help-23

[dense sierra]

yea

[rich moss]

I will try that

Nice I got f = 18

@dense sierra Thank you btw

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[dense sierra]

gald to help

[safe radish]

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[rich moss]

How do I find n

[dense sierra]

you can use the same way you did bro

[rich moss]

I tried but I got 6^2 = (5+n) x n

[dense sierra]

yes

thats correct

[rich moss]

my brother said that I have to plug into the quadratic formula

To get n= -9

and n=4

[rich moss]

(-b±√(b²-4ac))/(2a)

[rich moss]

@dense sierra

[dense sierra]

yep

n cant be negative so it must be 4

[rich moss]

Do I have to use (-b±√(b²-4ac))/(2a) to find the solution

[dense sierra]

yes

[rich moss]

But I could just do n^2 + 5n -36 = 0

n^2 + 9n - 4n - 36 = 0

[rich moss]

Proof: 9n - 4n = 5n. 9n x -4 = -36

[rich moss]

n(n+9) -4(n+9) = 0

cross n+9

Leaving me with (n+9) (n-4) = 0

I was just stick to this now and find the solution

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[wanton horizon]

In question number b am I supposed to used the differentiated equation or the the equation of the curve?

[frosty vigil]

the first curve

[wanton horizon]

First curve as in?

@frosty vigil

[frosty vigil]

the top one

[wanton horizon]

Oohh thanks for your help

[frosty vigil]

np

[wanton horizon]

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[kind dove]

how to do this using comparison test ?

[quasi bison]

do some root conjugate shenanigans on this to rewrite the general term as $$\frac{2}{\sqrt{n+2}(\sqrt{n+1} + \sqrt{n-1})}$$ then recognize that this is $\Theta(1/n)$ and so should probably diverge by comparison with $1/n$ (up to a constant) from \textit{below}

[flat frigate]

Ann

[kind dove]

can i do limit comparison in this case ?

if i remove constants from denominator i end up with 1/n

[quasi bison]

sure

if you want to do limit comparison then go for it

unless there's an explicit instruction that says to use something else

or to NOT use limit comparison

[kind dove]

when im doing limit comparison i can use any series ?

why we are comparing it with 1/n

[quasi bison]

presumably because that's the only thing you could reasonably go with for a limit comparison that wouldn't give you an inconclusive result

and because, again, your series is seen to be Theta(1/n), so that is what it makes sense to compare it to.

[kind dove]

what do you mean by inconclusive result ?

[quasi bison]

like if you get that a_n/b_n -> 0 but \sum b_n diverges

or a_n/b_n -> infty but \sum b_n converges

nothing can be said about a_n from that

[kind dove]

uhmm, ok thanks for explanation

.cloose

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[sacred quiver]

I have troubles differentiating this function

x is bigger than zero

[sacred quiver]

this is my work

<@&286206848099549185>

[safe radish]

@sacred quiver Has your question been resolved?

[sacred quiver]

okay

the biggest problem for me is to simplify this big fracture

[maiden jetty]

You can first simplify the numerator and denominator separately

[sacred quiver]

so for the numerator:

$\frac{1}{x^2}+\frac{1}{x^3\sqrt{1+\frac{1}{x^2}}}\newline \frac{x^2\left(x \sqrt{1+\frac{1}{x^2}}\right)}{x^2\cdot x^3\sqrt{1+\frac{1}{x^2}}}$

damn

[flat frigate]

Big xdddd

[sacred quiver]

$\frac{\sqrt{1+\frac{1}{x^2}}}{x^2\sqrt{1+\frac{1}{x^2}}}$

[flat frigate]

Big xdddd

[sacred quiver]

YOO I CAN

$\frac{1}{x^2}$

[flat frigate]

Big xdddd

[sacred quiver]

ok

nvm

i forgot one thing

$\frac{1}{x^2}+\frac{1}{x^3\sqrt{1+\frac{1}{x^2}}}\newline \frac{x^2\left(x \sqrt{1+\frac{1}{x^2}}+1\right)}{x^2\cdot x^3\sqrt{1+\frac{1}{x^2}}}$

[flat frigate]

Big xdddd

[sacred quiver]

$\frac{1+\sqrt{1+\frac{1}{x^2}}}{x^2\sqrt{1+\frac{1}{x^2}}}$

[flat frigate]

Big xdddd

[sacred quiver]

ok so

can i just divide by the square root

so i tried to simplify and i got

$\frac{1+\sqrt{1+\frac{1}{x^2}}}{x^2+x}$

[flat frigate]

