help-28

[molten wadi]

3 + b^2 = 4

b^2 = 4 - 3

b^2 = 1

b = 1

[spice orchid]

Correct

[molten wadi]

is 1 the missing side

oh

[spice orchid]

It is

So now sinθ is?

[molten wadi]

1/2

[spice orchid]

Well done

[molten wadi]

and sin^-1(1/2) = 45

[spice orchid]

You don't actually need to find θ

[molten wadi]

i just need the ratio?

[spice orchid]

The question you posted just asked for sinθ

[molten wadi]

oh

ty

so every time it ask for this i just draw a triangle

[spice orchid]

If it asks for (trig function)θ given (some other trig function)θ, then you're most likely going to want to draw a triangle yeah

[molten wadi]

ok ty

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[torn jolt]

Can I get some help to figure out the 2 left angles and lengths please

<@&286206848099549185>

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@torn jolt Has your question been resolved?

[torn jolt]

No

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@torn jolt Has your question been resolved?

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@torn jolt Has your question been resolved?

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@torn jolt Has your question been resolved?

[gritty rose]

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[untold kestrel]

help with this complex analysis problem

[untold kestrel]

if i have this contour with the singularities denoted by the points black, green, and blue

how can i integrate this with respect to a function?

my thought process is i have to break it up into three contours

[gritty rose]

just show the original problem

[untold kestrel]

alright wait, thank you for replying

here's the original problem

[gritty rose]

Do you know residue theorem

[untold kestrel]

yup I do

[gritty rose]

How far did that get you

[untold kestrel]

not far because I was thinking of breaking it up into three regions, one for each "petal" of the "rose"

but I cannot find a parametric equation for each "petal"

[gritty rose]

Seems fine

Why do you need a parameterization

[gritty rose]

You already have the poles/singular points here right?

[untold kestrel]

yes

would it be enough to just use C and not C_1 : petal 1 when evaluating a single singular point and then sum all the integrals?

sorry complex analysis has been beating me around this semester haha really sorry for the questions

[gritty rose]

Yes. That's part of the residue theorem

Read it again if you don't remember

[untold kestrel]

Alright alright thank you so much have a nice day

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[full forum]

@mellow kraken Has your question been resolved?

[sick night]

@mellow kraken i am not very keen on geometry but i think for this problem, my way to do it is to first transform the equations to a single function of f(x), which I believe you should get 3((1-x^2)^(1/2))+3, then do some analysis on the geometry, you will realize that its a circle center on (0,3) with radius 3, and from that point forward you do some trig and you should get the result soon

few hints: the distance between that unknown point to (9,0) is 9, bc the angle between that tangent line and the point to its center is 90 degrees

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✅

[sick night]

@mellow kraken u can use the arctan function to find the angle, and then deductively find the slope

some hints: do you see the angle at (9,0)?

and the line that you draw bw that point and (0,3)

it bisect the angle

bc the two triangles are congruent

so if you find the angle for any of the right triangle, which is very easy if you are allowed to use arctan function

which is arctan(3/9)

@mellow kraken take some time looking at the graph and analysis the relationships, let me know if you need more hints

slope=(change in y)/(change in x)

there are probably other ways to do without tangent, but thats why i said, if you are allowed to use tangent and arctan functions, it makes things easier

first you use the arctan(3/9) to find smaller angle, then you times 2, then use tan function to find the slope

and at the end, add a negative sign for its going downwards

what part do you not understand? @mellow kraken

do you see that if you can find the angle at (9,0), you are pretty much done

bc if you know that, you know the slope, and with slope you can find the coordinate of the unknown point

[full forum]

@mellow kraken Has your question been resolved?

[sick night]

@mellow kraken remember, tan(\theta) is defined by (change in y)/(change in x) which is exactly of the slope

hold on let me give a another way to present this

a quick sketch, so might be confusing, feel free to ask for any confusions

this way, is actually easier bc with the slope method, you have to solve another equation

@mellow kraken if you still have trouble with this, i suggest you to review the trig functions, what it does in terms of cartesian coordinates, if you know sin(x), cos(x) or tan(x) you should know the slope, etc

and all of these trig functions are interchangeable in the way that if you know any of them, you know all of them at once

what it is that i wrote that you don't understand @mellow kraken

remember what sine function does

what that is, is it transform that 3 to x coordinate of that point

[rare wraith]

So i'm a little bit stuck on this one

I'm not sure what to use

Question is find sina+cosa/sina-cosa if tana=2

[sick night]

if you are allowed to use arctan then you use plug the arctan 2 to the thing you are calculating, so i assume you don't want to use arctan? @rare wraith

[rare wraith]

no we are not allowed to use it

[sick night]

you can do this problem solving a differential equation but i think it is overcomplicating the problem

also i don't think you set up the right equation, and i think this is the one you want to solve, and it is really unnecessarily complicated for this problem

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[limpid ermine]

Whats the middle step before factorising it like that

[light sonnet]

You mean this step?

[limpid ermine]

After this

How did we get (2y-1)(2y+9)
What formula to use to quickly do it?

