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[dire cradle]

I am stuck on part (ii). I am not sure how to write the eₓ component of L. I thought that angular momentum was the cross product of position with momentum and was used in polar coordinates. I can’t seem to find anything in textbooks or online that I can apply here. Any help would be greatly appreciated, even if it was a nudge in the right direction ! 🙂

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[dire cradle]

<@&286206848099549185>

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@dire cradle Has your question been resolved?

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@dire cradle Has your question been resolved?

[wet kite]

@dire cradle what is this app?

[tacit arch]

Does this just reduce to a differential equation?

[dire cradle]

ah yes, thank you @tacit arch ! i solved the ODE and got to the answer 🙂

@wet kite im on google chrome, i just trimmed the screenshot

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[steady badger]

Okay i know its wrong but not sure how tf i did it wrong,
im trying to graph 4x-5y=-20
and im getting it right but when i graph im getting a y-intercept of +4 when it should be -4

[icy crane]

why should it be -4

[steady badger]

i put the standard form into a graphing calc and thats what i get

when i graphed it i switched it to slope intercept and get +4

[icy crane]

this line intercepts the y axis at y=4

did you use desmos?

[steady badger]

yeah

[icy crane]

ok. so you see clearly y=+4

[steady badger]

im so stupid! i missed the negative lmfaoo

[icy crane]

you don't even have to put it into standard form. just throw the equation 4x-5y=-20

there you go

[steady badger]

i knew it was something stupid

thank you

.close

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[manic dagger]

Im currently working on this problem although I'm not sure how I would find the mass or velocity.

It says the object of "unit mass", and since the mass is a constant I thought I could potentially just plug in a random number.

derivative of velocity is acceleration I believe, although in this case if I antiderive acceleration I would get a variable for within velocity and I'm not sure what it would be

this is the formula I am using from my notes

if someone could provide some direction on this that'd be awesome

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[manic dagger]

<@&286206848099549185>

[glass lichen]

$T=t_{\text{air}}+t_{\text{liquid}}$

[ocean seal]

Mosh

[manic dagger]

I understand that I will have to add those, although I'm not sure how I would find the initial velocity or math

mass*

[glass lichen]

unit mass means m=1

[manic dagger]

oh really! wait lemme check

[glass lichen]

Yes

Unit means 1

[manic dagger]

that helps out a lot I feel stupid for that one lol

[glass lichen]

however you just solve time in air, find initial speed for the liquid portion, then find time for liquid

[manic dagger]

would the initial velocity be zero?

since at t=0 there would be no movement

[glass lichen]

yes

$v_{0,\text{air}}=0$

[ocean seal]

Mosh

[manic dagger]

although for water it would be different correct?

or the liquid

since it already has the momentum from being dropped in the air

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[lone heart]

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[alpine sable]

I had to find total distance in [0,3] interval of function v= 4-2t

[alpine sable]

The reason I changed the signs from (2,3) in the second integral ( indicated by purple arrow) is bc the area is below x axis correct?

[split oriole]

Last integral (4-1)-(0-0)=3

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[knotty wren]

does anybody know how to solve this?

[blissful whale]

yes

[knotty wren]

can u help me figure this out?

i am taking a self assessment test and i am stuck

[blissful whale]

I assume they mean linear subspace

So this is what we need to proof

Do you know the definition of a linear subspace?

[knotty wren]

nope

[blissful whale]

Hm

a subset that is again a vector space

most things from the vector space inherit automatically

but one must check that for u, v vector in the subspace u + v is also one

and that scaling preserved it, i.e. if v is a vector in the subspace

λv is too

for any real λ. (I am assuming it's a real vector space)

[knotty wren]

i thikn i need to go back to the basics

i don't understand this

[blissful whale]

okay

is there something in particular I can help you with?

the word vector space is hopefully familiar?

[knotty wren]

no it's not i will watch a video first

thanks a lot tho

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[vapid crypt]

how to do this?

[lone heart]

@vapid crypt Has your question been resolved?

[bright hedge]

To prove it exists or to prove that it doesn’t?

I think it’s to prove that it doesnt

So choose ur paths carefully

The obvious choice first would be for y = 0

Then what would another path be that wouldn’t give you that answer?

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[lone heart]

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[cold mica]

can some one explain my confusion here. the math here makes sense but sin(-60)-sin(-60) = sqrt(3) so i just didn't cancel out the 2 so i end up getting sqrt(3)/(2pi) which is not the same as their sin 60 / pi coefficient

[cold mica]

oh actually i see my error now never mind

.close

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[zenith compass]

Can you teach me the method of finding parametric equations via point and the norm of the plane?

<@&286206848099549185>

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@zenith compass Has your question been resolved?

[zenith compass]

<@&286206848099549185>

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[jaunty glacier]

how to calc this

[alpine sable]

@jaunty glacier do you know how to calculate a magnitude of a vector

<a, 1> is your vector

What does it state on a graph?

[jaunty glacier]

it means its going up one

cuz the y is 1

[alpine sable]

And what about a

[jaunty glacier]

i dont know

oh is magnitude like length?