Big xdddd

[sacred quiver]

damn you still there

ty for helping

[sacred quiver]

this one is wrong

because i took out the x

$\frac{1+x\sqrt{1+\frac{1}{x}}}{x^3\sqrt{1+\frac{1}{x}}}$

this has to be right

[flat frigate]

Big xdddd

[sacred quiver]

and can i simplify this more,?

ok

so now trying to simplify denominator

$\frac{1}{x}+\sqrt{1+\frac{1}{x^2}}$

[flat frigate]

Big xdddd

[sacred quiver]

this was the denominator

i can write this as

$\frac{1+x\sqrt{1+\frac{1}{x^2}}}{x}$

[flat frigate]

Big xdddd

[sacred quiver]

ommg

it worked

but

yea it worked

[safe radish]

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[pastel hare]

How did this happen?

I understand the product rule
x dy/dx + y
But how did 2y * dy/dx come?

[haughty jetty]

Chain rule

[pastel hare]

I know the chain rule but... I don't see how it got applied here..

[haughty jetty]

Let f(x) = x² and g(x) = y. Then f(g(x)) = y²

[flat frigate]

Si Arya

[pastel hare]

hmm

[haughty jetty]

So the outside function (f(x)) is x² and the inside function (g(x)) is y

[pastel hare]

ah I see

Ok thanks. 👍

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[silk lintel]

How do you get that answer

e-(e^-1)-(1-0)

Answer should be e-(1/e)-1

Where does that -1 go?

[winter moon]

$$ e^{-0} = e^0 = 1 $$

[flat frigate]

jaydamani

[silk lintel]

Why does it say that it is 0 lol

Prob some wrong

[worthy hemlock]

Because 1 - 1 =0

[haughty jetty]

He/she miswrote

[silk lintel]

Ye that's what confused me

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[lean otter]

Hello. i have a qeustion to simplify some sqrts

im stuck on one part that says 3sqrt30
i wrote down 9 bfr but going over it i dont know how i got it.... becuase you cant simplify sqrt 30.

[haughty jetty]

,w factorization of 30

[flat frigate]

Results provided by WolframAlpha

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

Requested by Si Arya#9554

[haughty jetty]

√30 can't be simplified because the power of every prime factors of 30 is 1

[lean otter]

yes thats what i see now. but im very confused as to how i got 9 before

3sqrt20 - sqrt50 / 3sqrt30

^ that is whole question

i put 9 as demoninator.

[worthy hemlock]

Then you did it wrong

[lean otter]

yep

im just sooo confused as to how XD

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[thin bridge]

which one of these are you stuck on

[hot star]

last 6

on the left side

[thin bridge]

so everything except f(-1)?

[hot star]

yes

[thin bridge]

how did you find f(-1)?
and what's stopping you from applying the exact same principle to find the values of the 5 following questions

[hot star]

i just evaluated it and got -f but im not sure what to do

[thin bridge]

f(-1) isn't -f

[hot star]

oh

[thin bridge]

in the context of function notation
f(-1) isn't the product of f and -1

you're given info about f(x),
(as well as the other functions)

to determine f(-1), you can simply substitute x=-1 into
f(x) = 2x - x^2

[hot star]

1

[thin bridge]

how are you getting 1

[hot star]

1,1 by evaluating and simplifying

[thin bridge]

what's 1,1 supposed to be

[hot star]

not sure

[thin bridge]

tell me exactly how your getting the values you're stating
leaving nothing out
typing stuff in full context

evaluating and simplifying
you said you did some evaluating and simplifying
show me exactly what you did

[hot star]

using photomath to solve and learn

[thin bridge]

well avoid photo math (because its crap)

and try attempting the task i requested of you yourself

to determine f(-1), you can simply substitute x=-1 into
f(x) = 2x - x^2

[hot star]

C?

[thin bridge]

how are you getting C

tell me exactly how your getting the values you're stating
leaving nothing out
typing stuff in full context

[hot star]

f(1)=2x1-1^2

[thin bridge]

why are you subbing in 1

[hot star]

x=1

[thin bridge]

when the question is asking for f(**-**1)

f( minus 1)

and when i also requested that you sub in x= **-**1

although you did correctly determine that f(1) = 1,
that wasn't what the question is asking for

also don't use x for multiplication especially not when the variable x is present

[hot star]

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[lean otter]

its not 12/24 but 12/36=x/(x+36)

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@lean otter Has your question been resolved?

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[finite linden]

My question is shouldnt question e be n=300sin(pi/3)t+550?