[light sonnet]

Just factor in general

Whatever method you know

The "fast" part is just practice

[limpid ermine]

Only remember middle split

[light sonnet]

The more you do the better you'll become

[wide sundial]

you use your brain to look at what multiplies to get 9

it's either 1, 9 or 3, 3

you then deduce those numbers paired with a 4 doens't work

so you split it to 2, 2

and somehow wiggle your way into a +16

[light sonnet]

.

[wide sundial]

at least that's kinda how i do it

[limpid ermine]

Yeah not happening...
This is a question part of applications of integrations
I am only used to middle split one
The one in which here we should get -36y^2 and on adding 16y

<@&286206848099549185>

Ah got it

Thnx Frosst#1111

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[quiet hedge]

Solve for y please

[light sonnet]

You can use software

[quiet hedge]

I tried

Didn’t work

[light sonnet]

Instead of asking people to do it for you

[quiet hedge]

See I can’t do it

[light sonnet]

Also, that equation is slightly complex, what's the purpose for it?

[quiet hedge]

I need it for an assignment where I have to create a logo. I basically need intercept points and can’t think of any other way to find them

Tried some calculators online and says no intersections but am unsure why as

[light sonnet]

As mentioned, it's too complex to solve

[quiet hedge]

Ok

So should I approximate

Intercepts?

[amber sapphire]

is cos^5 or the inside angle?

[quiet hedge]

I think it’s cos^5

[amber sapphire]

okay

[light sonnet]

Like here's whatever the result is

[quiet hedge]

Wow

[amber sapphire]

[light sonnet]

x^60 power is crazy

[amber sapphire]

how did you even reach that power?

[light sonnet]

Idk, software solved it

[amber sapphire]

lol

what i would have done would be

[quiet hedge]

Sooo i think I’m just going to approximate

[amber sapphire]

cos(6Θ)= something cos(Θ)

then power 5

[light sonnet]

Maybe because they're trig functions so it's periodic so it does something weird every even power up to 60 then it repeats itself

[amber sapphire]

and then ig arctan to arccos and we would be free from any trigonometric thing

but it would have been a mess

[summer burrow]

It’s 4😂

[amber sapphire]

wut?

[light sonnet]

They trolling

[summer burrow]

Yes

[amber sapphire]

lmao i thought god himself entered the chat

[quiet hedge]

Lol

[light sonnet]

I mean, it was obvious that they were trolling because the original equation had both x and y, on both sides and they magically got a singular number, impossible

[amber sapphire]

wait i think sukkie wants intercepts or something

so i thought he gave that

[quiet hedge]

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[pastel helm]

how do we approach this ?

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[pastel helm]

<@&286206848099549185>

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[wind pendant]

Hey can someone help me on this

[light sonnet]

,rotate

[light sonnet]

What's the goal?

[wind pendant]

i have to find the base area of the prism

And I don’t even know what prisms are

😅

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@wind pendant Has your question been resolved?

[rugged harbor]

think of it as a rectangle with a corner cut off

How would you find it's area?

@wind pendant ?

Uhm

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@wind pendant Has your question been resolved?

[wind pendant]

a =l × b?

[echo lance]

that does give you the area of the rectangle, what about the area of the corner?

you can do $\text{area} = \text{area of rectangle} - \text{area of corner}$

[glossy valve]

Doggo

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[random solar]

could someone please help me understand how to figure this out

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@random solar Has your question been resolved?

[rough plaza]

Hello @random solar
What have you tried?

For me, I would start with n_0 =1, since log n is negative in (0,1)

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[astral mango]

Ampitheaters are often designed so that are fewer seats per row for rows closer to the stage. Suppose you are
sitting in row 22 in an ampitheater and notice that there are 117 seats in your row. It appears the row in front
of you has 114 seats and the row behind you has 120 seats. Suppose this pattern continues throughout the
auditorium.

[astral mango]

How to solve for the amount of seats in the first row

or 49th row

[solar geode]

Do you know about arethemtic sequences ?

[frail cobalt]

Arithmetic progression

[thin quartz]

don't be lazy; if you're so curious just go up to the first row and count the seats

since you already painstakingly counted the 117 seats of the row you're in

[astral mango]

yes, im not quite sure which formula to use tho..

recursive or explicit

[solar geode]

Use the explicit formula
Common difference is 3

Is the seat number starting from top or bottom

[astral mango]

do i have it set up correctly in the second photo?

@solar geode

[solar geode]

Yes

[astral mango]

so i just got -334

[solar geode]

Then it probably -3

The rows starting from the opposite side

[solar geode]

The common difference i mean

[astral mango]

AHH SHIT U RIGHT

wait it wouldnt make sense still

[astral mango]

i got positive 380 but that doesnt make sense that a1=380

since the progression of seats is lowering from 23-1

[solar geode]

No
Your confusing the numbering of seats again

Its reverse

[astral mango]

so this is the correct forumla to find the seats in #1?

row 1 would have 54 seats

[solar geode]

Ye
Row 1 is the front row
Which says in the question has the fewest seats

[astral mango]

so in theory i couldve set up that same formula but using row 21 instead of 23?

