help-0

[alpine sable]

like i don't understand the question at all

[gray isle]

are you familiar with the trig formula for area of a triangle?

[alpine sable]

a bit

but not too much

[wheat shore]

sir anyone know?

[gray isle]

there's a formula for it, do you know it?

it should be a simple yes/no answer

[vale wigeon]

@wheat shore this channel is occupied, please move.

[wheat shore]

;-; where?

[alpine sable]

no i don't know

the formula

[vale wigeon]

a channel that isn't occupied

[delicate umbra]

does anyone know what a phase line portrait is

[alpine sable]

I don't really understand that haha

whats cdot

[gray isle]

argh bot delay

[alpine sable]

oh

[gray isle]

is the (trig) formula for the area of a triangle

where a and b represent the lengths of two sides and C is the angle between them

[alpine sable]

ah ok

but how do i find two possible angles though

[gray isle]

first set up an equation (applying this formula)

[alpine sable]

yes done

[gray isle]

what's your equation

[alpine sable]

4.5 = 1/2 (3)(4) sin(C)

[gray isle]

wrong use of c

in your question, you should be using <A,A or <BAC

[alpine sable]

oh i should use A?

[gray isle]

be consistent with what's presented in the question

[alpine sable]

ok, I believe i should use A

[gray isle]

yes

and then isolate sin(A)

[alpine sable]

oh i should solve for A?

[gray isle]

well yes

A represents the angle they want you to find

[alpine sable]

yes yes

sin(a)=0.75

i think

i might've done sum wrong

[gray isle]

A not a

[alpine sable]

yes

am i correct? is it 0.75

[gray isle]

sin(A) will be 0.75

[alpine sable]

ok

[gray isle]

and then using inverse trig and apply properties of sine to determine the two solutions to that (for 0<A<180°)

[alpine sable]

wait. how do i use inverse trig

and apply properties

[gray isle]

by the time you're given this problem, you should have an idea of how to do those things

[alpine sable]

sorry

[gray isle]

you should at least know how to use inverse trig from right triangle trignometry

[alpine sable]

i don't know... i really don't

hoping you can help me

will mean the world

[gray isle]

have you done any trigonometry problems with right triangles before?

[alpine sable]

yes, but i vaguely remember

[gray isle]

eg would you be able to find an approximate value for x here?

[alpine sable]

um

dudeee omg

idk

[gray isle]

sin(x) = 3/7
so you haven't seen anything resembling:
x = arcsin(3/7) = sin^-1(3/7)
?

[alpine sable]

eh

i have a bit

anyways, how do i find the inverse trig

[gray isle]

do you at least understand what's been written there?

[alpine sable]

yes

vaguely

[gray isle]

those statements are contradictory

vaguely leans a bit towards not understanding

[alpine sable]

yes i don't understand

sin^-1(0.75) is the inverse right?

of A

[gray isle]

given a trigonometric ratio, the appropriate inverse trig function can be used to help determine possible angles

given that
sin(A) = 0.75
you can apply the inverse sine to both sides to get the value of A between 0 and 90°

A = sin^-1(0.75) will give you one of the values they want

[paper minnow]

How do you calculate the power of hypothesis testing when using chi square distribution? I found the critical chi square value under H_0 but I don't know how to find the power

[gray isle]

applying properties of sine,
sin(180° - A) = sin(A)
(and that you triangle could be acute or obtuse)
180° - sin^-1(0.75)
would be another possible value for A

[paper minnow]

I can find the power when using t distribution or normal distribution, but chi square just throws me for a loop

[gray isle]

(and you can use a calculator to get an decimal approximation for those angles)

[alpine sable]

ok

[grizzled ingot]

How would I find a general solution to a matrix when I havent been given a b vector, but three unique solutions for it

[vale wigeon]

i think you may have an issue with your wording

if there's three solutions then they aren't unique

[grizzled ingot]

uh so I have equations with 4 unknowns, but i dont have a given b vector

ohh oops

yh specific solutions is probably a better way to word it

[vale wigeon]

so you're considering a linear system Ax = b, and you're not given A nor b, but you're given three different values of x which you're told are solutions to it?

[grizzled ingot]

im given A

Do i just make my b vector (0, 0, 0)?

[vale wigeon]

no?

am i correct in understanding that you are given three vectors x1, x2 and x3 and are told they are all solutions to Ax=b

[grizzled ingot]

uh yep

[vale wigeon]

then b will be A times any one of those.

[grizzled ingot]

but like even in my matrix i have like unknown coefficients

[vale wigeon]

you have $Ax_1 = b$

[ocean seal]

Ann

[grizzled ingot]

yh

[vale wigeon]

and also $Ax_2 = b$ and also $Ax_3 = b$

[ocean seal]

Ann

[vale wigeon]

it would help if you showed the whole problem exactly as stated.

[grizzled ingot]

I dont exactly have the whole problem

$\begin{pmatrix}
-30 & q & r & s\
0 & 0 & -5 & t\
0 & 0 & 0 & 0\
\end{pmatrix}$

[ocean seal]

legend27

[grizzled ingot]

this is the matrix im given

[vale wigeon]

that's A?

[grizzled ingot]

yep

[vale wigeon]

okay

and what are your x1, x2 and x3?

[grizzled ingot]

$\begin{pmatrix} -2 \ -1 \ -2 \ 10 \end{pmatrix}$

[ocean seal]

legend27

[vale wigeon]

that's one of them

where's the other two

[grizzled ingot]

$\begin{pmatrix} -8 \ 4 \ -4 \ -15 \end{pmatrix}$

[ocean seal]

legend27

[grizzled ingot]

$\begin{pmatrix} 1 \ -7 \ 5 \ 30 \end{pmatrix}$

[ocean seal]

legend27

[vale wigeon]

okay

and what exactly are you asked to find here?

the unknown entries in A?

