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[carmine reef]

yeah so if you make the "domain" negative

change the 10 to a 0

then it goes counterclockwise

[carmine reef]

otherwise it doesn't

or clockwise i mean

counterclockwise is default

[quiet vector]

well i got both the + and - domains

even for those domains it is still ONLY rotated..?

right?

<@&286206848099549185>

[torpid mirage]

can you make it start from 0 to 10

it will show clearly how it flips orientation

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[mellow kelp]

U= 2x + y^3 -3x^2y find anaylatic function by Milne thomsan , method please tell me that answer is( 2z + i z^3 + c) or( 2z - i z^3+ C )

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[thick lynx]

Does anyone know of a short proof for the third law of Kepler that proves it for all elliptic orbits, not just circular ones?

[thick lynx]

Almost all of the sources I saw just proved it for circular ones, using 2 pi r as the circumference

[limpid turret]

Do you know vector calculus?

[thick lynx]

I'm giving a lecture/presentation to an 11th grade about this, so I would have to introduce them to it if we decide to use that proof

[limpid turret]

Most 11th graders don't know vector calculus

And the proof is not very short

[thick lynx]

I could go with proving it for circular orbits and saying that it holds for elliptical ones too, but that would annoy me a little

Is that what you would do though?

[limpid turret]

You gotta know your audience

[thick lynx]

They've had vectors, we've had calculus, but not vector calculus

[limpid turret]

Do they know vector calculus? Do they know regular calculus? Do they know polar integrals?

[thick lynx]

They don't know integration though

[limpid turret]

Do they know cross products and conservation of momentum?

[thick lynx]

They know differentiation and vectors up to scalar products

[thick lynx]

Nope to both

[limpid turret]

It will be out of their scope for now

So close yet so far

Real shame

[thick lynx]

Well, I could introduce them to it if it doesn't take very long

[limpid turret]

I'll find you the proof. Your call if you want to use it

[thick lynx]

I have 40-60 minutes to present the three laws of Kepler, so I can probably afford 10-20 minutes with introducing them to vector products, cross products or conservation of momentum.

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[limpid turret]

@thick lynx

[thick lynx]

Thanks a lot! Which book is that?

[thick lynx]

Alright

Thanks

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[peak jasper]

I really don't know where to start with these

[peak jasper]

I can see A has a horizontal asymptote, and B has vertical asymptotes, but I don't know how to answer the question

Should I try to recreate them in desmos?

[plain flame]

probably a good idea yea

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[peak jasper]

if anyone could help I'd appreciate it

[slender quiver]

ok looks like the 2nd and the 3rd is periodic so that means

[peak jasper]

I got this for A

[slender quiver]

ok

[peak jasper]

the unit is combinations of functions, if that helps

[slender quiver]

going to guess trig

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[lone heart]

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[junior warren]

i am having trouble proving this

i am working with expanding the left side

however you get

i need to somehow get that -ve sign out

but when squaring the abs value it would end up as a positive correct?

because the product is -2zw but it is in abs value

[alpine sable]

this is not true, even in the reals

0 = |1 + (-1)|^2 ≠ |1|^2 + 2|-1| + |-1|^2

[junior warren]

ah i see

so how would i want to start this?

[mortal trellis]

|x|^2 = x*conj(x)

[alpine sable]

I would decompose z = Re(z) + i*Im(z) and bash it out that way, but maybe someone else has a nicer idea

[marsh rapids]

|a+b|^2 = (a+b)(a+b)* = aa* + bb* + (ab* + ba*) = |a|^2 + |b|^2 + (ab* + (ab*)*)
You can see the difference here

[junior warren]

is that a property

that |x|^2 is x*conjx

[marsh rapids]

I hate having the add a backlash everytime I want to write a * damnit

[marsh rapids]

It's good to verify that by hand yes

[limpid turret]

This is not true

[alpine sable]

this is the parallelogram law, complex numbers version lol

[peak bough]

LaTeX is friend not foe

[marsh rapids]

Very common and useful property

[junior warren]

so i should start here?

[marsh rapids]

Please don't go and confuse them with hermitian products yet

[alpine sable]

I might confuse myself also in the process haha

[peak bough]

Does it work in general for any vectors of ℝⁿ with Euclidean norm?

[marsh rapids]

Yes

[peak bough]

[marsh rapids]

Just use bilinearity I Believe

[junior warren]

or should i do something like this

[marsh rapids]

Any euclidian space

[alpine sable]

that's already nasty

just brute force it by expanding

[junior warren]

so i have to use the conjugate right

[marsh rapids]

Which is precisely what bilinearity does

[junior warren]

to square

[alpine sable]

ah I thought you meant something else

[marsh rapids]

the zz* way is the fastest I believe. Now ofc if you expand it all with (a+bi)'s it should work but that sounds painful

[alpine sable]

like assume it's bilinear to conclude it's an inner product

[junior warren]

oh i see

[alpine sable]

inner products are so smooth

[junior warren]

so i would expand like this?