[solar geode]

Yes

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[stark pond]

hi, can someone help me on question c ?

[stark pond]

would you just sub in the values from part b ?

of x, i mean

got it

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[rigid niche]

Why are these answers not correct (this isn't a test)?

[faint mesa]

total divided by 4, they ran 4 miles not 3

[rigid niche]

That wasn't correct -- like I said, I'm confused because this seems like a simple thing in my head

[faint mesa]

I have a feeling they want you to give the final answer.

[rigid niche]

It's the opposite -- we're encouraged to type it out exactly as if we'd put it into a calculator, which is why I do things like 19-6 instead of 13

[faint mesa]

nevermind it would be 3 in the denom....

wth

[faint mesa]

do me a favour and swap the 6 and 19

i wonder if your teacher made it so that order matters

[pliant radish]

you're dividing time by distance rather than distance by time

that's why

he took 25 minutes to cover all 4 miles so his average speed would be 4/25 not 25/4

[faint mesa]

yea donk is right

my reading is bad

[pliant radish]

nothing to worry about, it's the little things that always get everyone

[rigid niche]

That works -- thanks!

[pliant radish]

np

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[ionic thicket]

How do I know the order in which u write the corresponding sides

Like ab/pq

= bc/qr

=ac/pr

How does it work

If this triangle is prooved simmilar

By aa

How will the corresponding sides look like

[pliant radish]

you see how AP and PB are on the same line, which is AB

CP and PD are on the same line which is CD

[ionic thicket]

K

Ye they are on the same line

[pliant radish]

so AP/PB = CP/PB = AC/BD

DPB = APC, PCA = PDB and PAC=PBD

[ionic thicket]

K got it

Thansk

Thanks

[pliant radish]

np

[ionic thicket]

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[bright rose]

how do you find h in terms of x?

[vivid coral]

Right

[pliant radish]

find the perimeter of the cuboid in terms of x and h

you know the perimeter is 640

[bright rose]

yea

but it syas

syas

says

h in terms of x

[pliant radish]

yes

because you won't find a numerical value for either h or x

P=4(L+W+H)

so 640=4(4x+x+h)

divide both sides by 4

160=5x+h

h=160-5x

[bright rose]

waitttttt

what is that

[pliant radish]

perimeter

[bright rose]

why is 4 there

[pliant radish]

because there are four lengths, four widths and four heights in a cuboid

[bright rose]

oh yea

alr

how does this work

when you divided both sides by 4

640/4

160

but how did you do the other side

[pliant radish]

yes

4 takes out the four

[bright rose]

yea

[pliant radish]

so you only have what's inside the parentheses left

[bright rose]

ok

wait

so

4x + x + h

[pliant radish]

yes

4x+x+h=5x+h

so 5x+h=160

subtract 5x from both sides

and you have h in terms of x

[bright rose]

ok I got i

it

[pliant radish]

yes

that's it

👍

[bright rose]

is this like

L times W times H

[pliant radish]

yes

this bit is built on the h in terms of x bit from before it

[bright rose]

why does it say in terms of anthing

tho

' in terms of x'

[pliant radish]

because you can't find a numerical value

you don't have the value of x so you can't find h or V as a number

but you can find the relation between h or V, and x

[bright rose]

alr

got it

[pliant radish]

like how you proved that h = 160-5x

if x equals 1 then h = 155

if x equals 2 then h = 150

and it goes like that

as for the volume bit

V=4(x^2)h

substitute h by 160-5x

[bright rose]

wait

why that

isn't it

V= lwh

[pliant radish]

yes

l = 4x, w = x and, h is h

so it'll be 4x^2h

h=160-5x

so it'll be 4x^2(160-5x)

if you distribute the 4x^2 over the parentheses you get V=640x^2-20x^3

[bright rose]

but this is the answer

[pliant radish]

they're equivalent

[bright rose]

if I write that

[pliant radish]

if you distribute 20x^2 over the parentheses you also get 640x^2-20x^3

[bright rose]

do you think I get the mark

in a test

[pliant radish]

if the automatic correction doesn't accept only one form of the answer it should be correct

unless your teacher manually corrects it

but i can't guarantee it if it's automated

but in terms of math as math, it's correct

[bright rose]

alr so here

you just sub 11 in all the x's

in our equation

right

[pliant radish]

yes

[bright rose]

ah

nah

your equation is wrong

Im pretty sure

I put it in the calc

and I got

70k or somthing

like that

and the answer is 50 k

what did you do wrong

[pliant radish]

hold on

https://cdn.discordapp.com/attachments/903486480985489478/983526334410141716/unknown.png ok this one returned 50820

let me check my answer

it returned the same number

[bright rose]

wait

I have to go now

my test

😐

bye

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[young storm]

How do I find x? probably really easy and my brain's just dying

[daring citrus]

similar triangles
GH/FH=KH/GH (pls double check this)

[young storm]

Yeah that's right lol Idk why I thought they weren't similar

How about this one?

