help-17

[velvet knoll]

😭

[teal vapor]

bro

[velvet knoll]

[livid horizon]

just substitute y as 3pi/4 in the equation

And tan x is given

Yes coz the final ans is ||2||

[velvet knoll]

😵‍💫

How?

[livid horizon]

Tan x is 1/3

[velvet knoll]

Wait so (1/3)+1/1-(1/3)

[livid horizon]

Ye ye

[velvet knoll]

Ohhhh

K

[livid horizon]

Ye ye

[velvet knoll]

I GOT IT

YES

[velvet knoll]

thx man

[livid horizon]

Nice

Np

[vocal sleet]

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[echo fox]

According to figure, find the passing straight lines
A AC and SO;
B AS and SB;
C AC and BD;
D SC and BD.

[safe wharf]

Did you mean finding the point passing through the lines

[echo fox]

maybe yes.. , i just need find passing skew lines (google trasnalte can make mistakes)

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@echo fox Has your question been resolved?

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[pale wagon]

if i use cylindrical coordinates to perform this integral would it be correct to say that rho*(length) goes from 0 to 2, theta from 0 to 2*pi and the other angel from 0 to pi/2

this is the figure

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[wary mantle]

Prove that if $X$ contains atleast three elements, then the group $(S, \circ)$ is not commutative.

[hard atlas]

I assume that B is supposed to be S?

[wary mantle]

Yep

[wary mantle]

B should be S

[hard atlas]

start slow

what about the specific case |X|=3

[wary mantle]

Well then X contains three elements

f o g is not necessarily g o f

[hard atlas]

can you give specific f,g for which they are different?

[wary mantle]

f: R -> R, x -> 1/2x
g: R -> R, x -> x^2

1/2x^2 vs 1/4x^2

Oh wait

[wary mantle]

We need bijective functons, right?

So my f and g wouldn't work

[hard atlas]

R contains quite a bit more than 3 elements

and yes, they need to be bijective

[river minnow]

It says that X contains at least three elements

Ah wait you mentioned just the case |X| = 3

Right

[hard atlas]

I asked about the specific case of having exactly three elements

[river minnow]

Nevermind

[wary mantle]

f: {1, 2, 3} -> {1, 2, 3}, x ->
g: {1, 2, 3} -> {1, 2, 3}, x ->

Uh

How would we create something not commutative there

[river minnow]

The identity map commutes with every other element, that won't get you a counterexample

[hard atlas]

for starters, why dont you list all of the elements of S

[wary mantle]

A piecewise function, maybe?

[hard atlas]

dont think of functions as formulas here

[wary mantle]

Yeah

That's why I'm saying piecewise function

[hard atlas]

for each element of 1,2,3 write down where it gets sent to

[wary mantle]

Like 1 maps to 3

2 maps to 2

3 maps to 1

[hard atlas]

ok so that function you could for example write down as 321

[wary mantle]

Well, actually that's just 4 - x

Ok, another would be

1 maps to 2

2 maps to 1

3 maps to 3

[hard atlas]

please no

[wary mantle]

ok

[hard atlas]

ok so that second map you could for example write down as 213

[wary mantle]

yeah

[hard atlas]

what would the composition of that be with the other one

[wary mantle]

so if we take f as the 321 one and g as the 213 one, then
f o g gives us
231

Right?

[hard atlas]

yes

[wary mantle]

Yep, so now g o f

That gives us

312

So not commutative

[wary mantle]

Well, we proved it for |X| = 3 now

But we need to generalize somehow

Maybe induction?

Yeah induction should work, right?

We take the same f and g

But just append n -> n

[hard atlas]

induction only works for finite sets

[wary mantle]

Oh

[hard atlas]

you would never reach something like Z or R

[wary mantle]

How? You can prove statements that go up to infinity with induction

[hard atlas]

well but at every step the statement is only ever for a finite number

but do you actually need induction for your idea?

your idea is basically, take these three elements and do the same as before. and for the other elements, do nothing

[wary mantle]

Well, map them to themselves, right?

[wary mantle]

We don't need induction, I guess

We can just state something like take f and g as above, then for every n >= 3, map to the same element

[hard atlas]

ok lets be more precise

if our X doesnt contain numbers then we cant use the exact same f and g

so lets say X has three elements x1,x2,x3

what do you want f and g to do

[wary mantle]

So |X| >= 3

[hard atlas]

yes

[wary mantle]

map x1 to x3, x2 to x2 and x3 to x1

for f

[hard atlas]

and f(x)=x for all other x

yes

[wary mantle]

then for g, map x1 to x2, x2 to x1 and x3 to x3

Yeah

We only need to pick out one value to show it's not commutative, right?

We don't have to show it for all 3

[wary mantle]

Yep, great

Yeah, so we start our proof by saying pick f ... and g ...

(f circ g)(x_1) is not equal to (g circ f)(x_1) since ...

Thus circ is not commutative

[hard atlas]

yes

[wary mantle]

Ok, now what about the |X| = 1 and |X| = 2 cases?

So for the |X| = 1 case

It needs to be commutative, right?

[hard atlas]

list all elements of S

[wary mantle]

x_1 needs to map to x_1

[wary mantle]

the only function in S is x_1 maps to x_1

So it's commutative

Right?