I dont get why theres 60

since 2pi divided by pi/3 is 6

[safe radish]

@finite linden Has your question been resolved?

[finite linden]

anyone?

[safe radish]

@finite linden Has your question been resolved?

[rough totem]

so actually pi = 180°. here it's ( sin pi/3 = sin 180/3= sin 60)

did u get it

[finite linden]

oh

i dont think i fully got it though

is it both right?

oh i get it now

so there is two ways

60 and pi/3

[safe radish]

@finite linden Has your question been resolved?

[shell raven]

I need to ask some basic electronics question doubts is there a server for that ?

[final halo]

#old-network

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[tulip gust]

Could you please explain the concept and the question? I have searched the web but it does not say

Thank you

[plucky elk]

http://homepages.math.uic.edu/~bpower6/stat101/probability examples.pdf

[tulip gust]

Thank you for the help :)

[plucky elk]

maybe a video is better?
https://www.youtube.com/watch?v=4kmGPVjnSPo

[tulip gust]

yes that helps alot

thank you :)

can I close the channel?

[plucky elk]

it's your channel. you're the boss

[tulip gust]

ok

thanks for the help

.close

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[merry spindle]

Can someone explain how/why they got (4)17 plz

[worthy hemlock]

Because all the sides are the same length

And there are 4 sides

[merry spindle]

Yeah ik where the 4 comes from by where did the 17 come from?

[worthy hemlock]

It shows it right above

Pythagorean theorem to find SR

[merry spindle]

Ohhh

Ok thank you

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[jade vessel]

S

[worthy hemlock]

What are you stuck on?

[jade vessel]

am I right?

[sharp sage]

use desmos

[jade vessel]

I feel like I'm wrong

[worthy hemlock]

Submit it and see

[sharp sage]

seems right tho

[jade vessel]

it's a practice quiz bro I'm tryna do the quesitons I'm having problems with

[sharp sage]

desmos will help u understand much better

[jade vessel]

kk thanks

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[marsh walrus]

well, whats f(2)?

[lean otter]

-5

[marsh walrus]

whats g(-5)?

[lean otter]

sorry i was in the restroom

[lean otter]

im not sure

2?

[marsh walrus]

well how did you find f(2)

repeat the process for g(-5)

if you cant, why not

whats different

[lean otter]

i looked what was next to it in paraenthesis

[marsh walrus]

so whats different now?

[lean otter]

-5 isnt there

[marsh walrus]

how do you interpret not seeing -5?

[lean otter]

i dont see 05

-5*

for g

[marsh walrus]

no i mean, what does this mean?

that you dont see it there

theyve defined g to be a relation involving those 4 coordinate pairs

but (-5, ?) isnt included in the definition

what does this mean in math vocabulary?

[lean otter]

not sure

[marsh walrus]

and i guess specifically why is it a problem if we want to compute, say, g(f(2))

well what do you call something that isnt included in a definition

why cant we do 1/x where x=0?

why do we require all sorts of special dealing with like

[lean otter]

undefined

[marsh walrus]

0!, or 0^0

yea

its undefined

g(f(2))=g(-5)

but g(-5) is undefined

[lean otter]

so if the pairs dont

[marsh walrus]

big problem

[lean otter]

if

a certain number

isnt in there

its undefined

[marsh walrus]

yea, if its missing from the domain

the function isnt defined there

usually with these, its implied i think that (x,y) gives x~y

so the domain is the first coordinate

well, the collection of first coordinates

if a number is not in the collection of first coordinates, the function is not defined there

so in your case, g is only defined for -3, -1, 0, 1, and 3

[lean otter]

thank you sir.

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[lean otter]

Can somebody explain me 40.21b?

[hexed geyser]

they've used the limiting sum of a GP formula in reverse here

wait

should be 1 - x + x^2 - x^3 + x^4 - ... right?

bruh

[lean otter]

Okay.

[lean otter]

Would you please elaborate this?

[hexed geyser]

yea sure i'll send formula

[lean otter]

Thank you 🙂

[hexed geyser]

if you choose a = 1 and idk sub r = -x you'll see what they've done

[lean otter]

Yes, understood.

[hexed geyser]

40.21b is wrong i think

i had a stroke reading it tbh

[lean otter]

Haha.

May I know why do you think it is wrong?

[hexed geyser]

it doesn't follow on from the first step

and doesn't lead to the next step

actually wait i think i see the mistake here

i think they wrote the wrong powers for fourth and fifth terms and forgot to write the "+ ..." to make it clear that this is an infinite series

[lean otter]

I see.

