help-38

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[ashen prairie]

Characterize all finite groups with trivial automorphism group.

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[dusky ocean]

Legit been on this question since yesterday

[dusky ocean]

supposed to find derivative

[past river]

Do you remember the product rule

[sweet hare]

use the product rule

[dusky ocean]

now what

[sweet hare]

you didn't do the derivative of x^3 - x correctly

[dusky ocean]

forgot the ^2

[past river]

Right. But then, sometimes you can‘t simplify further and you have to leave it at that

[dusky ocean]

b) is the answer

[past river]

Hm, then put everything on a common denominator

[dusky ocean]

do you know if i’d use the chain rule next?

[sweet hare]

the chain rule isn't needed here

[dusky ocean]

so do i just find the derivative of what i have left

damn so i’m cooked

[sweet hare]

you've already found the derivative

it probably wants you to simplify somehow

[dusky ocean]

i do

but i don’t know how to simplify this monstrosity

[sweet hare]

yeah it's messy

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[wraith hinge]

I have a question on my paper and need help confirming an answer it says write an equation in slope intercept form and gives me this question, a cell phones company charges a flat fee of 30$ month plus 0.45$ for each minuite over the limit

[wraith hinge]

So u use y=mx+b

Will it be y=30x+0.45

Or will it be y=0.45x+30

[strong fern]

pretty sure its second one

[wraith hinge]

How come the 0.45 comes first tho

[strong fern]

because its 0.45 per minute over the limit so its depending on the minutes over the limit

so thats the slope

so if you imagine y is the total cost

total cost = 0.45 * minute over the limit + fee of 30

[wraith hinge]

So like if there’s any questions that are similar to that should it always be the slope

[strong fern]

well you just gotta think about it like

what's contantly changing and what isn't

[wraith hinge]

Wait hold on I got it

I get what u meannn

[strong fern]

yeah

[wraith hinge]

Can u help me out with this?

I have to make it into a equation in slope intercept form

[strong fern]

well

first you see that when x = 0, y = -2

so what does that mean

[wraith hinge]

That y intercept is -2

[strong fern]

yeah

so y = mx + b

b is -2

now for the m

[wraith hinge]

Y=0x-2

[strong fern]

0?

[wraith hinge]

Wait I have to solve for m

[strong fern]

how do you calculate the slope of a line

do you know

[wraith hinge]

I probably know it but like I don’t get the way u mean

What is it

[strong fern]

it's just when you have 2 points on a line (x1, y1) and (x2, y2) we say that m = y2 - y1/x2 - x1

so pick 2 points from that picture

[wraith hinge]

Don’t in this question u have 2 choose 2 coordinate

[wraith hinge]

Ya ya this

[strong fern]

each row is a point

[wraith hinge]

Okay I choose 0,2 and 1,6

[strong fern]

so (-1, -10), (0, -2), (1,6), (2,14)

[strong fern]

-2

careful

[wraith hinge]

Mb

[strong fern]

yeah so we have

m = 6 - (-2) / 1 - 0

so thats 8

y = 8x - 2

do you understand?

[wraith hinge]

Let me show u what I got on my paper rq

[strong fern]

yep

8/1 is just 8

so that's your m in y = mx + b

[wraith hinge]

And then what happens

[strong fern]

thats your equation

y = 8x - 2

[wraith hinge]

It’s that easy?

[strong fern]

yeah

😄

[wraith hinge]

Damn I feel dumb

Wait how did we get the -2

[strong fern]

and btw try writing 6 - - 2 as 6 - (-2) it'll be wy easier

you wont make mistakes this way

[wraith hinge]

Alright thanks for the info

[strong fern]

look at the picture again

[strong fern]

when x is 0

y = -2

so thats the y intercept

b is the y-intercept

[wraith hinge]

And like this only works when x is 0 right?

[strong fern]

ofc

the function only intercepts y once

when x is 0

[wraith hinge]

Ohhh I get it now

[strong fern]

and if you wanna understand this a bit more

think of it as f(x) = 8x - 2

x is some value

if you do x = 0

f(0) = 8*0 - 2

f(0) = -2

so basically just dont get confused f(x) is the same as y when writing the function

[wraith hinge]

Alrighttty thanksss

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[low sand]

Can somebody tell me how to do this

[onyx jacinth]

So you have the integral now lets subtract by -2ln(a)

using the logarithmic properties
its the same as -ln(a^2)

the 2ln(3x+5)
can also be written as
ln( (3x+5)^2 )

When you subtract to logs you divide the inside so it becomes
ln( (3x+5)^2 / a^2 )

since they are both squared you can just square the entire fraction

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[low sand]

Thank u

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[lethal rose]

i dont understand the steps taken to make -(12/x^2) become -12*x^(-2), chould someone please explain