[grizzled ingot]

general solution for the system

[vale wigeon]

ah

okay, so that is a little simpler

simpler as in it takes less algebra

[grizzled ingot]

ahhh

[vale wigeon]

if you know your way around linear systems, that is. because there's a trick.

first, observe that our A is in row-echelon form

[grizzled ingot]

yep

[vale wigeon]

and there are 2 free variables and 2 pivots

[grizzled ingot]

yep

[vale wigeon]

therefore our solution space will have dimension 2

[grizzled ingot]

yep 2 parameters

[vale wigeon]

okay now you have three solutions given to you: $$s_1 = \bmqty{-2\-1\-2\10}, s_2 = \bmqty{-8\4\-4\-15}, s_3 = \bmqty{1\-7\5\30}$$

[ocean seal]

Ann

[vale wigeon]

i've renamed them from x to s to hopefully have less notational conflicts later on

[grizzled ingot]

yh thats fine

[vale wigeon]

so $As_1 = As_2 = As_3 = b$ (even if we don't know $b$)

[ocean seal]

Ann

[vale wigeon]

this also means that $A(s_2-s_1) = 0$ and $A(s_3-s_1) = 0$

[ocean seal]

Ann

[grizzled ingot]

right

[vale wigeon]

thus s2-s1 and s3-s1 are solutions of the homogeneous system Ax = 0

[grizzled ingot]

ahhhh

[vale wigeon]

and you know that general sol = particular + homogeneous sol

[grizzled ingot]

ahhhhhhh

i think i got it

[vale wigeon]

so you have your general solution as $$x = s_1 + \alpha (s_2 - s_1) + \beta (s_3 - s_1)$$ where $\alpha, \beta \in \bR$

[ocean seal]

Ann

[grizzled ingot]

so $x = a + (\textrm{ General sol to Ax=0 })$

[ocean seal]

legend27

[grizzled ingot]

yep

thank you so much for your help!!!

[alpine sable]

hellooo

may i have help

[grizzled ingot]

@alpine sable you're asked to find angle ABC, right?

[alpine sable]

yes

[alpine sable]

yes

[grizzled ingot]

Notice that ABC is a right-angled triangle, where BC would be our hypothenuse, right?

[alpine sable]

yuh i figured it out kinda

[grizzled ingot]

oh so you dont need help?

[alpine sable]

uh go dm

[grizzled ingot]

Ann sorry to disturb you once more, but would I need to find general solutions to Ax=0 or would specific solutions work as well?

[grizzled ingot]

I can explain it here

[vale wigeon]

well you can check manually that s2-s1 and s3-s1 are linearly independent

or at least they should be

[alpine sable]

i just dont' know how to use soh cah toa

lmao

[grizzled ingot]

uhh i havent gotten that far in linear algebra lol

[vale wigeon]

they aren't parallel

thus they span the solution set

because it has dim 2

[grizzled ingot]

ohh right

thank you so much

[grizzled ingot]

ahhhh

lemme find a helpful diagram for you

@alpine sable there we go

Ill post this again so that we dont have to keep scrolling up again

So now we have to find theta in this case

we have been given the opposite and the adjacent values, so what trig function would we use? @alpine sable

[alpine sable]

uh. I guess we would use toa

[grizzled ingot]

right

so tan

[alpine sable]

sorry for the late response

yes

[grizzled ingot]

na its allg

[alpine sable]

16/9 rigth?

[grizzled ingot]

so we know now that $\tan(\theta) = \dfrac{\textrm{Opposite}}{\textrm{Adjacent}}$

[ocean seal]

legend27

[alpine sable]

yes

[grizzled ingot]

so $\tan(\theta) = \dfrac{16}{9}$

[ocean seal]

legend27

[alpine sable]

yes yse

[grizzled ingot]

now to find theta, we need to take the tab inverse of both sides

[alpine sable]

yes

tan^-1(16/9)

[grizzled ingot]

$\tan^{-1}(\tan(\theta))=\tan^{-1}\left(\dfrac{16}{9}\right)$

[alpine sable]

i got 60.64 degrees

[ocean seal]

legend27

[grizzled ingot]

that seems to be the answer

[alpine sable]

could you double check just in case

[strange plover]

What is $\int e^{-x^2} dx$

[grizzled ingot]

yep i just did

[ocean seal]

256bitEncoder

[strange plover]

And why is it so special?

[alpine sable]

ight

dude, whats b

[grizzled ingot]

from what i know that that integrand doesnt have a anti-derivative

[strange plover]

Ohh that's makes sense with erf(x) function

[grizzled ingot]

yep

[alpine sable]

😦

legend can you friend me

lets go dms

for a sec

[grizzled ingot]

we can find the value if we add bounds @strange plover

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

@strange plover maybe you wanna read this, if you have the time

@alpine sable sorry man, im kinda busy rn, i have some uni things to do

[strange plover]

Yeah I will sure read it

[grizzled ingot]

maybe someone else in this server can help you

[alpine sable]

@grizzled ingot yo

[old fiber]

how the fock did they simplify

[alpine sable]

rip

idk what that is

at all

[jagged imp]

they factored an (x^2-1)^3 out @old fiber

also post a better picture next time lol

[old fiber]

oooooooooooo makes more sense

[old fiber]

lol ok

[jagged imp]

had to open that one in browser

[old fiber]

its still wrong though

when u factor that out

what about the 8x^2 times the second part

it wont be 9x^2 -1

[jagged imp]

$(x^2-1)^4+8x^2(x^2-1)^3=(x^2-1)(x^2-1)^3+8x^2(x^2-1)^3\$ Now factoring $(x^2-1)^3$ out, we have $\(x^2-1)^3(x^2-1+8x^2)$

[ocean seal]

111211211111221312211

[jagged imp]

and then group like terms

[strange plover]

We all know the generalized stokes theorem $\int_{\partial M} \omega = \int_M d\omega$
So the integral of differential forms over curved space is meaningful, but from tensor calculus I know a fact that taking integral of a tensor in curved space is meaningless so what's the point?