Thank you for clearing my doubt and helping me out 🙂 @hexed geyser

[hexed geyser]

np 😄

[lean otter]

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[royal ferry]

Hi! I'm really stuck on this statistics problem. I think I can figure out b and c, but a really stumps me.

[safe radish]

@royal ferry Has your question been resolved?

[royal ferry]

<@&286206848099549185>

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@royal ferry Has your question been resolved?

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@royal ferry Has your question been resolved?

[marsh walrus]

still around?

[safe radish]

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[safe radish]

Please don't occupy multiple help channels.

[fickle trail]

.close

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[rocky aspen]

why is the third one a contradiction? I dont understand why 5 dividing x^2 and then 25 dividing x^2 give a contradiction

[quasi bison]

x^2 = 0 mod 25 but 35 isn't 0 mod 25

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@rocky aspen Has your question been resolved?

[rocky aspen]

ohh right i completely forgot about the 35, thank you

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[safe radish]

@oblique nexus Has your question been resolved?

[oblique nexus]

<@&286206848099549185>

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@oblique nexus Has your question been resolved?

[final halo]

Can 5 vectors in R^3 be linearly independent?

[oblique nexus]

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[upper badge]

do you want to validate your selection? then yes.

[tribal fern]

okaii!

thank you

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[silk lintel]

How does 2x turn into 1?

On the top of the fraction

[sharp sage]

hmm

[winter moon]

Take a look at how dx changed to du

[quasi bison]

du = 2x dx bc of your choice of u

[silk lintel]

But how does it turn into 1 still

[simple current]

Read up on substitution for solving integrals.

This is a very fundamental concept.

[frosty grove]

1*du = du

[winter moon]

It doesn't it goes with dx as 2xdx becomes du

[simple current]

https://tutorial.math.lamar.edu/classes/calci/substitutionruleindefinite.aspx

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[wind siren]

Hey

[wind siren]

I found a0+a1x to be the mean of f_a with variance 1

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[wind siren]

Still in need

[wind siren]

<@&286206848099549185>

[plucky elk]

factor the exponent and identify the new mean and variance

[plucky elk]

share your work and someone can help

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[safe radish]

@spare tulip Has your question been resolved?

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[fickle abyss]

if i find the area under a graph for -x^2, would it calculate area outside or inside

[fickle abyss]

is it right that it's caculating the shaded area?

[plucky elk]

,w plot -x^2

[flat frigate]

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Requested by riemann#2529

[plucky elk]

i don't think this is -x^2, but whatever function it is, the shaded area is the correct depiction of the area from A to B

[fickle abyss]

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[soft glade]

Does the following limit admit a closed-form? limx→∞[8ex√xx+1(x−1)!−8x2−4xlnx−ln2x−(4x+2lnx)ln2π]

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[soft glade]

.reopen

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[soft glade]

.reopen

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✅

[soft glade]

Does the following limit admit a closed-form? limx→∞[8ex√xx+1(x−1)!−8x2−4xlnx−ln2x−(4x+2lnx)ln2π]

[woven anchor]

the problem is we can't read that

[stoic dune]

That's 100% a picture of the text

[woven anchor]

can't read as in you should write ^ for exponents, put proper brackets etc.

(or use latex)

[soft glade]

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[lean otter]

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[eternal sparrow]

Hello guys, I’m not sure how to find the first value of n

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[lean otter]

Can someone please check my answers for me

[lean otter]

This one too

[regal girder]

what is $R_1$?

[flat frigate]

Doggo

[lean otter]

Oh shit

[lean otter]

Is they right bro?

[regal girder]

I can't really help about this sorry

[lean otter]

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[mortal kernel]

These kinds of problems give me trouble :/

[regal girder]

what is the problem asking, formula for h?

if so, you can assume B != 0 and move everything but h to the other side

[mortal kernel]

Yes it’s for h

Move everything??

[regal girder]

multiply by 3, divide by B

again, this assumes B != 0

because we can't divide by 0

[mortal kernel]

I’m still confused

[lean otter]

Multiply both sides by 3, then, divide both sides by B, (just remember b not equal to 0 ), this way you isolate h

[mortal kernel]

So 3(v)=1/3bh ( 3) ?

[lean otter]

Thats the first step

But u didnt simply the rhs completely yet

[mortal kernel]

Yea the first step I got, idk how to write it afterwards :/

[lean otter]

Divide both sides by B

For step 2

[mortal kernel]

I mean like how do I write it out after I multiply by 3

3V = ??

