help-0

[tender geyser]

tho calling infinities as extremal values is a bit of a complicated notion, id be careful with that (but it probably wont be relevant to you anyways)

correect, its the same idea ^^

i can flip the graph above on its head and you can see the same ideas for max

np 🙂

[plush epoch]

Hello

Context: I'm interested in generating curves that intuitively describe the surface of a 3d object. For example: Creating a spherical spiral based on a sphere. A bit like if a 3d printer would print a sphere. I'd like to do this not only for spheres, but all kinds of object. So I'm wondering.

My question: I this a specific case of problems or algorithms that are known in Math? If so, what are they called? I'm asking because my googling wasn't very successful so far because I don't really know what I should be googling for. What I've found are "slicers", which translate a mesh into motor instructions for a 3d printer, which must tackle the same problem, so I assume. But I would prefer looking in to the broader topic first (if it exists), before I start reading slicer code, hoping to stumble upon their solution of this.

while I'm sure I'd find spiraly functions for many 3d primitives, I'd also be interested in doing different things than spirals, and of course even use meshes instead of primitives if that would be possible as well :).

[alpine sable]

h

[warm phoenix]

Okay so sine of $240^{\circ}$ ($4{\pi}/3$) is $-\sqrt{3}/2$

cosecant is 1/sine right? Or well 1/sin($\theta$)

So it'd be $1/ (-\sqrt{3}/2$) right?

So how does it become positive $2/\sqrt{3}$? Why does the sign change?

Or am I wrong?

[ocean seal]

FloppyDingo

[velvet pelican]

,w calculate csc(4pi/3)

[ocean seal]

Results provided by WolframAlpha

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

Requested by Frosty_#7210

[velvet pelican]

I think you're wrong

[warm phoenix]

This online circle test is a big oof then: https://i.imgur.com/TXX0LWG.png

No wonder I was confused. Ty

[ocean seal]

Please give me something to evaluate.
See ,help calc for usage details.

Jun 5, 2021, 6:26 PM

[rough spear]

[empty moth]

,calc csc(4π/3)

[ocean seal]

The following error occured while calculating:
Error: Undefined symbol π

[vale wigeon]

,calc csc(4*pi/3)

[ocean seal]

Result:

-1.1547005383793

[empty moth]

yh

csc(4π/3) = 1/sin(4π/3)

sin(4π/3) = -√3/2

1/(-√3/2)

= - 2√3/3

[tranquil tulip]

How do i do this?

without a calculator

[gusty gorge]

write it as $\sqrt[4]{9^3}$ and simplify from there

[ocean seal]

Saccharine

[gusty gorge]

Note that $9^3 = 3^6$

[ocean seal]

Saccharine

[sterile relic]

say u have a^(b/c) the resulting radical will be $ \sqrt[c]{a^b}$

[alpine sable]

why is this the case?

wtf bro

Yeah but no

if f(x) = 3x +5 and g(x) = x then they aren't multiples no?

yes but in this case we have 3x and x, one grows 3* faster

ohhh

thx

[wet fulcrum]

is the channel free?

[somber shard]

Did I do this right? Because they’re the same thing but different letters

Oh okay

[ocean seal]

itsDaman ඞ

[glass lichen]

factor

why did ln(x) become ln(1)?

"take e" is vague

Like I said "take e" is vague

[alpine sable]

how do i prove this?

[oak chasm]

@alpine sable Use the associative and commutative properties.

[alpine sable]

i don't know how

it's all new to me

[oak chasm]

@alpine sable OK, what's the commutative property say?

[oak chasm]

OK, what does the associative property say?

[ocean seal]

Chai T. Rex

[alpine sable]

yes

[ocean seal]

Chai T. Rex

[alpine sable]

yes

but i don't actually know how to prove this

[oak chasm]

These are steps in proving it.

The commutative property and the associative property are generally things you don't have to prove.

These are just uses of them.

[alpine sable]

i don't understand what proving looks like

[oak chasm]

It's where you use the axioms you know to transform things.

The associative property and commutative property are generally used as axioms.

[alpine sable]

this is basic math though

[oak chasm]

Right, even more so then.

You use the commutative and associative properties to transform the starting expression into the ending expression, step by step.

We've done a few steps that way.

That's how you do the proof.

[alpine sable]

but i don't understand what a proof actually looks like

we can say these expressions are all equal to each other but how does that translate to an actual proof?

[oak chasm]

That doesn't translate to one. That is one.

[alpine sable]

?

[oak chasm]

You're asking how to translate from that to a proof.

You don't need to translate.

It's already a proof.

[alpine sable]

Hey guys can someone explain why the probability of only one cube being 6 is 10/36?

cant wrap my head around this, thanks

[oak chasm]

@alpine sable If you want to be more explicit, you can give what you used on each step, as in https://mathbitsnotebook.com/Algebra1/LinearEquations/LEjustify.html.

[summer parcel]

Can someone check if the solution is correct?

Anyone?

[merry coral]

back

wow looks like you solved it

credit to @noble sinew

odds of flipping a total, n, number of heads before flipping a total of 3 tails

[ashen wave]

Can anyone tell me what the difference between each are

[glass lichen]

one is standard form, one is factored form and one is vertex form

[ashen wave]

.

[manic halo]

its the same equation, just written differently

[knotty sleet]

They're used for different things

[ashen wave]

what are the different things-

[knotty sleet]

Standard is standard, easy for differentiation

Factored can be used to find roots

Vertex is used to find extrema

[ashen wave]

wdym by roots?

[knotty sleet]

The x-axis intercepts

[ashen wave]

oh. wait I remmember talking about y=-b/2a what was that for?

[knotty sleet]

That's going from standard to vertex form

Basically if you are given standard form (which is most often the case), the 'b' in the vertex form is the -b/2a in standard

[ashen wave]

nvm I wasnt thinking straight

[knotty sleet]

Yh it's the line of symmetry

x=-b/2a, that is

[ashen wave]

ohhh so that would also be the x value for the vertex

[knotty sleet]

Exactly

[ashen wave]

oh ok tysm!

