilac flame]

just don't use x

or d

[ember ivy]

using u is also wrong because of uv formula, right?

uv - Svdu

[wanton rune]

u and v are placeholders

like, you could replace u and v with f and g in the intergration by parts formula and it would still be fine

[ember ivy]

okay

So far so good?

[wanton rune]

no

specifically note sum/difference rule

[ember ivy]

I know the sum difference 100% but that's not what I have here

[ember ivy]

this?

[wanton rune]

are you okay 😭

do you know what you are doing

[ember ivy]

the t goes up I just have to switch exponent to -2

which I forgot

[wanton rune]

no...

[ember ivy]

Why not

[wanton rune]

why would you switch the exponent

[ember ivy]

okay then I split the fraction into two and add them

?

[wanton rune]

write down what you mean

[wanton rune]

yes, you forgot an extra integral sign but yes

[ember ivy]

alright, brb

Is this okay?

[wanton rune]

no.

recall the definition of the antiderivative of 1/x

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[frosty hinge]

How do it reach to this answer?
(note: can't use partial fractions, cause that wasn't taught yet. So we use transforming the integrand methods such as adding zero, changing radical to power, perform indicated operation, add zero, use of equivalent forms, etc)

[tulip perch]

think about this:

last term, u can sub u = x^3+1 to solve the integral

[frosty hinge]

Thank you! finally solved it after 2 hours of searching online lol

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[dull temple]

(this is partial fractions lol)

[frosty hinge]

(really??)

[tulip perch]

it works with their restriction of adding 0

x^3-x^3 = 0

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[subtle canopy]

guys i wanna just make sure that my approach is correct now i am thinking to prove that e´->n can be reprsented by e´=>k n where =>k is number of transition steps and analoguly do the same for e->n and finally by transitivity prove the sentnce

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[hidden dew]

<@&268886789983436800>

[vagrant marsh]

oh my god

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[glossy steppe]

Having trouble isolating y

[robust wigeon]

There's no non annoying way to do it, you gotta multiply by y+1 on both sides, and then distribute out and collect every y term

[glossy steppe]

Ugh

Ok thanks

[robust wigeon]

At least to my knowledge*

[glossy steppe]

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[willow mortar]

Que significa conjunto ?

[urban copper]

set = conjunto

like a set from set theory

[solid kiln]

938c2cc0dcc05f2b68c4287040cfcf71

[urban copper]

notice A + B = C implies A ⊂ C and B ⊂ C

[solid kiln]

938c2cc0dcc05f2b68c4287040cfcf71

[willow mortar]

That’s the first thing I’d check

[urban copper]

I don't even know how to prove that statement, but I have struggled with a lot of exercises and sometimes u need to know this tricks

the "A + B = C implies A ⊂ C and B ⊂ C"

[willow mortar]

Check for what value of a T is included in H

[urban copper]

4-2(4) + a^2 = 0

4-8+a^2=0

[willow mortar]

Yup

[urban copper]

a^2 = -4 + 8

a = {2, -2}

[willow mortar]

Good

[urban copper]

answer is d)

[willow mortar]

So now you need to make sure that S and T don’t intersect

[urban copper]

oh, right

[urban copper]

yeah sorry

I was rushing ahead of myself

[willow mortar]

It’s ok

So test that intersection for both values of a

[urban copper]

when a = 2

S = <(4,2,4,4)>

T = <(4,2,4,4),(1,0,0,-1)>

so a = 2 is not valid

when a = -2

[willow mortar]

Yup

[urban copper]

S = <(4,-2,4,4)>

T = <(4,2,4,4),(1,0,0,-1)>

Nice!

[urban copper]

so answer is not d) is a)

[willow mortar]

You made sure S vector is not a linear combination of T vectors ?

Nevermind, the solution is not null so it has to be a

[urban copper]

,w rank {{4,-2,4,4},{4,2,4,4},{1,0,0,-1}}

[solid kiln]

Results provided by WolframAlpha

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Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[urban copper]

🤓

ty for the help

.solved

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[surreal temple]

what can i do when i need to find the diagonal of a quadrilateral but the diagonal is split into two isosceles

[pure fiber]

what data do you have?

[lone basin]

The diagonal is split into two isosceles?

[pure fiber]

I guess the diagonal splits the quadrilateral into two isosceles

[surreal temple]

yes

[pure fiber]

?