[alpine sable]

the second expansion is false

it's a minus sign after zz

[junior warren]

ok i fixed that

so now i can just add them

[alpine sable]

yep

no wait

there is another mistake

[junior warren]

yeah there is z* twice

if thats what you are referring to

i also fixed that

[alpine sable]

no |z-w|^2 is equal to (z-w)(z_bar - w_bar) not (z-w)(z_bar + w_bar)

[junior warren]

oh right

[alpine sable]

once you fix it just sum the two expansions

[junior warren]

im going to assume theres some property that says zz* is |z|

^2

[alpine sable]

zz* = |z|^2, not |z|

[alpine sable]

yep

[junior warren]

right

[alpine sable]

you're done

[junior warren]

is there an quick explanation for zz*=|z|^2

or is it just a property

[alpine sable]

if you write z=a+ib, zz* is equal to a^2+b^2, which is the distance |z|^2 (according to the pythagorean theorem)

[junior warren]

ah ok

makes sense

ty

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[analog nymph]

im not sure how to start (a) and (b)

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[analog nymph]

got it

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[neat sierra]

Im confused why they are equivalent?

Why the F(x) - x - 1 = x(F(x) - 1) + x^2F(x) is equivalent to the previous line

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[neat sierra]

<@&286206848099549185>

[alpine sable]

Hacker0

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[flat ridge]

They have the same terms except for x and 1 because the def of F(x) is that sum where n starts at n=0 and the input to the f(n) and x^n is just n but when the sum is written out the input to f(n) and x^n is n+2 while still starting at n=0 which means it skips the first two terms thus they are subtracted from F(x) to match.

for the x(F(x)-1) part because the input to f(n) is n-1 it removes the first term hence F(x)-1 and because there is a de-sync with the x^n function it is multiplied by x to shift the coefficents generated by f(n+1) back to the correct power

for x^2F(x) it is the same this time all coefficients are correct but it is off by a factor of x^2 because of x^(n+2)

@neat sierra

If you are confused why the sums are equal it is the same principal as:
if a = b+c
then a*d=b*d+c*d

where d is some power of x and you repeat the whole thing infinitely many times

[neat sierra]

Thank

I think I get it now

@flat ridge

[flat ridge]

np

[gaunt flower]

hi is anyone able to help with geometry

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[alpine sable]

What

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[ornate tundra]

I found the derivative in part a which turns out to be -1/sqrt(1-2x)

I know that you have to use tan(gradient)=angle for part b but what does it mean by the “positive direction of the x axis”?

[stark scroll]

I think it’s like that, starting from the positive x-axis going counter clockwise

[ornate tundra]

That makes sense

Let’s say the question is says from the negative direction of the x axis going clockwise, would it be this?

[stark scroll]

Yeah

But I think it is standard to start from positive x and go counter clockwise

[ornate tundra]

Yep

Thanks

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[mellow root]

what is the quotient group $(A_4 \times Z_2) / Z_2$ and what does it look like

[ocean seal]

ball

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[hallow dragon]

How to find VAT of 25% on 0,020 tons with a unit price of 1,560,00

[hallow dragon]

help

pls

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[vale wigeon]

does this mean:

I have 0.020 tons of really cheap drugs which cost $1560.00 per ton, and the VAT on them is 25%. How do I find the VAT?

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[dusky kraken]

I need some help with this

[dusky kraken]

I know that the origin is (2, -3)

Just not sure where to go from there

[vale wigeon]

no

the origin is (0,0)

(2,-3) is the center of your circle

[dusky kraken]

Yes that’s what i meant my fault

[vale wigeon]

consider that a tangent line at a point is perpendicular to the radius drawn to the same point

[dusky kraken]

Oh

Thats a lot more simple than I made it out to be

I need to find gdt of the radius to that point

And then find the inverse gdt for the perpendicular tangent

Thank you that was correct

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[bright remnant]

hwllo, how can I attempt to linearize this formula:

[bright remnant]

im trying to find the answer to this question

from what I know, capacitance is C=q/V

NVM

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[slow wraith]

can anyone help? im not sure on how to solve this

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<@&286206848099549185>

[copper rock]

We have the length of the minor axis (which is 2b). Use the formula c^2 = a^2 - b^2 to find the length of the major axis

https://courses.lumenlearning.com/waymakercollegealgebra/chapter/equations-of-ellipses/

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[alpine sable]

How's it equal that exactly?

.close

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[frank apex]

get values for m, g, theta?

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[long axle]

Is this what ur talking about

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[undone cobalt]

anyone know where the 5/2 comes from this problem?