[ocean seal]

256bitEncoder

[foggy onyx]

can someone help me. i gotta find the x of the given figure

[oak chasm]

@foggy onyx Do you know how similar triangles work?

[foggy onyx]

i forgot-

[oak chasm]

OK, similar triangles have all three angles the same, and the sides of one get multiplied by the same number to get the sides of the other.

So, let's look at the left side of the triangle.

See the two marks on each part of that side?

Like an equal sign?

[foggy onyx]

bo

nope

wait lol i dont understand

you mean the lines inside the triangle?

[oak chasm]

No, the marks on the left side of the triangle.

Do you see the left side of the triangle?

[autumn timber]

https://tenor.com/view/dog-doggo-cute-math-formulas-gif-17580986

im just watching

[foggy onyx]

its not an equal-

[oak chasm]

What isn't?

[foggy onyx]

oh wait

nvm i can solve on my own

sorry if i disturb you

[oak chasm]

Oh, OK.

[strange plover]

<@&286206848099549185> Hey!, Could someone please help me out?

[upper ember]

[strange plover]

We all know the generalized stokes theorem $\int_{\partial M} \omega = \int_M d\omega$
So the integral of differential forms over curved space is meaningful, but from tensor calculus I know a fact that taking integral of a tensor in curved space is meaningless so what's the point?

[ocean seal]

256bitEncoder

[oak chasm]

@strange plover Please read #❓how-to-get-help. @Helpers is only for use 15 minutes after posting a question, and you're not supposed to post a question to multiple channels.

[strange plover]

Sorry I wasn't getting an answer but I ensured that 15 mins pass and it basically passed in every channel. Sorry for inconvenience.

[sonic snow]

How do i get the answers?\

[tame stirrup]

You should learn about mid point theorem ... All the problems hera are based upon that

[sonic snow]

oh ok

ty

[oak chasm]

@sonic snow The top and entire triangles are similar triangles.

[sonic snow]

okok

[split fulcrum]

not sure what the integral is that I should setup is

[ocean seal]

Chai T. Rex

[vale wigeon]

\2?

$\sin(\theta) > \sin(2 \theta)$

[ocean seal]

Ann

Chai T. Rex

[oak chasm]

Ahh, thanks Ann.

[jovial mica]

Can someone explain how i got this wrong, CAS says the answer is -380x^18

This is the question.

[gray isle]

such bad notation in the notes

[jovial mica]

as in my answer or the question?

[gray isle]

as in they shouldn't have used those $\floor{\ }$ which indicates floor

[ocean seal]

ℝamonov

[gray isle]

what you did seems ok

[jovial mica]

oh alright, can you see what i did wrong?

Do you i think i used CAS wrong?

[gray isle]

yes

[jovial mica]

mmmm.

[gray isle]

it seems you used a multiplication sign between the first two terms

instead of a +

[jovial mica]

ahhhhh, yess i have.

Thank you so much!

[lone vapor]

hi can anyone help me out

[shut wadi]

what exactly is the problem?

[lone vapor]

i like understand the maths step to finding the answer to this

[shut wadi]

Do you know how to find the highest value in a matrix in excel/spreadsheet?

[lone vapor]

no but im trying wanna do this with just my pen and paper

[shut wadi]

So could you please elaborate a bit more on where you are exactly stuck?
What exactly do you not know to get an answer

[lone vapor]

with the total baseline added up for 12 months

thats my total kwh for the year

right?

[shut wadi]

that's not peak consumption though if i am understanding correctly

[lone vapor]

yeah this " peak consumption "

[strange plover]

Guys?

[lone vapor]

how to get it

yes?

[shut wadi]

as far as i understand it should just be the highest entry in the baselines

[strange plover]

Can normal one forms exist?

[lone vapor]

okay so how can i find peak consumption

[shut wadi]

find the biggest number in the baselines?

[lone vapor]

that will be august = 27.13

[shut wadi]

Good and now that you know the peak consumption, calculate how much march's baseline is compared to that in %

[lone vapor]

march is 10.64

[shut wadi]

so what percentage is that of 27.13?

[lone vapor]

ooo

27.17 x 100

2713%?

[shut wadi]

hmnn not quite

what percentage is 10.64 of 27.13

for example 5 is 25% of 20.

[lone vapor]

10.64/27.13 * 100%?

[shut wadi]

yes!

[strange plover]

Hey, there!

[lone vapor]

39.2%

this would be?

[shut wadi]

yes that should be correct

[lone vapor]

i did that to find out?

march consumption in percentage

[strange plover]

Hey guys could I ask a calculus related question?

[shut wadi]

this channel is still occupied

feel free to ask in another (unoccupied one) :)

[lone vapor]

hold on 256 , there is an hard equation question being solve atm

[strange plover]

Oh ok

[shut wadi]

I don't know, this was just what the question asked you to calculate

[lone vapor]

round to 2 decimal

is 39.22

wow

thanks

!!

thats correct answer

thanks glitched!