[sturdy haven]

it honestly depends on the info that you have

[surreal temple]

can we work with this

[empty orchid]

Is that isoceles?

[surreal temple]

yeah

[empty orchid]

In what manner?

[sturdy haven]

the sides are different lengths

what

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[proven jewel]

is my numbr 1 correct

[vague walrus]

cant see

zoom in

[proven jewel]

here

for no.2

[vague walrus]

oh not sure about the epsilon delta definition

[proven jewel]

can any1 help

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[stable wharf]

wrong channel last time but i’m missing one step cause i keep getting -sinx no matter what, can someone help with it

[safe loom]

1 - cos(2x) = 1 - [1 - 2sin^2(x)] = 2sin^2(x)

[stable wharf]

doesn’t the 1 not subtract the 1 inside

[safe loom]

1 - 1 + 2sin^2(x)

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[potent vault]

I have no clue how to do graphs literally my first skeychong question

[lone basin]

Find the y-intercept, x = ...-asymptote

And make sure you know which direction the graph goes

$4 - x > 0$ must be satisfied

[solid kiln]

King Leo

[potent vault]

X = 2?

OHH X = 4 so y = 0??

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[potent vault]

No

<@&286206848099549185>

[wraith hinge]

Can you show the question again?

[wraith hinge]

This thing is confusing

[potent vault]

Yh

[wraith hinge]

https://www.google.com/url?sa=i&url=https%3A%2F%2Fwww.youtube.com%2Fwatch%3Fv%3DbY0i9Kmuyt4&psig=AOvVaw1VySDmbONQneMWJD1WNRbU&ust=1738297855214000&source=images&cd=vfe&opi=89978449&ved=2ahUKEwiBw9SizpyLAxXEwwIHHTtqKaMQjhx6BAgAEBk

This can help...

[potent vault]

I'll watch it thx

[lone basin]

😭

$$4 - x > 0$$
$$4 > x$$

[solid kiln]

King Leo

[wraith hinge]

So...

x ≨ 4

x cannot be 4

@lone basin Just remember that x cannot be same either bigger than 4

[potent vault]

Does this look right

[wraith hinge]

Seem maybe

But I didn't know this is right or not

<@&286206848099549185> Common who can help this guy

[potent vault]

Alright

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[gray swallow]

In the quadrilateral ABCD AB 14 BC 48 CD 40 and AD 30 find the diagonal AC and the area of the quadrilateral if AC divides its area in the ratio 14:25

[gray swallow]

any suggestions on how i can find the diagonal?

[frosty slate]

Huh?

It said AC divides its area in the ratio 14:25

Find the area then

[zinc ginkgo]

Is it a trapezoid?

[gray swallow]

quadrilateral

[frosty slate]

quadrilateral but is it cyclic?

[gray swallow]

i dont think so

[frosty slate]

U need to check it then apply "bruh man gotta" formula

Its Brahmagupta's Formula

I cant remember the original name so I come up with a way to remember it by "bruh man gotta" then google for the original 😭

First find S = sum of all length / 2

[past widget]

You'll find that ABCD is cyclic

[frosty slate]

Then if (s - one side)*(s - other side) ... do for 4 sides, ...> 0, then its cyclic

A = sqrt((s−a)(s−b)(s−c)(s−d))

[past widget]

,texsp $\frac{14}{25} = \frac{[ABC]}{[ADC]} = \frac{14\cdot 48 \cdot \sin B}{30\cdot 40 \cdot \sin D} = \frac{14 \sin B}{25 \sin D}$

[gray swallow]

ill work on it

[past widget]

I think Blues' idea is more direct :o Use Heron's formula

,texsp $\sqrt{\frac{(14 + 48 + x)(x + 48 - 14)(x + 14 - 48)(14 + 48 - x)}{(x + 70)(x - 10)(x + 10)(70 - x)}} = \frac{14}{25}$

[past widget]

once you're done with this, you can apply cosine law on ADC and ABC to solve for cases B = D, and B + D = 180° and thus get AC

[gray swallow]

thanks

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[void current]

The widths (in meters) of a kidney-shaped swimming pool were measured at 6-meter intervals as indicated in the figure. Use the Midpoint Rule with n = 4 to estimate the area S of the pool if a1 = 18.6, a2 = 21.6, a3 = 20.4, a4 = 16.8, a5 = 15, a6 = 14.4, and a7 = 14.4.