[cyan seal]

The drawing is not well done

If the triangle is equilateral, it should be like this

[naive crystal]

Well median bisect each other. Since the length is 5, each part would be 5/2

[cyan seal]

It's not five, I think it's "S", from "side"

[naive crystal]

Oh ye, the "S" looks like 5

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[undone cobalt]

if its s then where does 3/4 comes from

[slender quiver]

i feel like something is really wrong

[undone cobalt]

That’s what I’m saying but that’s my professors work

So idk

[slender quiver]

oh thats a s

ok everything is good

[gray isle]

they're s

[slender quiver]

oh

[gray isle]

they subtracted s^2/4 from both sides
then combined like terms

[slender quiver]

no its good

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[sullen jungle]

The angle at corner C in triangle ABC is obtuse. Given that |BC| = a, |CA| = b (measured in units of length), and the area of the triangle is S (measured in square units), calculate and express the length of side AB in terms of a, b, and S.

[sullen jungle]

i got the answer AB = c = (sqrt(a^4 + S^2))/a

and it feels correct to me

however the right answer is supposed to be

[mellow grail]

Wdym by |BC|

[sullen jungle]

the distance from point B to point C

[mellow grail]

|| stands for abs Val no?

Ohh

Well u gon have to use herons ig

[slender quiver]

has mutli meaning

[sullen jungle]

like i know how to get the correct answer its just i dont know why mine was incorrect

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[sullen jungle]

<@&286206848099549185>

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@sullen jungle Has your question been resolved?

[sullen jungle]

ping me

[sullen jungle]

<@&286206848099549185>

[fast fiber]

.reopen

sorry

@crisp iron

[undone cobalt]

can you use L'hospital rule for -infinity/infinity?

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[sullen jungle]

this still

ping me

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[proven leaf]

This looks like a \textit{Heron's Formula} problem, which relates the side lengths and area of a triangle, Heron's Formula states:\

$A=\sqrt{s(s-a)(s-b)(s-c)}$ such $s=\frac{a+b+c}{2}$ or the \textbf{s}emiperimeter.\

For this this problem we have all the components to set up the problem and solve, we can easily relate one side to the two other given sides and the area! The only issue may be expanding that $s(s-a)(s-b)(s-c)$ once squaring both sides.\

This would be my way of doing it and it looks like it would end up with what the correct answer after solving a quartic, which albeit is never fun, but it would be my method of doing it; it's probably wrong given my track record but if you don't end up with the right answer don't be afraid to ping the helpers!\

[ocean seal]

XxMrFancyu2xX

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[granite ore]

how do i start this question

[gray isle]

can you find the expression for the gradient of the tangent at (x,y)

[granite ore]

uh

no i dont think so

[gray isle]

do you know the definition of normal here

[granite ore]

yes

yes

normal is line on the cruve

the tangent is the tangent of the normal

[gray isle]

no and no

[granite ore]

oh

[gray isle]

very poor descriptions of what they are

[granite ore]

haiz

[gray isle]

first look up the proper definitions of tangent and normal

[granite ore]

The normal line to a curve at a particular point is the line through that point and perpendicular to the tangent.

there0

haiz

[gray isle]

so the normal is perpendicular to the tangent,
do you know the relation between slopes of perpendicular lines?

[granite ore]

wha

uh

i know

m1 x m2 = -1

[gray isle]

can you apply that here

[granite ore]

ye

is it

x^2/x^3-3x^2+4

i think

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[granite ore]

help😭

.close

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[flat vale]

Why is e. T and f. F

[flat vale]

by any function im pretty sure they refer to any real-valued functions

wait i get part e but part f?

[lone heart]

@flat vale Has your question been resolved?

[marsh rapids]

The zero vector is unique

[flat vale]

yea but for real-valued function it will be at the origin right

[marsh rapids]

what does it mean for a function to be at the origin ?

[flat vale]

wait nvm got it

$\vec{0}$ is $y=0$ right

[ocean seal]

TheWhiteShadow

[flat vale]

@marsh rapids

[marsh rapids]

That's the zero function yes

[flat vale]

nice nice

.close

[lone heart]

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[marsh rapids]

And no other function is the zero vector

[flat vale]

yea cheers

[lone heart]

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[void agate]

Q.
An object X has mass m and velocity v.
A second object Y has mass 2m and velocity 2v.

What is the value of the ratio,

kinetic energy of y : kinetic energy of x?

(KE = 1/2mv^2)

The way I did this is by just inputting values to get the answer 8:1. I am wondering if there is an alternative way to do this using just algebra.

[ornate ginkgo]

$X = \frac{mv^2}{2} \
Y = \frac{(2m.(2v)^2}{2} \
\frac{Y}{X} = \frac{\frac{2m.(2v)^2}{2}}{\frac{mv^2}{2}}\
\implies \frac{Y}{X} = \frac{8mv^2}{mv^2} \
\frac{Y}{X} = \frac{8}{1}$

[ocean seal]

ColdTee

[ornate ginkgo]

Perhaps at your level this is the only way to do it

[lone heart]

@void agate Has your question been resolved?