[hard atlas]

yes

[wary mantle]

Ok for |X| = 2

We have the possibilities
x1 maps to x1, x2 to x2
x1 to x2, x2 to x1

Wait

We can map to the same element

So we have more possibilities, right?

Ah nvm

That'd break bjectivity

Yeah, those two

[hard atlas]

yes

[wary mantle]

So taking f and g as the same function obviously commutes, so take f as the first and g as the second

Then f(g(x1)) = f(x2) = x2

g(f(x1)) = g(x1) = x2

f(g(x2)) = f(x1) = x1
g(f(x2)) = g(x1) = x1

They commute

So for |X| = 2, it commutes

Right?

[hard atlas]

yes

[wary mantle]

Ok, so how'd we precisely define f

f: X -> X, right?

$f: X \rightarrow X, \quad x \mapsto \begin{cases} 3 & \text{if } x = 1 \ 2 & \text{if } x = 2 \ 1 & \text{if } x = 3 \ x & \text{if } x > 3 \end{cases}$

[hard atlas]

well if X =N then this works I suppose, sure

but you are overcomplicating this

its perfectly fine to write down f as we did

without this piecewise notation

[wary mantle]

Oh, well then replace x > 3 with "otherwise"

That'd work too, right?

[hard atlas]

yes

well and x1,x2,x3 instead of 1,2,3

[wary mantle]

Ah, yep

[wary mantle]

Wait, how do we define x1, x2 and x3

[hard atlas]

well just some three elements from X

[wary mantle]

Do we need the axiom of choice for picking them?
I guess not, because of min

[hard atlas]

no and no

your general set X doesnt generally have a min

[wary mantle]

Ah, true

[hard atlas]

X has three elements

hence we can take three elements

and call them x1,x2,x3

formally, we know there is an injection k:{1,2,3}->X

so we call x1 = k(1), x2=k(2), x3=k(3)

but thats just useless technical stuff

[wary mantle]

Thank you!

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[patent ridge]

Hi

[patent ridge]

Why is that wrong?

I am confused

Its actually e

[flat whale]

You're not taking the limit together

You're choosing to take the limit of the inside before the outside

And that's incorrect

[patent ridge]

So that would mean 0^infinity

[flat whale]

No

[patent ridge]

Which is a indeterminate point

(1+0)^infinity

[flat whale]

Yea

[patent ridge]

No

I just took the limit of the inside

And then took of the outside

[flat whale]

Yes you did

[patent ridge]

But it should be from outside to inside

[flat whale]

And that's incorrect

No

You have to take the limit together

[patent ridge]

So I am not allowed to sum up the terms?

[flat whale]

I don't know what terms you want to sum

[flat whale]

Try plugging in a few small integers for x

[patent ridge]

1+0 which is 1 and then to infinity is 1

[flat whale]

x=2,3,4

[patent ridge]

Dw I know this limit very well

[patent ridge]

But I never asked myself to do this first like I said: (1 + 1/inf)^inf = (1+0)^inf => 1^inf

And you said that this is incorrect

So

I have to take them together

[flat whale]

Right. For this reason

[patent ridge]

Hm I see

thanks

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[cinder mulch]

Find the radius of the sphere circumscribed around a straight prism, the height of which is equal to c, and the base is a right-angled triangle with legs a and b

[cinder mulch]

how to calculate this?

[vocal sleet]

@cinder mulch Has your question been resolved?

[cinder mulch]

<@&286206848099549185>

[bronze osprey]

so OC = OA = OB right

but also O is halfway vertically between C and C1

so the height of O above the base ABC is c/2

and O lies directly above the circumcentre of ABC using the Pythagorean theorem (if circumcentre = M, then CO^2 = CM^2 + MO^2 = AM^2 + MO^2...., but CO = AO = BO and MO is the same)

so find the distance from any one of the vertices to the circumcentre

and then the height is c/2

and then Pythagoras

it's not bad at all

[cinder mulch]

but how can the center of the circumcircle be M if it is O

[bronze osprey]

if you know what the circumcentre of a right triangle is

[bronze osprey]

the circumcircle of ABC

[cinder mulch]

I didn't understand anything at all

how can the height be c/2 if it is according to the condition c

[bronze osprey]

O is c/2 above the triangle ABC

are you even reading what I'm saying?

[cinder mulch]

How can I understand if you entered some letters and values that I need to represent

[bronze osprey]

you mean you can't do anything unless someone does all the steps for you

mam nadzieję, że nie zdasz matury

[cinder mulch]

You don’t know how to explain normally, and why should I look for anything further if the answer to this problem is C/2???

[bronze osprey]

[cinder mulch]

.close

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[brave tendon]

how do I get the answer every time I try it’s wrong

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[native vale]

i need help

[native vale]

A business executive travels 1120 miles in a corporate jet and then travels an additional 60 miles by
car. The plane ride took two hours longer than the car ride. The jet travels 8 times the speed of the
car. Find the total time for the 1180-mile trip.

thats the question

x, 8x

or (y+2), y

<@&286206848099549185>

please

<@&286206848099549185>

[urban edge]

Continue mathing it

Jk lol, are you familiar with d=rt?