[knotty sleet]

👍

[gloomy lintel]

The integral from 2 to 0 is the same as the one from 4 to 0 minus the one from 4 to 2 as we only want 0 to 2

[gloomy lintel]

No, how are you getting 31?

Ahhhhh nvm I’m dumb I thought the limits read 2 to 4

lollll

I should read the q next time

[elfin snow]

it is an octagon with vertices T, R, I, A, N, G, L, and E

oh

idk

[knotty sleet]

The area

Maybe

I've seen it used to mean area

Square brackets

Lol imagine naming an octagon TRIANGLE

[elfin snow]

lol

[somber spoke]

can someone help me wrap my head around this

im new to integration

if i did the integral of this graph between -1 and 7 then what bit would be shaded

would it be like this??

[vapid ermine]

@somber spoke

this area

you should use both graphic

[vapid ermine]

but if you say between y=x²-4x-4 , x=-1 x=7 , then it would be like this

[somber spoke]

Yes but I was just confused cus I didnt know what the area under the curve between -1 and 7 was

[prime ice]

Determine the coordinates of D
can someone help pls im new to this type of geometry

[vapid ermine]

you forgot using both of them so i think you know for now?

@prime ice first one 18-2x²=0

x²-9=0

[prime ice]

i finshed those sorry

[vapid ermine]

oh ok

[prime ice]

i need help for Determine the coordinates of D

[vapid ermine]

make equal them

[somber spoke]

I was trying to visualise them separately at first sorry I'm confused

[vapid ermine]

@prime ice did u found c=3?

the little one

[prime ice]

yes

that was question 2 i finshed that

[vapid ermine]

mx+3=0 mx=-3

x was 3

so m=-1

[prime ice]

yeah i got that

[vapid ermine]

ok then y=-x+3

and make equal them

18-2x²=-x+3

root are C and D

@prime ice 2x²-x-15=0

x=-2.5 x=3

-x+3= -(-2.5)+3=2.5+3=5.5

D is (-2.5; 5.5)

@prime ice did u found A or not?

if u didnt then

[prime ice]

yeah a is (0;18)

[vapid ermine]

ok so we found all of them?

[prime ice]

yeah

but can u explain this

[prime ice]

.

[prime ice]

from here

[vapid ermine]

i found x is -2.5 for D from 18-x²=-x+3

cause other root is for C

so i put x in a place in -x+3

we found m=-1

little c is 3 from y=mx+c

then mx+c=-x+3

[prime ice]

so c + x gives us the y coordinate

[vapid ermine]

c-x

cause m=-1

[prime ice]

so if i were to write it masthematically

how would i get the y coordinate

[vapid ermine]

just put root in X

our roots are -2.5 and 3

3 is for C

so only -2.5 can be for D

[prime ice]

ok

So like this

[vapid ermine]

yeah

[prime ice]

after they ask for this

sorry for troubling i just got exams comming up and im tryna prepare

[vapid ermine]

B was -3 0 right?

[prime ice]

yes

heres the thing

[vapid ermine]

so if new graphix is y=kx+b then m=k for paralel to dc

and m was -1 so k=-1

-x+b=y -(-3)+b=0

b=-3

then y=-x-3

[prime ice]

sorry im confused

[vapid ermine]

oh ok again then

İf they want line through B

and line is y=kx+b when we put -3 in x then y should be 0

cause Line is on B also

also they want paralel to DC line

and we found DC line is -x+3

if New line and DC are paralel then in y=kx+b and in y=mx+c k should be equal to m

idk why but it's just rule

so new line is y=mx+b for now

and also it is on B and B is -3 0 so we can put -3 in X

also we can put 0 in y cause line is on B

so y= x=-3 then for y=-x+b 0=-(-3)+b 0=3+b and b=-3

[vapid ermine]

k is -1 from his cause m also is -1

[vapid ermine]

so line is y=-x-3

[prime ice]

what does k have to do with any of this though

[vapid ermine]

cause they say new line should be paralel to DC

[prime ice]

im atry and do it myself and get back to u

give me a sec

[vapid ermine]

in my country time is 03:59 i have to sleep

[prime ice]

oh thanks alot man

have a good night

[vapid ermine]

if i could help, then i am happy and good bye

[wary stream]

It's not a rule, it's a known fact. Parallel lines have the same slope, while perpendicular lines have negative reciprocal slope

[prime ice]

what would i call the straight line through b

[wary stream]

What do you mean?

[prime ice]

Would I solve that like this ?

[prime ice]

.

[wary stream]

Yeah, that's right

[prime ice]

thanks

can someone explain how id go about doing this?

Draw the graphs of f and g on the same system of
axes.

[wary stream]

Just plot the functions on the same coordinate plane

[prime ice]

im lost sorry

[wary stream]

Same coordinate plane, both functions should be in the same plane like this

Just plug in values into the functions

And plot the points

[prime ice]

how i do this

[wary stream]

Pick numbers as x values, to get a y value

[prime ice]

could u please shoe me and example

[wary stream]

If you don't understand how to draw a graph of the function by plugging in x values, and getting a y value, I suggest looking at resources on the internet

[prime ice]

ok ima do this tomorrow

let me do something i understand better for now

So I done this but is there a rule I can say to say AC=BD

[gray isle]

from the midpoint alone, no

[prime ice]

and from this

[prime ice]

and is this fine?

[gray isle]

from the midpoint alone, no

[prime ice]

ok

was my calculation fine??