[marsh rapids]

It's linear in m and quadratic in v, so doubling both, it's multiplied by 2*(2)^2 = 8
But that's nothing more than analysing and understanding the formula

[lone heart]

@void agate Has your question been resolved?

[void agate]

This does seem to make sense on paper

But when I work it out i dont know how to get beyond this step:

[alpine sable]

can i just post my question here, is this how it works

[void agate]

Go to the 'available' category

[void agate]

I don't know how to get past this T_T

Yeah we can cross out the 0.5's but then what

[marsh rapids]

He already showed you

[void agate]

OH

Im a complete dunce

8mv^2/mv^2

D'oh

.close

[lone heart]

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[lilac temple]

So I'm kinda stuck here

What do i do next

[lone heart]

@lilac temple Has your question been resolved?

[lilac temple]

Okay so i got some help but what im trying to know is why did we do dx/dt = 0 and x+y=2???

This is the point where i dont understand

[sand pawn]

duh

@lilac temple

have you actually looked properly in the workings?

[lilac temple]

Wdym

[sand pawn]

read this line

very carefully

[lilac temple]

Yeah i took out dx/dt out of the line

[sand pawn]

It's like the root of the equation

either x'=0 or x+y=2

[lilac temple]

Wait so if its part of the equation or the root why did we do dx\dt =0 and x+y-2=0?

Isnt dx/dt part of x+y-2?

Or its root

[sand pawn]

'x' is a function of t

and 'y' is a function of 't' too

well, not necessarily

the derivative here only acts on certain components as shown therein

@lilac temple

[lilac temple]

okay so lemme get this straight if i take out a root from the equation and the line is equal to zero i act as if the root is a=0 and the components are b=0?

[sand pawn]

nah man

you gotta read FTA

@lilac temple

[lilac temple]

what does fta stand for

[sand pawn]

duh

😑

[lilac temple]

sorry im an arabic student so all this stuff is like foreign to me

[sand pawn]

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

read it @lilac temple

[lilac temple]

alright

Ill try to look at a yt video abt it l8r

Thanks for helping tho

.close

[lone heart]

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[sand pawn]

close it

ok

[lone heart]

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[cinder fossil]

Gradient Descent is a strategy applied to solve the minimization problem on the function C.
Let's suppose that we're trying to make a move Δv in position so as to decrease C as much as possible. This is equivalent to minimizing ΔC ≈ ∇C⋅Δv.
The size of the move shall be ∥Δv∥. We're trying to find the movement direction which decreases C.

The problem it's asking me to prove that the nudge direction is gonna be Δv = − η∇C.
Eta is η = ∥Δv∥ / ∥∇C∥.

It also suggests to take into account the Cauchy-Schwarz Inequality.

[cinder fossil]

By doing a simple substitution I got this

Have I solved it?
Is this the answer?
I'm quite confused

[lone heart]

@cinder fossil Has your question been resolved?

[cinder fossil]

<@&286206848099549185>

[lone heart]

@cinder fossil Has your question been resolved?

[cinder fossil]

.close

[lone heart]

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[alpine sable]

Right or wrong

[ocean hawk]

wrong

[lone heart]

@alpine sable Has your question been resolved?

[torpid mirage]

wrong

[alpine sable]

Ok look

So this wrong?

[torpid mirage]

okay so i don't know how you got there

but the rules

are like so

[alpine sable]

Yea

Im listening

[torpid mirage]

$\log(xy) = \log(x) + \log(y)$

[ocean seal]

Mohamed Mohsen

[torpid mirage]

$y\log(x) = \log(x^y)$

[ocean seal]

Mohamed Mohsen

[torpid mirage]

these are the two basic rules for logs

[alpine sable]

I do know the rules but where am ai wring

[torpid mirage]

well then you should use the rules

first things first

what is the question asking

[alpine sable]

I did

To

Convert it into a single log

[torpid mirage]

great

[alpine sable]

Great what

[torpid mirage]

now walk me through it

[alpine sable]

Ok

[torpid mirage]

what

[alpine sable]

Just look at my steps

Where am I wrong at

How am I not following the rules

Thats what am I asking for

[torpid mirage]

quite frankly i don't know how the second step comes from the first xD

the whole exercise is done in one step

[alpine sable]

Alright

[torpid mirage]

can you tell me how you got it ?

explain at least the first step

[alpine sable]

Log x -2 log x can be log x /( log x)^2 and +3 log (x+1) -log (x^2-1) can ve log (x+1)^3/ log (x^2-1) yeah?

[torpid mirage]

"Log x -2 log x can be log x /( log x)^2"

compare that to the rule

is it log(x)+log(y) = log(x)log(y) or is it log(x)+log(y) = log(xy)

do you see the mistake here

[alpine sable]

No

Wait

So youre saying

No..