[solid kiln]

faiyrose

faiyrose

[lethal rose]

right

[solid kiln]

faiyrose

faiyrose

faiyrose

[lethal rose]

thank you, it makes sense now

thanks

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[sudden drum]

!help

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[sudden drum]

hi i dont understand how to graph absolute values

so i know that the -6 is telling me to shift to the right by 6 units on the x-axis

the +2 is telling me that you need it goes up by two units on the y-axis

but i dont understand the factoring or multiplying by 5 part

youre suppose to multiply the slope by 5

but how do i know what the slope is

[pseudo bear]

the five is the slope of the graph, so the higher number the steeper the graph

[sudden drum]

oh

so for example

3|x - 2| + 3

the slope of this graph is 3

is what youre saying

[pseudo bear]

yup

[sudden drum]

ok

also

sqrt 17(17)

am i suppose to

take the square root of 17

then multiply that number by 17

or do 17 * 17 first then take the square root of that number

[pseudo bear]

$\sqrt{17} \cdot 17$

[sudden drum]

yes like that

im asking what the order of operationsi s

cause i forgot

[solid kiln]

caspar

[sudden drum]

yes

so do i do 17 * 17 first

[pseudo bear]

you take the sqare root first

[sudden drum]

or do i do square root of 17 first then multiply that number by 17

ok thx

[pseudo bear]

np

[sudden drum]

are u sure

cause i just did symbolab

and they told me to multiply by 17

well i guess u were suppose to multiply first lol

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[pseudo bear]

$\sqrt{17} \cdot 17 \neq \sqrt{17 \cdot 17}$

[solid kiln]

caspar

[pseudo bear]

$\sqrt{17} \cdot 17 = \sqrt{17 \cdot 17^2}$

[solid kiln]

caspar

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[quick hound]

need hint

i forgot what i did to do this

[frail palm]

remeber that the function of cos looks like this

[quick hound]

yeah?

wait

[frail palm]

Where its highest value is 1 and its lowest is 0

[quick hound]

so infinity?

not -1?

[frail palm]

Sorry yes -1

[quick hound]

so n= infinity

and cos -s 1

so 0

is 1

so its just n^2(0)?

idk man

wait

am i finding the sum?

not the value of teh last tierm

nah im lost

[frail palm]

okay, are you trying to find the upper bound and or the lower bound or are you trying to find the limit of the function?

[quick hound]

limit?

i think

to infinity

[frail palm]

Cause if its the upper/lower bound then cos has the values either +1 or -1

[quick hound]

yeah

where 5.3^2/2 from then

im triyng toreverse engineer that

i bet its some lopital shenanigns

[frail palm]

Well if its the lim of inf i would aproch it like this

and then apply L'hoptials

[quick hound]

ye

did it

ty

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[quick hound]

why does tihs not diverge

i mean conerge

nvm im dumb

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[vale blade]

Hello, I need to solve the given inequality here. However, I don’t understand how 0 is included in the first x solving case. The answer in the first section is: {0} u [1 ; +)

[shrewd vessel]

what have you tried/what was your approach when doing this question?

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[vale blade]

Left side is the part I’m asking about

Actually I see it why it is included now 😂

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[low sand]

Hello can someone tell me how to do this question

[vivid silo]

$\sin(3x)=3\sin x - 4\sin^3 x$

[solid kiln]

jolimath

[vivid silo]

$\sin x + 3\sin x - 4\sin^3 x = 4 \sin x - 4\sin^2 x$

[solid kiln]

jolimath

[vivid silo]

$4\sin^2 x - 4\sin^3 x = 0$

[solid kiln]

jolimath

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[low sand]

I don’t understand this part can u explain 😭😭

[vivid silo]

Wait -- which part? The initial identity?

[low sand]

The very first

[vivid silo]

You can break $\sin 3x$ down using the angle sum formula ...

[solid kiln]

jolimath

[vivid silo]

$\sin(3x) = \sin (x + 2x)$

[solid kiln]

jolimath

[vivid silo]

But if this problem is appearing here, I'm assuming the triple angle formula would be something you had access to prior to this problem ...

Oh -- I'm so sorry ...

I misunderstood the problem.

[low sand]

I don’t have access to the triple angle formula 😭

[vivid silo]

You're right -- my bad.

So, this is a proof.

First, I would take the left side, and I would write it as

$\sin x + \sin (x + 2x)$

[solid kiln]

jolimath

[vivid silo]

Do you know the angle-sum formula for sine?

And if so, do you know the double-angle formula for sine?

[low sand]

Yes

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[low sand]

.close

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[misty laurel]

hello

[misty laurel]

I am trying to do this maclaurin series limit

the answer is -1/3

I've been trying for over an hour but still no dice haha

i feel so dumb

[digital bolt]

can you just sorta maclaurin everything

[misty laurel]

?