[lean otter]

3v=bh

[mortal kernel]

Is it 3V = bh after I multiply?

[lean otter]

Yea

Genshin player moment😞

[mortal kernel]

Oh so it’s h=3v/b

I’m not good with problems with just variables in it

[lean otter]

Where b not equal 0

[mortal kernel]

Idk what that means

[lean otter]

Try to put b=0

And calculate the rhs

😈

[mortal kernel]

So it’s not h=3V/b ? Cause it is one of choices

[lean otter]

It is

But under the condition that b not 0

[mortal kernel]

That confuses me lol

[lean otter]

Yes thats why

Try putting b=0

Ane calculate what h becomes

[mortal kernel]

I’m still confused

[lean otter]

Well do you understand that

V,b,h are variables

[mortal kernel]

Yes that I understand

[lean otter]

And if we assume v,b,h can only take values of real numbers

Then technically you can assign h,v any value but b can be anything but 0

But you cant assign all of 3 of them random values at the same time unless for those values the equality holds💀

[mortal kernel]

See that sounds complicated to me

[lean otter]

😞 me too

[mortal kernel]

Especially if it’s only letters & no numbers

[lean otter]

Numbers are also letters in a way

[mortal kernel]

Oh yea 1

[lean otter]

Why is 2 written the way it is written

If i wanted i could have made the symbol 3 equal to the value 2

😡

[mortal kernel]

Lol okay that sounds more complicated to me

Hopefully I remember how to do it during my exam

[lean otter]

Oh no dont do this in an exam

😨

[mortal kernel]

I mean the original problem I needed help with

Thank you though!

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[spice raven]

yr 11 exams coming up, trying to understand most of the difficult question areas.

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[spice raven]

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[full fiber]

@shadow glade do you mind helping me again? my original question was part of an approach that i tried, but it didnt work. i want to know why my approach doesnt work

what i did was take the area of the small circle divided by the intersection

because the intersection is all of the points that it can be; a side with 1, and two ranging from 0 to 1

[shadow glade]

how about using law of cosines with c = 1? that's my initial instinct

[merry sleet]

i dont even get this, how is a random number a point in the plane

[shadow glade]

you would need gamma > 90

[full fiber]

side lengths

[merry sleet]

oh ok

[full fiber]

does this have to do with my diagram or is something else?

[winter pivot]

Hint: rewrite γ in terms of a and b

[shadow glade]

seems to be the trend in math teaching these days, they dont bother with precise wording, they should say "let these two numbers be the side lengths of a triangle...blah blah" you are correct in saying that these numbers shouldn't be equated with being the sides themselves

[winter pivot]

Ι personally don't see the problem but whatever

[full fiber]

^

[shadow glade]

it's not the worst : )

[merry sleet]

for an obtuse triangle, you have simply that if c is the longest sidelength and a b are the two other ones, a²+b²<=c².

In your exemple, 1 is always the longest side, so your condition should be a²+b²<=1

[winter pivot]

Do this

[winter pivot]

This is better actually

[shadow glade]

yeah thats a lot better

[full fiber]

im confused

[winter pivot]

I agree that the wording is imprecise now

What does between mean

[half cliff]

graph x^2 + y^2 <=1. what do you see?

[shadow glade]

lol

[winter pivot]

Is that an open interval or a closed interval

[full fiber]

inclusive pretty sure

[merry sleet]

doesnt matter, the borders have measure =0

[full fiber]

circle

[half cliff]

yes

yes

[winter pivot]

It's not a circle

[half cliff]

yeah but epiconan has the idea

[full fiber]

filled in circle

?

[half cliff]

pay attention to only one interval of the circle

0 < x < 1 and 0 < y < 1

because thats where x and y can be sleected from

[winter pivot]

Ball

[full fiber]

so quarter circle?

[half cliff]

yeah

[shadow glade]

you could think of the possible picks that are valid as the quarter circle in quadrant 1, and all possible picks as the 1x1 square from 0,0 to 1,1 so then take the ratios of the areas of those two for your percentage

[half cliff]

this is all the possible values of x and y

yeah what sooshon said

[shadow glade]

are you using mathematica for these graphics by the way or what is it?

[half cliff]

desmos

[shadow glade]

aha

[full fiber]

x and y are the two side lengths right?

that means x + y > 1

[half cliff]

that produces any valid triangle

but not necessarily an obtuse triangle

[full fiber]

obtuse would the purple

right?