[gray isle]

yes

[prime ice]

ok thanks

[glass lichen]

change of base formula

[gray isle]

probably

[merry coral]

its trying to calculate log(2)

then is confused af when you have a 6 next to it with no operator

[gray isle]

is there a comma button?

try alphalog(2,4)

[merry coral]

odds of flipping a total, n, number of heads before flipping a total of 3 tails

so apparently

this is equal to

odds of flipping at least n heads in n + 2 tries

how do you calculate this lol

[alpine sable]

anyone know fourier series

[flat vale]

When the question asks to answer to 2 decimal places do you use = or ≈

[merry coral]

depends on the teacher honestly

in my class
x = ~56.69
gets full points

[merry coral]

I assume ≈ is better in general tho

[flat vale]

alr thx

[blazing rapids]

i found the first three derivatives and their values

but im getting stuck trying to figure out how to convert this trend into a sum formula

bruh you couldnt have sent it in any of the other questions chats

i left the calc chat because you were getting help there

[alpine sable]

https://discord.gift/rztP3jyqJWsDvsd6
peepee poopoo

[steep briar]

Careful. Make sure you're doing a Taylor series and not a Maclaurin series since you're centered at x=1.

Do you know the difference?

[fading jay]

a series without minuses will obviously be more in modulus and it will converge to exponent-1, which means this will absolutely converge

[steep briar]

I ask because the equation you gave looks like a Maclaurin.

[fading jay]

and according to the Leibniz rule, the members of the series tend to zero and each next from the second is less than the previous one in absolute value

[steep briar]

Yeah, so what'd the skeletal Maclaurin series at x=1 look like?

(Idk why I added the adjective skeletal, don't think that's something special. The question just wants you to plug and chug the values from the table into the Taylor series at x=1, then you'll simplify by factoring the powers)

No that's a Maclaurin series, not a Taylor series

Yea

Did you try that originally?

One moment, I'll try to verify it's not a bugged question

[fading jay]

you may need to open the brackets

[steep briar]

It does

By any chance you know Binomial Theorem, it would hasten the work a lot.

[fading jay]

oops there near f '' '(1) should be (x-1) ^ 3, not (x-1) ^ 2

third derivative*

[steep briar]

(try to make due with this, it's especially useful in this question)

Could you show your attempt?

[steep briar]

And the third term should be 2! not 1! again, heh

[blazing rapids]

would anyone be able to help guide me in the right direction>

[rustic night]

Anyone know how to solve this?

[jade sail]

use proportions

[steep briar]

Try not to simplify terms, that question is mainly about finding the pattern of the summation.

[blazing rapids]

so leave it in the form all the way to the left

i know im gonna need a (-1)^n to show the alternating sign

[steep briar]

Yeah a bit, let me see if it's right. I'm used to it being as a fraction.

Also a 3ⁿ term, you see why?

[blazing rapids]

umm not really, i see im going from 3 to 6 to 18, which wouldnt be 3^n

wouldnt a 3^n trend be 3 9 27

[steep briar]

Yea, lemme see how you got the 18.

[blazing rapids]

looking at how i wrote it i screwed it up

because originally i wrote it as (1+x) not (1+3x)

lemme redo it from the start and see if i get the same thing

[steep briar]

gotcha

I'mma plug in what I got on Desmos to verify I'm not misleading you

[blazing rapids]

i got -3, 6, and -18 again

[steep briar]

You did? Strange.

[blazing rapids]

,rotate

[blazing rapids]

@steep briar am i doing something wrong

[steep briar]

I think your first derivative is wrong, it's missing the 3 coefficient from chain rule

[blazing rapids]

ah yeah so -3 (-2) * 3 for the second derivative

so 18

[steep briar]

Try to redo again keeping in mind what I said about chain rule since you seem to have the -2 a single derivative too early. Here's a really good piece that'll help you finish. I have to go so also see if someone else can help.

[light dirge]

Can any one give the answer

Plz

[blazing rapids]

wait this is going to sound really stupid, but shouldnt it be -1, 0, 1, 2 ,3

ive been doing -1, -2, -3, -4

[steep briar]

Where'd you get -1,0,1,2,3 from? 0.o

[blazing rapids]

doing the power rule

of 3(1+3x)^-1

the derivative of that should be -9(1+3x)^0

[steep briar]

oh no

[blazing rapids]

rip

trying to do series without knowing how to do basic derivatives

[steep briar]

It'd be -(3)²(1+3x)⁻² since you'd -1 from -1

[blazing rapids]

dang okay thanks

[blazing rapids]

ah this ones not bad

n=4 means you have 4 rectangles

can you please post this in questions #1 im tryna help this guy

@alpine sable so n=4 means you have 4 rectanges, since youre going from 0 to 2, you do 2/4 to get that each rectangle is 0.5 in width, or thats your delta x

so you have 4 rectangles 0.5 in width from 0-2, so now you figure out height

since they are right endpoint rectangles your rectangles height is determined by their right endpoints, so here if rectangle 1 is 0 to 0.5, your height is determined by the height of the curve at 0.5

like this

so to find the answer you do f(0.5) * delta x + f(1) * delta x + f(1) * deltax + f(2) * deltax

which is just [ f(0.5) + f(1) + f(1.5) + f(2)] * 1/2

[fast marsh]

I have a wierd magic the gathering related question that pertains to math

[blazing rapids]

yep, did you get the answer?

[neon dome]

Hi guys, hope you all are doing well.

I had a very basic question:

I wanted to determine ratio's between two numbers, that are high in value.

Now I know that the ratio for 10:2 would be 5:1, but can someone simplify this particular question for me:

Q) What is the ratio between 1.931 M: 945.79 K? (M = Million, K= Thousand)

Also do add the method for calculating the same.

[shell widget]

@neon dome You can write 1.931M as 1,931,000 and M as 945,790, now they are both in thousands

and just divide them

[hearty ravine]

guys whats 10 x 10

[warm smelt]

no clue

[wispy helm]

10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 ?

[hearty ravine]

thanks dude

[warm smelt]

reposting here

anyways i need help with 4.

[blazing rapids]

no

thats wrong @alpine sable

f(x) = sin(x)

so its sin(0.5) + sin(1) + sin(1.5) + sin(2)

*.5

type that into your calculator

yes

[thorn kindle]

@warm smelt let girl = 0 and boy = 1. 00, 01, 10, 11

[warm smelt]

i get it now thanks

[thorn kindle]

Good job!