[digital bolt]

i mean ok so it's just like

sin is basically like x + x^3/3

cos is basically like 1 - x^2/2

sin + cos is basically like 1 + x - x^2/2 + x^3/3

and then there's an expansion for log(1 + x)

but sin + cos is like 1 + something

[misty laurel]

I did all of that

[digital bolt]

ok what did you get

can you show your working

[misty laurel]

here’s one of my attemps

[digital bolt]

where did the log go

[misty laurel]

maclaurin

[digital bolt]

what does maclaurin say

what's maclaurin of log

[digital bolt]

right

so why do you only have one term

my guess is that you need an extra one maybe

you can probably get rid of the x^3 and the x^4

so just expand log(1 + x - x^2/2)

and then use two terms instead of 1?

maybe that works

[misty laurel]

i tried that

earlier

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[frozen prism]

how did they approximate s

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[wraith hinge]

What would be the formula to represent the following problem:

[wraith hinge]

A student with the number 1 and multiplied it by either 6 or 10. He then multiplied the result by either 6 or 10, and continued this procedure many times.

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[torpid cradle]

Hey everyone. I am not sure how to go about this question I have on a review assignment for my Algebra II class. The question is just:
"For all values of x for which the expression is defined, ( x^2 + 3x ) / ( x^2 + 5x + 6) is equivalent to" and some random multiple choice answers. I plugged the expression into my Ti-nspire cx ii CAS and it told me the right answer is x / (x + 2). Can someone walk me through solving this?

[burnt mulch]

Factor the numerator and denominator

[torpid cradle]

i am a dunce thank you lmao

[burnt mulch]

[torpid cradle]

.close

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[outer flicker]

if someone could explain it very simply thatd be nice

[idle meadow]

yo

[outer flicker]

yoyo

[idle meadow]

ill help you

[outer flicker]

u know how to do it right haha

because its mainly crytography

[idle meadow]

send me a dm

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[shrewd rivet]

Hello! i have questions regarding the metods to resolve this problem, they ask me to analyse the continuity of this scalar field in (0;0), they also tell me it is defined in (0,0) and it is 0. So my train of thought is that this is a bound field between 0 and 1; but is not being multipled by an infinitesmy, so i dont know if its continuous or not, i appreciate any help

[solemn schooner]

Try finding a path that approaches the origin where a isn't 0. Hint: ||Try looking at the term (x-y^2)^2||

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[scenic vessel]

this is all I know , how do i answer thsi question

[ionic pendant]

if they are not scalar multiples, then the only way for the two sides to be equal is for it to be 0 = 0

[scenic vessel]

ohhh ok

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[mossy wagon]

Can someone help explain how thy went from cos pi/2 x 3,5 - d = 1 to pi/2(3,5-d) = 2nPi

I dont get the 2nPi

[split chasm]

2n * pi is the general solution for
cos(this) = 1

[mossy wagon]

Oh okay thanks

I didnt know tjat

How about sin

Is it the same

[split chasm]

no

but similar

it would be helpful to consider the unit circle or their graphs

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[mossy wagon]

Oh okay thanks

Hello guys i do not understand i know we move 2pi/3 toe the right and we divide it but where did 3/4 comefrom

Wait nevermind i hot it

2pi/3 x 2/Pi = 3/4

Bur what did they do with the Pi on the right why is it remoevd?

[split chasm]

there's a factor of pi on each side
divide both sides by pi to simplify

[split chasm]

pi/2 * 3/(2pi) = 3/4
they multiplied both sides by 3/(2pi)

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[dense breach]

First 2 look right

Last 2, can't take the square root of a negative

[open fulcrum]

true

so which one is >= and which one is >

[dense breach]

maybe 3x^(9/2) is all I can think of

?

[open fulcrum]

couldnt u write that as 3(sqrtx)^9

or should i write as 3x^4.5

or 9/2

[dense breach]

Yeah you probably shouldn't but you could

4.5 maybe

[open fulcrum]

second one would be x^5/2 right

(x^[5/2])/3

[sweet hare]

no . 5/2 = 2.5

[open fulcrum]

this worked

[open fulcrum]

x^3 times x^0.5 = x^3-0.5 = x^2.5

[sweet hare]

f(x) = x^4

[open fulcrum]

oh so its 7/2 not 5/2

[dense breach]

Yeah

[open fulcrum]

whats the domain of (f/g)(x)

its not x =/= 0 nor is it all real #s

idk if its > or >=

[sweet hare]

plug in 0

[open fulcrum]

is 0^ any power undefined or 0

[sweet hare]

depends on the power

[open fulcrum]

7/2

but whats the general rule for what power = what

[mossy wagon]

Oh okay thanks

[sweet hare]

actually you should plug it in before simplifying

[open fulcrum]

i dont know what 0^3.5 is so i dont know if its x>0 or x>=0

when is it undefined and when is it 0

[sweet hare]

before simplifying

[open fulcrum]

did you ever have a math test and when you asked the teacher a question they just repeat the question back to you

and they act like it helped

anyways i figured out x>0 via t&e

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[ancient edge]

How do i integrate this

[hallow kite]

using a calculator

you can't find an antiderivative

using known calculus techniques

[ancient edge]

Why not?