[void current]

hey guys i dont rly kno how to do this problem, any formulas or tips ?

<@&286206848099549185>

<@&286206848099549185>

???

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[tiny horizon]

How do we differentiate this?

[tiny horizon]

Im thinking product rule or chain rule but im not too sure

Im confused what to do with the x^2, do we still reduce the power by 1 when we turn it into cos?

[void berry]

No you don't need to reduce it to one inside the cos just outside when you're doing the chain rule

[tiny horizon]

So it would be

Sinx^2 * 1 + x * cosx^2 * 2x?

so whenever there it is raised to a power it is separate to the sin or cos?

[void berry]

Yes I think that's correct

[tiny horizon]

Ah alr thanks

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[vague walrus]

chain rule should work yeah

[tiny horizon]

Ye I figured it thanks

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[wraith hinge]

hey

[wraith hinge]

isnt this wrong?

shouldn't it be x*y*x^-1

or im missing smth?

[sick zenith]

yeah thats a misprint it should be x*y*x^-1

[wraith hinge]

thnx

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[lilac lynx]

can someone explain how after using Pk the equations becomes that

[meager tulip]

its a simple substitution

u got this right?

so put the value of lhs in here

and u get
(k/2 +1) +1/(k + 1)

[lilac lynx]

yeah i get it until there

i dont understand where this comes from

[meager tulip]

broooo

it a simple sum

jus sum the fraction k/2 and -(k+1)/2

[lilac lynx]

yeah but where is -(k+1)/2 from

[trim jolt]

@lilac lynx Has your question been resolved?

[meager tulip]

i think this is not complete

is there a definition of P

oh my bad its there in 1st pic

so to answer ur question i think they substracted -(K+1)/2 to generate the form at end (1-k)/2(k+1)

then we know for sure that the this

is <= 0

which implies

why (k+1)/2 u can ask?

why not something else?

thats because we need to prove that as rhs in the final eqn

i agree this one is a short but a lil complicated way of doing the proof

so if u want u can look some other proof on this

[trim jolt]

@lilac lynx Has your question been resolved?

[empty orchid]

Oh no, the (using P_k) note is for the line adjacent to it. Not the line below it.

Weird part of the problem statement. I would have proved it like this:
Now $\sum_{i=1}^{k+1}\frac{1}{i}=\qty(\sum_{i=1}^k\frac{1}{i}) +\frac{1}{k+1}\le\frac{k}{2}+\frac{1}{k+1}+1\=\frac{k+2}{2}+\frac{1}{k+1}\ \ \cdots$

[solid kiln]

;(

[lilac lynx]

Ooo thanks i kinda see what i have to do

[lilac lynx]

Thanks ill try this aswell

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[empty orchid]

How would you go about starting with $\sum_{n=1}^{\infty}\frac{1}{2^n\cdot n^2}$? I remember (or misremember) seeing this in a Micheal Penn video once, but I completely forgot how to solve it. Any hints?

[solid kiln]

;(

[midnight vessel]

The terms get small very very fast.$\\$

$\frac12+\frac1{16}+\frac1{72}+\frac1{256}+\frac1{700}+\text{...}\\$

So I guess you could find the sum for the first 5-6 terms and then approximate it since the farther you go, the less difference it makes.

[solid kiln]

GoldBarley

[midnight vessel]

You can approximate it till 10 terms for a better accuracy I guess.

[vernal palm]

0.58224052646501250590265632015968003287921568838685734695129391486614285424

that's wolfram alpha for the first 100

[midnight vessel]

So it converges to ≈0.6

[amber python]

pi^2/12 - log(2)^2/2

thats wolfram alpha for the first infinity

[vernal palm]

yes but why

[midnight vessel]

So I guess it doesn't go past 0.58

[vernal palm]

it looks like the basel problem

[amber python]

it looks like it because it essentially is...

[vernal palm]

yes but it is also multiplied by 1/2^n for every term

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@empty orchid Has your question been resolved?

[past widget]

Consider: $$S(x) = \sum_{n = 1}^{\infty} \frac{x^n}{n^2}$$

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[empty orchid]

Is there a power series for this?

[marble wharf]

are you asking whether there is a closed form for the power series?

something something differentiate twice should probably work

[empty orchid]

Oh, ok.