[ionic jewel]

is this intentionally a troll problem

or did someone unironically give it to you

[fallen patio]

They did😀

[thorn kindle]

1 - 4/(x+1)

@fallen patio

[ionic jewel]

your problem is either a trick I dont know or a higher level of math

[fallen patio]

Wdym

[thorn kindle]

Ok. Do you know what a mixed number is?

Basic arithmetic but it's important.

1 2/5 is a mixed number

[shell widget]

$1 \frac{2}{5}$

[ocean seal]

Sup?

[thorn kindle]

Yes

[shell widget]

🙂

[thorn kindle]

1+2/5

An integer plus a fraction less than 1

[fallen patio]

Are you talking bout the qns i asked?

[thorn kindle]

Yes i am

[warm smelt]

nah not troll I just don't get how to do with with math

that's why I asked

[thorn kindle]

I am showing you a very important technique

[warm smelt]

like is there a formula to use?

[ionic jewel]

look up the "boy girl problem"

[warm smelt]

okay thanks

[ionic jewel]

theres two answers for the problem depending on your interpretation

[fallen patio]

Ohk...

[thorn kindle]

7/5 = (5+2)/5 = 1 + 2/5

[fallen patio]

But how do I solve this?

[thorn kindle]

(x-3)/(x+1) = (x-3+4-4)/(x+1) = (x+1-4)/(x+1) = 1-4/(x+1)

Now the variable only appears once. Start by taking a composition or 2

1-4/((1-4/(x+1))+1)

[fallen patio]

Um.....

[thorn kindle]

Im literally showing you how lmao

Is there something you dont get?

[fallen patio]

But I am not understanding

[fallen patio]

Can you explain this one?

[thorn kindle]

That is what i am explaining. Or do you not understand what the problem is asking at all?

[fallen patio]

I didn't understand at all

[thorn kindle]

Do you understand what a composition is?

[fallen patio]

No

[thorn kindle]

Why are you doing this problem then?

[fallen patio]

Because it was assigned to us as h.w, even tho we weren't taught about this.

[thorn kindle]

You're fucked then lmao

Wait for your teacher to explain

GL

[fallen patio]

Thanks anyway

[thorny charm]

If ∠𝑎 and ∠𝑏 are vertically opposite angles, where 𝑚∠𝑎 = (2𝑥−5)˚ and 𝑚∠𝑏 = (𝑥+6) ˚ , find 𝑚∠𝑎 and 𝑚∠𝑏.

[fallen patio]

Since both the angles are vertically opposite, they would have the same magnitude.
So,
(2x-5)=(x+6)
Solve this and you will get the values of angle a and b

[ionic jewel]

comedy gold

[final crag]

Could someone help me with this problem?

I believe its pigeonhole principle

[vale wigeon]

is that even true

cause i don't buy it

if the sum of all ten people's ages is odd there's no way to split them into two groups with the same total age, and nothing about the problem implies this can't happen

[thorn kindle]

You dont have to use all ten people though

[vale wigeon]

oh wait

misread

[thorn kindle]

Min of the sum is 2, max is 540

Idk if that helps

Actually i think you can reduce that max to 300

[ionic jewel]

intuitively this doesn't seem like it should work but the Pidgeonhole principle had a way of making absurd looking stats actually right

i don't see a way to apply it though

[vale wigeon]

let's see

uh

okay no my idea doesnt quite work

back to the mental drawing board

[thorn kindle]

The sum of n numbers leq 60 is equal to the sum of 10-n-k numbers leq 60 for n geq 1 and k geq 0

Yeah thats a bit tricky

[final crag]

uh i have another question

i could just ask my TA about that one

[vale wigeon]

okay show us the other question

this one feels kinda weird

[final crag]

that one might be worded wrongly or something

[vale wigeon]

no the wording is fine

[final crag]

this was the other question i was confused about

[vale wigeon]

a bit eccentric but fine

this one is way more straightforward

[ionic jewel]

okay i feel like the first problem is like infinitely harder

[vale wigeon]

two integers have an odd sum iff they have opposite parities

[final crag]

what does it mean by 1011 distinct integers?

[ionic jewel]

this one has the n/2+1 giveaway but i can't see anything like that in the other one

[vale wigeon]

...a thousand and eleven integers, all distinct from one another

[final crag]

oh i thought it was like some binary coding or something lol

[ionic jewel]

[final crag]

uh awkward

[vale wigeon]

so in order for no two of your integers to have an odd sum, your integers must all have the same parity

[final crag]

even then i prob still couldnt solve it if i wanted to ngl

[vale wigeon]

but there are only 1,010 odd integers and 1,010 even integers in your range

and you have 1,011 integers in your set

so there is bound to be at least one even and one odd

[final crag]

so theres 2 sets?

like one set is all odds and one set is all evens 1-1010?

[vale wigeon]

1-2020.

if you wish.

[thorn kindle]

Hmm, is there anything special about 10? Consider the list 60,1,30,2 of 4 people

[ocean seal]

itsDaman ඞ

[vale wigeon]

doesnt matter, you'll end up having to IBP twice anyway

[ionic jewel]

^

[final crag]

okay thanks

[ionic jewel]

i still want to solve the age problem

[ionic jewel]

where does this get us?

[final crag]

since i'm here, could i ask one last one?

[vale wigeon]

yes sure

let's all ignore itsdaman's kinda rude interruption

and let's all ignore the cesspool nature of questions-0

[thorn kindle]

There has to be a minimum number of people for which the statement is true.

Clearly not 4

My intuition tells me 6 but that could be wrong

[final crag]

okay last one haha

[ionic jewel]

you could add "4" to your set and it would still be true

[final crag]

its okay for the first question, i'll ask my TA on monday

[ionic jewel]

wait is this a log_2(n) solution

1,2,4,8,16...