[hallow kite]

you learn to recognize nonelementary integrals after a while, but you could also mentally try all your options

there isn't a particular u sub that makes this easier

integration by parts doesn't help, it's not really a product of two recognizable functions

it's not the right form for partial fractions

there's no other nice substitution like a trig sub that would help here

[ancient edge]

is nonelementary integrals like complex numbers

[hallow kite]

nonelementary means that you can't express the antiderivative in terms of standard mathematical functions

[ancient edge]

When do i learn that?

[hallow kite]

so stuff like trig, polynomials, rational functions, radicals, exponents, logs, etc

you don't really learn it in a calculus class

in a calculus class they teach you how to solve integrals that are actually solvable

but if they asked you to find an antiderivative of e^(x^2)

you wouldn't be able to no matter how hard you tried

[ancient edge]

would u sub

then int by parts work?

[hallow kite]

no, you can try messing around with different techniques if you want

but i'm telling you that no matter what you come up with, it's impossible to express the antiderivative

using normal functions

people eventually came up with algorithms to determine if antiderivatives are nonelementary or not

but that's beyond the scope of this

the integral you gave is one of those integrals that ends up being nonelementary

if the square root wasn't there it would be easy

when you try using something like wolfram alpha to integrate this, you get a nonelementary antiderivative

this is defined in terms of a special function called the hypergeometric function

but now you're really getting beyond the scope of stuff

that doesn't mean you can't evaluate a definite integral, it just means you can't plug in the bounds into a nice antiderivative to find the answer

but a calculator can use different numerical approximations to estimate the integral

https://en.wikipedia.org/wiki/Nonelementary_integral

hope this helps

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[blissful laurel]

how do I find the particular solution

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[hallow kite]

one method to find the particular solution here would be the method of undetermined coefficients

assume y_p = Ae^(-4t)

you can compute y' and y'', substitute those into your differential equation and you'll ultimately get an equation that you can solve to find A

[blissful laurel]

i did that but i got 0=4e^(-4t). so im a ltitle stuck unless i made a mistake somehwere

[hallow kite]

hm

y_p = Ae^(-4t), y' = -4Ae^(-4t), y'' = 16Ae^(-4t)

then substituting in gives 16Ae^(-4t) - 32Ae^(-4t) + 16Ae^(-4t) = 4e^(-4t)

what the hell

0 = 4e^(-4t)

hmm

ohh

your guess should actually be y_p = (Ax^2 + Bx + C)e^(-4t)

just trust here

if y_p = Ae^(-4t) was a valid guess then this should also work in theory, you'd just have the other coefficients disappear and be redundant

but trying with this, you get y' = (2Ax + B)e^(-4t) - 4(Ax^2 + Bx + C)e^(-4t) and y'' = 2Ae^(-4t) - 4(2Ax + B)e^(-4t) - 4( (2Ax + B)e^(-4t) - 4(Ax^2 + Bx + C)e^(-4t) )

i just realized i was using x here but imagine those are ts

then simplifying gives y' = e^(-4t) * (-4At^2 + (2A - 4B)t + (B - 4C))

[alpine saffron]

Isn't this equation not homogeneous?

[hallow kite]

it is nonhomogenous

okay i've gotten lazy so i'm using wolfram now

if y = (At^2 + Bt + C)e^(-4t), then y' is

and y'' is

when you plug these into the original diff eq you get

hmm

what is happening

ohh

you get 2Ae^(-4t)

after you plug all of that in and everything simplifies

so 2Ae^(-4t) = 4e^(-4t)

2A = 4, A = 2

i think you can just let the B and C terms be 0 since they don't matter

so y_p = 2t^2 * e^(-4t)

and you can check this since y' = 4te^(-4t) - 8t^2e^(-4t) = e^(-4t) * (4t - 8t^2) and y'' = -4e^(-4t) * (4t - 8t^2) + e^(-4t) * (4 - 16t)

then plugging those into the diff eq

you get exactly 4e^(-4t)

so your particular solution is indeed 2t^2 * e^(-4t)

another way to see why this would be the case is the following

for the same reason that you build up the exponential in the general solution when you get a repeated root in the characteristic equation

like since -4 is a repeated root, you got c1e^(-4t) + c2te^(-4t)

if your particular solution was a constant multiple of e^(-4t) or a constant multiple of te^(-4t), it would just be absorbed with the general solution

so your guess for the particular solution needs to be something like At^2 * e^(-4t) for it to provide something unique from the general solution