$S’’(x)=\sum_{n=1}^{\infty}\frac{n(n-1)x^{n-2}}{n^2}$

[solid kiln]

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

[empty orchid]

$S’’(x)=\sum_{n=1}^{\infty}x^{n-2}-\sum_{n=1}^{\infty}\frac{x^{n-2}}{n}$

[marble wharf]

you have to be a bit careful where the sum starts

[solid kiln]

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

[marble wharf]

after you differentiated it once you can actually recognize the series

[empty orchid]

Hmm.

$S’(x)=\sum_{n=1}^{\infty}\frac{x^{n-1}}{n}$

[solid kiln]

;(

[empty orchid]

This must be log, right?

Just shifted by one?

$S’(x)=-\frac{1}{x}\ln(1-x)$

[empty orchid]

Interesting.

[limpid dawn]

great point

[empty orchid]

So now we can do $-\int_0^{\frac12}\frac{\ln(1-x)}{x}dx$ and we are done.

[empty orchid]

Hm?

[amber python]

,w expand log(1 - x)/x at x=0

[solid kiln]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by tearfox

;(

;(

[empty orchid]

Yeah, this is pretty easy to work with.

Power sum, Fubini’s Theorem.

$\int_0^{\frac12}\sum_{n=1}^{\infty}\frac{x^{n-1}}{n}dx$

[solid kiln]

;(

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[plain storm]

r(x)=(1/25) (-x+35)^(3)+5 = blue
f(x)=5 sin((nπ/40) x) = orange

intersect/ root = (40; 0)

what is n and how do i make it so its tangent to blue

[limpid dawn]

https://www.desmos.com/calculator/gu54ko4gio
tangent

If you work with cleaner numbers your computations are clearer

[plain storm]

yeah thats why i changed them

and reopened

[limpid dawn]

you can also set contraints where it should be defined

[plain storm]

i dont understna

d

[limpid dawn]

it's a desmos link

[plain storm]

yeah

why is a the derivative divided by 5

[limpid dawn]

a is stretch factor

so i calculated the derivative of sine in terms of a and wanted to find such a so that the slopes are the same (aka they are tangent in (40,0))

the result was

a = -slope/5

[plain storm]

so you put the derivative of t(40) equal to the derivative of q(40)?

and solved in terms of a

[limpid dawn]

exactly

[plain storm]

ok so for t'(40) i got 3

is q'(x)= 5acos(ax) ???

[limpid dawn]

wha

[plain storm]

what

[limpid dawn]

are u doing this by hand and you also flipped the functions

the slopes are not that clean

[plain storm]

wdym 😭

[limpid dawn]

q'(40) = -2.38 something

[plain storm]

i changed the numbers

[limpid dawn]

lisa you are dealing with cube roots and stupid trigonometric values use a calculator

[plain storm]

i am i am

[limpid dawn]

what do you want to do now?

make it steeper?

[plain storm]

no i wanna see how you did it 😭

[limpid dawn]

t(x,a) and q(x)
t for trigonometric and q for qubic polynomial

you want them to be tangent at (40,0)

so

[plain storm]

oh

[limpid dawn]

t(40,a) = q(40) and t'(40,a) = q'(40)

[plain storm]

yeah

wait ill

write it down

[plain storm]

what do i do now💀

[limpid dawn]

it was -5sin(a(x-40))

so

[plain storm]

why -40

[limpid dawn]

-5cos(a(x-40))

[limpid dawn]

because you sine to be at (40,0)

t(40) = -5sin(0) = 0

[plain storm]

ah ok

[limpid dawn]

so in your final calcs

q'(40) = t'(40,a)
3 = -5acos(a(40-40)) = -5a

[plain storm]

what how

[limpid dawn]

cos(0) = 1

[plain storm]

ohh

[limpid dawn]

are you still trying to build a roller coaster or are you doing calculus

[plain storm]

so a is -3/5

[plain storm]

rollercoaster

[limpid dawn]

yea should be

[plain storm]

ok thank you

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[rancid salmon]

Is this correct

[robust wigeon]

!original

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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 dirge]

-b²/a at the end

and what happened here

the c doesn't get distributed onto b²

and the b doesn't get distributed onto -b

when you bring them into the fractions

[rancid salmon]

Thanks for catching that i was in a hurry and made a mistake i was wondering why it was wrong

[rancid salmon]

How is it not orginal

[neon dirge]

np

[rancid salmon]

Btw for this problem

[robust wigeon]

what. That's a command. Look at what the bot said

[rancid salmon]