[thorn kindle]

4 people do not always satisfy the claim is the point I'm making

[ionic jewel]

wait im an idiot

[ionic jewel]

yes i was giving a 5set that doesn't satisfy the claim too

[thorn kindle]

Mm good point

[ionic jewel]

wait maybe I'm not an idiot, does that 1,2,4 work?

i don't actually see a way to sum them to the right amount

,w log base 2 of 60

[ocean seal]

Results provided by WolframAlpha

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Requested by 兔子#6562

[ionic jewel]

you could do 5 people if that's true

[sterile beacon]

For which x applies the difference |x-7| < 2?

Guys?

[thorn kindle]

Applies the difference

It is JEE hours

Time for me to log off

[sterile beacon]

At least i am not helper who is warned

[final crag]

huh

so wait 1,2,4 works here?

how so?

[sterile beacon]

'

[ionic jewel]

it doesn't matter I was thinking about it wrong

[tawny cedar]

this question has had me screatching my brain for a while

[ionic jewel]

I'm not sure how to form it into 9 or fewer groups that would enforce the group sum property

[sterile beacon]

@ionic jewel what makes you trustworthy

[ionic jewel]

take the derivative of both and set them equal to each other i guess

[sterile beacon]

Wow so smart Damien

[ionic jewel]

,w derivative of tanx

[ocean seal]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by 兔子#6562

[ionic jewel]

oh yeah i knew that

2 cosx = 1/cos^2x

cos^3x=1/2
cos(x) = 1/cbrt(2)
x =0.653

[sterile beacon]

Well i guess is should cry now

[ocean seal]

itsDaman ඞ

itsDaman ඞ

itsDaman ඞ

[ionic jewel]

what is this step?

that's not a pythagoran identity but ok

and i meant the next step

where you decided to do different things to each side

no you didn't

you cancelled a cos(x) and a 1/cos(x)

by your math cos(x) = sec(x) is true for all x

[ocean seal]

itsDaman ඞ

[ionic jewel]

that is right this time

[ocean seal]

itsDaman ඞ

$cos^{3}x =\frac 12$

[ionic jewel]

what

just cube root both sides to get rid of it

then take the arccos of both sides to get rid of cos

wait

does that actually work?

[thorn kindle]

(Cos(x))^3

Trig powers are confusingly notated

[ionic jewel]

^

i almost jebaited myself there

,w sqrt(2)/2 = 1/cbrt(2)

[ocean seal]

Results provided by WolframAlpha

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

Requested by 兔子#6562

[ionic jewel]

you can't just say "thing is basically equal to other thing"

those are not equal

yes

i really hope you meant $\sqrt[3]{2}$

[ocean seal]

bunny

[acoustic shadow]

Can someone please help me. Please dm me if you want to help me. I really need someone to help me.

[tight locust]

let me define a concept i will call the "class" of a function. 0.1x, x, and 999x are all the same class. 0.1ln(x), ln(x), and 999ln(x) are all the same class. etc etc. what is the largest "class" of function whose infinite reciprocal sum still diverges?

O(f(x)) basically if you are familiar with that

x^3 + x and x^3 + x^2 would also both be in the same class of x^3
basically the fastest growing term, ignoring coefficients

[alpine sable]

can some check if i did correct answer or no

[jagged imp]

Ye

You did

[alpine sable]

was it correct

[thorn kindle]

Ye

[alpine sable]

okay

[tiny oyster]

👍

[prime ice]

Guys pls check this

Did I do this fine

[subtle smelt]

looks good

[thorn kindle]

SUS

[subtle smelt]

have you learned about vectors

[jagged imp]

when the

[thorn kindle]

AMOGUS

[subtle smelt]

cyan sus

[prime ice]

i have not no

[subtle smelt]

oh ok

[thorn kindle]

Vectors are sus!!!

[subtle smelt]

[prime ice]

what grade is vectors

[alpine sable]

@thorn kindle where have i seen you before

[ionic jewel]

you proved like 4 things that each individually would prove they are collinear that's definitely enough

[subtle smelt]

it's about 10-11 in my country?

[prime ice]

ok thanks

[ionic jewel]

i learned vectors in 7th? grade but obviously results vary wildly

[prime ice]

oh im in 11 so maybe ill learn later in this year

[thorn kindle]

Vectors are independent of transformation

You can't really use them in this problem

[ionic jewel]

i think real vector math comes up in linear algebra and or calc3 so that might be late highschool or somewhere in college depending

although they might briefly come up in algebra I or II (or even a "precalc" I suppose)

[subtle smelt]

Doesn't $\overrightarrow{AB}=4\overrightarrow{BC}$ proof this?

[ocean seal]

Cyantist

[ionic jewel]

yep

you could use vectors if you wanted

[subtle smelt]

ok

[ionic jewel]

[thorn kindle]

Lol nvm then

[ionic jewel]

okay back to this problem

how to solve

it would be more efficient to spend like 20 minutes writing a bot to check all the possibilities @ebon badge

modern problems require modern solutions

each possible sum requires a unique solution I think

[strong furnace]

if we assume a set has k units in it then the total number of possible sets with non-common units would be 2^(10-k)-1 right?

[ionic jewel]

where did you pull that out of

[strong furnace]

adding them up gives some kind of binomial sum

[ionic jewel]

2^k -1 seemed closer

[strong furnace]

if you have k objects taken

then some object out of 10-k need to make up the other one

and for that number of ways of choosing r would c(10-k,r)

but r!=0

so 2^(10-k)-1 sum of c(10-k,r)

[ionic jewel]

and then how does this help us

[strong furnace]

I am just thinking out loud but I think I have the idea

[alpine sable]

quack

i got muted

on home work help

[ionic jewel]

I think 7 is the breaking point

i don't think you even need to have 10 people

[strong furnace]

I am stuck this is not going anywhere for me -.-

[ionic jewel]

I think i got something

using your math

maybe

the first number can be any of 60

the second number can be any of 59

the third number can't be either of the first two numbers or their sum, so only 57 choices

so it's the number of combinations? of previous elements

[strong furnace]

so 60c10 ways of having non-trivial people

[ionic jewel]

then you can sum those up and it'll be greater than 60 or whatever

[subtle smelt]

so we have 1023 methods to choose a group from the 10 people, and the sum of their ages can be 1 at minimum and 600 at maximum?