I dont get it

Can I ping

[torpid mirage]

sure

[alpine sable]

@torpid mirage

[torpid mirage]

look

at the two pics

[alpine sable]

Yea

[torpid mirage]

$\log(x)\log(y) = \log(x) + \log(y)$

[ocean seal]

Mohamed Mohsen

[torpid mirage]

can you spot the difference

[alpine sable]

But isnt log x/log y = log x- log y

[torpid mirage]

it isn't

that's the point

[alpine sable]

But theres a rule

[torpid mirage]

this is wrong

this is correct

[torpid mirage]

is this image

the same as the rule on your board ?

[alpine sable]

Hol up my internet is slow the pics aint loading

[torpid mirage]

ok

[alpine sable]

The first one?

[torpid mirage]

are you sure ?

look at them closely

are they exactly the same

spot the difference

[alpine sable]

.close

[lone heart]

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[torpid mirage]

I hope you got your answer

[lone heart]

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[torpid mirage]

.close

[lone heart]

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[strong dirge]

what is a counter example to C

[modern sedge]

f mapping every element of S to a single element of T I guess

[outer lark]

straight line through {1,2,3,..} so 1 -> 1 , 2->1 , 3->1 etc

as opposed to a bijection which could be 1->1 , 2->2,3->3,....

[strong dirge]

thats no 1-1

or is it

[outer lark]

it is

[strong dirge]

but every element maps to the same element

[outer lark]

yes

sorry

thought you were responding to the other message

that is the counterexample

the second one is the bijection

[strong dirge]

orite

so what about (abc)

(a,b)(b,c)(c,a)

This is bijective

and the sets are the same size

[keen plinth]

[peak bough]

When it says it is bijective, it means that it needs to hold for any chosen mapping from S to T

[keen plinth]

a -> b, b -> c, c -> c is not bijective

[peak bough]

It says (c, a) not (c, c)

[keen plinth]

i am giving a counter example

[peak bough]

Oh right. Excuse my brain fart

[keen plinth]

their claim is the same as saying "If x is a real number, then x must be 0"

"what about x = 0, then x = 0"

[modern sedge]

bijective => same size, but not same size => bijective

[keen plinth]

doesnt show anything for all of the other x in the room

[strong dirge]

ok i understand now

[outer lark]

good luck on the exam soon jmp

[strong dirge]

yes

u too

Powerset of Q is countable?

[peak bough]

I don't think it is

Powerset of N is not countable, so my intuition says it would be the same for Q

[strong dirge]

how is C correct

[peak bough]

I think in general you can say that there exists $|Y|^{|X|}$ mappings $f : X \to Y$. I don't know if that also count for $Y$ not being finite, but it would make sense in regards to $| \mathbb{N} | = | \mathbb{Q} | = \aleph_{0}$ when you consider that the cardinality of powerset of the Naturals are $2^{\aleph_{0}}$.

[ocean seal]

Mikkel

[strong dirge]

ye

[keen plinth]

Q itself is countable, you can construct a bijection with N

[strong dirge]

So the closest one to 2^N is the power set of Q

[keen plinth]

you can also construct a bijection between N -> {0, 1} functions and the powerset of N

[strong dirge]

nice

(5 choose 2)(3 choose 2)(1 choose 1)=30

[lone heart]

@strong dirge Has your question been resolved?

[lone heart]

@strong dirge Has your question been resolved?

[lone heart]

@strong dirge Has your question been resolved?

[strong dirge]

.close

[lone heart]

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[proud elk]

Why not 0 --> π/2 and that area * 5

[oak yacht]

if you draw the part of the graph in 0 to $\frac{\pi}{2}$

[ocean seal]

Mudkip

[oak yacht]

you'll probably notice the problem with it

try it in a graphing calculator

[proud elk]

?

[oak yacht]

it's polar so graph it in polar

[proud elk]

uh

isnt that the thing up there

[proud elk]

^

[oak yacht]

yes, what part of the given graph would 0 to $\frac{pi}{2}$ give

[ocean seal]

Mudkip

[proud elk]

top right quadrant

[oak yacht]

mhm

and if we multiplied that by $5$

[ocean seal]

Mudkip

[proud elk]

5 loops

[oak yacht]

we would have more than just 1 of the loops

since the first quadrant includes a bit of the 2nd loop

however, there is a way to do the solution with the method you're thinking of

[proud elk]

hm now im confused

[oak yacht]

instead of splitting it in $4$, you split it into the $5$ for each loop

[proud elk]

cuz i see that

[ocean seal]

Mudkip

[oak yacht]

so you know the whole thing is $2\pi$ yes?

[ocean seal]

Mudkip

[oak yacht]

sorry I'm way to used to just typing everything in latex

[proud elk]

lol

[proud elk]

yep

[oak yacht]

so to split that into $5$ you would divide

[ocean seal]

Mudkip

[proud elk]

uh

by 5

[oak yacht]

yes because 5 loops

[proud elk]

then?