!close

[strange plover]

Sorry but I didn't find any unoccupied channels that were active. Is this one free now?

[alpine sable]

yeah free

[strange plover]

So the question is that how can we take the integrals of one forms over a curved space.

[jovial mica]

Can someone please check if i have done this properly?

[lament wyvern]

looks right to me

[jovial mica]

alright, thank you.

[glossy brook]

Could someone explain why integers are not in rational numbers??

[strong furnace]

but integers are in rational numbers

[glossy brook]

So the site is wrong?

[strong furnace]

I think it is the notation

sets do not belong to other sets

they are subsets of other sets

[lament wyvern]

is the set of all integers a rational number

[glossy brook]

Ah

so $A \in B$ is never true for any sets $A$ and $B$?

[ocean seal]

time

[lament wyvern]

you can have a set of sets (e.g. the power set)

you can't have a set of all sets

but $a \in A \subset B \neq A \in B$

[ocean seal]

PrettyPrincessKitty FS

[vale wigeon]

you need to know the difference between $\in$ (element-of) and $\subseteq$ (subset)

[ocean seal]

Ann

[vale wigeon]

$x \in \bQ$ means ``$x$ is a rational number''

[ocean seal]

Ann

[glossy brook]

alright

[vale wigeon]

and $\bZ$ refers to the entire set of integers. not ``an integer'', the entire set of integers as a single unit.

[ocean seal]

Ann

[vale wigeon]

and that certainly is not an element of Q.

it's true that $\bZ \subset \bQ$ though.

[ocean seal]

Ann

[glossy brook]

thank you very much

[strange plover]

Hey, there! Is Angular momentum different from a vector quantity?

[bleak ibex]

As far as I know it is neither scalar nor vector

[sly mantle]

classically, angular momentum is a vector

[blazing zinc]

uhh

[oak chasm]

@strange plover According to the following page, angular momentum is a vector: https://courses.lumenlearning.com/boundless-physics/chapter/vector-nature-of-rotational-kinematics/.

[manic glade]

need to solve this in C

i dont think i even really understand the question

[oak chasm]

What does the line above z mean?

[limpid spade]

It means adjugate

Forgot name but it's a - bi instead of a+bi

[oak chasm]

Oh, complex conjugate.

[limpid spade]

Yes

[ocean seal]

Chai T. Rex

[manic glade]

i mean, thats just a different way to write it

im not sure that leads to a solution

[strong furnace]

it should lead to a solution

can you show your work after this ?

[manic glade]

there is no work i genuinely dont even understand what is being asked

[strong furnace]

they want you to find out complex numbers z which satisfy the equation z^3=z'

[manic glade]

looks like someone else solved this with trigo

[strong furnace]

that is another way of doing it

[noble sinew]

it's asking for all solutions for z where that is true. Would probably recommend multiplying both sides by z and solve that and/or writing in polar form

[manic glade]

Z^4=|Z|^2?

[noble sinew]

yep and then we have z=|z|e^(i\theta)

[manic glade]

e?

[strong furnace]

cis(theta) can also be represented by e^i(theta)

[manic glade]

i need to go over the textbook again i guess none of this sounds familiar

[jagged imp]

no you don't you probably haven't learnt it

the e^(itheta) they're talking about is just your cis(theta)

[manic glade]

dunno what cis theta is either : (

[fervent grail]

Pls help

[strong furnace]

cis(theta) is cos(theta)+isin(theta)

[manic glade]

(i do see it in the picture i posted )

[fervent grail]

I know the ans is b

But how ???

[strong furnace]

use a different channel this one is busy

[fervent grail]

Can anyone help ??

[manic glade]

oh.

[fervent grail]

Ok

[noble sinew]

can also be done with expanding (a+bi)^3=(a-bi) then remember both real and imaginary part must be equal. So we can write 2 equations with 2 unknown and solve that (more tedious than what I suggested but works)

[manic glade]

im going over the chapter for this again ;x

wait so when using polar coordinates

do i need theta or tan(theta)

[noble sinew]

Expanding (a+bi)^3 gives us:
(a+bi)^3=a^3-3ab^2+i(3a^2b-b^3)
For z^3 to be equal to z conjugate we must have real and imaginary parts equal, so
a^3-3ab^2=a and 3ba^2-b^3=-b

Then solving above gives the solutions.

[noble sinew]

theta

[marble otter]

can someone please help with this q

[lament wyvern]

can you make an equation for the length of the rectangle

[vale wigeon]

@marble otter channel taken please move

[marble otter]

oh ok

[vale wigeon]

(unless scapeprof and corylus are done)

[manic glade]

i might have followups but its going to be sporadic as im reading through this i wouldnt claim the channel for this

maybe just to confirm this is ok

[vale wigeon]

i'd advise against asking in #help-0 if you can at all avoid it

[strange plover]

Oh guys actually I found Angular momentum to be a bivector totally unfamiliar to me

[manic glade]

me?

[strange plover]

I meant that the term "bivector" is not familiar to me

[manic glade]

how exactly do you go from tan(theta) to theta?

im noticing in all the examples in the book its assumed to be something known and obvious

i goes directly from calculating the tan to choosing which of the theta options are correct

[glass lichen]

arctan..?

[manic glade]

the heck is that

oh wow this thing is never mentioned

[glass lichen]

the inverse of tan..?

$\tan{x}=a\implies x=\tan^{-1}(a)=\arctan(a)$

[ocean seal]

moshill1

[manic glade]

the wiki is kind of loaded is there a simple calculation for it? like how tan theta is y/x

[glass lichen]

the button on your calculator. . .