@native vale

[native vale]

yeah

to an extent

[urban edge]

So we need to make two equations

[native vale]

right

[urban edge]

Lets start with the car

d = rt

We know d

[native vale]

yeag

[urban edge]

So lets plug that in

[native vale]

1180=

[urban edge]

The car

[native vale]

x

[urban edge]

What is the distance traveled by the car

[native vale]

60 miles

[urban edge]

Ok so that is our d

60 = rt

We dont know r or t so those will be our variables to solve for

[native vale]

k

[urban edge]

Now for the plane ride, we know that the plane travels 8 times faster

So what does that do to r?

[native vale]

it makes it 8 times larger?

oh

smaller

[urban edge]

Yes, in other words 8r

[native vale]

division

uh

oh

[urban edge]

Do you know what r means?

[native vale]

rate

[urban edge]

Yes, in other words speed

[native vale]

yeah

[urban edge]

So the speed would be 8 times faster, so you use multiplication

[native vale]

right

[urban edge]

then the plane takes 2 hours longer, so what does that do to t?

[native vale]

t+2

[urban edge]

Yup

[native vale]

(t+2)

[urban edge]

So we have a second equation

[native vale]

oh

[urban edge]

Can you try making it

[native vale]

1120=(8r)(t+2)

[urban edge]

Nice

So now we have a system of equations
60=rt
and
1120=(8r)(t+2)

[native vale]

i guess

[urban edge]

So lets mess around a little bit with the second equation

We can start by dividing by 8

What does the equation become?

[native vale]

140=r(t+2)

[urban edge]

Cool

Now distribute the r

[native vale]

140=rt+2r

[urban edge]

Nice! We have an rt, we know what rt is because of the first equation

[native vale]

oh yeah

[urban edge]

We can replace the rt

To get?

[native vale]

uh

sorry a friend of mine is messaging me

give me as econd

ok

hm

oh

whered we get rt

60=rt

[urban edge]

Yup

[native vale]

with that we can find r

140=60+2r

[urban edge]

Yes

[native vale]

80=2r

r=40

[urban edge]

Good job, now you can find t

[native vale]

A business executive travels 1120 miles in a corporate jet and then travels an additional 60 miles by
car. The plane ride took two hours longer than the car ride. The jet travels 8 times the speed of the
car. Find the total time for the 1180-mile trip.

[urban edge]

Now be careful, this isnt the t that will solve the problem

[native vale]

mkay

60=rt

60=40t?

[urban edge]

Yup

[native vale]

t=1.5

[urban edge]

Ok, so thats the time the car ride took

We can use the information in the problen to figure out the plane ride and then add them together

[native vale]

right

wait what

[urban edge]

Yeah t=1.5

But

Reread the problem

Before that

A little further

[native vale]

oh

1.5+3.5

[urban edge]

Yup

[native vale]

wow

thank you

[urban edge]

Youre welcome

[native vale]

.close

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[prime umbra]

i don't think u are supposed to do double derevative here because it seems really complex

but i will post my work in a sec

i am having trouble making the relative max be simple enough

[pale perch]

i have no idea what youve done there

[prime umbra]

my favorite hello

oh i found the derevative

[pale perch]

im not sure your derivative is correct

[prime umbra]

and then i tried to make it equal to zero so that i couldfind a point

ok

let me look

i did it again i think it is correct

what did you get?

[pale perch]

aha youre good, i was mentally discombobulated

but the stuff after that, you completely lose me

[prime umbra]

yea

idk what to do

like i tried to separate it then find square root

[pale perch]

if that fraction =0
the only possibility is that the numerator=0

-4x^2+4=0

[prime umbra]

well i was setting it too zero

like if zero is the intercept

honestly idk

tryna find absolute min now

[pale perch]

you just have to solve this really

youll get both your x values from it

[prime umbra]

ok let me try

okokokok

nioce

plugged in the answer and got it

could you exsplain why i only needed to olve the top one more time?

ok looked at it again i think that makes sense

[late ingot]

multiply both sides by the denominator, multiplying the side with 0 stays 0

so you are left with 0 = numerator

[prime umbra]

yea ok

thanks guys

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[tropic birch]

I completely understand everything up until g for which I am completely stuck. I can't get anywhere with this. Any help would be great. (This is a practice test question)

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@tropic birch Has your question been resolved?

[tropic birch]

<@&286206848099549185>

[torn trench]

Bruh

Lol

[tropic birch]

.close

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[ionic dagger]

i have 1 attempt left

ik its pythagorean but i dont wanna risk another one wrong

cause its my last attempt for full cred

[sand hedge]

whats 30 degrees?

[ionic dagger]

well C

D is 90 and A is 60

degrees

wait i think i got it

NVM

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[pure talon]

Anyone want to double check my work? https://www.desmos.com/calculator/uzopetr3fn
Goal was to find the maximum distance you would be able to see standing inside a rotating space habitat,with a roof. I wanted to get an idea about sight lines in an oneill class space hab.

[pure talon]

Inputs at the bottom of the list. The rest is calculated

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[tidal cedar]

can someone please explain how to solve this?