Probably similar too

But i'm not sure

@daring citrus

[daring citrus]

apparently this is a thing
idk how this formula comes ¯_(ツ)_/¯

[young storm]

alright then lol

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[molten notch]

d1+d2= or d1=d2?

[molten notch]

i am not sure which formula to use

[tidal citrus]

your image is too small to read

also, which of these two problems are you asking about?

[molten notch]

both

how about this

[tidal citrus]

ok

you want an equation for the distance between aircraft 1 and aircraft 2, in terms of time

we have the distance the first aircraft is away from the base, and we have how fast both are going.

what can we do with that?

[molten notch]

find out the time?

[tidal citrus]

that is our goal. What's the formula for the distance between the aircraft?

[molten notch]

d1+d2=200 km

r1 t1 + r2 t2 = 200 km

[tidal citrus]

well, that depends on what d1 and d2 is

we want our distances in terms of time

[molten notch]

so r1 would be 450km/h

[tidal citrus]

and involving the velocities of the aircrafts

[molten notch]

450 km/h * x = 350 km/h * (x - t1)

that's what i got

[tidal citrus]

maybe

it's early for me

I wrote something else

[molten notch]

what is it?

[tidal citrus]

what is x here?

[molten notch]

the time

as in how long

[tidal citrus]

the distance the first aircraft is away?

[tidal citrus]

well then what is t1?

[molten notch]

there's no mention for time

so i can't use x - (the time)

[tidal citrus]

it says "how long"

before the aircraft meet

that is the problem we want to solve

[molten notch]

yeah i know that

but how do substitute for t2

[tidal citrus]

at the highest level, if the distance aircraft 1 is from the base is d1
and the distance aircraft 2 is from the base is d2
then we want to find

[molten notch]

nono

[tidal citrus]

$0=d_1(t)-d_2(t)$

[molten notch]

d=rt

[glossy valve]

Disorganized

[tidal citrus]

where d_1 and d_2 are functions of time

[molten notch]

so d becomes rt

d1 = r1t1

[tidal citrus]

which aircraft is d1?

for you

it is arbitrary but we must be consistent

[molten notch]

the one travelling at 450km/h

[tidal citrus]

ok

yes

how about the other

[molten notch]

d2 would be for the one travelling at 350 km/h

[tidal citrus]

right, and what would that formula be?

that expression

?

[molten notch]

r1 t1 + r2 t2 = 200 km

wait

[tidal citrus]

think only of that plane, don't involve the other

only worry about d2

[molten notch]

i see

[tidal citrus]

almost there...

?

[molten notch]

yeah

[tidal citrus]

it's all in the first sentence.

* How far away from the base does that plane start and
* How fast is it going?

[molten notch]

thanks

i got it

what's your answer btw

mine is 1/4 hour

[tidal citrus]

same

[molten notch]

question no. 7 is the same formula?

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[maiden arch]

I have to give the exact value and not a decimal approximation. I’m confused on what to do with the square root of 55 .

[eager violet]

Why did you subtract 3²?

[last lynx]

thats just wrong everywhere

[maiden arch]

… I have no answer to that, is it supposed to be 73?

[last lynx]

pythagoras?

[eager violet]

√73, yeah.

Also, on that second line, that shouldn't be H²

You already took the square root

[maiden arch]

Oh ok

[flint moon]

The square root of 73 isn't rational so you will have to keep it as √73

for the cos(Theta) just use the definition

[maiden arch]

Ok that was my next question

Thanks so much

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[hallow yarrow]

So we have a spinner with 6 divisions(1,2,3,4,5,6) each with probability 1/6.

What is the probability of getting (1 and 2 and 3) is after spinning the spinner in 10 times?

Find the probability of getting on (1 or 2 or 3) is after spinning the spinner in 25 times?

Find the probability of getting on 1 is after spinning the spinner in 5 times

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[wary nest]

Getting HTH and HHT are equiprobable after 3 coin flips. But I keep flipping it, I'm more likely to get one of these combinations than the other. Why is that?

[wary nest]

this is confusing to me since P(H) = P(T) = 0.5

so how can i get one combination than the other

confused as why

[stuck rampart]

Because randomness
Its very unlikely you get the same
Like if you roll 10 dice it's very unlikely you role the same number of sixes as you roled ones

[wary nest]

uncertainty

[stuck rampart]

Ye

[wary nest]

oh so thats the answer?

[stuck rampart]

Idk
I just know the basic idea behind it
Idk how you would construct a mathematical proof

[wary nest]

ah

[fathom saddle]

It's unlikely they'll both happen together, but that's not what we mean when we say "one is more likely than the other".

If the question means "keep flipping and immediately stop when you get this string of 3"
Then I get the odd feeling that HTH becomes more likely. I'll think on it.

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[wary nest]

hmu if u think of something

[fathom saddle]

Oh, figured it out? Cool.