[alpine sable]

<x, y> means going x units on the x axis, and y units above

[alpine sable]

Precisely

[jaunty glacier]

ahh

[alpine sable]

It's the length of the vector from it's tail to the head

@jaunty glacier you'll see, a triangle will be formed

[jaunty glacier]

so a should be 2 right

[alpine sable]

No

[jaunty glacier]

oh wait

[alpine sable]

Magnitude of a vector is usually represented as $||\vec{v}||$

[ocean seal]

Pencil

[alpine sable]

Now

@jaunty glacier what would <a, 1> mean on a graph now

You told about 1 correctly

What about a

What does it denote

On a graph

[jaunty glacier]

the x axis

[alpine sable]

Elaborate please

[jaunty glacier]

a denotes how much the vector is moving along the x axis

[alpine sable]

Correct

If you're moving a units on x axis, and then 1 units upwards, how would you find displacement

The shortest distance

From the starting point

Till the ending point

@jaunty glacier i would prefer u to draw a rough sketch of this

To get an idea

[jaunty glacier]

am i supposed to apply pythagoras theorem or is there another way

@alpine sable

i thought if a is 2 the displacement would be 2

sorry my pc is potato..

sry for late reply

[lone heart]

@jaunty glacier Has your question been resolved?

[white jolt]

@jaunty glacier dude is a +-root3

Yeah a is root 3

[jaunty glacier]

Ahh

I see

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[alpine sable]

found the solution to part 1, but i am stuck on part 2

[alpine sable]

i think i need the reciprocal of the expression we found in part i

but what do i do after? my brain doesnt work atm

mark scheme kinda useless too

[wanton junco]

Well
the expression you found in part 1 is the derivative

so just plug in the points

do you see it ? @alpine sable

[alpine sable]

hmm

so points being 0?

or -4

[wanton junco]

which one you're referring to ?

you have 2 points here

But if your derivative is with respect to y, then it should be evident which coordinate to plug in

[alpine sable]

oo

[wanton junco]

ngl this is the first time i've seen such a take on derivatives lol

not including multivariable stuff obv

[alpine sable]

mannnn i dont get it 😢

multivariables sounds scary

this is what the mark scheme says

and like - im big dumb

[wanton junco]

yea seems right

what don't you get ?

[alpine sable]

basically all of part 2

[wanton junco]

part 2 is literally finding the slope at a point using the derivative

[alpine sable]

oooooooooooooooooooo

[wanton junco]

Consider the graph was reversed,
so you'd have y=(x+4)ln(2x+3)

and I told you to find the gradient at the intersection points with the axles

how would you go about it ?

[wanton junco]

this is part 1 ?

[alpine sable]

And 2

Wait I should've made it clearer

[wanton junco]

wait

Ohh

[alpine sable]

I'm hopeless 😭

[wanton junco]

this is what you had in mind of doing ?

[alpine sable]

Yeaaa

[wanton junco]

the reciprocal thing you did is all wrong

this is not at all how differentials work lol

[alpine sable]

ohh shoot

oh dang

[wanton junco]

you're marking the rate of change of a function in respect with y

not x

so all you need to do is plug in the y coordinate

[alpine sable]

OH WAIT

i think i u nderstand

I GOT IT

TY

YESS

it was so easy but my brain could not comprehend it

[wanton junco]

Notice that the linear equations of the tangents for this function would be in this form :

x=my+c

where c is the intersection with the x axis

and m=(d/dy)x(y_0)

[alpine sable]

oo yeas

i understand now

TYY

have good day - might come back with a different question later

lol

.clsose

[wanton junco]

np, gl !

[alpine sable]

tyy

.close

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[lone heart]

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[alpine sable]

in the answer key it says (a) y = 4.96x + 68.2 (3 significant figures)

I got something different

which is correct?

[gusty gorge]

I get y = 3.767x + 73.876

[alpine sable]

that's what I got

why does the answer key say differently?

[gusty gorge]

might be that they goofed one of the numbers

[alpine sable]

maybe

can another person double check?

I just want to be sure

[noble sinew]

yes its false, you can just compare the MSE of the two fits

34 vs 42 so the answer key is wrong

[alpine sable]

what is the MSE?

[noble sinew]

mean square error

[alpine sable]

sorry how do I check it?

[noble sinew]

by applying definition, but there is no reason to even check it

any software will do it correctly

and give you the least squared error fit

[alpine sable]

alright thanks

.close

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[sudden crater]

Any idea how to find the complementary function for the second differential equation in Explore 24.1?

[sudden crater]

Wait

I should probably substitute u= dy/dx and treat it like a second order equation

Would that work?

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@sudden crater Has your question been resolved?

[lone heart]

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[lime wadi]

sup y'all!

does anyone here who knows how to solve this?
thank you in advance ✨

[gray isle]

do you know the ivt?

[lime wadi]

n o

[gray isle]

try looking it up

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@lime wadi Has your question been resolved?

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[covert agate]

tag help after 15 mins

[alpine sable]

Sorry

[open folio]

What have you tried so far?

[alpine sable]

I tried corresponding angles with the 82

[wary stream]

It's pointless to delete a ping

[open folio]

What are you saying corresponds with 82?

[covert agate]

it isn’t

at least in my experience

[alpine sable]

Yeah, that part of angle d is 82

[open folio]

Chromium it means its a midsegment

[alpine sable]

Yeah

[open folio]

Which angle d is 82?