[oak yacht]

that would be the angle where the first loop ends

so you could do $5\cdot\int^\frac{2\pi}{5}_0\text{thefunction}\dx$

[ocean seal]

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

[oak yacht]

whoops

[proud elk]

what

[oak yacht]

I messed up the dx

[proud elk]

i see but what about this

the working there

i dont seewhy itsays the curve completes at π

[oak yacht]

because at pi/2 you have the first quadrant

[proud elk]

although there are 5 loops till π hmm

[proud elk]

uh

[oak yacht]

and the first quadrant has the first loop and then half the 2nd loop

so multiplying by 5 would essentially give you 7 1/2 loops of area

I also don't get the 0 to pi

hmm

pretty sure it should be 2pi

[proud elk]

here from 0 to 2π you get 5 complete loops

but 0 to π/2 gives you 2 loops + 1/2 as u said

yeah i guess it's an error

[oak yacht]

you should probably be graphing it in polar to see it

[proud elk]

i dont think i should spend too much time on this question anyway

first time im seeing a weird problem like this

thanks mr potato

[oak yacht]

🙂 np

[proud elk]

.close

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[brave bolt]

Hello, guys ....what is the distance formula for lt and lon? (latitude longitude)
I have this --->
=acos(sin(lat1)*sin(lat2)+cos(lat1)*cos(lat2)*cos(lon2-lon1))*6371 (6371 is Earth radius in km.)
But the result is not correct ...
I find these resources --->
https://community.powerbi.com/t5/Desktop/How-to-calculate-lat-long-distance/td-p/1488227
https://en.wikipedia.org/wiki/Haversine_formula

[tired trout]

Hello, I am currently in high school and I would love to practice math more frequently. There topics such as Probability, Trigonometric functions and etc I would love to get better at. Is there any interactive websites where you can learn math in a fun way? I mean for example there is doublingo for language learners, I thought there could be something like that for mathematics

[oak yacht]

aops alcumus is a good one

however it's spelled

competition math problems are your friend for probability

[strange hound]

@brave bolt You probably want Haversine formula. Here is the formula in code, you can reverse engineer it.
https://stackoverflow.com/questions/27928/calculate-distance-between-two-latitude-longitude-points-haversine-formula

[tired trout]

Is there an app for learning math, that you would recommend? (Highschool lvl)

[strange hound]

@tired trout Try a Coursera course, there are dozens.
https://www.coursera.org/learn/introductiontoprobability

There is also EdX. Here you get huge bragging rights to say "I learned Probability at Harvard!"
https://www.edx.org/course/introduction-to-probability?index=product&queryID=9f575a886e7cc5e205b80cb0a1d64bf7&position=4&v=1&linked_from=autocomplete&c=autocomplete

If you can afford it, yes, pay for the certificate. Sometimes you can transfer them to count towards a degree, and they look amazing on a resume.

[lone heart]

@brave bolt Has your question been resolved?

[tired trout]

Thank you so much!

[pulsar laurel]

I find learning from textbooks and youtube is the best way for me to learn maths. Idk how much it might help you but I've learnt a lot from tl maths on YouTube.

[lone heart]

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[ionic sedge]

can someone help me on this

[sour dove]

on what?

okie dokie. What have you tried so far? What is the setup?

[ionic sedge]

hold on let me send a picture

[sour dove]

kk sounds good

[ionic sedge]

Idk if I did it right

[sour dove]

looks good to me

nice work!

[brave bolt]

thanks... I am not tested it yet.. but I think it works

[ionic sedge]

Thanks for checking!

[sour dove]

yep np, but you did it yourself so nice work!

[ionic sedge]

so the constraint equation that x and y satisfy is x+y=100?

these are the other questions part of the main one. i bielieve i have every one of them answered

[sour dove]

yeah you're looking for two folds: that x + y = 100, and that xy is a maximum.

So you have x + y = 100, which equivocally you get y = 100 - x. Then we plug that into x*y to get x(100-x), which is a parabola that is concave down! The maximal point in a concave down parabola will be its vertex, which is also the critical point!

yep looks like you got all that covered 🙂

[ionic sedge]

thanks for checking

.close

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[tardy nest]

I'm trying to figure out trigonometrics and I think a worksheet I was given is incorrect and it's confusing me. In the second section of cos(30), why does the substitution turn into sin? Is the worksheet incorrect or am I missing something because I don't know why cos was turned into sin. Please help this is driving me into an electric bath.

[slender quiver]

looks incorrect

[tardy nest]

Thank you bless you

[tardy nest]

Also I have another question if you don't mind answering

[slender quiver]

sure

[tardy nest]

Tyty give me one minute

Oops give me one second

[last ether]

Well there's the 30-60-90 theorem

[tardy nest]

I understand that so I'm given this right triangle here

[last ether]

Yeah

[tardy nest]

And it's asking me to find the sin cos and tan for 60 degrees

[last ether]

Mmhm

[tardy nest]

Wouldn't all the answers be the same

[last ether]

No

sin(60) is the same as cos(30)

if you use your 30-60-90 theorem

Which side is opposite from the 60° angle?