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[young aurora]

Sorry I need help again

Im still a little confused

Do i find the area of the square above?

[young aurora]

wait for this one I just add the sides?

the last one I had to mutiply and split them into two

so its 128?

ty

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[honest heath]

Does this mean that, for example, $2^{1/6}$ is not constructible?

[twin meteor]

goobybalooby

[honest heath]

it does doesnt it? cuz the characteristic polynomial is $x^6-2$

[twin meteor]

goobybalooby

[honest heath]

.close

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[icy spear]

Find the largest integer m such that P(1) - 2P(7) + P(13) is divisible by m.

[icy spear]

For all P in Z[x]

Can I argue linearity and just consider the form x^n

[vocal sleet]

@icy spear Has your question been resolved?

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[lilac plaza]

Hello

[lilac plaza]

Not sure what to do after this

[gaunt sparrow]

Well first off be wary of the bounds of integration when you change variables.

[lilac plaza]

How does the intervals change?

[gaunt sparrow]

If u = cos(x) then you integrate from cos(0) to cos(pi/2)

[lilac plaza]

Like this?

If so, how does the second line interval change?

[gaunt sparrow]

Yes, although it might be easier to refer to them as their more common names, 0 and 1.

For the second one I think you set u = tan(theta) right?

[lilac plaza]

Yes

[gaunt sparrow]

So theta = arctan(u)

So theta ranges from ?

[lilac plaza]

tan(1) - tan(0)?

[gaunt sparrow]

arctan(1) - arctan(0)

Well no need for minus actually

arctan(0) = 0

arctan(1) = pi/4

So theta ranges from 0 to pi/4.

[lilac plaza]

I'm confused why it's arctan

[gaunt sparrow]

I think the antiderivative is right at the end, so just plug in the bounds as per the theorem of integral calculus says

You go from a u-integral ranging from 0 to 1.

So u = 0 for the lower bound

And u = 1 for the higher bound.

[lilac plaza]

Ohhh I see

Then do I just sub back in u

[gaunt sparrow]

Then you apply a substitution resulting in theta = arctan(u)

[lilac plaza]

And calculate the difference?

[gaunt sparrow]

No need for that

You already have the bounds.

[lilac plaza]

Does it result in the same value though?

[gaunt sparrow]

Yes.

[lilac plaza]

Oh okay

[gaunt sparrow]

If you needed the antiderivative of the original integral, then you would need to undo the subs.

But since it's a definite integral and you kept track of the bounds, it's alright,

[lilac plaza]

But doesn't tan(π/2) = undefined?

[gaunt sparrow]

pi/4.

Not pi/2.

[lilac plaza]

Ohh right

Thank you

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[warm wraith]

studying for physics under Work, power and energy

im so confused as to why I’m wrong 😭

ive searcged for solutions online and I dont get why their solution is diff

like why do other have 15f/4

[vocal sleet]

@warm wraith Has your question been resolved?

[warm wraith]

<@&286206848099549185>

🙏😭

[visual oracle]

are both answers incorrect?

[warm wraith]

based on this

i dont get where the 15f/4 cane from

or why there was no use of cos180 for the Ff

[visual oracle]

cos60 is 1/2

[warm wraith]

is that diff from directly putting it in the calcu as (2250)(7.5)cos(60) ?

i keep getting 8437.5 instead of 843.75

[visual oracle]

oh yeah they made a mistake

[warm wraith]

so i’m correct? :DD

what about the Frictional force tho

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[subtle lava]

I have a room with a wall that has a length of 174 in. and I have 4 acoustic panels that are 24 in. wide. On that wall, I want to hang those 4 acoustic panels evenly, so I need even spaces between each panel, including the spaces between each panel closest to each corner of the wall. How could I figure out the spaces I’ll need between each panel?

[ancient knoll]

draw a diagram

[last siren]

And determine the known values

It makes more sense when you stop trying to visualize it in ur head

[floral pike]

this is probably the most fun a math learner at your level with have with Math.
I also recommend you make a sketch and do some algebra 🤓

it's not very hard, I promise

draw all the horizontal distances, including the unknown distance.

you can label that whatever you like ('x' or 'd')

@last siren post the sketch and we'll help you convert it to Algebra

[last siren]

I dont have anything on me to sketch with 😆

[floral pike]

where you at

[subtle lava]

It seems so much more simple than my brain is trying to make it 😂

[last siren]

Im on a holiday lol

[subtle lava]

I’ve done way harder math than this. I’m just letting it confuse me a bit

[floral pike]

oh, sorry @last siren

I'm @ing wrong person

[floral pike]

did you make the sketch?

this is actually how all mathematicians solve unfamiliar problems

(modeling problems)

[subtle lava]

Not yet. You’re saying sketch the room out how it will look?

[floral pike]

(if you consider "sketches" and "tables" to be roughly the same thing)

[floral pike]

just do a quick-and-dirty sketch of all the distances between and including frames, across the room

label the length of the room as well

whole room --> long rectangle
4 panels evenly spaced --> ok, 4 rectangles inside that big rectangle
label the unknown distances. Use the same variable for distances that are equal.
label the known distances that are relevant in the HORIZONTAL direction

should take a few seconds to draw

[floral pike]

wow, that's pretty

[subtle lava]

😂

[floral pike]

now label the distances between vertical line segments

you said the the panels were 24 inches across

and that the unknown distances are all the same length

so that the panels are spaced evenly from the side walls and each other

[subtle lava]

So write 24 in. above each panel ?

