help-0

[broken pivot]

Start with an equilateral triangle and split it into two parts

[plush valve]

I forgot how to calculate the leg

[plush valve]

so 6sqrt3?

[broken pivot]

No; notice how the hypotenuse of [the given 30-60-90 triangle](#help-0 message) is 2, not 1

[plush valve]

I mean this triangle, and from what you said i think 3sqrt3

[broken pivot]

✅

[plush valve]

thx

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[spring lodge]

.

A

[spring lodge]

What is

[( cos {2x} ) ] ^2?

[broken pivot]

$\cos^2(2x)$

[ocean seal]

King Leo

[spring lodge]

Sure?

[broken pivot]

Youre trying to simplify this?

Or remove the double angle?

[spring lodge]

Lemme rewrite on a board

I don’t get this

I got

Nvm

Stupid me

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[zinc haven]

do the notations defined in the proof's one direction have a scope beyond the direction?

[zinc haven]

particularly, can i use the fact that e_1, ..., e_n is a basis of W in my <= direction without having to redefine it

this is the whole proof if it's helpful

i think the <= is wrong

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[zinc haven]

i just need to replace the f's with Se's

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[earnest lance]

I need help I keep getting 5.2 but my friend gets 6.2 can someone double check

[prime badge]

5.2

[alpine sable]

its 5.2

[earnest lance]

ok

ty

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[neon remnant]

Trying to figure this out but I feel like im not getting close to the answer

[fierce cipher]

what is lim(g(x)) when x -> 3

[neon remnant]

I tried factoring and whatnot and got 6

[fierce cipher]

what is g(3)

[neon remnant]

are you asking if i were to replace k with 6 and do 6(3)? That'd be 18

[fierce cipher]

no

im just asking what is g(3)

[neon remnant]

0?

I mean replacing x with 3 will give 0/0

[fierce cipher]

no, you replace it at the 2nd function

[neon remnant]

k(3)

[fierce cipher]

yep

so 6 = 3k

that's the requirement of a function to be continuous at that point

[neon remnant]

I'm a bit confused. Is that just the end of it?

[fierce cipher]

yep

[neon remnant]

How does 6 = 3k make it continuous?

[fierce cipher]

requirement of continuity is lim g(x) at x -> c = g(c)

[neon remnant]

So then what exactly is the value of k? 6? or 2 since 3x2=6?

[fierce cipher]

ye it's just 2

[neon remnant]

Oh okay. Damn this is mind boggling

I think I understand though. Thank you for the help

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[ebon widget]

If some mixed strategy nash equilbria of a game played on a finite set of size n convergence to a sequence S as n approaches infinity, does that imply the sequence S is a nash equilibrium for the same game played on a countably infinite set? For example, if [(\tfrac{1}{2}, \tfrac{1}{2})] is a nash equilibrium for a size 2 set, [(\tfrac{1}{2}, \tfrac{1}{4}, \tfrac{1}{4})] is a nash equilibrium for a size 3 set, and [(\tfrac{1}{2}, \tfrac{1}{4}, \tfrac{1}{8}, \dots \tfrac{1}{2^{k-1}}, \tfrac{1}{2^{k-1}})] is a nash equilbrium for a size k set, does this imply that [(\tfrac{1}{2}, \tfrac{1}{4}, \tfrac{1}{8}, \dots)] is a nash equilibrium for the game played on a countably infinite set?

[ocean seal]

Wife Receiver

[normal bone]

Is this tuple a mixed strategy for a particular player?

[ebon widget]

yes

[ebon widget]

in the game I was analyzing it was a mixed strategy for all three players

[normal bone]

I see

[ebon widget]

My question is more general though

the game I was analyzing was the game described in this video https://www.youtube.com/watch?v=9vdElgo-xNM

[normal bone]

I would say it is plausible, but there may be a tricky counterexample lurking

[ebon widget]

yeah I would say that as well, I was hoping for a proof or a counter-example

[ebon widget]

I thought maybe compactness could apply here but I don't know much about translating game theory problems into first order logic

[normal bone]

Yes, I don’t know how to prove this sort of thing

[ebon widget]

thanks anyway

[lone heart]

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@ebon widget Has your question been resolved?

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@ebon widget Has your question been resolved?

[ebon widget]

<@&286206848099549185> Anyone familiar with statistical game theory?

[alpine sable]

No?

[charred flint]

just guessing but it should be, if the leftover probabilities tend to zero and a nash equilibrium even exists in the infinite case

like if you reward the highest number chosen instead it's bad

[ebon widget]

yeah that was a counter-example I came up with

but the sequences don't converge

[charred flint]

apparently there's theory about continuity of nash equilibria if you change parameters

[ebon widget]

oh interesting

[ebon widget]

Is this related to compactness or is it unrelated?

because going from finite solutions to infinite solutions felt like compactness

[charred flint]

no clue, I just know compactness is part of some existence condition for infinite choices

[ebon widget]

thanks

[lone heart]

@ebon widget Has your question been resolved?

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@ebon widget Has your question been resolved?

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@ebon widget Has your question been resolved?