[tardy nest]

Ohh so it would be the 30 degree one which I have no clue which side that sife belongs to

I don't know if it's the hypotenuse or the adjacent leg

[tardy nest]

Why is the sin of 60 the same as the cos of 30

[last ether]

because the side opposite from the 60° is the same side adjacent to the 30°

In fact if a + b = 90°, sin(a) = cos(b), and sin(b) = cos(a)

[tardy nest]

Hm okay that makes sense

Okay thank you

.close

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[jade prawn]

how do I integrate (arcsinx)^2

[jade prawn]

Chain rule?

[minor needle]

IBP

[rancid token]

with respect to dX?

you can use integration by parts

$\int U dv = UV - \int v du$

[ocean seal]

Black Cat

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@jade prawn Has your question been resolved?

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[ionic sedge]

can someone help me on this

[ionic sedge]

this is my work, im not sure if im right

[ocean hawk]

if the second deriv is positive, what is the concavity

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[ocean hawk]

it isn't asking you for the range of the frequency, so you have it right

it states it in the question...

[sour dove]

What is the range of the numbers of people per reservation

[lone heart]

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[peak mist]

hello

[peak mist]

which way to solve it is correct?

[high rapids]

Solve what

[peak mist]

this way

or this way

[sly mantle]

@sly juniper post in an unoccupied channel, see #❓how-to-get-help

[peak mist]

@high rapids this

.close

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[meager dagger]

Hi, I have to prove that z^7 - 6z^3 + 4 = 0 -> |z| <= 2

[naive valley]

what tools do you have available?

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[steady basin]

how did the numerator go from there to there

[naive valley]

looks like they just multiplied it out

[steady basin]

i did that and got this

[naive valley]

well for one thing, you incorporated the e^(i theta) into the first factor, they kept it separate

[winter bane]

PING ME PLS

[naive valley]

ok, the last two terms can be simplified

e.g. the third term:

2 (1/2)^n = (1/2)^(n-1)

[steady basin]

like this?

[naive valley]

looks fine so far, almost the same as the solution screenshot you showed before

now just do 2(1/2)^n = (1/2)^(n-1) in the third term and you're done

[lone heart]

@steady basin Has your question been resolved?

[steady basin]

in this part here, how do they get 2costheta - 1?

[naive valley]

looks like they took the 1st and 3rd terms from the previous line:

$$Re(2^{n+1}e^{i\theta} - 2^n) = 2^n Re(2e^{i\theta} - 1) = 2^n (2\cos(\theta) - 1)$$

[ocean seal]

Bungo

[steady basin]

yep ive got it

thanks bungo

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[ionic sedge]

can someone help me with this

[ionic sedge]

this is a multi part question

first question is "what is the diferential dA"

[naive valley]

thoughts on what the answer should be?

looks good

yep looks right

[lone heart]

@ionic sedge Has your question been resolved?

[ionic sedge]

@ocean hawk what do you think

[ocean hawk]

looks right

[lone heart]

@ionic sedge Has your question been resolved?

[lone heart]

@ionic sedge Has your question been resolved?

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[ionic sedge]

.reopen

[lone heart]

✅

[ionic sedge]

here is my work:

can a helper check if i did this one right

@ocean hawk

[ionic sedge]

<@&286206848099549185>

[ionic sedge]

.close

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[ionic sedge]

.reopen

[lone heart]

✅

[carmine reef]

probably

A = pi*r^2

dA/dr = 2pi*r = 2A/r

dA/a = 2 dr/r

never seen that before but interesting

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[hasty shale]

this is the power rule

why is the derivative of 10x^2 is 20 if we apply this rule?

because it would be 2[10x]^1 . 10

20 * 10 = 200?

[surreal crystal]

d/dx(10x^2)= 2*10 x^(2-1) = 20x

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@hasty shale Has your question been resolved?

[vale wigeon]

"why is the derivative of 10x^2 equal to just 20?"

is that the question you intend to ask?

@hasty shale

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[vale wigeon]

👻

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[zenith olive]

hi

[zenith olive]

7 please

.close

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[noble notch]

i don’t know where to start proving for n=k+1

for Q10

[lone heart]

@noble notch Has your question been resolved?

[frank apex]

get from prevcase (n + 1)(n + 2)...(2n-1)(2n) to nextcase (n + 2)...(2n + 1)(2n + 2)
what manipulation can you do to the first to get the second

that's on one side
if the same manipulation works on the other side then you're done

[noble notch]

i’m not sure

how to get

from prev to next case

wait

is it

divide the first case by (n+1)

and then sub into the next case

[frank apex]

well first case divided by (n + 1) does not get you to next case

but you're on the right track

[noble notch]

do you need to turn the previous case into the next one?

[frank apex]

yeah

by multiplying/dividing/adding/etc some things

[noble notch]

i’m not sure how to get it exactly

could multiply the first case by (2k+1)(2k+2)?