[covert agate]

i saw AB and BC

genius

[open folio]

Lol it's fine

[alpine sable]

The right one I guess? The one not inside triangle DBF

[open folio]

Do you know how to write angles?

[alpine sable]

Yeah

[open folio]

Then do so

Because I have no clue what you are saying

[open folio]

What makes you say so?

[alpine sable]

I thought it was corresponding

[open folio]

Well... is it though?

[alpine sable]

No

Actually wait

[open folio]

Ok I'll wait

[alpine sable]

Isnt angle DFB just 82? Because of alternate interior angles?

[open folio]

That would assume that FDE is actually 82.

[alpine sable]

Ohh right

Then I have no idea

I’m completely lost

[open folio]

You haven't tried anything with the 26?

[alpine sable]

Idk what to try with it

[open folio]

Your parallel lines are DE and BC so you can use those

Well what congruence statements can you make with the 26?

[alpine sable]

It’s corresponding

I believe

[open folio]

Yes

Also you may have overlooked something in the directions

"DF bisects angle BDE"

So can you post what you have down now?

[open folio]

Ok what do you think you should do now?

🙂

[alpine sable]

Find angle ADE

[open folio]

Yes

Exactly

[alpine sable]

I got 72

[open folio]

Yes

Ok now what should you do next

I think you got this now

[alpine sable]

Would I do 180-72 to get angle d, then divide by 2?

[open folio]

I don't know would you? 🙂

Keep going

[alpine sable]

Isn’t angle F and 72 corresponding?

The right side of F

[open folio]

No, also it's not necessary. What about that angle bisector?

[alpine sable]

Ummmm

180-72 to find one side of the bisector, meaning both sides of the bisector are equal?

[open folio]

72 is the angle ADE, you know that by definition of a bisector that angles BDF and EDF are congruent

[alpine sable]

Right

[open folio]

So what do you have now?

[alpine sable]

I have the same. I’m very confused

Sorry

[open folio]

Can you post a picture?

[open folio]

What is confusing you?

[alpine sable]

How to find angle BDF and EDF

[open folio]

You just said, they are congruent

I don't want to give away too much

[alpine sable]

Is jt alternate interior angles with 26?

[open folio]

No

Stop it

Lol

||They are angle bisectors, so just divide (180-72) by 2||

You cannot set up parallel lines cut by a transversal when the lines aren't parallel. DF and AC aren't parallel, you won't be using 26

[alpine sable]

Ok

So would I do 180-72 divided by 2

I think I said that earlier

[open folio]

Yes but I wanted you to figure it out on your own 😦

[alpine sable]

Ahh ok

Thabk you

[open folio]

Did you find the angle?

[open folio]

But its DFB you are finding?

[alpine sable]

Yeah

[open folio]

You are like 2 steps away

Can you see how?

[alpine sable]

Do I do 54+x= 180 and solve?

[open folio]

You are still missing angle B

So that won't work

You must get angle B first to use that

Can you see how to get B?

[alpine sable]

No, but is F 54 due to alternate interior angles ?

[open folio]

Yeah...

Lol

I was following with your idea

Because it also works

[alpine sable]

Ok

Thabk you sm

[open folio]

Sure

[alpine sable]

I appreciate it

[open folio]

Type .close if you are done

[alpine sable]

Ok thx

.close

[lone heart]

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[young spear]

Which channel is focusing on Digital signal processing?

[wary stream]

In my opinion, none, because that's a lot more related towards engineering vs a topic in math

[wary stream]

Because DSP is all about Fourier transforms and signal processing

[young spear]

Damn it, so no one can help here

[wary stream]

I mean, I know DSP, it's been a while

[young spear]

uhhh well idk just a simple question if you don't mind.

[wary stream]

Post it

I'll see what I can do

[young spear]

So i wanted to know how this guy got this figured out. How he got sin out that

or i can do one better tbh...

i'm tryna figure the steps for this question @wary stream

[wary stream]

I'm still looking

Give me a second

[young spear]

ight

[wary stream]

I actually don't recall this

Sorry

[young spear]

that's alright

it is hard

i don't blame you

thank you anyway

[wary stream]

I have notes, but it's been about like 2 years since I've dealt with it

Like I have these formulas but don't recall how to use it

[young spear]

https://youtu.be/fFyjc6ozy10?t=350 this guy said something about factoring which i just didn't understand how he just could do that

would you happen to know if there are any other discords that would be experts in DSP?

[wary stream]

You can look up engineering/stem related ones

[young spear]

Thank you so much for your help (thanks for trying at least) . I appreciate it alot

.close

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[whole wadi]

Anyone know how to tell which one is greater or lesser than the other (the fast way)?

[bleak ridge]

4*6^2 > 5^2

[lone heart]

@whole wadi Has your question been resolved?

[lone heart]

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[spark reef]

for part c would i evaluate the limits of f(x) and g(x) or f’(x) and g’(x)

because since F’(x) already has the derivatives wouldnt evaluate the anti derivatives for F(x)?

or is it just the second derivative

[lone heart]

@spark reef Has your question been resolved?

[indigo jetty]

limits of f'(x) and g'(x)

[white jolt]

I think so if it's continuous at 2 then dy/dx at 2 should be +

And it's also differentiable so it'll be greater than or equal to 0

[spark reef]

so i just use the two functions given at the top to determine it?