I know but why did u do jt

[rancid salmon]

Nvm i figured it out

[robust wigeon]

Because we can help you much easier if you show the original problem so we can understand the context

[rancid salmon]

That problem was wrighting it in terms of v

Not much else i coulda put

[robust wigeon]

Idk what "it" you're refering to so that for example

[rancid salmon]

Wrong react

But

[robust wigeon]

lol

[rancid salmon]

Ig that makes sense

[neon dirge]

impossible to retract :}

[robust wigeon]

It's fine tho looks like lunatic was able to help you regardless in this case

[rancid salmon]

For the problem v is volicty over time and h is height over time where x is time and the others are variables and I'm just trying to prove you can allways simplify down after substitution and cancel out the v and be left with just v^2 so you can allways inverse

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[nova spire]

!status

[trim jolt]

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

[vapid idol]

I dont know where to begin

[nova spire]

ok

so do you know that with uniform probability, the probability can be computed as:

(number of satisfying outcomes)/(total number of outcomes)

how many total outcomes are there?

meaning in how many ways can we choose three points on this grid?

[vapid idol]

9.8.7 = 504??

[nova spire]

here's the problem with what you wrote

I believe your intuition might be "we choose the first point, 9 choices, then we choose the second point, 8 choices, etc..."

[vapid idol]

yes

[nova spire]

however

how do we distinguish

"first point top left, second point top middle, third point top right"

from "first point top right, second point top middle, third point top left"?

[vapid idol]

I dont get it

[nova spire]

basically

[nova spire]

those two

[nova spire]

give you the same set of 3 points

[vapid idol]

yeah

[nova spire]

even though those are two DIFFERENT ways to count it

[vapid idol]

oh right

that makes it harder

so how do we do it

[nova spire]

Well

Just don't order the way you choose your points

And instead choose the three of them simultaneously

Isn't there a type of number

That tells you exactly how many ways to simultaneously choose k things

Among a total of n things?

[vapid idol]

like fraction or what do you mean

[nova spire]

it's a number

Here's an easy list to see if you get it

"how many ways to choose 0 things out of 0 things in total"? 1, you just don't pick anything

more generally, "how many ways to choose 0 things out of n things in total"? 1, just don't pick anything

same thing for choosing n things out of n things in total, you have 1 way to do it, just pick everything

eventually

if you list it like a triangle

it starts looking a little something like this

do you have it now?

[vapid idol]

sorry but no

[nova spire]

combinatorial numbers?

pascal triangle?

doesn't ring a bell at all?

[vapid idol]

no, i have never heard of those ever

[nova spire]

wow

$nCk = \frac{n!}{k!(n-k)!}$

[solid kiln]

rafilou is not not born in 2003

[nova spire]

well that's the formula for combinatorial numbers

[vapid idol]

is the task supposed to be this difficult?

i dont understand anything

[nova spire]

ok at least tell me this

if you have n things in total

how many ways are there to order them?

in a line

how many ways to rearrange n objects

for example, take digits 1,2,3,4

how many ways to arrange them in a line

for example 1234 is a valid way

2314

2413

etc...

[vapid idol]

okay lemme see

24?

[nova spire]

yes

that's 4!

4 * 3 * 2 * 1

in general

how many ways to rearrange n objects?

answer: n!

which is n(n-1)(n-2)...* 1

[vapid idol]

okay

[nova spire]

so

[nova spire]

the problem with your answer there

is that taking "first point, second point, third point"

is that ANY rearrangement of positions 1,2,3

will give you the SAME set of 3 points

so for example say I focus on how to obtain the 3 points on the top row

[vapid idol]

theres one way only?

[nova spire]

I could have started with first point = top left

second point = top middle

third point = top right

[vapid idol]

okay

[nova spire]

that's already one way to obtain the top row, using your method of counting dots in order

but ANY rearrangement of first, second, third

will give me the same 3 points

so top right first, top left second, top middle third is another way

of getting the exact same thing

[vapid idol]

i get it

[nova spire]

my question is: in how many ways can I get the exact same 3 points?

with your method

[vapid idol]

6?

[nova spire]

yes!

so when you compute 9*8*7

each set of 3 points

is actually counted 6 times!

[vapid idol]

ohh

[nova spire]

when we would like to count them exactly once

so as to not overcount

[nova spire]

so how do we fix your result?

[vapid idol]

divide 504 by 6?

[nova spire]

exactly!