[ebon widget]

<@&286206848099549185> anyone else familiar with nash equilibriums? If not I’ll close the channel.

[ebon widget]

F

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[spice dawn]

excuse me what and how

[spice dawn]

this dont make any sense

how am i supposed to figure it out without any equal sides or parallel lines :(

[opal portal]

the sum of all exterior angles is ..?

[spice dawn]

uh

idk

[opal portal]

take a guess

[safe meadow]

Remember interior + exterior = 180

[spice dawn]

interior + exterior is 360 tho

right

or no

[safe meadow]

U sure g

[opal portal]

focus on my question

[spice dawn]

1440?

[opal portal]

no

[spice dawn]

oh

[opal portal]

its always 360

for any polygon

[spice dawn]

oh i thought you meant all the exterior

[ocean fossil]

That is all the exterior

[opal portal]

if you sum up all the exterior angles it will be 360

[spice dawn]

how so

im bad at math :(

[opal portal]

you can imagine sliding all the edges to all converge on one point

[ocean fossil]

This is defined as exterior angle

[spice dawn]

ohhhh

[ocean fossil]

Whoops I didn't realize the background was see through

[safe meadow]

Or

If you have 5 side

Then

You will have 5 interior and 5 exterior right

Interior + exterior = 180

5 * 180 = 900

The sum of the interior is 180(n-2)

So we get 180n-180(n-2)

We get 360 for any n

[spice dawn]

ah

[opal portal]

anyways you need to find am expression for the sum of the exterior angles

[spice dawn]

ah ok ill come back when i do

i think i can get this

[noble gyro]

The sum of the exterior angles of any polygon is 360°, regardless of the number of sides.

[opal portal]

• 72

[noble gyro]

Guh

46

[spice dawn]

ok so i did some calculations and i got (180-x) + (180-2x) + (180-3x) = 396

im stumped

im just trying to figure out x rn so its easier

[safe meadow]

6x + 72 + 46 + 98 = 360

Solve for x

Here

[spice dawn]

24?

ok i fact check it is 24

i can now solve on my own thanks

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[keen glade]

why the denominator do fractions I'm so confused

[safe meadow]

Could you explain more clearly

[keen glade]

Why does negative numbers make it fractions

[safe meadow]

Negative in exponent ??

[keen glade]

Okay for question one

Like

[safe meadow]

Ohhhhh

Here here

[keen glade]

How they get 1/16 and 1/4

[ocean whale]

Because $a^{-n} = \frac{1}{a^n}$

[ocean seal]

CaptainNova22

[keen glade]

ohhh

[safe meadow]

True

[ocean whale]

Exponent rules

[keen glade]

wait whats the ab^x thing again

How do I find key points

[ocean whale]

It just looks like the key point is the y intercept

[keen glade]

Oh okay

For 5

How do I get 3/16

[ocean whale]

When you plug in -2 for x?

[keen glade]

Yes

[ocean whale]

So if you plug in -2 for x, you get 3 * 4^-2

[keen glade]

Yes

[ocean whale]

Appl the exponent rule I gave you

To simplify 4^-2

[still loom]

Help guys

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[keen glade]

Uhmmm is it 1/16

[ocean whale]

Yes

[keen glade]

Kaboom

[ocean whale]

Then you have 3 * that

[keen glade]

hmmm

3/16 😛

Woah that exponent rule rocks! 😼

Question, what's a and b again I lowk forgot

[keen glade]

uhh

Hmmmm

so what's key point?

[lavish seal]

15 no. Plz

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[fickle musk]

put z^2 = t

then set discriminant as less than zero

[ocean whale]

As I said before, it looks like a key point that was used in you're worksheet, is the y intercept

[keen glade]

Okay sorry I forgot 😭

[ocean whale]

This is not their channel so please don't add to the clutter

[keen glade]

Okay thanks brochacho 🤑 I'm gonna attempt the homework 😼

.close 🤑

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[lavish seal]

I am doing that thing only but ans is not matching. .😭

[fickle musk]

hol up

is that z complex number

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[lavish seal]

Yes

[fickle musk]

not sure then

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[digital pecan]

The purple line has an argument of 5pi/6

[broken pivot]

Yooure using trig on this?

[digital pecan]

The trig isn’t working- I’ll explain

The x intercept of -1 is horizontally 2 away from the blue dot

Thus if we form a triangle, we will get something like this

However, this would make the vertical distance to the blue dot 1 (using sine rule or special triangle or whatever) which it isnt

[broken pivot]

Ok i see what youre trying to say

[digital pecan]

As you can see in DESMOS, it’s a bit further than 1

[digital pecan]

Yeah, I’m so lost why it works out like this

[broken pivot]

Can you give the equation of the line?

[digital pecan]

arg (x +iy-1 +2/sqrt3i)

[broken pivot]

Idk what arg means 🙃

Bye

[digital pecan]

Can anyone else help me

[broken pivot]

You can ping @ Helpers in 2 mins

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[alpine sable]

A: if Quant A >, B: if Quant B >, C: if equal and D: if no conclusion

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[vale ginkgo]

Hello, can someone please help me

[vale ginkgo]

i just don't know how the answer is 16.05

I found the original sum of the values 15.6 x 20 = 312

since jessie mistyped

i did 312 - 10.6 + 1.6

which is 303

so i did 303/20 = 15.15

[lone heart]

@vale ginkgo Has your question been resolved?