[wary nest]

no

just closing the channel

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[hallow yarrow]

we have a spinner with 6 divisions(1,2,3,4,5,6) each with probability 1/6.
Find the probability of getting on 1 is after spinning the spinner in 5 times

[hallow yarrow]

so prob of getting 1 is 1/6

how do we find what will be the probability of getting 1 in 5 spins

[fathom saddle]

I assume you mean getting "at least one 1" in 5 spins?

Then it's 1 - (1 - 1/6)^5

Easier to calculate the probability of never getting a 1, then take the complement of that

[hallow yarrow]

oh

then how do we consider cases where it is or condition

For example, prob of getting "at least one 1 or 2 or 3" in 25 spins

we spin 25 times, what is the probability that we got at least one 1 or 2 or 3

[fathom saddle]

Probability of doing that in a single spin is 1/2.
Probability of getting it at least once in 25 spins is 1 - (1 - 1/2)^25

[hallow yarrow]

can you please explain how it is 1/2

there are three options , getting 1, 2 or 3

[fathom saddle]

The probability of getting a 1, 2, or 3?

When all options are 1, 2, 3, 4, 5, 6?

That's half the options.

[hallow yarrow]

in this case the answer would become ~1

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[iron pecan]

Hi I'm looking for a function that takes integers as input and output integers divisible by 2 and 3. How does such a function look like?

[spice orchid]

Depends what other requirements you want for the function

The function that sends every integer to 6 satisfies what you wrote, but i expect that isn't exactly what you're after

[wild sleet]

they just mean filter

sequence to shorter sequence

ask some programming server tbh

[iron pecan]

The original question is "what is the percentage of integers divisible by 2 and 3". My idea was to create a function and then to differentiate it and find the limit

[low bloom]

or just divisible by 6

n(Z) isn't defined

so calculating a percentage of an infinite set is probably not defined

[fast peak]

n(Z) may not be defined, but you can definitely do a limiting process

[iron pecan]

Sorry i meant divisible by 2 OR 3

[low bloom]

alright

[fast peak]

something like $$\lim_{n\to\infty} |{x\in {0, 1, \ldots, n}: x \text{ is divisible by } 2 \text{ or } 3}|/n$$

[glossy valve]

Denascite

[iron pecan]

Yep i think so

[low bloom]

1, 2, 3
2/3 divisible by 2 or 3
4, 5, 6
2/3 divisible by 2 or 3
7, 8, 9
2/3 divisible by 2 or 3

[fast peak]

you can also think of it from the other perspective: which numbers are not divisible by 2 and 3

1/2 of all numbers are not divisible by 2 and 2/3 of all numbers are not divisible by 3

[low bloom]

https://youtu.be/QBsFWaxsOV0

[iron pecan]

Oh that's neat clever

[low bloom]

finding fractions involving infinity is weird 🤷‍♂️

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[rugged harbor]

I was playing with desmos, and I tried to plug in some functions to the mod() function (to be clear, mod returns the remainder of a division, it's not the absolute value).
mod( f(x), g(x) )
The function has a very special graph: there are a lot of curves and their shape depends on f(x), but their range depends on g(x). I can see why that's happening.
In some particular cases it looks like the curves of the mod function are the same function, but translated.
My question is: is there a way to extend all the curves that mod makes past the values of g(x) to make a graph?
Do you have any resources to recommend to learn more?

[queen sun]

what do you mean past g(x)?

you mean that mod(f(x),g(x)) will plot a line past the field of g(x)?

[rugged harbor]

something like this

[queen sun]

yeah

impossible

you know how mod works?

[rugged harbor]

Yes

[queen sun]

then you should know it can generate only values less or equal g(x)

[rugged harbor]

Yes, I know

[queen sun]

besidea

the thing you juat draw

iis not considered a function

[rugged harbor]

Yes, I know

That's fine

[queen sun]

since for 1 argument x you have 2 values y

[queen sun]

if ypu want a result like you have shown, then it won't be just 1 function, but all bunch of them

just a matter of glueing and stretching

[full forum]

@rugged harbor Has your question been resolved?

[queen sun]

you won't get any analytical reduction here

solution will be numeric

you need something better than deamoa for this

[rugged harbor]

What do you mean?

[queen sun]

you wanted to enlong your lines of g(x)

so you need all points such that mod(f(x),g(x)) = f(x)

[rugged harbor]

Wait, g(x) is fine, I want to enlong all the curves that mod( f(x), g(x) ) makes

[queen sun]

this ain't easy task

yeah

so need all points where mod and f connect

because thoae are atarting pointa.of every new function

[rugged harbor]

The start of every function might be describable by a progression

[rugged harbor]

I've experimented a bit, either it's a constant or it's too hard to find

So yeah, no progression I guess

[full forum]

@rugged harbor Has your question been resolved?

[rugged harbor]

"In some particular cases it looks like the curves of the mod function are the same function, but translated."
Apparently I was wrong

[spice orchid]

Out of curiosity, what lead you to believe that?