[strange plover]

@Corylus $tan^{-1}(x)$ basically gives the angle that helps u to create a tangent line with length tan(x)

[ocean seal]

256bitEncoder

[strange plover]

R u fine with the solutions u got?

If not then just mention.:)

[manic glade]

oh no i dont have the slightest inkling of understanding about any of this

right now trying to understand how you go from tan theta = -sqrt(3) to theta=2pi/3

[glass lichen]

knowing exact values or calculator and knowing what quadrant

[manic glade]

tan^-1 of -sqrt(3) gives me -60 on my calc but i dont even know if im doing it wrong or its a different way to write it

[strange plover]

It actually has a intuitive meaning but is often calculated on calculator as it is just a compositional inverse of a function.

[glass lichen]

yes that's the quad4 angle, 300degrees or 5pi/3

[strange plover]

@glass lichen is right

[manic glade]

quad4?

[glass lichen]

quadrant 4

[manic glade]

how do you get that from -60

[glass lichen]

cause it went 60 clockwise

[manic glade]

oh its backwards circle from 360

man this would have been so much easier if they actually taught the relevant information

[glass lichen]

$Q_4: \theta\in\left(\frac{3\pi}{2},2\pi\right)=\left(0,-\frac{\pi}{2}\right)$

[ocean seal]

moshill1

[strange plover]

If u want some more explanation u can go to #geometry-and-trigonometry

[manic glade]

ok and how do i go from this to pi?

[glass lichen]

convert between deg and radians

[manic glade]

@_@

[glass lichen]

60 degrees is pi/3

[strange plover]

Yeah

Radians r pretty cool

[proper hare]

I need help

With fathers

Would mean a lot

[manic glade]

can i convert on the calc as well?

[glass lichen]

yeah. .

[strange plover]

Use trig identities

[proper hare]

So I think I got the last one

But the first one I think I wasn’t supposed to get this answer

So I’m wondering how it’s supposed to even look

[strange plover]

Such as $1 - sin²(\theta) = cos²(\theta)$

[manic glade]

what button is that? ;x

[ocean seal]

256bitEncoder

[glass lichen]

No fucking clue, idk what calc you have.

[proper hare]

Kinda lost tbh

[strange plover]

And inverse trigonometric functions

[proper hare]

May I see how you did the first one

Yeah

[strange plover]

Get $\alpha$ and $ \beta$ by $sin^{-1}(\theta) and cos^{-1}(\theta)$ then ur good to go :)

[ocean seal]

256bitEncoder

[proper hare]

So in this case

It would be...

How did I get 0 🤦‍♂️

I definetly am doing something wrong

I just want to se show you did 1 so I can do the rest

[strange plover]

Oh okay

$\alpha = sin^{-1}(40/41)$
$\beta = sin^{-1}(12/13)$

[ocean seal]

256bitEncoder

[proper hare]

So do I write this way as an answer for example: (12/13) or so I literally write the whole thing

[strange plover]

Wait I am telling the vals

[proper hare]

And the whole thing I mean the whole a=sin-1 (40/41

[strange plover]

Of alpha and beta

[strange plover]

Nah

Ofcourse not

[proper hare]

Yeah I was kinda confused why lol

[strange plover]

Alpha approximately is 77.3 degrees

[proper hare]

So for Cos (a+b) it would be 77.3?

Because I thought I needed radical

[strange plover]

Beta is approx 67.3

[alpine sable]

LRLRLELELRLRLRL3LR

REAL OVLE

[strange plover]

Cos(77.3 + 67.3)

This should be 1

[proper hare]

I understand!

Thx

[strange plover]

Welcome

[strange plover]

:)

[alpine sable]

LRLRLRLRLRLRLR

[jagged imp]

<@&268886789983436800> le epic trolls come to raid

[proper hare]

So 144.6

@strange plover

[strange plover]

So cos of it is -0.99 approx

[proper hare]

Wait that’s the answer?

Not 144.6?

[strange plover]

144.6 is alpha + beta

Isn't it?

In a u have to take cos of it

[proper hare]

So 143.61 or just -0.99 as the answer?

[strange plover]

-0.99

Of a.)

[proper hare]

Ohhh

Do you mind showing how u got that real quick

I wanna solidify this

[strange plover]

From the unit circle

And accuracy by calculator

[proper hare]

Oh literally like from 0.97560 and you round

From the exact form of 40/41

[strange plover]

Oh sorry my unit circle had a mistake

I didn't see the rads and degrees form

It's -0.809

[proper hare]

Is it -0.322

[strange plover]

U r working in radians?

Or deg?

[proper hare]

Which should I use

[strange plover]

U can't get a value that low because 144.6 is real close to 180

[strange plover]

The one that's listed

If it isn't then insure to put degree symbol

[proper hare]

So can you show a way of how u did it all in order for the first one so I don’t miss anything

[strange plover]

Ok

First u can take the cos inverse of 40/41

That is radians I believe

??

[proper hare]

0.5605

[strange plover]

R u using radians

Oh my god

[proper hare]

0.00978

[strange plover]

These r radians not degrees

I can't use inverse trig functions

U should use radicals I believe

[proper hare]

Idk then

[strange plover]

Radicals r essential for this question

I am a physics student so I work with degrees

Where I can use inverse funcs

[strong furnace]

you can use inverse functions with radians as well

[strange plover]

What

Really?

[strong furnace]

why would inverse functions not work with radians

[strange plover]

Oh u mean to convert into degrees

Radians*

Te get it

But @proper hare has a bit different problem can u help him out

I am not great at trigonometry

[strong furnace]

arcsin(1)=pi/2 , arcsin(1/2)=pi/6 arcsin(sqrt(3)/2)=pi/3

[proper hare]

Hey I did that one I’m good on that one guys!