[blissful laurel]

would u add the particular solution with the complimentary solution to get the general solution, so y = C_1 e^(-4t) + C_2 te^(-4t) + 2t^(2)e^(-4t)

[hallow kite]

yes

to better understand why i had to build up the power in front of the exponential

https://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx

read this and specifically scroll down to example 9

[blissful laurel]

ohh okay thank you

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[stone cradle]

i need help

[stone cradle]

how to do this

@dull temple

@alpine saffron

[alpine saffron]

$g(x)=e^{-x}f(x)\implies g'(x)=e^{-x}(f'(x)-f(x))\implies g''(x)=e^{-x}(f''(x)-2f'(x)+f(x))$

[solid kiln]

otheol

[alpine saffron]

Wait

[stone cradle]

wtf

[alpine saffron]

Are you sure that the problem is not asking for $g''(a)$?

[solid kiln]

otheol

[stone cradle]

yeah that is the problem

[alpine saffron]

Because if $f$ has an inflection point at $(a,g(a))$, then $f''(a)=0$

[solid kiln]

otheol

[stone cradle]

that is the question

my teacher wrote it idk how to do it

[alpine saffron]

^

[stone cradle]

its to find f''(a) in terms of f(a), f'(a)

[alpine saffron]

I mean

$0=0f(a)=0f'(a)$

[solid kiln]

otheol

[stone cradle]

ik the question is not wrong

but how would we do it

[alpine saffron]

For a function to have an inflection point, its second derivative needs to be 0

Since $f$ has an inflection point at $a$, $f''(a)=0$

[solid kiln]

otheol

[stone cradle]

Idk ill do it later

Thanks tho

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[drifting turtle]

Hi, I am currently a 15 years old student but my physics teacher taught me a bit of Lagrangian mechanics, which involves some calculus and partial differentiation which I have not come across. This problem is just about a simple pendulum where the position (vector) is given by some function in terms of time. I am trying to figure out the Lagrangian (kinetic - potential energy) of the system, to obtain the equation of motion. I know this is physics, but all it involves is some mathematics.

I attempted this problem by looking up relevant YouTube tutorials, but somehow I am missing a "cos(theta)" in the last part of the question.

Please can anyone spot my mistake, or provide the "proper/best" way to solve this question?

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[drifting turtle]

I suspect that I might have differentiated the position vector wrong, but I am not sure what is wrong with it

[drifting turtle]

<@&286206848099549185>

[undone forge]

you're 15 and learning lagrangian mechanics? TF

[drifting turtle]

not really

a mathematical physics undergrad / graduate (i am not sure) came to our school

and he taught me this

he derived formula for the Lagranian for double pendulum, and then i generalized the formula for n pendulums

I got done gcse specificatoin and half of a-level, all i knew was newtonian mechanics

but i just tried really hard to understand uh this pendulum mechanics thingy

and im struggling to picture the derivatives of angle (with respect to time), because angle has no dimenisons (and it is not dependent/ interms of time)?

@undone forge if you can offer a bit of insight beyond just the question above I would be extremely grateful

also at what age is this thing taught typically???

although im doing like this level stuff, in maths lessons i am still learning relatively simple things, like currently im self learning invariance/invariant lines and matrix transformations

i also didnt know how to differntiate sinx or cosx before doing this pendulum thing

i suppose there are some things that I am suppoed to first know in between the euler lagranian formula and Force(net) = ma

this is my expression for the lagranian of n-pendulums

[drifting turtle]

really need some help because i have basically no teacher, cus my classmates are still learning like simultaneious equations and quadratic equations

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[fast tendon]

.reopen

o

[fast tendon]

.close

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[wraith hinge]

Hey

[wraith hinge]

Got lamda and mu

Just stuck on the last part

Part c

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[wraith hinge]

Help?

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[frank merlin]

last one is A^TA

the first one is AA^T

why does the last one have a a.m more for 0

it should be the same for SVD

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[wraith hinge]

has anyone ever tried to integrate the binomial theorem for a known or unknown n

[wraith hinge]

how would it be done

[nimble stone]

the integral of a sum is a sum of integrals

so you can just integrate the term inside the summation with respect to your variable

if thats what youre asking

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[quick hound]

uh

ye ty

why do they compare it to 1/n^0.75?

is (ln(n))^2=sqrt(n)?

in the scale in which they grow?

i guess it does

is there a test for ln(n)^2?

like p series test

stfu lil bro

give me 2 bitcoin and ill buy your course

love how the number doesnt even work

amazing

[quick hound]

is there a test for this like p series test for 1/n where n must >1 for it to converge?

oh the bot deleted his messages

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<@&286206848099549185>

If anyone can help before I go to bed

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[cedar goblet]

Can anyone help me check if p(t) = t/3 is a particular solution for a)?