[rugged harbor]

I looked at it, look at the x+ side only

[rugged harbor]

At this point I'll just start finding equations of the parabolas on the x+ side, there might be some patterns

f(x) = x², g(x) = x
1° parabola: x² = 0 →that's f(x)
2° parabola: x² - x = 0
3° parabola: x² - 2x = 0
4° parabola: x² - 3x = 0
5° parabola: x² - 4x = 0
...

Well, now that's a nice pattern

What functions should I try next to have the best possible results for a generalisation?

if anyone is following, feel free to recommend functions

Another nice thing from the previous result. If we change the coefficient (let's call it n) of x in g(x), the only thing that changes is that only the curves that start form an x divisible by n stay. *there's a reason behind this, I found it

[queen sun]

that's something compleyely different from what you initially wanted

[rugged harbor]

Well, I wasn't at all expecting that haha

[queen sun]

...?

[rugged harbor]

I'm looking for nice patterns to understand how mod(f(x), g(x)) builds it's curves

[queen sun]

it's just f(x)

that has been cut into little pieces

in the size of g(x)

[rugged harbor]

[queen sun]

ffs

I use paint

ok I see your point

hold on a sec

you cheated there a bit

because in example 1 your f(x) = x

and in example 2 your g(x) = x

see that it aligns completely with initial function for first interval?

same thing happens with your example at parabola

[rugged harbor]

Yes

[queen sun]

the next interval starts at first zero of f(x)

in your example next interval started at point that connected f(x) and g(x)

from this you can see that next interval is stretched

OH I KNOW

i know how can you look at that

imagine that this a game of snake

but your map is only in the area between y=0 and y =x axis

if your function (snake) passes trough it, it will land on the opposite side

this is why I told you this is sliced:

[rugged harbor]

but the curve transforms in a weird way

[queen sun]

you sure those little pieces won't make initial function when put together?

[rugged harbor]

Do you mean if I translate them upwards?

[queen sun]

yeah

hmmm

I got an idea

[rugged harbor]

If that's the case, if I consider an interval on the x axis (starting form the start of a curve, ending at the start of the next one) in mod(f(x), g(x)), Δy of the start and end points should be the same if I did the same process considering the same interval in f(x)

But that's not always true

[queen sun]

IT WORKS

crap

linking desmos doesn't work same way as wolfram

[rugged harbor]

You have to save the graph

[queen sun]

nvm

I just made print screen

see?

it alligns perfectly

[rugged harbor]

Try with the next interval

[queen sun]

ok

works as well

altough

you need to translate it by a value of a function

so first translation was by 1

so 1^2 = 1

next was

2

so 2^2 = 4

[rugged harbor]

but that's not f(x) cut into little pieces

[queen sun]

how it is not?

it's not deformed

[rugged harbor]

I mean mod(f, g)

[queen sun]

yeah it is cut into little pieces of x^2 in this case

I just translated them if they align meaning they are not deformed

meaning all of them make up for entire function

[rugged harbor]

That's mod(f, g) with f and g translated

[rugged harbor]

.

[queen sun]

not always true huh?

because g(x) = x is trival example

try other boundaries

and other images

the one with sin(x) is a bit tricky

[queen sun]

.

[rugged harbor]

I don't think this path we are following helps to understand how to extend the curves

[rugged harbor]

<@&286206848099549185>

got any ideas?

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@rugged harbor Has your question been resolved?

[rugged harbor]

I spent 5 hours on this, should I just give up?

[rugged harbor]

Yes, I give up

Wild problem

Thanks to everyone who helped! ^^

bye

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[rugged harbor]

.reopen

[full forum]

✅

[rugged harbor]

With f(x) = x³
the function that describes all the extensions is y = x³ - nx, where n is a natural number (and 0 included)

And with f(x) = x^4
the extension function is y = x^4 - nx, where n is an integer

[gritty rose]

Do you have a summary of what your question is

[rugged harbor]

in desmos there's this function, mod(a, b), it gives the remainder of the division a/b.
If you plug in functions (I used f(x), g(x)) it has some weird properties.
If you graph it, it makes a lot of curves, most of which stop at g(x), because the remainder won't be bigger than g(x)
I'm trying to find a way to understand how those curves behave

[kind jay]

Try f=sinx

[rugged harbor]

I have some great news ^^
a lot of functions' curves are describable by using
f(x) ± kx

the domain of k depends on the function

I don't know how to generalise it

Also it behaves super weird with functions with vertical asymptotes nevermind, it works just fine

any suggestions?

[noble surge]

Huh

What's the question?

Seems like this convo has been going for some time

[rugged harbor]

.

[rugged harbor]

It really has lol

[noble surge]

Hm

The behaviour will be in terms of a summation

f(point) - sum(previous intersections with g(x))

Does that answer it?

The cursor resets every time it hits g(x), but the accumulator (f(x)) still yields numbers

[rugged harbor]

That doesn't seem to work with points -1 < x_p < 1

[noble surge]

Very much true, it doesn't work on sinoidals?