[half cliff]

yes

[full fiber]

because it satisfies both obtuse rule and triangle inequality rule

ok

[shadow glade]

ah, true

i like this solution

[half cliff]

all that is left to do is to find the area of the purple part, because all the possible outputs can be represented as a square with area 1

[winter pivot]

Ok so now let's set this up as a double integral

[shadow glade]

lol

[half cliff]

do you really need to?

[shadow glade]

i think it's a joke

[half cliff]

its just a circle and a segment...

[shadow glade]

pi/4 - 1/2 right?

[half cliff]

and the problem asks for a final answer of 4p

so just π - 2

[full fiber]

thanks this makes sense

[half cliff]

np

[shadow glade]

oh that was originally meant as a solution to this problem? interesting

[full fiber]

i tried to take the circle divided by the intersection

[full fiber]

yes

[shadow glade]

what was your reasoning behind that idea? i dont quite follow

[half cliff]

wait were you trying to find the area in between the 2 circles?

[full fiber]

yes sooshon and some other ppl helped me in another channel. i thought that would lead to the solution to this problem but it didnt

[half cliff]

because you do have a correct approach(the drawing)

[full fiber]

so, the intersection represents all of the possible triangles right?

[half cliff]

well

[full fiber]

because the radii are 1, so any point inside of them will be between 0 and 1

[half cliff]

trying to find the set of all triangles that work is a different problem

while this might seem annoying, you can consider one half of the intersection

really bad drawing

[shadow glade]

im just really confused by the whole idea

😄

[full fiber]

my reasoning is that the set of all triangles would be in the denominator of the problem (bc it's asking for probability)

and then to find the ones that are obtuse, i take the ones inside of the small circle

or does all of this not make any sense

[half cliff]

i thought you were trying to find the area of the intersection of circles

[full fiber]

we found that

[shadow glade]

we did all that earlier lol #help-16 message

[full fiber]

[4pi - 3sqrt(3)]/12

[half cliff]

you don't need to worry about cases of triangles for finding area

[shadow glade]

he didnt present it in the context of this problem back then, just as a geometry problem

[full fiber]

the probability that it is a obtuse triangle is:
area that leads to obtuse / total area

[shadow glade]

but why does that intersection of the 2 circles represent the probability that it is obtuse?

[full fiber]

do you mean the small circle?

[half cliff]

there isn't a small circle in the diagram

[full fiber]

intersection of two circles is total area right?

[shadow glade]

well what area are you claiming represents the probability that it's an obtuse triangle

and what total area are you referring to?

[full fiber]

sorry for being unclear

here

blue circle = obtuse area
red intersection = total area

[shadow glade]

and the argument behind that is what?

behind it isnt obvious to me 😄

[half cliff]

are you talking about all the triangles with its base as this line?

[shadow glade]

ah ok im kinda starting to see how you were thinking about this

[half cliff]

i think you might be trying to say that the if the 3rd vertex of such a triangle is within(not on) the blue circle, you are right

but does the red area really represent all triangles, regardless of what type?

[full fiber]

the two center points in the two triangles are two points of the triangle, and are a distance of 1 away. anywhere inside the red intersection refers to a possibility for a triangle, side lengths wont exceed 1.

the reason the blue circle represents obtuse area is that um....i know if it is on the blue circle it is 90, and it seemed intuitive that if i go inside then it will become obtuse

[shadow glade]

yeah im buying this now

[full fiber]

yes yes yes

[shadow glade]

i think

[full fiber]

mb

[full fiber]

all triangles in the bound of this problem, because side lengths 0 to 1

[shadow glade]

wouldnt the solution be just half of the small circle though, since the side lengths on the bottom of the circle would just be reflections of the side lengths on the top right?

like these two triangles would have same side lengths

[full fiber]

wdym

[shadow glade]

but i guess ratios of areas would be same if youre looking at just half or whole thing so never mind, doesn't really matter

[full fiber]

yeah

[half cliff]

i think one problem is that all points in each circle don't equally map to the red square/quarter circle in the plane

which messes up the areas

[full fiber]

is it missing area or have extra area?

[shadow glade]

by the way do you have access to the correct answer to this problem? was our previous solution correct?

[full fiber]

yes the answer is pi - 2 and you guys got it correct

in fact the solution presented (on aops) is the same as yours

[shadow glade]

im intrigued by this two circle thing idea though

it would be quite interesting and seems like it should work

[half cliff]

send me the link to the aops problem, I still don't see why the red area isn't 2/3 * (pi - sqrt(3))

or at least show a screenshot

including the original problem

[half cliff]

wait that was the original problem?