[cedar goblet]

I've calculated the first and second derivates, p'(t) = 2at + b and p''(t) = 2a then just subbing them into the x'' + 3x' = t equation

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[white frost]

Can someone tutor me tonight I have a civil engineering exam tommorow pls help I need to learn stuff like bending moment diagrams, sheer force diagrams etc

[white frost]

I'm super dead unless I learn this stuff

I'm willing to pay for tuition tonight

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[wraith hinge]

I can't understand why this one question is using pi/4 in the answer key.

[wraith hinge]

The question is asking to find the volume of the two lines sqrt(x) and x^2 if the area between them were circular disks

I understand everything except for why it's using pi/4

[frail yarrow]

pir^2=pi(d/2)^2=pix1/4xd^2=pi/4d^2

It's just a weird way of putting the area of a circle in terms of diameter

@wraith hinge

[wraith hinge]

I'm sorry, I don't quite understand

Wait, it's diameter, yeah.

So the area is really this:

And the /2 gets turned into a 1/4

Alrighty then

[frail yarrow]

yep

[wraith hinge]

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[orchid wagon]

In a circus, there is a trick consisting of 6 different birds balancing (see diagram). Given that:

• The weight of each bird is a natural number from 2 to 9kg
• No bird has the same weight
  What is the weight of the bird B?

[orchid wagon]

is A + 12D = B + 6F correct? and not sure what to do either

[worn mist]

uhhhh

i don't get it

would bird A = C+D

[orchid wagon]

what

no lol

[vagrant marsh]

should be 3A and 3B

[orchid wagon]

oh right that matters too

but rn i have like 4 variables

[vagrant marsh]

you also know C=3D

and E=2F

[orchid wagon]

yeah thats how i got the equation

[vagrant marsh]

yeah but you can take some guesses and solve from there

[orchid wagon]

$A + 4D = B + 2F$

[orchid wagon]

oh so its just guesswork?

[vagrant marsh]

yeah pretty much

[orchid wagon]

a bit of guessing gets me A = 8, B = 6, D = 2 and F = 3

thanks!

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[vagrant marsh]

@orchid wagon B = C = 6

also you get 16=12

[orchid wagon]

.reopen

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✅

[orchid wagon]

oh what

[orchid wagon]

thats what would happen if i assume this?

[vagrant marsh]

yes

[orchid wagon]

fucj

oh i found the problem

dumb me

[vagrant marsh]

(A,B,C,D,E,F)=||(2,6,9,3,8,4)|| i think

[orchid wagon]

A = 4 not 8

either way both case yields B = 6

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[surreal nebula]

Is this right

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[surreal nebula]

Is it to big

Cuz this is the problem

I keep thinking I neeeded to subtract 1 first before proceeding

Ur not a dumdum

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[spring burrow]

how am i supposed to do this with no t variable

[knotty quest]

no t variable just do integral dt = t

[spring burrow]

oh wait so would the integrand be cos(t^6)?

[knotty quest]

no no

ull be getting

cos(s^6)*(t)

from 0 to s^5 for t

[spring burrow]

OH

that maeks more sense

oh that makes a lot more sense actualyl tahnk you

[knotty quest]

npp

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[unkempt tinsel]

I have no idea where i should start here

[unkempt tinsel]

I checked my class notes and it didnt help me and i am really confused

[alpine saffron]

One thing we can do here is check for a potential u-sub with the denominator

[unkempt tinsel]

yea and how do i do that?

[modest yarrow]

Do a variable change

U= x^2+16

Du= 2x *dx

[unkempt tinsel]

so it would be u/(x^s+16)

?

[crimson pebble]

That hints that you need to do a preliminary step first

[unkempt tinsel]

i am getting more confused

[modest yarrow]

Substitue x^2+16= u

[crimson pebble]

Consider splitting the fraction into two

Such that you can apply the u-sub Amelia mentioned in one of them

[unkempt tinsel]

oke i will try

thx

.close

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[weak maple]

yo

[weak maple]

how to do it

?

[full dock]

,w integrate x^2/(1+x^8)

[solid kiln]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by waterrbeam

[full dock]

ggs

[modest yarrow]

(x^8)=(x^3)^(8/3)

U=x^3

Than du=3x^2 dx

Maybe can work not sure

[full dock]

it probably doesnt work based off of the answer alone

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@weak maple Has your question been resolved?