9*8*7/6

[nova spire]

which is also 9C3

"how many ways to simultaneously choose 3 objects out of 9"

so now that this is out of the way

total amount of outcomes = 84

[nova spire]

so we got one of the two things needed for this formula

we now have to count how many of those outcomes are "satisfying"

meaning how many form a line

[nova spire]

I think it will be easy counting them using the graph

[vapid idol]

8?

[nova spire]

there's not that many

yeah 8

3 vertical + 3 horizontal + 2 diagonals

[vapid idol]

ok

so its 8 out of 84?

okay tysm

[nova spire]

don't forget to simplify to simplest terms and then you'll be done

[vapid idol]

alright

2/21

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[sweet turret]

how do i start part C

[sweet turret]

whats the relation of normal vector and plane equation

[red mountain]

notice that z=1+2f(x,y)

[sweet turret]

yes

and how do i link those two

[gilded pulsar]

For a surface defined by the equation 0=f(x,y,z) the tangent space is given by Kern Df. Since the gradient is the differentials dual vector, for a given tangent vector v <v, grad f> = 0

[sweet turret]

wha

[gilded pulsar]

So grad f is the surfaces normal vector

[sweet turret]

whats grad

[gilded pulsar]

The gradient

[sweet turret]

so basically partial derivative of x,y,z is the normal vector

[gilded pulsar]

The gradient

Which is the vector made up by the partial derivatives

[sweet turret]

den how do i find the normal vector

is the gradient vector basically normal vecotr?

[gilded pulsar]

I just said if you have a surface given by f(x,y,z)=0 then grad f is a normal vector

[sweet turret]

oh lmao

okok

thank you

luv u

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[woeful gorge]

I did eveerything but got k=1/3 but my textbook says k=1/15

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[woeful gorge]

.close

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[barren tinsel]

I have to check if the function is descending in R

for me it is

but the answer says no

i dont understand why

[lethal valve]

is that arctanx??

[barren tinsel]

@lethal valve Yes

[limpid dawn]

can you explain why

[barren tinsel]

I've drawn it

the |arctan| is descending lal the time

all*

and then the -log is of course descending always

@limpid dawn i can show u my drawing

[lethal valve]

for me it is also yes 😭😭

[barren tinsel]

maybe the answer is wrong?

@limpid dawn what do u think

[limpid dawn]

i don't think so

[barren tinsel]

alright

so the function is descdending in R

[limpid dawn]

|arctan(x)| = -arctan(x) since it's defined for x <= 0

so derivative -1/(1+x²) < 0

-log_2(x) = -1/(xln(2)) < 0 too if x > 0

[barren tinsel]

OOH i may know why

maybe I cant type that is descdening in all R because those are two different intervals

and when writing down the intervals we dont sum them

we like seperate them by a comma?

what do you think

so it is more like the function is descnedning in
(-infinity;0>, (0;infinity)

therefore it is not descdending in all R

am i right or is it a bunch of nonsense

[limpid dawn]

R = (-oo, 0] U (0, oo)

[barren tinsel]

but when we like write down those intervals while writing down the monoticity we cant sum them

we seperate them by a comma

[barren tinsel]

like i did here

[limpid dawn]

if it descends in "(-infinity;0>" and "(0;infinity)" then it does descend everywhere?

[barren tinsel]

i mean yes

but maybe the two functions have to be connected

to make it in all R

damn mb i will just ask my proffesor

[limpid dawn]

ok

i have one clue

it's not continuous in x = 0

therefore it's not differentiable in x = 0

so technically you can't say what the monotony is in x = 0

[stable crescent]

hello, whats the question OwO

[barren tinsel]

@stable crescent

[stable crescent]

descending as in it monotonically decreasing?

[barren tinsel]

Yes

[barren tinsel]

i mean you kinda can tell because when you draw the function u see it

[limpid dawn]

yea but mathematically no

that's most likely why

[barren tinsel]

alright I will jsut wait for @stable crescent to look at it and then im gonna close the thread and just ask my proffesor 😂

[stable crescent]

Since its a piecewise-defined qustion, you can just study monotony over each interval where its defined

obviously it isnt continuous at 0

so its useless to study it at 0, so study it for x <0 and for x > 0

[limpid dawn]

decreasing would also be equivalent to a < b => f(a) > f(b), but take a = -1 and b = 0.1 then f(a) = 0.78 and f(b) = 3.3 so it would contradict f(a) > f(b)

[barren tinsel]

okay guys one more question haha.
Why not all elements in this sequence are not positive?

one more thing I was thinking about a lot too

[stable crescent]

it isnt continuous

[limpid dawn]

i know.