[full fiber]

yes

it's from alcumus so quality is not that great

[half cliff]

oh so your strategy was to do the 2 intersecting circles

[full fiber]

yes

[shadow glade]

this is what we got for red area in our last channel

[full fiber]

although the incorrect answers i answered were something else

if i were to do blue/red i would get this:

[half cliff]

i think the way you randomly pick triangles ends up being different than picking the sides separately

[full fiber]

hm

how come

[half cliff]

as i said before, the red circle doesn't equally map to the red square

[full fiber]

do you know that because the areas are different?

are you saying that if i did it correctly, the red area shold simply be 1

[half cliff]

take the set of triangles such that x<1 and y<1(red circle as well as red square)

[half cliff]

no, the red area in the circle woudn't be 1

[full fiber]

quick question whats the equation to make a square on desmos

[half cliff]

oh

i convered up something to avoid confusion

[full fiber]

ik im doing something wrong

[half cliff]

i just put x>=0 as a filler equation

that colors everything

[full fiber]

ohhhh i see

[half cliff]

back to what i was saying

if you geometrically try to create the set of triangles by using circles, it won't work because the drawing causes x and y to be dependent on each other

if you use the coordinate plane, x and y won't be dependent on each other if you try to calculate probabilities

at this point what i am saying is getting less and less comprehensible for me, so I won't try to go into the details(i have the intuition that it is true, but i don't know a rigorous proof)

[full fiber]

but each circle has it's freedom to be any length they want though

how would they be dependent

for every single length of circle one there is every single length of circle two

[half cliff]

well ok

when you are using the circles, you are randomly generating the angle

not the sides

and going from sides to angle doesn't produce a perfect linear mapping

so areas get messed up

[full fiber]

very interesting

[shadow glade]

i dont know about that, this feels like it should work

[half cliff]

its like trying to compare cos(x) and x ig?

when you take cos(x) some output areas are more dense than others

[shadow glade]

i dont know why you think you are "generating an angle", every point in the diagram corresponds to some chosen length of one side and a chosen length of the other side

[half cliff]

true

[shadow glade]

i dont see it as fundamentally different than the previous solution

[half cliff]

i wish i could animate like 3blue1brown lol

[shadow glade]

😄

[half cliff]

if you take 2 pieces of the red circle and map them to 2 pieces on the coordinate plane, those 2 pieces could have different area in the plane

[full fiber]

@half cliff if you were to first pick the length of first side, you will have a circle of different possibilities. then, when you make the third side, you can make any side length for the third one

[half cliff]

yes

[full fiber]

doesnt that mean that they arent dependent

[half cliff]

it does cover all points

[half cliff]

sorry for that

that was bad reasoning

but now i realize whats really going on

[full fiber]

then what is the problem? i dont fully understand

[half cliff]

should i make a diagram in desmos?

it will take me a while

[full fiber]

your choice

[shadow glade]

now im playing with geogebra website : )

they have a nice straightedge and compass construction tool

[half cliff]

bruh desmos doesn't allow nested lists

looks like i have to do this on python matplotlob >:(

[full fiber]

just making sure that this solution is actually wrong, the area of the blue circle is pi/4 and area of red intersection is [4pi - 3sqrt(3)]/6

right?

and probability is just blue/red

[full fiber]

what are you trying to do with lists?

[half cliff]

im trying to show how all the points get mapped from the red circle intersection to the square

[full fiber]

could it be possible that the red part is correct, but it's actually the blue circle that is wrong?

because my reasoning for that was very weak

[shadow glade]

im willing to bet theres a way to solve it using this idea, but we just aren't seeing something

but im gonna call it a night for now, if you ever solve it using this please @ me because i find it interesting

good luck : )

[full fiber]

good night

[safe radish]

@full fiber Has your question been resolved?

[half cliff]

this is a bad graph

[full fiber]

what are your conclusions?

[half cliff]

but its pretty clear that these points won't perfectly map to these points:

the reason why you get different areas is because some parts of the circle have different point densities, so taking the area of some part circle will actually contain more or less possibilities than it should, while taking a section of the square does correspond to reandomly choosing x and y

i think the circle is euqivalent to choosing the angle ut im not sure

either way it gets messed up

@shadow glade

the red points in both graphs represent x and y from 0 to1 (intervals of 0.1)

[full fiber]

i see

[full fiber]

what do you mean by probability densities

[half cliff]

i meant point densities

in other words there are more points in that area

[full fiber]

oh ok.

why is there that gap in the graph?