[indigo jetty]

yes

[spark reef]

okay i was just confused because it asks for differentiability at F(x) and the derivatives are in the piecewise function for F’(x)

[indigo jetty]

basically if F(x) is differentiable at x = 2, the limit of F'(x) as x tends towards 2 must exist

therefore the left and right limits of F'(x) as x tends towards 2 must be equal

[spark reef]

ohhh okay i see

yeah it’s not differentiable

thank you

[indigo jetty]

👍

[spark reef]

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[tame hound]

I need to find CD. Why is this calculation wrong:

12/20 = 40/AC
AC = 40*20/12
AC = 800/12
CD = 800/12 - ((12*40)/12))
CD = 320/12

[echo socket]

Why're you dividing AD by AC

It instead should be 12/20 = CD/AC

[tame hound]

De/ab = dc/ac

[echo socket]

Yes

[tame hound]

Oh DC ≠ 40, AD is 40

Is that where the problem is

[echo socket]

Yeah

[tame hound]

12/20 = CD/AC

[echo socket]

I'd recommend starting with this 20/12 = (CD + 40)/CD

[tame hound]

12AC = 20CD

[echo socket]

Because AC = 40 + CD

Yeah you can go that way as well

[tame hound]

How do I find Cd here

In my method

[echo socket]

12(40 + CD) = 20CD

Just solve for CD here

[tame hound]

20/12 = (CD+40)/CD
AC = 40+CD
12(40+CD) = 20CD
(480+12CD)/20 = CD

Hmm I’m supposed to solve this without a calculator

[echo socket]

You can do it by hand

480 + 12CD = 20CD
480 = 8CD
from your last line

[tame hound]

24+0.6CD = CD
24 = 0.4CD
24/0.4 = CD
60 = CD

[echo socket]

Yeah CD = 60 is the answer

[tame hound]

Okay thanks

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[whole wadi]

Anyone know how to tell which one is greater or lesser than the other (the fast way)? (Had to repost because I was gone)

[whole wadi]

Is the easiest route to write it like this?
(I) 9 + 4*36 - 25
(II) 9 - 4*36 + 25

There can be scenarios where it's more abstract with variables like a, b, and it says ”5>a>4” and ”6>b>5” and it's written similarly

[bleak ridge]

You know 4*36 is more than 25

In one of them you're adding big and taking away small

In the other you're taking away big and adding small

Which one would be bigger

[whole wadi]

The second one is a negative (4*36) and the first is a positive, and we also know 6^2 > 5^2, so I think the first one will be greater

np

Does any of these apply in this scenario?

[bleak ridge]

Yeah

c = 3^2

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[pale kestrel]

9 = 10-1 might help

[median mauve]

😮

[median mauve]

how does that help

[pale kestrel]

you write numbers in base 10

[median mauve]

you means this ?: {10-1, (10-1)^2, (10-1)^3, (10-1)^4, ..., (10-1)^N}

[little drum]

||ping me if you've got a clue how to do it|| I found a necessary condition {N log 9} > log 9 but that doesn't help at all

[pale kestrel]

like that.

i think we need the big triangle

pascal

[bleak ridge]

Pascal

[median mauve]

[bleak ridge]

$\floor{\frac{x}{10^{\floor{\log_{10} (x)}}}$

[little drum]

$\floor{\frac{x}{10^{\floor{\log x}}}}$

[ocean seal]

Ansh

[bleak ridge]

Yes that

Log being log base 10

[little drum]

doesn't help either, the best it does is give you ${k \log 9 } > \log 9$ where k is the exponent on 9 such that $9^k$ has the leftmost digit 9

[ocean seal]

Ansh

[little drum]

for example consider 9^22

so the fractional part of it is .99 > log 9

which verifies with the fact that 9^22 has leftmost 9

[bleak ridge]

,w plot y=floor(9^x/10^(floor(log(9^x)/log(10))))

[ocean seal]

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[bleak ridge]

Graph go up and down

[little drum]

Lol, the periodicity isn't defined either

I bet the question intends on doing something with the N and M together 🤦‍♂️

[tacit arch]

I like big triangles and I cannot lie, you other polygons can't deny

[pale kestrel]

play around the calculator

its fun

[little drum]

riemann gave leftmost digits of upto 9^100 terms lol

[bleak ridge]

Hold on

[little drum]

zero periodicity.. even if it looks like 9, 9^22, 9^44, 9^66, 9^88 follow the trend, you'd be hugely devastated when you learn that 9^(176) is a ...