[floral pike]

sounds good

that's a horizontal edge

[subtle lava]

And they’re each 48” height, but not that it matters

[floral pike]

no, not really

in fact, you need the constraints you provided for the horizontal distances

to solve this problem

you didn't provide the vertical distance and that space you probably want to be different anyway

(unless you want them close to the ceiling)

[subtle lava]

Well I have an a-frame shaped room (above my attic). So just the angled part of the wall is where I will want to hang the panels on and they’ll need to be centered on the wall height-wise

[floral pike]

then you already provided the proper constraint by saying the horizontal distance you are filling is 174 inches across

[subtle lava]

Oh okay

[floral pike]

did you mark up your sketch yet

[floral pike]

ok so what I would also have added

is double-arrows between the vertical edges

|       d      |
|<------------>|
|              |

as blueprints have

[subtle lava]

You’re saying like this?

[floral pike]

yeah

with the variable above

[subtle lava]

Okay cool

[floral pike]

see how I have the 'd' above the arrows?

[subtle lava]

Yep

[floral pike]

perfect.

now, write the algebraic equation for the TOTAL WIDTH of all these segments (knowns and unknowns)

think: how many d's? how many 24's?

[subtle lava]

Is that v not important that I wrote?

I can erase that

[floral pike]

oh, I mean 'd' could have been 'v'

it was whatever you wanted

[subtle lava]

Cool, I wrote v before I know we’d use d lol

[floral pike]

np

what is the equation then

[subtle lava]

Hm…

[floral pike]

^

[subtle lava]

lol oh goodness my brain hasn’t done equations in awhile

[floral pike]

^

[subtle lava]

does it start with d =

?

[floral pike]

answer my questions first

how many 24's?

count them

[subtle lava]

Ohh sorry I missed that question

5

[floral pike]

no

how many 24's?

in the picture

@subtle lava

[subtle lava]

4

Sorry you asked d’s at first

[floral pike]

$4\cdot24$

[twin meteor]

Disorganized

[floral pike]

it's ok I had a hunch

how many d's then?

[subtle lava]

Oh no worries. Sorry 😅

5 d’s

4 of the 24’s

[floral pike]

$5d + 4\cdot24$

[twin meteor]

Disorganized

[floral pike]

what is the total width of the room?

in the picture

[subtle lava]

174 in.

[floral pike]

the total length of these distances added together

yeah

$5d + 4\cdot24 = 174$

[twin meteor]

Disorganized

[floral pike]

simplify and solve for d

(the dot means multiply)

[floral pike]

nice

[subtle lava]

Look correct?

[floral pike]

yeah

[subtle lava]

Cool ^-^ so until I can hang the panels with the hardware that is coming in the mail on Tuesday, I am putting tape up to square out where each panel should hang. So I can start by coming 15.6” over from the corner of the wall?

And taping there?

[floral pike]

yeah

now, you said that your wall is what, trapezoidal?

you said the side walls are angled or something

are you haning the panels on the angled wall or the trapezoidal part

[subtle lava]

Here’s a pic of the wall shape. We’ll be hanging the panels on the angled walls, not the lower straight parts of the wall

[floral pike]

right

are they going on the rectangular sides?

[subtle lava]

They’ll be going longways like in the picture I drew. 48” long, standing up that way

With 24” width

[floral pike]

I'm more of a "shapes" guy

since the long walls are in the picture

I'm guessing the walls are rectangular

so you got nothin' extra to worry about

[subtle lava]

Yes they are rectangular, but at an angle

[floral pike]

(except I guess how close the panels might be to equipment)

[floral pike]

dunno if that really matters for your purposes

[subtle lava]

Yeah for our room size we decided to go with the thinnest panels we can to save some space for wiggle room

[floral pike]

you may want to put them up as high as you can

[subtle lava]

Oh good point. I wonder if it will look silly not being center though

[floral pike]

I'll bet it gets hot up there. You don't leave heat-sensitive equipment up there when it's hot, right?

[floral pike]

nah

they're centered horizontally

it's very Zen

[subtle lava]

Nope. Our equipment is monitored quite often. And the room stays pretty cold unless it is summer

[floral pike]

you can fit 2 more of those on either side

or are you just doing one wall

[subtle lava]

We’re going to put some panels up on the other wall too. Probably less, just enough to dampen behind the speakers

[floral pike]

ah, ok

if you had 6 panels, the spacing would be 4.285714 (repeated)

(30/7 inches)

[subtle lava]

And I just thought about it.. we’ll probably have to not hang them up very high so that the sound coming from the speakers will have a good amount of panels within it’s path to bounce into

[floral pike]

ah, k

[subtle lava]

Oh you mean 6 on the one wall (back wall across from the desk) ?