[ionic jewel]

I am fairly sure my binary approach does work. The numbers 1,2,4,8,16,32 can sum to any number (0,60), so you can't iinclude any other number

[strong furnace]

this problem is boiling down to how many different groups of people you need to choose so that you have atleast 2 groups of non-common people group

which is much more easier I guess

[ionic jewel]

there's no way to include the 7th number

[thorn kindle]

2 people can share the same age

[ionic jewel]

no

[strong furnace]

can you make why you can't include a 7th number more clear

[thorn kindle]

Can they?

[ionic jewel]

no

becuase they could be your two groups

[thorn kindle]

The wording is not clear on that

[strong furnace]

that would be a very trivial problem don't you think?

[ionic jewel]

then it all falls apart

all of them can be 60

qed

[thorn kindle]

It doesnt all fall apart. Thats not a given

[strong furnace]

yeah you can have {x},{x} as the two common groups

[thorn kindle]

But it is a possibility

[strong furnace]

and thats it

[prime ice]

yoh this all sounds like gibirish to me lmao

[strong furnace]

if their were two equal aged people they would make up 2 group that have equal sum

so that would be very trivial

[ionic jewel]

the numbers given can uniquely sum to any number in 0 to 60, so if you add another number, you can set it in a group against whatever sums to it, making it break the problem

[thorn kindle]

The wording says nothing about whether or not multiple people can share the same age

[strong furnace]

they don't have to sum to (0,60) they just have to be in(0,60 )

sum can be from (6,600)

[thorn kindle]

If they do that's the trivial case, so moot point

[ionic jewel]

no you misunderstand

you cannot add another number to the set I gave

[thorn kindle]

Sum can be from 2 to 300

Or rather 3 to 300

No sharing apparently

[subtle smelt]

yes every number can be expressed as sum of powers of 2

[thorn kindle]

Oh wait no sharing

That makes computing the max a bit harder

[ionic jewel]

also the difference in two people's pairs of ages can't be the same for obvious reasons

[strong furnace]

I don't really understand what you are going for why are you considering powers of 2 for ages or am I misunderstanding what you are trying to do , can you explain what you're trying to do because its not hitting my head in the right way

[ionic jewel]

so starting at 1,2, then only way to continue picking the lowest numbers is the 2^n sequence

[strong furnace]

yeah continue

[subtle smelt]

oh I'm replying to nyann

[ionic jewel]

I don't know how to prove the 2^n series is the optimal way to express ages

I was merely showing that 1,2,4,8,16,32 is a "full" set such that you couldn't add a 7th element without breaking the sum rule

I suspect that it is the optimal way and 6 is the largest possible set, but i don't see a way to prove it

actually

i think i do know how to prove it

using Pidgeonhole actually if I'm right

[strong furnace]

I see what you are trying to do and if you're successful can you write the whole proof in what we would say a formal way?

[ionic jewel]

no because im awful at proofs but I can write it in an understandable way

lets see if this works

We know that each number we add must not be able to be summed to by any previous numbers in the set. Starting at 1 2, the next number we can add is 4 (which is 1 + 2 + 1). After that we get (1 + 2 + 4 + 1) = 8, and so on.

[strong furnace]

I am sorry was not paying attention what was this supposed to be?

[ionic jewel]

This isnt what i wanted to go for

and isnt a formal proof

ugh

the ohter one fell apart as I was typing it

[ionic jewel]

I think this one does happen to be correct anyways

i just have to show its true for any two starting numbers

why do i feel like this is going to be something super trivial

[subtle smelt]

I mean, {1, 2, 4, 8, 16, 32} can sum to (0, 60)

[strong furnace]

I mean I have something which deals with worst case I think this might have a solution

[ionic jewel]

that just means you cant add anything to that set, it doesnt mean there isnt a set thats better than that

[strong furnace]

what I am going for is something like this , if we have 2^k-1 possible groups for k selection we just need to find how many groups do we need to select such that no 2 groups have anything non-common at max

[ionic jewel]

i might be tired but I dont know what that means

i get the 2^k-1 part

[strong furnace]

I actually messed that up really bad

ok so here's where I am at

there are 1023 possible groups

and 594 possible values

so we split them as the worst case

594 and 1023-594=429

so 429 groups must have a common group

[ionic jewel]

oh

yes

[strong furnace]

if I can show that no matter how many groups we select that have no non-common group

and it is less than 429

then by pidgeonhole

two non-ocmmon need to be

common

idk what I am saying anymore

[ionic jewel]

no

you already did it didnt you

[strong furnace]

but these 1023 also include groups which have common elements

[ionic jewel]

oh do they?

[strong furnace]

we can choose any group out of 10

that is why the split

[ionic jewel]

$\sum_{n=1}^{10} {10 \choose n}$

[ocean seal]

bunny

[strong furnace]

yeah thats what I went for

[ionic jewel]

1023

lmao

of course

[strong furnace]

that is how I got the number tho

[ionic jewel]

oh i thought you did 2^n-1

didnt realize they were the same

[strong furnace]

that is how I got 2^n-1

binomial sum

[ionic jewel]

oh lmao

i got 2^n-1 a differnet way but thats ok

so we have 1023 groups possible groups, and only 594 possible sums, that means there have to be repeat groups on certain sums

$2^n-1 > 594$

[ocean seal]

bunny

[strong furnace]

yeah that is where I am at

but from here you need

[alpine sable]

plz

[strong furnace]

the minimum number of groups that need to be selected

such that we have atleast 2

non-common element group

[ionic jewel]

that would support the theory that there is a 9set out there somewhere if im thinking about this right