[stable crescent]

thats not the same f , though

[limpid dawn]

?

[barren tinsel]

damn guys I will just go about asking my proffesor thank you

.close

[trim jolt]

Channel closed

Closed by @barren tinsel

Use .reopen if this was a mistake.

[stable crescent]

@barren tinsel it isnt decreasing nor increasing

[barren tinsel]

understood

[trim jolt]

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• Show your work, and if possible, explain where you are stuck.
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[unkempt kindle]

Hey there, I'm currently preparing for my abstract algebra exam and I worked on a question asking if 3Z/(6) is a field, or in other words, if (6) is a maximal ideal in 3Z, which it should not be (i guess), because (3) is a "bigger" ideal than (6). Is this reasoning correct?

[mighty prism]

Right, (3) contains (6) so (6) is not maximal

[unkempt kindle]

Ah okey, thanks 🙂

.close

[trim jolt]

Channel closed

Closed by @unkempt kindle

Use .reopen if this was a mistake.

[trim jolt]

Available help channel!

Send your question here to claim the channel.

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

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[junior viper]

is m = 60 correct?

[lethal valve]

nope... its not

[junior viper]

oh

what s it

[lethal valve]

should try solving again

[junior viper]

help#

[lethal valve]

try again yourself

[junior viper]

im stick

stuck

[kindred pier]

how did you get 60?

[lethal valve]

you can use the cross multiplication method

[junior viper]

whats that

wha t did i do wronh

wrong

[kindred pier]

you did m=20L, not 20L³

[junior viper]

but i already cubed it before

[kindred pier]

and you need to cube it again

[junior viper]

thast how i got 8

oh

[kindred pier]

m is proportional to the cube of L

[junior viper]

so its 20 cubed

[kindred pier]

20 is the constant of proportionality, not L

[junior viper]

o

so its 3 cubed

81

i mean 27

sorry

im stupid

27 x 3 then

which is

[kindred pier]

we all make simple mistakes

[junior viper]

81

[kindred pier]

why are you multiplying by 3

[junior viper]

uh]

wait

i dunno

this so conffusing

so is m 81

[kindred pier]

Remember, you started with $m\propto L^3$. You correctly wrote the equality as $m=kL^3$. You were given $M=160$ when $L=2$, giving you $160=k(2)^3$, you solved for $k$, getting $k=20$. So, your proportionality is $M=20L^3$. So, what is $M$ when $L=3$?

[junior viper]

what the f

[solid kiln]

SWR

[junior viper]

is it not 27 x 3 then

which is 81

[kindred pier]

where are you getting 3?

[junior viper]

from l

L

[kindred pier]

then where are you getting 27?

[junior viper]

L cubed

[kindred pier]

then what happened to k?

[junior viper]

what was k

20

oh yeha

20 x 27

540

yoo

thats so difficult

[kindred pier]

I think you just got lost in a the sauce a bit

[junior viper]

for sure

how is this even maths man

[kindred pier]

I mean, you are doing arithmetic

I don't know what else it could be if not math

[junior viper]

why is the porportional sign a fish

[kindred pier]

why is any symbol anything?

[junior viper]

whats the point of direct proportoo

[kindred pier]

wdym?

[junior viper]

proportion

like where would you use it

[kindred pier]

a lot of places

Mostly science and research

[junior viper]

oh

[kindred pier]

Proportionality is a good way to determine correlations of two events

Like, how much longer does your commute take based on how many cars are on the road

How many people are born each year given the current total population

How fast does a virus spread given how many people are already infected

[junior viper]

how would you even work that out

thats next level

you need to make your own factors

[kindred pier]

Take several data points, and see if they make a trend

[junior viper]

thats socomplex

can i try anoter question

another

one sec

lemme work this out

is it (A) = 50X

and B) 17500

[kindred pier]

no to both

I think you got something backwards]

[junior viper]

nooo

is the mistake /10

i dont remember why I divided by 10

oh

[kindred pier]

dividing by 10 leaves you with 1=50k

[junior viper]

/500?

which gives 50

so it was right

o

50 x 10 is 500

[kindred pier]

$\frac{10}{500}\ne50$