[pale kestrel]

ah i see a pattern

prove it works though

[little drum]

what pattern? 👀

[pale kestrel]

idk i think i do

just stare

it probably relates to log

theres slowly more repeats

thats the pattern

wait no

noo why is it

665444

Near the end

[little drum]

imagine riemann somehow managing a typo towards the end

[pale kestrel]

maths is broken

the pattern is all the digits are between 1 and 9

[bleak ridge]

[9, 8, 7, 6, 5, 5, 4, 4, 3, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 9, 8, 7, 7, 6, 5, 5, 4, 4, 3, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 9, 8, 7, 7, 6, 5, 5, 4, 4, 3, 3, 3, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 9, 8, 7, 6, 6, 5, 5, 4, 4, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 9, 8, 7, 6, 6, 5, 4, 4, 4, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 9, 8, 7, 6, 6, 5, 4, 4, 3, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 9, 8, 7, 6, 5, 5, 4, 4, 3, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 9, 8, 8, 7, 6, 5, 5, 4, 4, 3, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 9, 8, 7, 7, 6, 5, 5, 4, 4, 3, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 9, 8, 7, 7, 6, 5, 5, 4, 4, 3, 3, 3, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 9, 8, 7, 6, 6, 5, 5, 4, 4, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 9, 8, 7, 6, 6, 5, 4, 4, 4, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 9, 8, 7, 6, 6, 5, 4, 4, 3, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 9, 8, 7, 6, 5, 5, 4, 4, 3, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 9, 8, 8, 7, 6, 5, 5, 4, 4, 3, 3, 3, 2, 2, 2, 2, 1, 1, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan, -nan]

Programming can't handle big number

[pale kestrel]

we have an answer

[bleak ridge]

Cool pattern though

Looks like ocean waves

[little drum]

(@_@;) what if the leftmost digit is 0

[bleak ridge]

[pale kestrel]

thats the answer

the leftmost digit is always 0

[bleak ridge]

it's true

[little drum]

why would they suggest $9N$ has M digits and not $9^N$ has M digits

[ocean seal]

Ansh

[little drum]

looks pretty fishy to me

[pale kestrel]

inequality time

[little drum]

$\floor{\log (9N)} + 1 = M$?

[ocean seal]

Ansh

[pale kestrel]

try inequalities not floors idk

ill have a go at this when i get up

[little drum]

$M-1 \geq \log(9N) > M - 2$

[ocean seal]

Ansh

[median mauve]

here is a table (maybe useless maybe helpful):
https://www.online-python.com/ZynGvzKc2p <-- magic python

[little drum]

🤦‍♂️

[pale kestrel]

could test this q in other bases

[bleak ridge]

,w plot y=floor(9^floor(x)/10^(floor(log(9^floor(x))/log(10))))

[ocean seal]

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[bleak ridge]

Nice

[little drum]

Find the number of solutions to ${k \log 9} > \log 9$ for $k \in [0, N]\cap \bZ$

[little drum]

okay got it

I mean -_- got the graph riemann.. How about you use the riemann brain here

[pale kestrel]

can we try the binomial approach?

is it that much of a dead end

[tacit arch]

nah i just suck at combinatorics

[lone heart]

@median mauve Has your question been resolved?

[median mauve]

[little drum]

wait for shuri.. shuri's defo gonna get the ans.

[pale kestrel]

thats a lie.

sleep feels better rn

[lone heart]

@median mauve Has your question been resolved?

[drifting hull]

as i found out when searching in the internet to get some ideas: This Problem was Problem 13 of the AIME 1990, you can find solutions there

[tacit arch]

wtf is this solution
Since $9^{4000}$ has 3816 digits more than $9^1$, $4000 - 3816 = \boxed{184}$ numbers have 9 as their leftmost digits.

[ocean seal]

riemann

[median mauve]

😮

https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_13

[median mauve]

not same but similiar

[drifting hull]

just replace 4000 with N and 3816 with M...

[median mauve]

how many digits are there for 9^N

[drifting hull]

i guess that the 9N in your setting is a typo

[median mauve]

you mean it is 9^N

[drifting hull]

yes

[median mauve]

and answer is N-M-1?

[drifting hull]

yes

[median mauve]

I don't get why it is N-M-1

[drifting hull]

did you read the solution on aops?

[drifting hull]

there is a longer one, too

Let's divide all elements of $T$ into sections. Each section ranges from $10^{i}$ to $10^{i+1}$ And, each section must have 1 or 2 elements. So, let's consider both cases.

If a section has 1 element, we claim that the number doesn't have 9 as the leftmost digit. Let this element be $9^{k}$ and the section ranges from $10^{i}$ to $10^{i+1}$. To the contrary, let's assume the number ($9^{k}$) does have 9 as the leftmost digit. Thus, $9 \cdot 10^i \leq 9^k$. But, if you divide both sides by 9, you get $10^i \leq 9^{k-1}$, and because $9^{k-1} < 9^{k} < 10^{i+1}$, so we have another number ($9^{k-1}$) in the same section ($10^i \leq 9^{k-1} < 10^{i+1}$). Which is a contradiction to our assumption that the section only has 1 element. So in this case, the number doesn't have 9 as the leftmost digit.

If a section has 2 elements, we claim one has to have a 9 as the leftmost digit, one doesn't. Let the elements be $9^{k}$ and $9^{k+1}$, and the section ranges from $10^{i}$ to $10^{i+1}$. So We know $10^{i} \leq 9^{k} < 9^{k+1} < 10^{i+1}$. From $10^{i} \leq 9^{k}$, we know $9 \cdot 10^{i} \leq 9^{k+1}$, and since $9^{k+1} < 10^{i+1}$. The number($9^{k+1}$)'s leftmost digit must be 9, and the other number($9^{k}$)'s leftmost digit is 1.