[frank apex]

and do this

exactly

if you put that all together it will simplify

[noble notch]

ok

ty

.close

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[jade prawn]

How do I differentiate (arcsinx)^2

[limpid turret]

!show

[lone heart]

Show your work, and if possible, explain where you are stuck.

[limpid turret]

!status

[lone heart]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[jade prawn]

1

[limpid turret]

Do you know how to differentiate arcsin?

[jade prawn]

Yea

[limpid turret]

Do you know how to differentiate x^2?

[jade prawn]

1/(1-x^2)^1/2

[jade prawn]

of course

[limpid turret]

Do you know chain rule?

[jade prawn]

Yeah

[limpid turret]

That's all the ingredients then

Just gotta put them together

[jade prawn]

2x/(1-x^2)^1/2

?

[limpid turret]

Not quite

We'll convert (arcsin x)^2 into the form f(g(x))

What is your f(x) and what is your g(x)

[jade prawn]

f(x) = arcsinx and g(x) = x^2

?

[limpid turret]

You sure?

[jade prawn]

No

arcsinx and arcsinx?

[limpid turret]

Which function is inside the other here?

[jade prawn]

f(x) = x^2 and g(x) = arcsinx

[limpid turret]

Yes

So in terms of f and g, what is chain rule

[jade prawn]

idk man

[limpid turret]

You don't know derivative of f(g(x))?

[jade prawn]

f'(x) * g'(x) 😭

[limpid turret]

Not quite

You should review chain rule if you're unsure

Very important

[jade prawn]

f'(g(x)) * g'(x)

[jade prawn]

I remembered

2arcsinx/(1-x^2)^1/2?

Is this it?

[limpid turret]

Looks perfect.

[jade prawn]

Ok question isn't finished yet might be back

Ok now I need to find the second derivative

Quotient rule?

[limpid turret]

Yes

[jade prawn]

I hope there is nothing finnicky about this one

[limpid turret]

Just standard stuff. A little tedious but nothing you can't do

[jade prawn]

I did it but it looks a little messy

(2+2arcsinx(x(1-x^2)^-1/2))/1-x^2

[lone heart]

@jade prawn Has your question been resolved?

[limpid turret]

I'll trust you

I can't mental math it

[jade prawn]

It was right, thanks so much

I've now found a question that makes that question look like a joke

[lone heart]

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[void agate]

How do I go about solving this in the fastest way?

I first thought 'trial and error' just keep guessing numbers and see if they meet the inequalities and lie on y = 0.5x, but that seems rather inefficient.

Is that the only way?

Found the answer (8,4) through trial and error by plugging values into y=0.5x and checking if they match the inequalities.

Still wondering if theres any faster ways.

[frank apex]

can you graph those lines on graph paper

[void agate]

Nope

[median oar]

8,3?

Is that the answer

[void agate]

(8,4)

[median oar]

Wait 8,4 also satisfies?

[void agate]

(8,3) does not since it must lie on y = 0.5x

[median oar]

Oh

It doesn’t lie on the line

Right

Ok there’s a smart way to do it

[median oar]

I forgot this condition

[void agate]

All good bro

[void agate]

I'd love to hear it I'm stumped lmao

[median oar]

So look at the intersection of the horizontal and vertical line

That’s (9,2)

[void agate]

Yep

Point C (9,2) got it

[median oar]

You can by inspection see that the closest integer point to 9,2 that could possibly lie within the triangle is 8,3

8,2 is not greater than 2

9,3 is not less than 9

[void agate]

Ahh

[median oar]

8,3 is the closest point to 9,2 in the direction of the triangle

[void agate]

Thats a smart way to think

[median oar]

That’s why I said 8,3

[void agate]

And then you just account for 0.5x

So go up 8,4

[median oar]

If 8,3 isn’t in the triangle then no point are

Now

8,3 is not on the line so ignore it

Next closest integer pairs

7,3 and 8,4

7,3 doesn’t satisfy the line

8,4 does

8,4 is your first solution

[void agate]

Yeah I got it thanks for the help @median oar

.close

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[median oar]

It seems to suggest there’s only 1 solution so you could end there or prove that the next set of points don’t lie on the line and/or are not in the triangle

[lone heart]

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[median oar]

Also since each set of points you consider are equidistant from 9,2, and each set is further than the last, if at some set it is not in the triangle, all subsequent set of points will also not be in the triangle

[median oar]

.close

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[lone heart]

@ashen kernel Has your question been resolved?

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@ashen kernel Has your question been resolved?

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@ashen kernel Has your question been resolved?

[lone heart]

@ashen kernel Has your question been resolved?

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[woeful wind]

what does it mean for a y or x value to be bounded?

the context is there is are two exponential solutions, one in x and y, which are simulatenous

does it just mean it does not tend to infinity

as t approaches infinity

[modern sedge]

yes, I think so. It means it doesnt tend to +-infinity.