[strong furnace]

9 set would be 511 so that eliminates any set < 10 out of the equation

[ionic jewel]

. doesnt this prove it

[strong furnace]

I don't think it does

[ionic jewel]

dont we know they are all unique groups

or is that the hangup

[strong furnace]

they are not unique groups

they are unique in a way

but they share common element

[ionic jewel]

ah yes

quite the problem

[strong furnace]

with at least 1023-2^(10-k)+1 group

k is the number of elements

[ionic jewel]

for the first group we choose n people where n <10, then the second group chooses m people where m <= 10-n

seems like some kinda janky double sum here

[strong furnace]

I will stop typing and think for some time because I have not been thinking ever since I got to this point

[strong furnace]

I got this

[strong furnace]

from this

[subtle smelt]

wait

if they share common element

[ionic jewel]

$\sum_{n=1}^9\sum_{m=1}^{10-n} { 10 \choose n} {10-n \choose m}$

[ocean seal]

bunny

[subtle smelt]

just eliminate them?

[strong furnace]

its in the question

[ionic jewel]

is this what you simplified nyann?

[strong furnace]

nah , what are you going for?

[ionic jewel]

2 unique groups

[subtle smelt]

wdym

[strong furnace]

ohh total number of possible ordered pairs of 2 unique groups?

[strong furnace]

they need to be (non-common) its in the problem statement

wait if that is the value

[ionic jewel]

unordered

but i got a huge number

57002

[strong furnace]

still smaller than c(429,2)

[ionic jewel]

wait

when i run it on 6 i get 602

thats slightly too high isnt it

since we know theres a legal 6set

[ionic jewel]

this implies theres 602 ways to make groups, and only 600-1 possible sums, so there should be dupicates

so theres something wrong here

when i run it on 3 I get 12, which seem like too many groups by a bit

well i guess not

its exactly right

[subtle smelt]

I mean if we found two groups A, B that have the same sum of ages but share common elements
then we can choose A\B and B\A, whose sums are the same and share no common elements

[strong furnace]

yeah makes sense

so that ends it thenn

[ionic jewel]

so whats the soln

[strong furnace]

we don't need to find the worst case

if we have 2 sets with common elements with same sum

and remove the common elements

[ionic jewel]

was I right about 7 being impossible as well?

,calc 2^7-1

[ocean seal]

Result:

127

[ionic jewel]

huh

im still confused then

[strong furnace]

any set less than 10 would be impossible

[ionic jewel]

{1} clearly works

[strong furnace]

how

[ionic jewel]

like I mean whats the lowest integer you could replace 10 with in the problem for it to still be provable/true

[strong furnace]

10 is the lowest

since we want 2^n-1>594 for it to repeat

[ionic jewel]

so for 9 people you couldnt always split them into two groups such that the sums are even?

ah

i see

okay but that implies the existance of a 9set that works, which I have a hard time figuring out what they could be

[strong furnace]

maybe 9 is not so trivial

maybe not every value between 6 and 600 can be allowed

wait

I had the wrong thing from the start

possible values of subset were 1 to 600

[ionic jewel]

hm?

oh possible sums?

[strong furnace]

so 600 possible values for groups

[ionic jewel]

well not actually 600

only 300

[strong furnace]

yeah 300

[ionic jewel]

,calc 2^9

[ocean seal]

Result:

512

[ionic jewel]

,calc 2^8

[ocean seal]

Result:

256

[ionic jewel]

hm

an 8set?

I doubt it

perhaps a 7set and an 8set dont exist due to a non-pidgeonhole related rule

[strong furnace]

is there no open problem thing , I want to try to solve this further and with more clarity but right now I have to leave :/

[ionic jewel]

i can dm it to you if you want to save it like that

[strong furnace]

that would be great 🙂

[ionic jewel]

ok

[strong furnace]

maybe we can discuss if we get any further results?

[ionic jewel]

of course

[prime temple]

This empty right??

Can someone help me please

I am studying for my SAT and I came across this

Its a quiz I am taking and this is the question I got wrong and I cant figure it out

[uncut hull]

@prime temple If I understood the question well, I believe it is C. When you "move" a function up by 2 units, you need to replace x with x + 2. Similarly, if you move it down, you replace it with x - 2.

Now, if it is transformed to move 4 units towards the left, you need to add 4. If the function is transformed towards the right by 4 units, you need to subtract 4.

Hence, your function should be $g(x) = (x + 2)^2 + 4$

[ocean seal]

Nethiriel

[prime temple]

Ohhhh

Now I get it

Thank you soo much

[mint bronze]

and this

?

[uncut hull]

@mint bronze So, the chimney is comprised of two squares and its area is 1/8 in^2. Thus, the area of one square is 1/16 in^2. Now, this means that each side of the square is 1/4 of an inch.

To calculate Δ LMN, you need to use the formula area = 0.5 * height * base of the triangle.

[prime temple]

Can you help me with this

??

@uncut hull ??

[mint bronze]

@uncut hull so whats the amswer

[prime temple]

Or @mint bronze

[uncut hull]

I am not here to solve it for you, I am only here to help you. It is fairly easy to find it now that I've explained how.

[mint bronze]

just this one i promise

[prime temple]

Can you explain mine please

I’ll just for you guys to finish

Take your time

[uncut hull]

Calculate it first and send me the answer, then I'll tell you if it's correct or not.

[forest knot]

Bro just search it on google "Sphere surface area formula", plug in the numbers. You should be good.

[prime temple]

I tried but googled doesn’t explain it well

And my English is kind of bad

[uncut hull]

I will explain it for you.

[prime temple]

Thank you

[uncut hull]

The volume of a sphere is V = 4/3 * π * r^3. From there you will find the value of r, which is the radius of the ball. In order to find the surface, you use surface = 4 π r^2**. And that's it.

[prime temple]

Ohhhhh

Can I try to solve it and you tell me if I am right?