There are total 4001 elements in $T$ and 3817 sections that have 1 or 2 elements. And, no matter how many elements a section has, each section contains exactly one element that doesn't begin with 9. We can take 4001 elements, subtract 3817 elements that don't have 9 as the leftmost digit, and get $\boxed{184}$ numbers that have 9 as the leftmost digit.

[ocean seal]

Alexander42

[median mauve]

is that same idea as the first one?

[drifting hull]

that just the answer without any proof, the one i posted is more informative

[median mauve]

i feel i understand

[drifting hull]

👍

[median mauve]

ty

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[tribal temple]

Solve tan (x/2) = 2.1 in the interval for 0 < x < 360

[tribal temple]

Does any1 know how to do this question

[little drum]

the 3rd solution is more feasible but okay.

[little drum]

If you're allowed calculator, you can use inverse trigonometric functions appropriately.

arctan(tan (x)) = x ; for -pi/2 < x < pi/2

[drifting hull]

i only read the second one and it was fine, so i didn't look at the third one

[little drum]

You'll feel pretty disappointed when you'll read it though

||read on your own risk||

[drifting hull]

its basically the same argument, just formulated a bit different

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@tribal temple Has your question been resolved?

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@tribal temple Has your question been resolved?

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@tribal temple Has your question been resolved?

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[jolly parrot]

I got 7.5 = x but im not sure

5.3x-4.1x+2 = 3.6x - 1.6
1.2x-1.2x + 2 = 3.6x-1.2x - 1.6
2+1.6 = 2.4x-1.6+1.6
18/24 = 24x/24
x = 7.5

My bad I found my mistake

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[split oriole]

The whole angle sum is 360

Hope this can help

[alpine sable]

ohhh okay tysm

.close

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[cunning trout]

How would I draw the diagram?

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[lone heart]

@hybrid flare Has your question been resolved?

[hybrid flare]

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[slim sequoia]

hello

[slim sequoia]

i need help on this

*these

[white jolt]

3 a is easy right?

B1 is quadratic graph ..

[slim sequoia]

dang

what's 3a then ?

[white jolt]

What's the basic structure of a quadratic equations?

X^2-(a+b)x+ab

[slim sequoia]

ax2 + bx + c = 0

[white jolt]

Lad

Theres also x^2- (sumofroots)x+ multiplication of roots

[white jolt]

By this 3a Is very easy

Just do additional and multiplication of the roots of required quadratic

[slim sequoia]

ok 1 sec

what's the answer then ?

[white jolt]

Dude can't u even do that?

[slim sequoia]

mb bro im in a hurry but thanks

yes

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[meager fiber]

Hello I need help with game theory

[meager fiber]

I checked for PNE's and I got none so I think thats ko

ok

But I have no clue how to solve for mNE's with a matrix that is not 2x2

Normally when I solve for an mNE, I first calculate expected payoffs by doing: EU1 = prob[outcome1] * U1(outcome1) .... for all outcomes

but how can I assign probabilities to something like this??

[charred flint]

uh what do v and P stand for

[meager fiber]

v from VNM?

[charred flint]

yea

[meager fiber]

you can ignore that its some acronym for von neumann or something

but it wont change anything

P is pure

for pure nash equilibrium

[charred flint]

the equations should be the same with more variables

[meager fiber]

Heads and tails game for example I would calculate expected payoff by doing: EU1 = prob[HH] * U1(HH) ... for all outcomes
EU1 = xy*1 + ... etc

normally probabilities are like x and (1-x)

[charred flint]

yeah now it's p, q, 1-p-q

[meager fiber]

for player 1
and for player 2 its y and (1-y)

[meager fiber]

this is 3 probabilities so that would be for player 2 correct?

and for player 1 it would be like x, y, z, 1-x-y-z?

[charred flint]

oh right

yeah

[meager fiber]

hmm

ok ill try just expanding it out

and see what happens

also idk if you would know

what is the difference between these 2 methods??

I have no clue what he means by "some proper dimension all best mixed response"

[charred flint]

idk what "proper dimensions" or SC theorem mean sorry

[meager fiber]

alright np

I looked through my class notes and found nothing on it

and nothing on google

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[hollow carbon]

If a or b or c is equal to zero is it still a quadratic function or equation?

[little drum]

:p it's still a quadratic

but I'm not aware of the convention. I learnt that it's a quadratic with one root at infty. or smth like that

[pale kestrel]

a = 0

not a quadratic

[little drum]

it is

[pale kestrel]

not.

You are suggesting all linears and constants are quadratics

[little drum]

exactly one root at infty.

[pale kestrel]

They're not by the conventional definition

[little drum]

we're talking about quadratic-

and how coefficients affect the quadratic

[pale kestrel]

The convention is to define quadratics with non-zero x^2 coefficient.

[little drum]

Given a question: ax² + bx + c = 0 has exactly one root at infinity. Find all possible values for "a", are you gonna suggest it's not even a quadratic?

[pale kestrel]

???

Thats not relevant

I am not denying there are conventions besides the one I stated.

But what I said is how it's taught at almost all high schools

No need to confuse people.

[little drum]

"If "a", "b", "c" equal zero, is it still a quadratic?"

:o I feel like I'm being relevant here!?

[pale kestrel]

And I replied, if a = 0, it is not.