Now this one I need help

If I can get both these I can do the rest

It’s just these 2 are more difficult than usual

[jagged imp]

those look a lot like the format of the compound angle formulas ay

[strong furnace]

they sure do

[strange plover]

True

[jagged imp]

just a passing observation

[proper hare]

Could someone help with these 2? Would mean so much 🙂

[finite holly]

try using the formula sin(A-B)=sinA.cosB - cosA.sinB

[proper hare]

Is it this

[strange plover]

Yeah

[finite holly]

correct..

[proper hare]

For both?

[jagged imp]

exact answer

[proper hare]

?

[finite holly]

yess

[jagged imp]

.8660 is not exact

[proper hare]

Then what is it

[vague coral]

with fraction

[proper hare]

?

[jagged imp]

the bottom box wants sqrt(3)/2

and the top one wants the single trig function you evaluated to get sqrt(3)/2

[proper hare]

But I’d it’s not what I said for decimal idk

[strange plover]

$sin(2\pi/3)$

[finite holly]

ie sin(2*pi/3)

[proper hare]

What is the exact decimal then

[ocean seal]

256bitEncoder

[jagged imp]

there is no exact decimal that will give you sqrt(3)/2

exact form is sqrt(3)/2

[strange plover]

It's irrational

U wanna say this?

[proper hare]

I mean top box

[glass lichen]

please use actual latex with the bot 🥺

[proper hare]

What is the top box supposed to be

[vague coral]

The value is irrational, there is no exact decimal

[strange plover]

Yeah u can leave it as frac

Bottom is gonna use same method

[proper hare]

Wait so both answers are square root 2/2?

Square root 3/2

[strange plover]

Yeah there r same equations

Just different representation isn't it

?

[proper hare]

So this?

[jagged imp]

[strange plover]

Bro

[jagged imp]

the top one says to write the trig function you used to get sqrt(3)/2

[strange plover]

1st box is sin(3π/4)

[proper hare]

Mad confused

I got it

[jagged imp]

lmao

[strange plover]

Man first orn says

To express as sine

[jagged imp]

,w sin(3pi/4)

[ocean seal]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Sneaky#0928

[strange plover]

Guys

Pls see the first ques

Says to write ans as

Sine of a expression

[jagged imp]

but thats the wrong expression to write the sine of

is my point

[strange plover]

2π/3

I am an idiot typo psycho sry

And then u have to write fac

Fraction

In second one

[proper hare]

Wait really

[strange plover]

Yeah it says that express it as sien and the expression should be fraction or integer

[alpine sable]

Can someone assist me in 4 if anyone is free?

[proper hare]

Is this right?

Jus checkin

I think box 1 isn’t but everything else is

[woeful pulsar]

erm... it looks off

do you know the pythagorean identities?

[strange plover]

Tan = sin/cos

Sec = 1/cos

[proper hare]

So that’s the top?

Like this look

Sin t for first box

[woeful pulsar]

not really, have you seen $\sin^2(\theta)+\cos^2(\theta)=1$?

[ocean seal]

Element118

[proper hare]

That’s what I have to put in the first box

So this is what I got

[woeful pulsar]

erm... no

[proper hare]

?

What is it supposed to be

[woeful pulsar]

don't just blindly plug stuff in

[proper hare]

I’m not

[woeful pulsar]

you can get two more equivalent identities by dividing $\sin^2(\theta)+\cos^2(\theta)=1$ by $\sin^2(\theta)$ and $\cos^2(\theta)$, try it out

[proper hare]

Like put that as box 1 or actually do it

[ocean seal]

Element118

[woeful pulsar]

of course you need to try it out

not just plug into box 1 blindly

[proper hare]

I get 0

This is really off

What is it supposed to come out as

I am definetly missing something unless I get what you get

[bleak ibex]

Okay, I got the answer, but idk how to explain it

[woeful pulsar]

it's supposed to be equal to sec t - cos t

Let's try this exercise: Divide both sides of $\sin^2(\theta)+\cos^2(\theta)=1$ by $\sin^2(\theta)$

[ocean seal]

Element118

[woeful pulsar]

You'll get an equivalent form of the Pythagorean identity

also try dividing by cos^2 theta

[proper hare]

So on the top I was just supposed to do COs t because

[woeful pulsar]

erm... I don't get what you mean

Okay, basically you are supposed to find an expression equivalent to tan^2 t

to keep equality

[bleak ibex]

@woeful pulsar can I dm u to show what I did?

[woeful pulsar]

hmm, I guess

have you tried it out yet? @proper hare

the exercise dividing by both sin^2 and cos^2

[carmine lion]

I don't get what this question is asking of me

What's with the let theta

[glass lichen]

,rotate

[carmine lion]

ignore the writing on the right

[glass lichen]

prove that if $f(x)=\sin^{-1}(x)$, then we have the fact that $f(-x)=-f(x)$

[ocean seal]

moshill1

[carmine lion]

aren't we supposed to use substitution

and not function theory

or are we allowed to use that

[woeful pulsar]

function theory?

[carmine lion]

like

for a function to be odd

[glass lichen]

I just wrote f(x) to save me writing \sin^{-1}()

[carmine lion]

-f(x)=f(-x)

[glass lichen]

. . .