[weak maple]

why there is cos

: <

[full dock]

thats ur question and not why the answer is 999 pages long

??

😂

[weak maple]

i mean idk what happend there tbh

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[jagged wharf]

Suppose we throw a die until we get a 6. Then the outcome space is $$\Omega={1,2,\ldots,6}^\mathbb{N},$$i.e. an element of $\Omega$ is a sequence $\omega=(\omega_1,\omega_2,\ldots)$ My book says the $\sigma$-algebra on $\Omega$ is the smallest $\sigma$-algebra containing the collection $$\mathcal C={\omega\in\Omega :\omega_1=i_1,\omega_2=i_2,\ldots,\omega_n=i_n},$$ where $n\geq 1$ and $i_1,i_2,\ldots,i_n\in{1,2,\ldots,6}$, however, it doesn't stipulate that $i_n=6$. Am I right in interpreting $i_n=6$ always? Also, is $\mathcal C$ closed under finite intersections?

[solid kiln]

Philip

[marble wharf]

it could make more sense to say that i_1,..,i_n are in {1,..,5}. depending on how you wanna model stuff

[jagged wharf]

but how could i_n be in {1,...,5}? We end the experiment if we get a 6, which is on the nth roll or never, right?

[marble wharf]

well we could say that the n+1th roll was a 6

which we dont need to write down

because its implied by the sequence ending after n

I dont know how your book wants to continue to build the condition of the sequence "ending" after the first 6 into the model

so far C is just Omega itself unless they give more restriction on what the i_1...i_n can be

[jagged wharf]

well, it wants to construct the random variable $$X(\omega)=\inf{j:\omega_j=6}.$$ Also, what I wrote above was a bit misleading regarding $\mathcal C$. The $\sigma$-algebra on $\Omega$ is the smallest $\sigma$-algebra containing all sets of the form $${\omega\in\Omega :\omega_1=i_1,\omega_2=i_2,\ldots,\omega_n=i_n}.$$

[solid kiln]

Philip

[jagged wharf]

where n>=1 and i_1,...,i_n are in {1,...,6}

[marble wharf]

aka the set of all sequences which start the same

for a given start

[jagged wharf]

hmm I'm not sure, I think all finite sequences, over all combinations of i_1, ... i_n in {1,...,6} (where my interpretation was that i_n is always 6), e.g. if n=1 we have {(6)}, if n=2 we have {(1,6)},{(2,6)},{(3,6)},{(4,6)},{(5,6)}, and so on, does that make sense?

[marble wharf]

no

they are still elements of Omega

which contains infinite sequences

so eg we have the set of all sequences which start with (1,2,3,6,2,3)

or the set of all sequences which start with (1,1,1,1,1,1,1)

or all which start with (5)

[jagged wharf]

ok, I have to think about this for a moment 🙂

[jagged wharf]

are you saying the set of all sequences which start with (1,2,3,6,2,3) is in the sigma algebra? I don't understand how this can be an event when we're supposed to stop after 6?

[marble wharf]

yes its in the sigma algebra, no its not an event. I dont know right now why your book chose this model

maybe its cleaner later on

sometimes its better to allow more

[marble wharf]

ah ok yes. by allowing more sets now, describing the probability measure is at least very simple

for now we just arent concerned with the condition about the 6

that will come when we consider the preimages of those random variables X

I have to go

[jagged wharf]

shame, but thanks 🙂 🙏

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[stone panther]

Using the diagram on the right, determine the displacement x(α) of the piston, taking the origin of the axis
axis (Ox) at the point where the volume is minimal and α = π. Hint: use the algebraic height y, the length z and the angle π.
length z and angle γ to help you when starting from x + z = b + R, but at the end these quantities should no longer
the formula.
Deduce the volume V = S × x variable between volume Vmin and volume Vmax as a function of α.

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[stone panther]

<@&286206848099549185>

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@stone panther Has your question been resolved?

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@stone panther Has your question been resolved?

[stone panther]

<@&286206848099549185>

[stable karma]

What’s your question?

@stone panther

[stone panther]

@stable karma I can't express the volume only as a function of alpha

maybe someone has an idea

[stable karma]

It’s a cylinder with radius d/2 and we have to find the value of x as a function of alpha

👆🏻

here you can find the value of x

[stone panther]

No i must express V

But yes it's the same thing

Because V depend on x

[stable karma]

Yes

[stone panther]

Yes but x,z, b and R can't appear in relation

[stone panther]

I noticed that here

@stable karma

[stable karma]

Can’t appear in relation? but you yourself have written the relation which is correct

?

[stone panther]

Which relation ?