Like are all triangles quadrilaterals? You could consider them to be such if you like, but usually people do not.

[little drum]

mhmm. Got you.

@hollow carbon

In algebra, a quadratic function, a quadratic polynomial, a polynomial of degree 2, or simply a quadratic, is a polynomial function with one or more variables in which the highest-degree term is of the second degree.
So by definition, it doesn't remain a quadratic.
||however, there are questions regarding one or both roots of a quadratic being at infinity in which case, you need not confuse the curve with a quadratic||

[hollow carbon]

Oh sorry guys didn't mean to cause an argument.. thanks for all the help!!

[little drum]

was a positive one, dw. hope you got your answer

[pale kestrel]

A debate, just making sure everyones on the same page 🙂

[lone heart]

@hollow carbon Has your question been resolved?

[lone heart]

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[opaque totem]

Can anyone help, i can solve it using tree diagram but is there a faster way? my answer is B btw

[vale garden]

How many choice in the first digit

[opaque totem]

2

3 and 6

[opaque totem]

then how?

<@&286206848099549185>

[inner fable]

Looking at it 1 sec

[opaque totem]

okay

[inner fable]

https://www.ck12.org/c/probability/permutations-with-repetition/lesson/Permutations-with-Repetition-BSC-PST/

https://www.youtube.com/watch?v=hXkxh459nsE&feature=emb_logo

The zero also can't be in the first spot I assume? or is 0366 considered a 4 digit number?

[lone heart]

@opaque totem Has your question been resolved?

[opaque totem]

Ohhh

why 3 broo?

wait

[inner fable]

Why 3 what xD sorry 0366 was a random example

[opaque totem]

Ohhh

[inner fable]

Was just wondering if a "0xxx" counted as a 4 digit number

affects the question

[opaque totem]

Ohhh LMAO

erhh

soo

in counting

the first one only 2 possible right?

so 2

[inner fable]

Sure

if we're saying it cant be 0

[opaque totem]

yess

butt for next one

should i say

3 or

6?

[inner fable]

definitely 3

[opaque totem]

so 2 X 3 X 3 ?

X 3

wait

noo

2x 3 x 3 x 2

waitt

i confuse

[opaque totem]

u know how?

[inner fable]

It's technically 3 but if you want to calculate the whole thing you shouldn't look at it like that

[opaque totem]

Ouuu

[inner fable]

You should do it like the link I sent you, using the permutation equation for the top, then dividing the bottom by the repeating numbers

but I have to think about how to remove the 0 from the calculation

[opaque totem]

Ohhhh

i think shall watch it first

is it fine if i dm u ?

later

?

@inner fable

[inner fable]

Yes 🙂

[opaque totem]

Ahhh okeoke tyyy

.close

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[solemn tinsel]

can someone help me understand why the remainder is -x-2 and not x+2 ?? havent done polynomial long division in a while

[pine kettle]

$\frac{-x-2}{x^2+x+2}$ is the remainder

[ocean seal]

sills

[solemn tinsel]

oh yeah my bad i forgot the whole polynomial is the remainder

however why is the numerator negative

[pine kettle]

The numerator is just left over terms right?

Or whatever couldn't go into the denominator

[solemn tinsel]

yah

[pine kettle]

and $x+2$ is a part of $x^2+x+2$ so it has nothing to do with the numerator

[ocean seal]

sills

[solemn tinsel]

okay i think i get what u mean

can u give me a minute

and show u how i did the division real quick

so i did that

and i understand why the remainder is f(x)/g(x)

but why is the answer negative

rather f(x)

why does the (x+2) become -(x+2)

[pine kettle]

It doesn't

It's just leftover

Let me try this

[solemn tinsel]

im sorry if this are stupid questions but ive genuinely forgotten the gist of this haha

[pine kettle]

Yeah I don't know how to do this either

But let me try it

[solemn tinsel]

alright thank youu

im trying to look up videos but rlly havent gotten anything out of them :(, i dont understand whats special abt this case

[solemn tinsel]

did u do x^2 ÷ (x^2+x+2) or (x^2+x+2)÷(x^2)

[pine kettle]

I did your problem so the first

[solemn tinsel]

yeah i see that thank you

however i think i understand what my error was

[solemn tinsel]

EXACTLY

i divided it in the wrong order now i see that

thank you so much this was a silly mistake of mine

[pine kettle]

I want to explain box method 🥲

But I don't know how to

[solemn tinsel]

yeah this is a way of division ive not seen before

well long division yk

[pine kettle]

It's a way of multiplying polynomials that you can also use to do division

Since division is just multiplication backwards

[solemn tinsel]

ohhh i see

seems a lil more efficient that way ngl

[solemn tinsel]

so organized omg

[pine kettle]

So in the box method for multiplication, a 1st polynomial times a 2nd polynomial = the product polynomial, which is inside the box

If you start with the product polynomial inside the box as the product already there, and you know what one of the polynomial is, you can work backwards to get the 2nd polynomial

There we go, thats what I wanted to explain

[solemn tinsel]

ohhhhh i seee

thats actually so easy to understand omg

ty i had never heard of that method before

i'll start applying it in my problems, its a less messy than the regular method

ty sills!! i hope u have a good day/night :))

.close

[lone heart]

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[pine kettle]

@solemn tinsel If you didn't notice, it works for all the terms and stuff, ok bye

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[ripe sedge]

hi need help on this question

[blissful whale]

Just go ahead and post the question. 🙂

[ripe sedge]

ok

one min

there we go

@blissful whale

hello

[blissful whale]

I am going to bed soon so I can't walk you through it step by step.