[floral pike]

yeah

well, either

[subtle lava]

Hmm, we have more panels. It would just depend on how many would get the job done

That would cover more that’s for sure

[floral pike]

I don't know too much about accoustics

[subtle lava]

They say you want to cover up as much as you can to dampen the room, but sometimes it’s good to leave a little bit of the natural room sound in there too (more useful for a high-end recording studio building that is highly treated though)

[floral pike]

that's cool

I guess it depends on how heavy you're gonna go with the post-effects

if people really want to go crazy they probably want recordings with the least amount of echo or interference with other sounds

[subtle lava]

Yes, exactly! Which is why we mostly want to deaden the room.

Because we know our home studio space isn’t the best sound naturally for a room, so we use lots of our plug-ins to help with that in post usually

[vocal sleet]

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[waxen hawk]

how to transform 1+cos140-isin140 into polar form

[vocal sleet]

Please don't occupy multiple help channels.

[waxen hawk]

and whats the point to tranform it into polar form

[waxen hawk]

that's the question on my textbook

[tidal dock]

i would convert cos140-isin140 into cartesian form first, add one, and then into polar form

the 1+ ruins a direct conversion

[waxen hawk]

I suppose the verb "convert" has to be followed by "into" rather than "to"?

[waxen hawk]

sounds like a good idea

[tidal dock]

sure

[waxen hawk]

yet, you dont know the value of cos140 degree

IMO, you cannot convert it into cartesian form, as you dont know the value of these trig function.

[tidal dock]

hm

[vocal sleet]

@waxen hawk Has your question been resolved?

[waxen hawk]

<@&286206848099549185>

[keen depot]

https://www.whatsapp.com/channel/0029Va8tObJE50UnrZc5c13h

[waxen hawk]

wdym

[gaunt sparrow]

I don't think there is a nice way to do it to get an exact answer.

Just write it in terms of cos(140) and sin(140)

[keen depot]

t'inquiete mon reuf

[waxen hawk]

Yet, the question is asking to convert it into polar form

[gaunt sparrow]

Can you not invade channel for shitpost, petal?

[gaunt sparrow]

You can still convert to polar form but keep it in terms of cos(140)

[waxen hawk]

I suppose polar form bears the identity of rcos(theta)+isin(theta)

[gaunt sparrow]

a = 1+cos(140), b = -sin(140) and you have a + b i.

[waxen hawk]

that is not polar form, as the "a" have a constant with it

[gaunt sparrow]

The point is you make it polar.

If you had x + yi how would you convert it to polar?

It's the same thing here.

[waxen hawk]

Im not sure if it is viable without a computer

[gaunt sparrow]

As I said, your answer will be in terms of cos(140) and sin(140), as they both have no closed form.

[waxen hawk]

It can be converted into the polar form tho, using trig formulas

[waxen hawk]

but i think that is horrible and unintuitive

I will show the process

the only intuitive part is that cos(2theta)=2C^2 (theta)-1

of which the negetive 1 can offset the positive 1 on the given expression

[gaunt sparrow]

Aren't you forgetting the +1 in the beginning?

[waxen hawk]

i aint?

emm

[gaunt sparrow]

In any case, I don't think this will get you anywhere since 2pi/9 is a messy angle, and even using double angle formulas will keep you stuck with 20 or 10 degree angles that also don't have a closed form.

[waxen hawk]

But that's what the questions asked for

[gaunt sparrow]

Why not just leave it with the trig functions though? This is far too intricate to be the intended goal of the problem.

$$x + iy = \sqrt{x^2 + y^2}(cos (arctan (\frac{y}{x}) + i sin (arctan (\frac{y}{x}))$$

[twin meteor]

Azyrashacorki

[waxen hawk]

what about the extra 1 in our expression

[gaunt sparrow]

x = cos(140) + 1

[unborn wind]

We can write it as cos0+isin0-cos40-isin40 so x is 2sin^2(20) and y is -2sin20cos20

Idk if this helps

[gaunt sparrow]

Yeah you would've gotten something like that at the end when simplifying the magnitude of the polar form.

But it doesn't help, just do $arctan(\frac{\sin(140)}{\cos(140)+1})$ for the angle (or start from your "simplified" expression.

[twin meteor]

Azyrashacorki

[unborn wind]

The answer comes out as 70

[gaunt sparrow]

Oh it actually does yes.

I guess we can find that $\frac{\sin(x)}{\cos{x} + 1} = \tan(\frac{x}{2})$.

[twin meteor]

Azyrashacorki

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@waxen hawk Has your question been resolved?

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[vast shale]

i dont know what this is called and i know its like 8th grade math but its 3 am and i dont know how to do this🙁🙁

[sweet birch]

algebra, simplifying, or algebraic manipulation

what you want to do is get all of the stuff like things * n on one side and all things like <just numbers> on the other

so first of all, look at the right side. you have $6n$ and you also have $-8n$. maybe we can combine those two?

[twin meteor]

PrettyPrincessKitty FS

[vast shale]

okay

ive got it down

to

6n - 1 = 8 + 15

do i just do the inverse operations thing again to move the 1

[sweet birch]

yep!

[vast shale]

thank u

very much

[sweet birch]

(you can get $n = number$ here, so make sure not to give up at $6n = number$!)