[stable karma]

x+z = b+R

[stone panther]

Yes

[stable karma]

But you don’t have to use this relation in this question

[stone panther]

"But at the end these quantities should no longer the formula"

So i in the final formula i can't have one of these 4 elements

[stable karma]

In the final formula you will only have R and alpha in the place of x

[stone panther]

But R i can't

[stable karma]

R is constant here

[stone panther]

I think that "quantites" include all of 4

He doesnt talk about variable only

[stable karma]

It can’t as x and z are variable and we have to find volume in terms of alpha, so it can’t have x or z

[stone panther]

Yes it's the problem

[stable karma]

Let’s put this aside, and tell me what’s the volume of the cylinder

[stone panther]

Base * height

?

It's piR² * height

[stable karma]

Yes

[stone panther]

Base * height 🙂

[stable karma]

And what’s the value of R and H here

Base area

[stone panther]

x+z i think

[stable karma]

Only x

[stone panther]

Oh yes

And H = d/2 * pi as you said

[stable karma]

r= d/2

[stone panther]

Yes

Sorry No H base

[stable karma]

H is height

[stone panther]

Yes sorry

But the problem is for express x

[stable karma]

And x is height or displacement, which is also 🟰

.

[stone panther]

Tan(alpha) = b/z+x

So x = -z + b/tan(alpha)

[stable karma]

No

[stone panther]

I can express also funcrion of R

[stone panther]

But the probleme is always here

Cos(aplha) = y/r

[stable karma]

Hmm

[stone panther]

Ah no

Sin

Sorry

[stable karma]

The value of x in this image will only be in terms of R and alpha

[stone panther]

Yes ok

I will see how to express R with vmin and vmax maybe

[stable karma]

You don’t have to, just focus on x and R and alpha,

[stable karma]

Get it?

If you have doubt, you can ask

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[stone panther]

thanks

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[wraith hinge]

isn't XO = O-X?

apparently they have it as X-O=XO

not sure if I'm missing something

[shrewd vessel]

prob just a mistake

[wraith hinge]

ah okay, thanks

.close

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[proper fern]

These are correct?

[proper fern]

(Question was answered on the other one so I closed it and made this one with a new question)

[drifting sand]

First one looks good

second one, you have a mistake in the 2nd row

[proper fern]

Really?

I divided 416432 by 375

[drifting sand]

Population density is population over area

Divide 416 432 by 21 110

It's not 375

[proper fern]

375 is a set answer

Im trying to find out the Area in the second row

[drifting sand]

Yeah to work out the area

[proper fern]

Where it is highlighted

No no the population density is what they give us to solve the question

[drifting sand]

Yes

[drifting sand]

If you do this, you find that you calculated the density wrong

[proper fern]

I didnt calculate the density

To find Area you do

Population / Population density

416432 / 375

[drifting sand]

not 21110

[proper fern]

The 2 isnt part of the equation

[drifting sand]

Oh shit

[proper fern]

2 is the question number

[drifting sand]

Yeah my bad it's good to hen

Then

[proper fern]

Alright thank you

[drifting sand]

The 2 confused me a bit

[proper fern]

Ah

I see

Can you answer one last question for me?

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@proper fern Has your question been resolved?

[proper fern]

Alright

I was working it out

Is this correct?

[peak perch]

This looks correct to me

[proper fern]

Alright thank you

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[serene timber]

Hello, i am a little confused since there is both arccos and arcsin, not just one.

[alpine saffron]

You can use the property that $\arccos x=\frac{\pi}{2}-\arcsin x$

[solid kiln]

otheol

[alpine saffron]

And vice versa

[serene timber]

Ah

Thank you very much

[serene timber]

Is this correct?

[alpine saffron]

It asks for the answer rounded to three places after the decimal

[serene timber]

Yeah, i got 0.9596 something

Therefore 0.96

my process was:

arcsinx + 3(pi/2 - arcsinx) = 1

then -2arcsin(x) + (3\pi)/2 = 1

,tex x = \sin(\frac{1-\frac{3\pi}{2}}{-2})

[solid kiln]

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

[serene timber]

thus, 0.96~

[alpine saffron]

There is something off about this question

,calc -(1-3*pi/2)/2

[solid kiln]

Result:

1.8561944901923

[serene timber]

Hm

Thats odd

[alpine saffron]

There are no solutions for $\arcsin x=1.8561944901923$

[solid kiln]

otheol

[cosmic meadow]

That equation has no solutions

[alpine saffron]

As you can see,

[serene timber]

Interesting

I can ask the instructor tomorrow

[cosmic meadow]

The “if there are multiple solutions” got me

[alpine saffron]

If we extend the function, we do get that x could = 0.96

Which matches your answer

It would probably be best just to ask your instructor though

[serene timber]

Hm okay

Thank you

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