That's above my paygrade

[rugged harbor]

haha, very nice observation btw

[rugged harbor]

It's not about sinoidals, I think it's strictly for points -1 < x_p < 1, because it doesn't work on x³ + 4 in that range too

[noble surge]

Won't work for points beyond that range too if it doesnt work on that range

Maybe it's a property only from kx² type equations?

[rugged harbor]

f(x) = f(x) - kx works fine with polynomial equations and trig functions

[rugged harbor]

Oh wait, I read intersection, not intersections

That's true, but some of the curves are left out

Well guys, we did a great job, but now it's time to take a break

Thanks to everyone who helped, I really mean it

Bye!

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[earnest rover]

Hello I cannot algebricaly find the equation of the intersection between a sphere and a paraboloid (that would be a curve) can anyone help me

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[earnest rover]

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[thin quartz]

just set the equations equal to each other

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[idle marsh]

i got 8x from finding the derivative of 4x^2

but how do i derive ln x

[hexed dagger]

😑 , haven't you studied product rule?

[idle marsh]

whats that again? i might have forgotten

[hexed dagger]

and basic differentiation formulas which are needed to be remembered

,w product rule

[glossy valve]

Results provided by WolframAlpha

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[safe trench]

we advice you to review differentiation before attempting this question

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[idle marsh]

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[ruby parcel]

https://cdn.discordapp.com/attachments/903481109420048385/985602256332288090/unknown.png

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[ruby parcel]

how do i find the cor ordinates of this equation

<@&286206848099549185>

[sacred fog]

!15m

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[ruby parcel]

i had to remake the channel because it was at the bottom , my bad

[torn jolt]

find derivative and then find where it's 0

[ruby parcel]

what do you mean find where its 0

so i find the sderirivative

and make it equal to 0?

[torn jolt]

yes, a stationary point is a point where the derivative is 0

[ruby parcel]

so once i've made the derirvate equal to 0 what do i do?

[rugged harbor]

The geometrical meaning of that is that you are finding a point where the tangent has m = 0, so it's an horizontal line

Solve for x and find that point where the derivative is 0

That's what I'm assuming, since I can't see the question anymore

[chrome idol]

solve for x

[ruby parcel]

Right so i should solve for x

but in this instance i got my answer as

how do i solve for x is numbert with negative power

[chrome idol]

write it as a fraction

$x^{-1} = \frac{1}{x}$

[glossy valve]

Kurama

[chrome idol]

same thing

[ruby parcel]

i got 1 over 2x squared

so my entire equation so far is

[chrome idol]

your derivative is wrong

the 2 shouldnt be in the denominator

the x^{-2} only effects the x term

[ruby parcel]

oh wait

yea your right

2 over x squared

[chrome idol]

now set that to 0 and solve for x

that will give you the x coordinate

[ruby parcel]

right

2x + 2 / x^2

how will i solve for x

will i need to times both sides by x ^ 2

[chrome idol]

couple of things you can do

try to make the 2 terms into 1 fraction

would be what I try first

[ruby parcel]

right i'll try doing that now

thanks for the help btw

[ruby parcel]

what other ways could i deal with this because what i worked out is totally not correcy 😢

[chrome idol]

$2x + \frac{2}{x^2} = \frac{2x^3+2}{x^2}$

[glossy valve]

Kurama

[ruby parcel]

oh right

how do i simplify this equation now

🤔

since were trying to find x

[chrome idol]

You set it to 0

[ruby parcel]

😂 😂

[ruby parcel]

once set to 0 will i have to move

everything onto the other side

[hot herald]

why everything

[ruby parcel]

wait i mean only 2 since were looking for x right

[chrome idol]

find x: $\frac{2x^3+2}{x^2} = 0$

[hot herald]

ideally there'd be a step before that

[glossy valve]

Kurama

[chrome idol]

$\frac{x}{2} = 0$ how would u find x here

[glossy valve]

Kurama

[ruby parcel]

you would times both sides by 2 right?

[chrome idol]

yeah, so its the same here

just solve for x

[ruby parcel]

i times both sides by x^2 first

and then the pictures shows the rest

[chrome idol]

yeah

or you can factor 2 so you get $2(x^3+1) = 0$

[glossy valve]

Kurama

[chrome idol]

Then divide 2 on both sides

youd get the same

[ruby parcel]

thats a solid point

didnt even see that

[ruby parcel]

so x = -1 , 1 or 2

[chrome idol]

why 2?

and why 1?

plug x = 1 or x =2 , it doesnt equal 0

so only -1 works

[ruby parcel]

bro is albert Einstein now 💀

[ruby parcel]

i'll try it

[ruby parcel]

because x^3 + 1 x can be 1 or -1

because it is cubic

[chrome idol]

$\sqrt[3]{x}$

[glossy valve]

Kurama

[ruby parcel]

wwe can make 2 = 0 here too right

[chrome idol]

????