[twin meteor]

PrettyPrincessKitty FS

[vast shale]

ik that part but thank u

have a good night

.close

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[vast shale]

kind of stupid question, but how exactly do i tell if this a function or not, and when i do find that out how do i explain why it is and how it is

[vast shale]

i dont understand graphs

[waxen hawk]

there's a identity for it

[vast shale]

What do you even need to do here

[vast shale]

so youre supposed to define wether the graph would work as a function

[waxen hawk]

draw vertical lines and see if they have only one intersection points

[sweet birch]

A function has one and only one output for each input

[vast shale]

A single variable function

[vast shale]

how do i tell the difference between output and input in a graph is basically what im asking🙁

[waxen hawk]

draw vertical lines and see if they have only one intersection points

[vast shale]

Is a circle a double variable function?? I just realized this..

[sweet birch]

Input is typically the x axis

[vast shale]

okay

well that helped

i know how to do this now

thank u

again

[vast shale]

@sweet birch ?

[vast shale]

it specifically says on the assignment that using the line test wont get me credit

i accidentally cropped it out

mb

[waxen hawk]

wdym by wont get you credit? the assignment tells you not to use the "vertical line" method?

[vast shale]

Probably yes

[vast shale]

it says that if i put that as the reasoning i wont get credit

not that i cant use it

but i need to explain it using other terms

[waxen hawk]

You have to explain the rationale of this method

[vast shale]

dawg idrk

i skip this class usually

[waxen hawk]

functions can only have one output when putting in a input or more

[waxen hawk]

it basically reflect the definition of function onto the geometric meaning

[bold hound]

we write f(x)=y
where f is our function
here we can easily see that x is the input and y the output
when we graph stuff, the x axis (horizontal) is thus our input and the y axis (vertical) the output

to check if something works as a function we look at how a function is defined:
a function is well defined if for each x, if there is only one y value associated with that x value
mathematically speaking we could write it like this:
f(x)=y1, f(x)=y2 => y1=y2
meaning there is no x value which has 2 y values

if we write our function in tuples like this:
f(x)=x^2 becomes {..., (0,0), (1,1), (2,4), (3,9), (4,16), ...}
then this would mean that there must not be two tuples (x,y1), (x,y2)

[waxen hawk]

this implies functions will not have more than two of intersection points with vertical line x=k, where k belongs to real numbers

[vast shale]

i know how to do it in that form😭

i am just not good at graphs

wait let me cook

[calm pecan]

imagine a point on the graph

that point has coordinates (x, f(x))

you know f(x) can't be different for same x

[bold hound]

a quick rule of thumb:
follow the graph and forward (in x direction)
if the graph at any point moves backwards, then that is not a function

that would be the case in the circle example

[calm pecan]

so, if a graph has two y coordinates at the same x coordinate, it aint a function

if you want to dumb that down, if you were to draw a line parallel to y axis

and you find 2 points from the graph on such a line

its not a function

[vast shale]

10 {(-4,5) , (-2,2) , (0,-1) , (2,2) , (4,5)}
11 {(-4,0) , (0,4) , (4,0) , (0,-4)}
13 {(-6,-3) , (-4,2) , (-1,2) , (1,6) , (4,6) , (6,4)}

okay that took literally forever bc im on phone

[calm pecan]

okay well

[vast shale]

theyre all functions

i think

am i right

[calm pecan]

are those given or what

[vast shale]

no

[calm pecan]

no

[vast shale]

dman it

[calm pecan]

where did you even get those

[vast shale]

the graphs

theres no way im actually that dumb

[vast shale]

this makes more sense now

thank u

i didnt see that before

.close

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[spring tapir]

this isnt a unique solution is it?

because solving this analytically gets down to bnSin(npix) = 0. there are inf solutions there right

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[spring tapir]

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[quasi cloak]

why is this wrong?

[quasi cloak]

i used this rule

[sharp lynx]

are you given that x is a constant or is it related to y somehow?

[quasi cloak]

there is no further information given

i thought that it is a constant

[sharp lynx]

the main issue is that the derivative of x^2+1 with respect to y is not 2x

thank you chartbit but I can handle this myself

[quasi cloak]

so instead of 2x i have to put x^2+1?

[sharp lynx]

the derivative of x^2+1 with respect to y is not x^2+1, no

[quasi cloak]

how do i get the derivative with respect to y?

[sharp lynx]

you said yourself that x is just some constant

[quasi cloak]

so it is just x^2?

[sharp lynx]

the derivative is not x^2, no

[quasi cloak]

i am confused, can you give me another hint?

[sharp lynx]

if x is just some constant unrelated to y, is x^2 also just some constant unrelated to y?

[quasi cloak]

yes

[sharp lynx]

what about x^2+1? is it also just some constant unrelated to y?

[quasi cloak]

yes

[sharp lynx]

what is the derivative of a constant?

[quasi cloak]

0

[sharp lynx]

so what is the derivative with respect to y of x^2+1?

[quasi cloak]

0

[sharp lynx]

there lies the fix to your solution

[quasi cloak]

yes, thank you very much

[sharp lynx]

now note that you don't even have to use the quotient rule for this

(as it would help with the fact that your final answer is unsimplified)

since x is just a constant, you could have just started by rewriting it as $\left( \frac{x}{x^2+1} \right) y^5$ and just differentiate one little thing

[twin meteor]

Steakanator

[quasi cloak]

so then (x/x^2+1) would be my constant and i would only have to create the derivative of y^5, right?