[fierce cipher]

yeah new mean should be 15.15

only the smallest data got changed

[vale ginkgo]

yeah but my answer sheet says 16.05

thats why i'm confused

[violet jetty]

It becomes 16.05 if the value 1.6 is removed and replaced with 10.6. The wording is pretty bad but if the mean = 15.6 case was produced with the erroneous dataset (ie. with the mistyped 1.6), the "real" mean with the correct data (10.6) will become 16.05

Its a wording thing so I wouldn't think of it much, at the end of the day you succesfully understood the intent/technique of the problem, which is what actually mattes

[vale ginkgo]

ahh

i see

thank you very much

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[static wagon]

how do i do question 2?

im stuck on what to do next

is it right to do 4(3M)?

since we substitute 4^k - 1

[fierce cipher]

substitute n = k + 1 then you make it so that the term (4^k - 1) appears

[opal moss]

n = k+1

[static wagon]

but i already substituted k+1

[fierce cipher]

then do the latter

make (4^k - 1) appear

[static wagon]

havent i done that as well

4^k x 4 - 1

u can rearrange it that it becomes 4 x 4^k - 1

[fierce cipher]

LMAO

NO

the whole (4^k - 1) term

that is not even 4^k - 1

[static wagon]

then can u guide me through it??

[opal moss]

You could do it this way: You have 4^k−1=3m
so 4^k=3m+1
Now multiply both sides by 4
4^(k+1)=4(3m+1)=12m+4=12m+3+1=3(4m+1)+1=3n+1

then u could go like

4^(k+1) = 3n + 1

4^(k+1) - 1 = 3n

[static wagon]

but i already have that

[opal moss]

wait um

[static wagon]

lol

[opal moss]

you're showing its divisible

[static wagon]

yes

[opal moss]

here

see how it equates to 3n

[static wagon]

yea

[opal moss]

u dont have that in the question

[static wagon]

yea?

[opal moss]

thats what induction is

wym yea

[static wagon]

this is a different type of induction

look at question 1

that’s the process

[opal moss]

my dude

thats the same thing here

you just have to prove that 4^(k+1) is divisible by 3

doesn't always have to substitute 4^k in the second step

[opal moss]

or theres another way

slightly weirder way but it would work with what you're asking for

[static wagon]

what is it

[opal moss]

4^(k+1) -1 = 4^k . 4 -1

and then 4 = 3+1

[static wagon]

yep

[opal moss]

4^k(3+1) - 1

3(4^k) + 4^k - 1

3(4^k) + (4^k - 1)

3(4^k) + 3m

3(4^k + m)

[static wagon]

=3n

ahhh

that makes more sense

i was looking for that solution

tyty

[opal moss]

weird way tho 😭

[static wagon]

it's how they want me to solve it

[opal moss]

okye understandable

[static wagon]

thanks again

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[radiant hedge]

Why am I wrong about uniqueness?

[radiant hedge]

Is it because it's homogenous?

[lone heart]

@radiant hedge Has your question been resolved?

[vale crag]

What condition do you have for uniqueness ?

@lean locust

Oops misping lol

[radiant hedge]

this but first order

I said it's unique because the lesser order is given

[vale crag]

Yeah your equation aint linear second order tho here

linear being the most important issue here

You don't have another more general theorem for first order ?

[radiant hedge]

I found this when I looked around a little

[radiant hedge]

That's why I asked this as a follow up question

[vale crag]

What's the theorem you have tho ?

Even if the equation is autonomous you could still apply it if we're thinking of the same thing

[radiant hedge]

this is Picard-Lindelöf but I'm not sure if I fully understand it

[vale crag]

Yeah

You haven't stated the theorem yet

You just posted a 1st order cauchy problem

[vale crag]

But sure ask away

[radiant hedge]

How do I apply it?

[radiant hedge]

Do I need to find f here?

[vale crag]

Yeah

[vale crag]

Well the theorem tells you the solution is unique under some conditions for f

Check them that's all

[radiant hedge]

How do I check them when idk what f is?

[vale crag]

Well the RHS is f

f(t,y) = cube root of y^2 -3y +2

[radiant hedge]

So do I put t in place of y and see if they're equal?

y(0) = 2

2^1/3 =/= 0 So not unique?

[vale crag]

No?

[radiant hedge]

What does f(t,y) mean? It's my first time seeing a function take 2 variables

[vale crag]

Well now you're initiated

Yes a function which depends on two inputs

[lean locust]

?

[vale crag]

Misping

I typed vol and it switched to you

[vale crag]

Derivative of y is allowed to depend both on time and the current value of y, that's how you can interpret it

But in your question it doesn't depend on time

[radiant hedge]

So is it the same as f(y) here

[vale crag]

Sure

[radiant hedge]

So shouldn't it be unique in the neighborhood of every t value?