[woeful pulsar]

not always

only for odd functions

[carmine lion]

yes

i know

this property only works for odd

[glass lichen]

the question is to verify that something is odd

[carmine lion]

yes, but part (a)

doesn't ask u to verify

the whole of question 7

[woeful pulsar]

so all we know is sin is odd

[glass lichen]

part a is: Use the fact that sine is odd to show that $\sin^{-1}(-x)=-\sin^{-1}(x)$

[ocean seal]

moshill1

[carmine lion]

yes

oh

but can't you do it using substitution

[alpine sable]

Can anyone help with an accounting question?

[carmine lion]

channel occupied

[glass lichen]

you can start by showing that $\sin(-\theta)=x$

[ocean seal]

moshill1

[glass lichen]

which just stems from the definition of theta in the question and the oddness of sine

[carmine lion]

im left with this

[alpine sable]

any help?

[carmine lion]

$\sin^{-1}(x)=-\theta$

[carmine lion]

channel occupied

[ocean seal]

abe

[glass lichen]

$\theta = \sin^{-1}(-x)\implies \sin(\theta)=-x\implies x=-\sin(\theta)\ x=\sin(-\theta)\implies \sin^{-1}(x)=-\theta$

[ocean seal]

moshill1

[glass lichen]

in use

[carmine lion]

ok so the first step

let $\theta=\sin^{-1}(-x)$

[ocean seal]

abe

[carmine lion]

right

[glass lichen]

yes, then isolate for x

[carmine lion]

$\implies \sin(\theta)=-x \ $
Now, $\sin(-\theta)=-\sin(\theta)\$
$\implies \sin(-\theta)=x \ $
So, $-\sin(\theta)=x \ $
$\therefore \theta = -\sin^{-1}(x)$

[strange plover]

Yea

@carmine lion r u gud in calc I got a question if u don't mind

Related to stokes theorem

[glass lichen]

dont ping random users. Just ask the question in an unused channel and be patient

[strange plover]

Oh ok

[glass lichen]

IN AN UNUSED CHANNEL

[strange plover]

Like $\int_M d\omega$

[glass lichen]

In case you only read half of what I wrote

yes

[strange plover]

Oh sry ❤️

[ocean seal]

abe

[carmine lion]

is that working good

[glass lichen]

No

[carmine lion]

huh

[glass lichen]

cause you did -sin(t)=sin(-t)=-sin(t)

[carmine lion]

where

[glass lichen]

which doesnt progress you anywhere

and the therefore statement came out of nowhere

the Now, line

[carmine lion]

whats wrong with that

[glass lichen]

the line after that doesnt progress you towards the answer

cause it undoes what you did

$\theta=\sin^{-1}(-x) \implies x=-\sin(\theta) \implies x=\sin(-\theta)\implies -\theta =\sin^{-1}(x)$

[ocean seal]

moshill1

[glass lichen]

condensed it, so obviously in the final answer it would be flushed out

[carmine lion]

wait but from ur working

the final answer will be

theta = -sin^{-1}(x)

[glass lichen]

yeah

[carmine lion]

but i don't want that

[glass lichen]

yes you do

[carmine lion]

i want -sin^{-1}(-x)

[glass lichen]

what????

[carmine lion]

oh wait nvm

[glass lichen]

No you dont

You'll have that $\theta = \sin^{-1}(-x)$ and that $\theta =-\sin^{-1}(x)$

[ocean seal]

moshill1

[glass lichen]

so $\sin^{-1}(-x)=-\sin^{-1}(x)$

[ocean seal]

moshill1

[glass lichen]

QED

[carmine lion]

$\qed$

[ocean seal]

abe

[carmine lion]

ty

this channel is now open

and to whoever asked me

i have not even learned calculus yet

so don't bother pinging me

ur wasting both of our time

[woeful pulsar]

what have you tried?

[safe sable]

I don’t know where to start.

[woeful pulsar]

find a less active channel which doesn't have a message just one minute ago

[woeful pulsar]

do you know what slope is?

[safe sable]

Yes 1/6

[woeful pulsar]

and what a slope of 1/6 means?

the concept of slope

not the value provided

[safe sable]

Yes that’s the slope of the line

[woeful pulsar]

what does it mean to have a slope of 1/6

it's about how y changes when x changes

[ocean seal]

abe

[carmine lion]

do you notice anything here @safe sable

[safe sable]

Would I do 1-y=1/6(5-x)?

[carmine lion]

$\frac{1}{6}=\frac{y-1}{17-5}$

[ocean seal]

abe

[carmine lion]

right?

[safe sable]

Not sure.

[carmine lion]

you're given the value of the gradient

and if you know the gradient formula

then you can solve for an unknown coordinate

right?

[safe sable]

So then I would take 17-5 and multiply it by 1/6?

[carmine lion]

$m=\frac{y_2-y_1}{x_2-x_1} \text{ or } \frac{\Delta y}{\Delta x}$

[ocean seal]

abe

[carmine lion]

in your question

they give you m, y1, x2, x1

[safe sable]

correct

[carmine lion]

you can easily find y2 as well now right?

[safe sable]

Let me think it through one second

I’m not sure

Wait

So 12=6(y-1)?

[carmine lion]

y should be 3

[twin raft]

hey can someone help me decipher a problem, i dont even understand where to start

[safe sable]

Then i divide by 6 on each side, to get 2=y-1

[twin raft]

yes

[safe sable]

Do I move the y to the other side and change the sign?

[twin raft]

no just move the -1

u wanna get positive signs on both sides to make it easier

[safe sable]

Ok, so 3=y then correct?

[twin raft]

yep

[safe sable]

Thank you.

[twin raft]

well done c:

yo @carmine lion u got some time?