[uncut hull]

Sure, go ahead.

[prime temple]

I cant find r

[uncut hull]

Then try harder.

The channel is open for anyone else who needs help.

[surreal linden]

someone help me with the domain and range (interval notation) for the following equations https://www.desmos.com/calculator/kflj14wpnu

[obsidian crane]

are the options wrong ?

I'm getting

$x^2 - y^2 = 1/4$

[ocean seal]

King_ftw

[spare fern]

,w x√2-x/√2=1/√2

[ocean seal]

Results provided by WolframAlpha

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

Requested by The Godfather#1908

[spare fern]

You have made some calculation error somewhere

[obsidian crane]

I don't understand what happened here

I used the following method

[ocean seal]

King_ftw

[obsidian crane]

and

[ocean seal]

King_ftw

[obsidian crane]

e = eccentricity

c = distance from vertex to the focus (from that I understand)

[spare fern]

Where did you get that???

[obsidian crane]

So half of 1/√2, that is : 1/2√2

[spare fern]

This I mean

[obsidian crane]

It's a property of hyperbolas.

[spare fern]

Equation of directrix is x=a/e

Which means it's a/e distance from origin

Focus is ae

Hence the distance between directrix and focus is given by

ae-a/e

[obsidian crane]

oh

let me try

thanks

[sage brook]

hi i solved a graph problem but im making a mistak and can find it

can you help if i send my ss

there are 3 line on the left . I multiplied them with 120 and 100. And found their common point. No problem until here.

but my intersection point of 3rd looks like this

what is my mistake can you tell me pls

[alpine sable]

Is this statement true or false ?

[bitter kite]

my brain hurts

[alpine sable]

And how do we get to know ?

[bitter kite]

is r a radius?

[alpine sable]

No no

It is algebra

[bitter kite]

oh

[alpine sable]

Rings

[bitter kite]

do you have any ideas in my problem tho?

[alpine sable]

This one right ?

[bitter kite]

yes

[alpine sable]

You can cover the entire shade region in 3 squares

Can you see that ?

[bitter kite]

i can cover the 5cm by 5cm square

but i dont know the rest

[alpine sable]

Ok see here

On the top of that 5×5 square

You can put another 5×5 square

You see that ?

Then it will cover the entire shaded region.

You getting the idea ?

[bitter kite]

its still loading

i know some parts that i can cover with the 5 by 5 square but what about the rest of the shade?

[alpine sable]

Now you see ?

[alpine sable]

.

[weary cipher]

Maybe try counting for the triangle's length?

[bitter kite]

the shade in that square can be part of the 5 by 5 square tho

[alpine sable]

Here we have a big rectangle. Can you see ?

[bitter kite]

i can see it

[alpine sable]

Ok

[bitter kite]

oh i get it

[alpine sable]

You sure ?

Find are of big rectangle first

[bitter kite]

i have my ways but the words cant come together

i cant explain it

[bitter kite]

50cm2

[alpine sable]

Then subtract the area of the half triangle

*half white triangle

From the big rectangle

[bitter kite]

how do i get the area of the white triangle-

[alpine sable]

The white rectangle below

10×(10+5)

[bitter kite]

thats 150

[alpine sable]

Divided by 2

From the big rectangle

Subtract the area of those two triangles

*two white triangles

You get the area of the shaded region

The white triangles are marked 1,2

And they are right angle triangles

So finding area is easy

(1/2)×(base×height)

[bitter kite]

how do we get the left side of triangle 2, then

[alpine sable]

Smaller side is 5

Do you see that

[bitter kite]

triangle 2 is right triangle, right?

[alpine sable]

Yes

[bitter kite]

so that means axbdivided by 2

[alpine sable]

Bigger side is 5+5=10

[alpine sable]

That is the formula for finding area of right angle triangle

[bitter kite]

for triangle 2, 10x5 divided by 2 is 25. is area of triangle 2, 25cm2?

[alpine sable]

Yes

You get the idea ?

I have to go

Quick

[bitter kite]

you can go now

[alpine sable]

Ok

[bitter kite]

i know it, thanks

[alpine sable]

You welcome

[bitter kite]

sorry if i dont answer your question, im not kinda use in algebra

[plucky crow]

i dont get how they leave aremainder of 1

dont they leave a remainder of 1/p1+2+...

lets take 4 for example

thats 1 more than a prime

4 = 3+1

4/3 = 1+1/3

the remainder is 1/(the prime which is in this case 3)

[elfin snow]

well prime numbers can't be directly next to each other

and if X is composite (since it is made up of primes) then a number +1 is likely prime

I think

idk what the remainder argument is tho

[plucky crow]

3 *7 = 21 + 1 = 22#

not prime

but not the point

i dont get the remainder thing

[elfin snow]

yeah idk

[plucky crow]

how is it a remainder of 1

[pallid trout]

X is the product of all primes + 1. X cannot be prime (it's not part of the list of finite primes). If X isn't prime then it must be divisible by a prime number.

[plucky crow]

yes i get that part

what i dont get is the "it will leave a remainder of 1"

[minor heath]

adding 1 makes the result not a multiple of any prime

if it were to add to 2, then it could be a multiple of 2, which is a prime

[plucky crow]

again i get that part

[pallid trout]

The second part if there is a prime divider

Then it would divide (X-(product of all primes))

So it would divide 1

Since X - product of all primes = 1

[minor heath]

the least prime number (from the finite number of primes) is 2

you factor that out from the product

this become 2(the rest of the primes)+1

now divide it by 2

this also applies to all primes

[plucky crow]

(the rest of the primes) + 1/2

?

[minor heath]

no what is 2(the rest of the primes)+1 mod 2?

lets say that the rest of the primes equal n

it becomes 2n+1

[pallid trout]

You're complicating this a lot more than it needs to be lol

[plucky crow]

2n+1/2 = 2n/2 + 1/2 = n + 1/2

[minor heath]

because it had to