[vale crag]

At some point you should actually state picard-lindelof

You're just saying whatever now

[radiant hedge]

Oh wait I only read the big parts

[vale crag]

The not formula is the important part

[radiant hedge]

It says it should be Lipschitz continuous in y

[vale crag]

Yes

[radiant hedge]

What's "Lipschitz continuous"?

"a Lipschitz continuous function is limited in how fast it can change"

So f'(y) should be bounded from above?

[vale crag]

For some constant K, |f(x1)-f(x2)| <= K |x1 -x2| for any x1 x2

But yeah if f is differentiable, it means the derivative is bounded

[radiant hedge]

it's not differentiable since the limit is different in 0+ and 0-

Is that right?

[vale crag]

Sure it's not differentiable

But non differentiable functions can be lipschitz continuous also

Say abs(x)

Also you have to be careful, our f is differentiable at 0

You just looked at the cube root, but you forgot what's inside it

[radiant hedge]

oh sorry I meant 2

because y(0) = 2

[vale crag]

Yeah exactly

You're interested in what happens around y=2

[radiant hedge]

How do I tell if they are lipschitz continuous if they are non differentiable?

[vale crag]

You check the more general condition here

That's the actual definition

[radiant hedge]

Can K be anything here?

[vale crag]

Well something that doesn't depend on x1 x2

Any constant

[radiant hedge]

Can it be inf?

Or a really big number

[vale crag]

A real number constant

Infinity isn't a real number

It could be super big tho yes

f(x) = 849159242894x is lipschitz continuous for example

[radiant hedge]

I'll try it on this one

[vale crag]

But yeah our f'(y) is unbounded around y=2

[radiant hedge]

|f(1)-f(0)| = K|1|

[vale crag]

You can't just check two inputs

It has to work whatever the two inputs are

x1 and x2 are arbitrary

[radiant hedge]

Wouldn't it always work as long as K is big enough?

Since they are both positive numbers

oh wait I think I got it

x1 - x2 should increase faster than f(x1)-f(x2)?

[vale crag]

f(x1)-f(x2) should be roughly as big as x1-x2

Not too big compared to it

[radiant hedge]

but for example if f(x1)-f(x2) is 999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999

and x1-x2 = 1

wouldn't it still work with K = 999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999

[vale crag]

Well the problems might happen when x1-x2 is super small yeah

[lone heart]

@radiant hedge Has your question been resolved?

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[warm quarry]

Doesnt MVT say that f must be continuous on [a,b] and not only (a,b)?

[robust geode]

Here it's a weird application of the MVT

It's not on f, it's on f'

I think

Wait

[warm quarry]

this is what the theorem says

[robust geode]

Welp if that's what you have as the MVT then yeah

My bad for that

Well your function is differentiable

[warm quarry]

i know but both must apply right?

[robust geode]

Yeah I guess it's just your professor who didn't want to write too much

But with all rigor it should have been

f from [a,b] to R, continuous in [a,b], differentiable in (a,b)

But talking about being continuous in a point a invokes the idea that there is some space left on both sides of a

But as you can see, on the negative side of a, there's none

So that's where there is some technical terms used where we say that the function is continuous on the right side, or ... right-side continuous I think ??
The same way, you would have right-side differentiable

But it might not be really important for the purpose of that course you're having

[warm quarry]

aaaah okay

ill just let it be then

thank you :)

.close

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[peak epoch]

$\int_{0}^{\infty}\frac{t^p \ln{t}}{1+t^2}\dd{t}$

[ocean seal]

OHHELLNAH

[peak epoch]

Any ideas?

[broken pivot]

What is p

[peak epoch]

$p \in (-1,1)$

[ocean seal]

OHHELLNAH

[peak epoch]

calculate for p

[broken pivot]

?

[broken pivot]

Based on?

[peak epoch]

what?

Thats the problem

[broken pivot]

How can you simply solve for p

[peak epoch]

yeah you can

[broken pivot]

!original

[lone heart]

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.

[peak epoch]

That is the original problem

solve for p\in (-1,1)

[broken pivot]

Are you supposed to find for what values of p the integral diverges?

[peak epoch]

Its a gamma, beta function problem

[broken pivot]

Oh then idk

[lone heart]

@peak epoch Has your question been resolved?

[tacit arch]

what does "solve for p" even mean

do you mean find what p the integral converges?

or evaluate the integral in terms of p?

[little drum]

The question is to compute the integral

[tight pier]

Maybe Feynman works

[lone heart]

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[warped topaz]

Im not really sure how to do this one

I think I know the first steps but im kind of just guessing

Do I unfactorize (idk if thats a word)

Then get the derivative and set it equal to zero

[tacit arch]

expand?

[warped topaz]

Yeye

Thanks 😅

[tacit arch]

yes use FOIL multiple times

(a+b)(c+d) = ac + ad + bc + cd

[tardy stag]

imo it's easier not to do that

but you can

[warped topaz]

What would you suggest

Cause I was just guessing anyways hehe

[tardy stag]

what's your plan afterwards?