<@&286206848099549185>

"for a quadratic function describes y=f(x). y'=0 when x=3. Find the highest and lowest values between 1<= x <=6"

so i understand that when x=3 there is either a minimum or maximum point

idk what to do here

[harsh swallow]

what do you know about the maximum/minimum of a quadratic

lets take x^2

[twin raft]

that the dy/dx is 0

hmm

[harsh swallow]

what x

is its maximum

[twin raft]

3?

oh nvm its 6

in this interval

english is hard

uh wait

is it just 6 is maximum, 1 is minimum and 3 is middle point

"extremepoint"

[harsh swallow]

for this quadratic function

the maximum of the entire function is at x =3

or minimum

the extrema

this means that the function is moved by 3

because the normal quadratic is x = 0

is the extrema

does that make sense?

[twin raft]

erm

it probably does but not for me

[harsh swallow]

that's fine

i probably explained it badly

when you have a quadratic

[twin raft]

yep

[harsh swallow]

think of the shape

it has 1 point where the derivative = 0

you agree

?

[twin raft]

its either a U or a n

yep

[harsh swallow]

alright

[twin raft]

shapewise

[harsh swallow]

we know that that point

is at x = 3

[twin raft]

yeps

[harsh swallow]

this is in the interval we want

[1, 6]

[twin raft]

alright

[harsh swallow]

so we know for a fact that that is either a minimum or maximum

[twin raft]

yep

[harsh swallow]

because it is the global minimum or maximum

in the entire complete function

so if it is in our interval

it is also the minimum or maximum

you agree?

[twin raft]

yep

[harsh swallow]

alright

so at x = 3 we have one

what else do we know about quadratics

we know they're symmetrical

[twin raft]

yep

[harsh swallow]

this means that if the dy/dx = 0 point of the quadratic is at x = 3

[nocturne gazelle]

is this a channel where i can ask about runtime and correctness of an algorithm?

[harsh swallow]

we're busy

one sec

the symmetry is around that point

[twin raft]

ok so its highest or lower point is at dy/dx = 0

[harsh swallow]

so we know it goes out from that point

yes

the other point that we want

lets say for the sake of understanding that our dy/dx = 0 point is the maximum

alright

for clarity

it can be either maximum or minimum but for clarity it is maximum

we know that the function goes down on either side of that point

you agreE?

[twin raft]

yea but

my book says minimum and maximum

not either

which is incredibly confusing

im guessing both minimum and maximum is the same then

[harsh swallow]

because we know it is symmetrical and goes down in both direction from x = 3

in our interval

[1, 6]

we need to find the x value that is furthest away from x = 3

because that would be the most down

does that make sense?

[twin raft]

yea

[harsh swallow]

what do we do to find the furthest away from x = 3?

[twin raft]

so if the maximum is x=3

then both x1= 1 and x2 = 6 is minimum

?

[harsh swallow]

no

because

|3 - 1| is not |3 - 6|

[twin raft]

ah

[harsh swallow]

holdup lemme show a graph to explain

[twin raft]

so 6?

[harsh swallow]

exactly

[twin raft]

since its 3 away while 1 is 2 away

[harsh swallow]

yes

so in your interval

[1, 6]

the maximum and minimum are x = 3 and x = 6

[twin raft]

gotcha

well holy shit my book did not prepare me for such a problem

thanks

[alpine sable]

Where does the sum come from ?

[harsh swallow]

@twin raft

this is one of the possible graphs that fits into the problem

[twin raft]

i see

[harsh swallow]

we have the interval [1, 6]

with the dy/dx = 0 point at x = 3

see how the x = 6 is the minimum in the interval

[vague coral]

telescopic sum i guess

[woeful pulsar]

the same induction that is used to show f(n)=n^2 can be used to show that sum

[twin raft]

ye i understand now

tyvm

im at the part of the mathbook where each problem takes like 10-15mins cause i have to draw graphs and calculate functions by hand its so annoying

[alpine sable]

gotcha

that helps

[harsh swallow]

geogebra.org/classic type in :
a = 1
b = 1
if(1<=x<=6, a(x-3)^2 + b)

that is the generalized version

play with a and b

it shows you that it doesn't matter what a and b are

x = 3 and x = 6 will always be the answers

[twin raft]

i c

[harsh swallow]

anyway

[twin raft]

tyty

[hidden cliff]

help

[alpine sable]

@woeful pulsar I think istill need to be clarified on how they get to n(k^2 - 1 + n)

[glass lichen]

solve $\dv{y}{x}=\frac{-2}{25}$

[ocean seal]

moshill1

[hidden cliff]

how

[glass lichen]

algebra

[woeful pulsar]

yeah so you have f(1)=k^2
can you calculate f(2)?

f(2)=f(1)+f(1)+2

and you can continue f(3), f(4) and so on

[alpine sable]

what does this

yeah , ohh

ok

even mean

i get it now

thanks element

[brazen pivot]

it means you need to find out the asked angles for given polygons

[alpine sable]

and how can you do that

i missed 1 day of school and they’re on a different unit

[radiant anvil]

hi, when exactly is sum of averages different than average of sums? when variables are not independent?

I forgot

[woeful pulsar]

context?

[radiant anvil]

sorry, just analyzing elapsed time in a workflow. An analyst calculated the total average time of workflow cases as the sum of the averages (of all cases) of times on each stage of the workflow. I was challenging whether we should also calculate the sum of times of all steps and then average them. I think I recall it's not always the same.

[carmine kayak]

im a dumb bitch

[woeful pulsar]

yeah, that makes a bit of sense

what if one step is performed 100 times and another step is performed 10 times>

[woeful pulsar]

what do you know about differentiability and continuity?