[sharp lynx]

indeed

[sharp lynx]

note that if you cancel the (x^2+1)s in this solution, you get the same thing

[quasi cloak]

wouldn't it be (5xy^4)/(x^2+1)?

[sharp lynx]

it would

[quasi cloak]

but 5y^4 * (x/x^2+1) isn't the same or am i wrong?

[sharp lynx]

it certainly is the same thing, just written differently

[quasi cloak]

ah yes i see, because i can write 5y^4 as 5y^4/1 and then multiply it

then the denominator would stay x^2+1 and the numerator would be 5xy^4

[sharp lynx]

yep

[quasi cloak]

this method could potentially save time in an exam

[vocal sleet]

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[cosmic kayak]

How do I solve this? I don't even know how to start

[dull bear]

seems like you'd probably want the Lambert W function for it

[fathom ridge]

Yea

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[finite hatch]

Hello

[finite hatch]

Is question 3b right?

I thought tan was like opposite over adjacent

So its 0.5/0.86

[flat whale]

,calc tan(330 deg)

[twin meteor]

Result:

-0.57735026918963

[flat whale]

,calc -0.5/0.86

[twin meteor]

Result:

-0.58139534883721

[flat whale]

Missing a minus sign

[finite hatch]

Im confused

So is it wrong then

,calc 0.86/-0.5

[twin meteor]

Result:

-1.72

[finite hatch]

Ohh its my approximation!

BTW why is it negative? Shouldnt both of them be negative so they cancel out

Sine in the 4th quadrant and cosine in the 4th quadrant

Oh...

Cosine is positive and then the sine is negative!

Ok thank u so much

😅😅

.close

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[elfin igloo]

Conditional Expectation

[elfin igloo]

I don't understand what E[Y|X] is, here.

I know it's the Conditional Expectation of Y, given X.

But... given X is what?

It's not E[Y|X = x], it's just E[Y|X]

[flat whale]

https://www.probabilitycourse.com/chapter5/5_1_5_conditional_expectation.php

[elfin igloo]

Right, so I'm supposed to interpret E[Y|X] as a function of X

?

[flat whale]

Yes

[elfin igloo]

OK. So in the image I pasted, I minimize the E [ ( E[Y|X] – f(X) ) ² ]...

...or, in words, I minimize the expected value of the square of what I get when I take a function that takes X inputs and returns Y inputs — which is a model I assume to exist in the real world — and subtract from that function my prediction function, which takes X inputs and outputs predicted Ŷ outputs...

...I minimize that when my prediction function f(X) is equal to the assumed-to-exist-in-the-real-world function E[Y|X]

?

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[elfin igloo]

.close

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[bright nacelle]

Can someone explain to me part d , I don’t understand it

[obtuse radish]

,rotate

[obtuse radish]

this is someone else's work?

[bright nacelle]

No we did this in class

I forgot how we solved it

[obtuse radish]

dont some stuff look awfully similar around this part

@bright nacelle im not gonna solve this for u

contribute dammit

[bright nacelle]

What do you mean

All of them are “x”

Thst just how i write in french

[obtuse radish]

dont a lot of these thinhs look similar?

like they might cancel out?

maybe?

[bright nacelle]

You mean one positive one negative

?

If you take a (x-1) in the numerator as common

You cancel it with the denominator

<@&286206848099549185>

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[vocal sleet]

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[eager elbow]

The half-life of stronium-90, Sr, is 25 years. This means that half of any amount of Sr will decay after 25 years. Look for an expression m(t) after t years of a mass of 24 mg of Sr.

[eager elbow]

need help

i dont know how to find the expression

he looses 12 mg at the moment 25 years or its progressiv

[dark kiln]

it's progressive

[eager elbow]

ok

[dark kiln]

it's just exponentiation things

[eager elbow]

i though it lost 12 mg at the year 25

[dark kiln]

it would mean it has to lose 12.94% in 5 years for example

,calc 0.8705^5

[twin meteor]

Result:

0.4998548119289

[dark kiln]

5 years happens 5 times, results in half remaining

[eager elbow]

how did you find 12.94

[dark kiln]

,calc (1/2)^(1/5)

[twin meteor]

Result:

0.87055056329612

[dark kiln]

5 years is 1/5 of 25

this is in 12.5 years (1/2)^(1/2)

,calc 0.7071 * 0.7071

[twin meteor]

Result:

0.49999041

[eager elbow]

wait my brain is lagging

[dark kiln]

if you lose 29% twice it adds up to half

,calc 1000 (1 - 0.29)(1 - 0.29)

[twin meteor]

Result:

504.1

[dark kiln]

this is because 0.71 is squre root of 0.5

[eager elbow]

i understand nothing sorry

there are alot of value

[dark kiln]

i don't think there's like anything to understand, it's just what the calculator shows

it solves the problem, somehow

we found how much is lost after 5 years, and 12.5 years

[eager elbow]

but i dont know where the 0.8705^5

come fram

from

[dark kiln]

so 25 years is 5 periods of 5 years