[warped topaz]

I dont really know... I know that the values for t I get would be the values where C crosses the origin

But I dont really know how to get that into a line

Oh wait

Getting the derivative and setting equal to zero would give me the slope right?

Or the tangents

and with the tangent and the point I could make the equation of a line

No?

[buoyant saddle]

why to zero?

dy/dt / dx/dt

[tardy stag]

your curve is in blue and the line you want is in green

[warped topaz]

Yeye

[warped topaz]

I thought thats what you do to get the slope of a tangent at a point? Take the derivative and set it equal to 0

[buoyant saddle]

no that’s to find critical points for relative max/min

L(x) = f(a) + f’(a)(x-a) is the tangent line

[warped topaz]

Oh oops

[buoyant saddle]

you need the derivative at the point of interest

in this case it’s the origin

notice that at the origin both x and y are zero, what value of t makes both x and y zero

(they have a common factor)

[warped topaz]

t = -3/7

[warped topaz]

Was the equation written this way already factorizied just so that I could see this easily?

[buoyant saddle]

yes it was

[warped topaz]

Ah I see now

[buoyant saddle]

even lined the factors up for you

[warped topaz]

So we take the derivative of -7t-3 with respect to t?

[buoyant saddle]

why?

[warped topaz]

Idk I thought thats what you meant

Oh wait

I get it now

We sub in t = -3/7

Right?

[buoyant saddle]

sure but that’s not the function

[buoyant saddle]

i mean you already know the x,y coordinates

you hold off on plugging in until you get the derivative

[warped topaz]

Ah ok

[buoyant saddle]

we were told it was at the origin to begin with

[warped topaz]

So we take the derivative of x(t) and y(t) with respect to t first

[buoyant saddle]

you still need f’(a)

yep

[warped topaz]

whats f?

[buoyant saddle]

i was referring to the equation i wrote earlier because i was too lazy to say the derivative evaluated at the origin

[buoyant saddle]

^

[warped topaz]

Oh haha

[warped topaz]

Wait can you explain this please

Where does the f(a) + f'(a)(x-a) come from

[buoyant saddle]

well you know that the equation of a line in general is of the form y = mx + b

m is the slope

so if we want a line to be tangent to some curve then it must satisfy two conditions, firstly it has to pass through the point and secondly it has to have the same slope of the curve at that point

[buoyant saddle]

f’(a) serves as the m here

and we write (x-a) so that when evaluated at a that part goes to zero and we’re left with f(a) meaning it goes through the same point

this is just point slope form

reorganized

y - y_1 = m(x-x_1)

[warped topaz]

1 moment im just rereading a few times to understand 😅

[buoyant saddle]

if you evaluate it at x = a then the (x-a) bit is zero

leaving you with f(a)

[warped topaz]

Only because it goes through the origin, right?

[buoyant saddle]

no this is the general form of any tangent line

it’s just point slope form

maybe you’d like an example

[warped topaz]

yeye pls 😅

[buoyant saddle]

let f(x) = x^2, say we wanted to find the line tangent to f at x = 4

well in order for a line to be tangent to a curve at a particular point then it must have the same slope as the curve at that point while also having the same y value at that point so that they "touch"

well the slope at x = 4 is just f’(4)

f’(x) = 2x so f’(4) = 8

so we want a line with a slope of 8

but we also want a line that touched the curve at x = 4

but f(4) = 16

so we need a line that passes through (4,16) with a slope of 8

let’s call the line L(x)

then L(x) = 16 + 8(x-4), notice that the slope of this line is just 8 but it’s clear when we write it in this form that it also passes through (4,16) since letting x = 4 yields 16 + 8(4-4) but 4-4 is zero so we just get 16

and we can generalize this for any curve

say we wanted to find the tangent line to some curve at x = a

well then the line must have a slope of f’(a)

and it must pass through (a, f(a))

hence L(x) = f(a) + f’(a)(x-a)

letting x = a will always make the f’(a)(x-a) portion zero leaving us simply with L(a) = f(a) which is what we wanted

[warped topaz]

Ok I think I get it but I will reread a few times to make sure I understand

So 1 moment again

Charbit 🥰 ❤️

Ok I think I get it now

Like if f(x) = x^3 and x = 2

f'(x) = 3x^2

Subbing that into f(a) + f'(a)(x-a)

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

8 + 12(x-2)

[buoyant saddle]

yep

[warped topaz]

so the equation of the tangent is y = 12x - 16

Alright I got it, thats a neat trick

Thats called point slope form?

Or whats the name for it

[buoyant saddle]

point slope form is y - y_1 = m(x-x_1) or y = y_1 + m(x-x_1)

[warped topaz]

Ah sry

[buoyant saddle]

y = mx + b is just idk slope intercept form i think

[warped topaz]

Ah oke

[buoyant saddle]

because b is the y intercept

[warped topaz]

You are the only other person I met that also uses these letters 😅

[buoyant saddle]

americans

[warped topaz]

I grew up with m for slope and b for y intercept

but ok

Original question

x = x(t) and f(x) = y(t)?