[ripe sedge]

ok

can i ask someone els

[blissful whale]

Essentially you have to translate the english sentances to equations

[ripe sedge]

e

[blissful whale]

and then solve the equation system

[ripe sedge]

@alpine sable

[blissful whale]

If the distances from Adam to mia is called A, from Mia to Cindy M and so forth

"Adam was stadning 80m for Cindy"

corresponds to A + M = 80

[ripe sedge]

ok

one min

[blissful whale]

Since the distance from Adam to Cindy

[ripe sedge]

ok

[blissful whale]

is the sum of the distances from Adam to Mia and fro mMia to Cindy.

[ripe sedge]

and now that mia is 3x fo a+m=80

or soemthing like that

right

[blissful whale]

You just have to carefully look at what is stated

and use the right distance variable.

Going to bed now though, good luck 🙂

[ripe sedge]

ok

@alpine sable

[lone heart]

@ripe sedge Has your question been resolved?

[lone heart]

@ripe sedge Has your question been resolved?

[ripe sedge]

.closde

.close

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[wet spindle]

what's wrong here?

[sage violet]

whats sze original problem statement

[wet spindle]

also, pq>0

value of $p_1q_1+p_2q_2$

[ocean seal]

hyperlix26

[sage violet]

oh you forgot

210/pq = 31-pq

210 = 31pq - (pq)^2

[wet spindle]

ohh

yeah

right

alright I have no idea what to do now

[sage violet]

turn it into a quadratic

solve for pq

@wet spindle and zen finally manipulate

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@wet spindle Has your question been resolved?

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[wet spindle]

oh right

[wet spindle]

wait

[lone heart]

@wet spindle Has your question been resolved?

[wet spindle]

alright found it

$p_1q_2+p_2q_1=58$

[ocean seal]

hyperlix26

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[frosty flume]

did I solve for y^k correctly?

[frosty flume]

it was from this stackexchange

c2 = m * y^k
y^k = c2 * m^-1

nvm its correct

.close

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[frosty flume]

now how do I solve for y^k

im sure my y^k is wrong

idk how to manipulate mod p

anyone even here

[supple oasis]

yo

can you help me solve this out

idk the steps

@frosty flume

[frosty flume]

what

came to get help but got asked for help

idk btw im not good at math

[supple oasis]

oh

[gusty gorge]

if p is prime, then it works mostly like normal multiplication, so you're actually right

[frosty flume]

i just calculated and (c * m^-1 mod p) does not give the correct value for y^k

meaning that 2nd statement is derived wrongly

unless i calculated wrongly

[gusty gorge]

probably calcuating m^(-1) wrongly?

[frosty flume]

i dont think shifting mod p it remains mod p on the other side

[gusty gorge]

what do you mean by that?

[frosty flume]

p = 3
yk = 10
m = 5
c = (m * yk) % p  # 2
result = (2 * m**-1) % p # 0.4, this should be 10

result = yk

[gusty gorge]

m^-1 is not 1/m here

using python to compute m**-1 is not correct

[frosty flume]

i dont get it

if i remove mod p from both

and do the same thing

I get the correct answers

[gusty gorge]

okay but that's not how it works

look up "modular multiplicative inverse"

[frosty flume]

i dont see how

this is correct

[gusty gorge]

are you familiar with modular arithmetic?

[frosty flume]

like 0.5%

[gusty gorge]

people usually write something like

[frosty flume]

LOL

sorry

[gusty gorge]

$y^k \equiv cm^{-1} \pmod {p}$

[ocean seal]

Saccharine

[gusty gorge]

and this means both sides are equivalent mod p

now the thing about the numbers modulo a prime p is that they actually form a finite field

which basically means things like multiplicative inverses exist

and a multiplicative inverse $m^{-1}$ of $m$ is an integer such that $mm^{-1} \equiv m^{-1}m \equiv 1 \pmod{p}$

[ocean seal]

Saccharine

[frosty flume]

🤯

my head cant process this

lemme do another example without mod p

[gusty gorge]

okay stop

[frosty flume]

yk = 10
m = 5

c = m * yk  # c = 50
yk = c * m**-1 # yk = 10

this is correct

[gusty gorge]

"without mod p" is completely different

[frosty flume]

oh

[gusty gorge]

a modular multiplicative inverse is not the same as the standard inverse

in the field of the real numbers, obviously the multiplicative inverse of something like 2 is 1/2

because 2 * 1/2 = 1

but that doesn't work in the integers mod p

say we're looking at the integers mod 5

the modular multiplicative inverse of 2 is 3

[frosty flume]

so if these two sentences are consistent, given the values of c, m and p, how can I solve for y^k? I can't just "plug" them in

[gusty gorge]

because $2\times 3 \equiv 1 \pmod{5}$

[ocean seal]

Saccharine

[gusty gorge]

I think you should read up on the basics of modular arithmetic