2 = 0

[ruby parcel]

2 = x*

wait so im gona test it now

and sede

[chrome idol]

dont worry about the imaginary numbers

[ruby parcel]

bro got the krabby patty formula 💀

[ruby parcel]

thanks bro it worked

i finally understand this

finally

thanks to @ivory palm @eternal maple

and ofc the goat @chrome idol

@sacred fog We appreciate your 24/7 server 😃

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@close

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[ruby parcel]

there we go

cheers mate 🍻

good bye 😢 👋 🤝

🤝

i got the questions

😃

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[ruby parcel]

😂 im actualyl in year 12 but last year was insane

cus all the papers got leaked and stuff

[eternal maple]

what did i do😆

[ruby parcel]

bro @sacred fog shoulda been the exam board senior 💀

[ruby parcel]

bro you just helpful

[eternal maple]

thanks

[ruby parcel]

exactly bro

illuminator spectating this chat rn i bet

you take care too bro

add?

[sacred fog]

I was making food

[sacred fog]

exam board senior?

not a mod

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[late olive]

If I have a 50% chance to receive either box A or B, and box A may contain either 1 apple or 1 pear and box B contains 1 banana, then what is my chance of getting a pear?

I'm trying to learn probability stuff just for myself, I love math but never learned this stuff 😄

By my knowledge, I should have a 33% chance to get a pear, or any fruit.
But it feels like my math is wrong here and there is something much weirded going on in the background..

I've seen some videos about probabilities with questions that seemed super obvious but in the end they would multiply different levels of probability or something like that and then answer to this question would turn out to be 0.5 x 0.5 = 25% chance to get any of the fruits..?

And then some people explained that answer would be 16.5% because first you have a chance to receive one of the boxes (0.5) and then you have 3 different outcomes (0.33) and by multiplying these two levels you get 0.5x0.33 = 0.165..?

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[late olive]

I got a little explanation that helped me think a bit more clearly on this stuff, basically each level of probability should sum to it's total chance..?

Like box A has 50% sum chance, meaning that apple+pear should not exceed 50% sum.

Am I moving in the correct direction with this knowledge? 😄

[dusty lava]

Following because I'm interested

[late olive]

o/

So far my new answer will be:

• apple = 25%
• pear = 25%
• banana = 50%

[dusty lava]

I'm wondering if it's law of total probability

Bayesian shenanigans

I was never good at this stuff

[late olive]

I only learned basic probability of one level, like 1 dice example 😄

[late olive]

Oh no, stuff like this makes my brain break, never understood these formulas

guess it's calculus level or something

I only know that symbol in the end should be "sum" 😄

[dusty lava]

P(pear)=P(box A)P(pear given boxA)+P(box B)P(pear given box B)

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[cobalt dirge]

Yes, that is correct.
Intuitively, you can also think of it as cutting a rod of length 100 into segments. First you cut it in the middle into two pieces, one piece is A and the other is B (banana). Then you cut piece A in the middle into two segments denoting apple and pear. Now the length of each piece is the total probability of seeing that item.
When you take half of something you multiply by 1/2 (cutting the rod in half). Taking half of a half is 1/2*1/2=1/4.

[dusty lava]

This also is 1/4

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[late olive]

Like this then?

[cobalt dirge]

Yes. Probabilities must sum to 1.

[late olive]

So, with this next example, answer would be:

• apple = 20%
• pear = 20%
• banana = 60%

[cobalt dirge]

Correct.

[late olive]

What if box A will contain 3 apples and 1 pear? And we can only get 1 fruit.

That means my chances for fruits are:

• apple = 30%
• pear = 10%
• banana = 60%

[cobalt dirge]

Again assuming that apple and pear is equally likely in box A.

[late olive]

Yeah ^^

[cobalt dirge]

If box A contains an apple 75% of the time and a pear 25% the time, yes.

[late olive]

Okay.. I think I'm getting it 😄

I will make one more harder example and see if I will get it correctly, may I mention you when I'll be done? 🙂

[cobalt dirge]

go for it

[late olive]

Thanks, will be back soon!

@cobalt dirge

Green % are my answers 😄

Those are approximate of course, 3.3% should be 3.4% so it would sum up to 100% in the end .w.

[cobalt dirge]

Correct.

[late olive]

Great! Thanks, I am pretty confident in my basic knowledge of this type of probability now 😄

[cobalt dirge]

When working with probabilities it is often useful to work with fractions instead of percent. This way you don't have to worry about rounding and just need to check that everything sums to 1.

[late olive]

Yeah that is true, but for user interface they wanna see percentages and not "3 1/3" 😄

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[kindred osprey]

Which test would I use for this?

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[wooden knoll]

i would start by figuirng out where the vertical asympototes are

[torn jolt]

I tried to do so using x + 5pi/8 = pi/2 + pi(n)

But it’s literally not working for me

[wooden knoll]

,,x+\frac{5\pi}{8}=\frac{\pi}{2}

[glossy valve]

epiphonically

[wooden knoll]

,,x=\frac{\pi}{2}-\frac{5\pi}{8}

[glossy valve]

epiphonically

[wooden knoll]

,,x=\frac{4\pi}{8}-\frac{5\pi}{8}

[glossy valve]

epiphonically