[buoyant saddle]

no

wdym f(x) = y(t)?

[warped topaz]

Ah yeah sry

What is f(x) then?

[buoyant saddle]

nothing

just x(t) and y(t)

all that matters here

[warped topaz]

Then how can we use f(a) + f'(a)(x-a)?

[buoyant saddle]

same principle

we have a point, (0,0) so "f(a)" will be zero

and our line is of the form y = mx

since it goes through the origin

take a to be zero

[warped topaz]

Oh yeah true

[buoyant saddle]

and f(a) to be zero

then find dy/dx

[warped topaz]

For which sry?

[buoyant saddle]

the parametric curve

dy/dx = dy/dt / dx/dt

[warped topaz]

(-10t - 3)(-7t-3)(4t+4) = 280t^3 + 484t^2 + 240t + 36

[buoyant saddle]

i mean you don’t have to expand fully but yes

you could’ve used a neat trick

[warped topaz]

(-7t-4)(-7t-3)(5t-1) = 245t^3 + 196t^2 + 11t - 12

[buoyant saddle]

well actually nevermind we don’t have t = 0

so go on then

find dy/dt and dx/dt

[warped topaz]

This differentiates to 735t^2 + 392t + 11

[warped topaz]

This differentiates to 840t^2 + 968t + 240

Oh wait other way around

[buoyant saddle]

yep

then let t = -3/7 or whatever it was

[warped topaz]

So its (735t^2 + 392t + 11)/(840t^2 + 968t + 240)

Subbing in t = -3/7 gives 77/72

Thats the location of the tangent right?

[buoyant saddle]

if that’s correct then it gives the slope

y = mx

m is the slope

[warped topaz]

How come? why would t give us the slope?

t gives us the coordinats, no?

[buoyant saddle]

dy/dx = dy/dt / dx/dt

you found dy/dx

then you plugged in the point of interest in terms of t

[warped topaz]

Ok I think I get it

So its just y = 77x/72

[buoyant saddle]

sure

not going to check the calculation

but yes

[warped topaz]

Yep 😄

I think I get it now

Thank you so much for explaining youre a really good teacher!!

❤️

.close

[lone heart]

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[buoyant saddle]

https://tenor.com/view/ratatouille-remy-eating-grapes-eating-grapes-gif-19543638

[buoyant saddle]

thank you sir

and you’re welcome

[tacit arch]

that ratatoiulle gif has too many resolutions

[lone heart]

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[tiny rivet]

I don't understand :((

How'd that happen 😭😭

[tacit arch]

$\sin^4(\th) = \sin^{2+2}(\th)$

[ocean seal]

riemann

[tacit arch]

,tex .exp prod

[ocean seal]

riemann

[north rover]

Note that $0\le\sin^2(x)\le 1$.

[ocean seal]

;(

[tiny rivet]

Ohh okay

Because it's less than 1 the sq of the sq will be even lesser

Okay, thank youu

.close

[north rover]

👍

[lone heart]

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[formal grotto]

i thought i was doing this right but i keep getting it wrong

i seprated to get 2sqrt(y)dy = tdt

integrated, plugged in "y", to find y(1)

[tight pier]

show ur work

[tight pier]

I can't tell what u tried to do, but it seems you forgot the +C?

[formal grotto]

no i had a +C

in the problem y(0)=1, so my C was 1

which i included in my final answer

[tight pier]

how did you get C=1

if t = 0 then the right side contains C which is equal to 2(2/3)(1)^(3/2) = 4/3

[formal grotto]

what? if t = 0 everything is multiplied by 0 which makes everything except C go away

[tight pier]

,, 2 \cdot \frac{2}{3} y^{\frac{3}{2}} = \frac{t^2}{2} + C

[tight pier]

2nd line of yours

[formal grotto]

i am solving for C when everything is simplified

solving C at that stage wont help because C will change

[tight pier]

no

it's an equation

you can also get C immediately here

[formal grotto]

bro what

[formal grotto]

i technically can, but it will change

[tacit arch]

initial conditions are usually used to solve for C

[tight pier]

she found C but 1 seems wrong

[formal grotto]

let me redo this

[tight pier]

ok I see

[tacit arch]

this is incorrect

[tight pier]

you redfined a new constant prob

[formal grotto]

probably...

[tight pier]

,, y = \sqrt[3]{\frac{3}{2}t^2+\frac{3}{4}C}^2 = \sqrt[3]{\frac{3}{8}t^2+\tilde C}^2

[tacit arch]

[tight pier]

well yes

[tacit arch]

just plug in y(0) = 1 here

[tight pier]

if t = 0 then y = 1

makes your new constant 1

[formal grotto]

uhh so why does mine not work again

[formal grotto]

your c should turn into my c at the final form right

im still confused...

[tacit arch]

you should have a +C right after integration

$\int f(y) dy = \int g(t) dt \Ra F(y) = G(t) + C$

[ocean seal]

riemann

[tacit arch]

F' = f and G' = g

[tight pier]

but it should still work out

[formal grotto]

i thought i can do a +C in the end, because that C can encompass all the simplifying

[tacit arch]

sure after 10 minutes of algebra and 25% chance of making a mistake

[formal grotto]

erm....

i should ask my TA...

[tight pier]

expuck me

anywy what is y(1) supposed to be

[formal grotto]

no clue

i'm on my last try on prairie learn 😭

[tight pier]

,w (4/3)y^(3/2)=1/2+4/3

[ocean seal]

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[tight pier]

,w simplify (3/8+1)^(2/3)

[formal grotto]

what the sigma

hold on

[ocean seal]

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[tight pier]

see same

whether you redefine the constant

or not

but yes one thing!

putting +C in the very end is wrong

[tight pier]

because here for example it's not outside but inside when you solved

for y

[formal grotto]

alright i can see how it can be easier

thanks for the help guys but since its my last attempt ill confirm my answer with a ta anyway

[tight pier]

did you get that?

[formal grotto]

no 😭 ill retry later i have to go to class

[tight pier]

ok good luck

[formal grotto]

thanks!

.close

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[lethal aspen]

Help, I want to solve this integral. (Solving this by hand seems like monumental task, but maybe there is trick I am not seeing.)

[lethal aspen]

$\int^1_0 dx \frac{3x^3-x^2+2x-4}{\sqrt{x^2-3x+2}}$

[ocean seal]

Rei

[mental python]

please write the dx at the end and not at the front of the integral ty

[ocean fossil]

[lethal aspen]

I stopped doing that a while ago. Just a preference.

[mental python]

filthy

anyway, try factoring

[lethal aspen]

I tried pretty much everything in my repertoire already, you can factors the sqrt to (x-1)(x-2) which doesn't really help you at all.

So my first approach was to factor it to (x-3/2)^2 - 1/4 and then substitute, but this leads to an absolute mess of terms.

[mental python]

you can factor the numerator aswell

hint: ||(x-1||

[lethal aspen]

I mean I can do that, but it won't really help solving the integral no?

[mental python]

partially it will

[lethal aspen]

I can get it into the form:

$\int_0^1 dx \frac{(x-1)(3x^2+2x+4)}{\sqrt{(x-1)(x-2)}}$

[ocean seal]

Rei

[lethal aspen]

How does this make it any easier?

[mental python]

you can get rid of x-1 in the denominator now

[lethal aspen]

That looks a bit nicer.

[lethal aspen]

Ah sadly I think there is no way to solve this nicely. I think numerical methods will have to do :D

Ty for your input though.

[mental python]

where did you get this integral from??

,w int (3x^3-x^2+2x-4)/sqrt(x^2-3x+2)dx

[ocean seal]

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[mental python]

ew

[lethal aspen]

Yea.

[mental python]

i mean, it is possible

[lethal aspen]

No one does this per hand in less than a day :D I thought I was missing some nice subsitution trick or something. But yea, maybe I wasn't.

[winter light]

Maybe u² = x - 2 might work

But I'm not so sure

[lethal aspen]

I can't make it work :/

[mental python]

perhaps u^2=x^2-3x+2

[lethal aspen]

Why did you pick this one?

[mental python]

because then the denominator becomes u

[winter light]

But didn't we already cancel x - 1?

[mental python]

yes

but i think it doesnt actually help

lol

but its bcuz i kinda cheated since i looked at the answer

[winter light]

Ahn that's why lol

[mental python]

and there i see sqrt(x^2-3x+2) occuring twice

which gives a hint that we should use that as u-sub

[lone heart]

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[paper bramble]

Hello I need some help

[paper bramble]

I dont understand why my answer is wrong

it seemed like a pretty straightforward question but maybe my udnerstanding is off

could someone help me understand why the answers are wrong

[lone heart]

@paper bramble Has your question been resolved?

[tacit arch]

are we supposed to know the data

[paper bramble]

thats all the question gives

im so lost

no data

[tacit arch]

what's given above the question

[paper bramble]

thats it

any ideas?

[tacit arch]

your answer should just be in terms of the x_i then

use the formula for cumulative density

[a] should be easy

[paper bramble]

yea but I put that in and i got it wrong

[tacit arch]

the answers you put are not in terms of the x_i

you put numbers

x_i are unkowns

[paper bramble]

wait what do you mean

[paper bramble]

I'm a little lost

I thought since we have 18 observations they would all have equal individual prboabobilties

and then since its increasing we just add as i increases for x_i

so the first has 1/18

and the third has 3/18 (cdf)

[tacit arch]

this is not given

[paper bramble]

which part, the fact that n = 18 or they have equal individual probabilities

because it does say n = 18

if I dont have their individual probaitlies then how do I answer :(

im stoopid

[lone heart]

@paper bramble Has your question been resolved?

[paper bramble]

nevermind figured it out

dumb question by the prof

[lone heart]

@paper bramble Has your question been resolved?

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[deft glen]

.close

.reopen

[pallid briar]

I'm playing a factory building game with a lot of really wonky crafting times, and despite being able to do calc 3 with ease I'm struggling to even figure out how to identify the building-to-building ratios.

My buildings:

1. Produces 1 x every 5 seconds
2. Consumes 10 x every 5.4 seconds

How many of Building 1 do I need to satisfy one Building 2?

[pallid briar]

Like this is really simple and I know it but I just cannot get my brain to work on it

[safe meadow]

I mean the second building

Make 100 every 54 second

The first building make 1 every 5 second

[woeful granite]

you can convert this problem to equalize the ratios of creation/consumption, where a is the number of building 1's
a(creation rate) = (consumption rate)
rate = #x / (second interval)

this can be rearranged to solve for a:
a = (consumption rate)/(creation rate)

[limpid turret]

dimensional analysis can be really helpful for stuff like this

[lone heart]

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[pallid briar]

.reopen

[lone heart]

✅

[pallid briar]

Thank you!

[pallid briar]

And you too, thanks!

.close

[lone heart]

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[lone heart]

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[formal owl]

hi

[formal owl]

so i lowkey need help w my math its due in like 3 hrs

it shouldnt be hard hard

it js my brain wont brain

😭

[buoyant saddle]

what kind of dog do you have

[formal owl]

pomsky

[buoyant saddle]

oh wow

[formal owl]

loll

[buoyant saddle]

i knew someone with one of those i think

it was either that or a husky

[formal owl]

thats cool

husky n pomsky r basically the same thing

[formal owl]

mind helping me w these

[formal owl]

😭

[formal owl]

nun

[buoyant saddle]

yea beautiful dogs for sure

i have a golden retriever

[formal owl]

cuteee

i love golden retrievers

[buoyant saddle]

man you probably don’t know how because you didn’t watch the videos

what is true of inscribed quadrilaterals?

in a circle

[formal owl]

to lazy tbh

i did like 50 pages of my ap euro packet

[buoyant saddle]

yuck

[formal owl]

my hands gunna fall off

[buoyant saddle]

never took ap euro

took ap world and apush

[formal owl]

our school forces us

[buoyant saddle]

that was enough for me

[formal owl]

geniunly tho

hate it

HELP ME THO

[buoyant saddle]

cant stand DBQs and LEQs

[buoyant saddle]

^

[formal owl]

jeez i got a F on mine

i mean a D

[formal owl]

idk bro

[buoyant saddle]

opposite angles are supplementary

[formal owl]

dam

180

so dyk how i would be able to solve it

[buoyant saddle]

that wasn’t enough?

[formal owl]

na

im shit at math bro

dont blame me

ANY other subject i be good in

plus these we havent learned its like we have to do a certain percentage every week lol

i gotta do 7 of these

😭

would it be 95

[winged pawn]

there's two ways to do this

they're identical so i'll only explain one

[formal owl]

i js did a dif thing

lem send it

[winged pawn]

uhhh

okay

[winged pawn]

Find their slopes and compare

if they're parallel, their slopes equal

[formal owl]

would it be neither

[winged pawn]

if they're orthogonal i.e. perpendicular, their slope products should be -1

did you check if they're either of those?

[formal owl]

i think

[lone heart]

@formal owl Has your question been resolved?

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Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[obtuse berry]

can somone please explain how this is the answer

[fierce cipher]

substitute x=4

[twin nimbus]

not quite.

[fierce cipher]

oh wait nvm

that point is not continuous lol
it's still the same thing tho but, just not the same thing, i mean it's complicated

[twin nimbus]

@obtuse berry what is the limit of f(x) as x->4 for starters?

[alpine sable]

ap euro is such a bad class omg

[obtuse berry]

of the graph in the picture?

[twin nimbus]

yes, because that graph in the picture is f(x)

[obtuse berry]

5

[twin nimbus]

that is correct

So if you do 2 cos(5) what do you get?

[obtuse berry]

ohhhh

shoot

i see

yeah i understand now

thanks

.close

[lone heart]

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Use .reopen if this was a mistake.

[twin nimbus]

you're welcome 🙂

[formal owl]

I HATE IT OMG

[fierce cipher]

pretty funny because you can accidentally get this right without knowing f is not continuous at x=4

(if you dont overthink that f(4) = 3)

[alpine sable]

what unit?

[formal owl]

6 i think

[alpine sable]

oh were pretty even then

we just finished 6

[formal owl]

its like industrial stuff

[alpine sable]

i still dont know the congress of europe but y

k

omg

👏

we're gonna get 5's

[formal owl]

no

[alpine sable]

yes

[formal owl]

failing this class fs

[alpine sable]

clutch up then

[lone heart]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[mint peak]

My braincells are fried

[dusty kayak]

you can’t use a calculator or sum

[tardy steppe]

you can divide each of these up into simpler shapes, then you can figure out those shapes' areas much easier than trying to find the area of the entire shape

for example

take this one

do you see any simple shapes that you might be able to cut this into?

[tardy steppe]

yeah, so the left triangle is going to be the hardest part of this shape i'd say, so lets leave that for last

can you figure out the area of the top triangle?

[mint peak]

Is it 42.5?

I haven't done this in a long time

[dusty kayak]

nah

wait