[half cliff]

idk

[half cliff]

i think i messed up something in my code

either the intervals are not good enough or whatever

but i think most of those points are "non"-triangles

so don't worry about that

[full fiber]

i think the densities are the same though

[half cliff]

focus on the points not stuck on the x axis

some are slightly more packed than others

[full fiber]

like where

[half cliff]

middle vs top

[full fiber]

i dont rlly see a difference but ill just take your word for it

but even if there is, why does it matter

[full fiber]

can you elaborate on this

[half cliff]

do you understand what it means to map one set of points to another?

[full fiber]

no

[half cliff]

...

than i don't really know what to do

hope sooshon can explain the concept

its getting late for me too

[full fiber]

well thanks for your effort

good night

.close

[safe radish]

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[half cliff]

one final thing. there are more points to help now. in the circle graph, the points aren't evenly spaced, but they are in the square one(the graphs are stretched). you only get the true probability if you calculate the area of a set where possibilities are "evenly spaced". As you take the limit to 0 between for the space between adjacent points, it should be the same for all points when you use an x and y axis, but not with the circle method.

[half cliff]

@full fiber

bye

[full fiber]

tysm

good night

[half cliff]

gn

[safe radish]

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[somber spire]

Hi, I know that the number of compositions of an natural number is 2^(n-1). Given an array = 1,1,2,2,4,4,...,2^k,2^k (k = 0,1,2,...,26). What is the number of compositions of an natural number representing by this array ?

[delicate aspen]

the way you have phrased this is not super clear

send a photo or a screenshot of the question please

is this array supposed to be a composition of the some number?

[somber spire]

Sorry, let's me write again.

For any non-negative k = 0,1,2,3,…. There are exactly 2 coins of value n = 2^k. For a given M, how to calculate the number of different ways that can represent M with the coins?

For example, M = 6 then there are 3 ways {1,1,2,2},{1,1,4}, and {2,4}

[delicate aspen]

So, you're saying that there are exactly 54 coins in total

just to make sure i understand

[somber spire]

No sorry, I calculate wrong. As M is up to 10^18

[delicate aspen]

well 2^26 is very big

[somber spire]

So I am trying to find a recursive way to calculate this by a program.

[delicate aspen]

the first thing you should do is determine what the biggest coin you can use is

oh wait, recursive?

[somber spire]

Yes

[delicate aspen]

you want to start doubling M and looking at how the partitions change

lets look at M = 3

M = 3 only has 1 way, {1,2}

so for M = 6, we take double of those coins, {1,1,2,2}, but we can also pair up the two 2s to get {1,1,4}. now that there are 2s available, we can also pair up the 1s

i think actually starting with odd numbers is going to be the best

because we have to use exactly one copy of 1

then our hand is forced

for example, M = 7

we need to make 6 with at most 1 copy of 1

i might now be being very clear but i hope im getting some of my idea across

[somber spire]

Yeah I get it somehow

[delicate aspen]

i have a feeling there is something special happening at 3 and 7

*1,3 and 7

this is really interesting problem

[somber spire]

feel like I get something but still not

[delicate aspen]

theres actually more recursion happening

my best advice is to start writing down all the partitions for values of M starting at 1, and pick an ordering that you write them in so that you have a system, and dont miss any

[somber spire]

ok I will try it

Thanks

.close

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[remote flint]

i’m not exactly sure what “From a point 35m from the base of the building and level with the base” means

[hexed geyser]

level with the base basically just means that its on the same horizontal plane

[simple current]

Exactly what it says on the tin.

[hexed geyser]

like its not on a hill or smth

[simple current]

Walk from the base of the tower to a point 35 m away.

And that point is at the same height as the base of the building.

[obsidian quartz]

im gonna have a math test soon, can anyone help me?

[hexed geyser]

#❓how-to-get-help

you gotta read this before asking for help

[obsidian quartz]

aight

[remote flint]

also, what does angle of elevation of the top is 72.5 mean

[hexed geyser]

lmao he answered ur question before you even asked it

[shadow glade]

: )

[misty tartan]

Bro someone pls help

[remote flint]

thanks

[misty tartan]

Me

[remote flint]

.close

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[somber spire]

.reopen

[lean otter]

@violet rain whats the question?

3x isn't multiplied by 5

.close

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[safe radish]

@hollow trout Has your question been resolved?

[safe radish]

@hollow trout Has your question been resolved?

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[stray ridge]

How to calculate the base instead of the exponent using logarithm?

[quasi bison]

,rccw

[quasi bison]

i can't read the LHS of this

$g \log(\frac{1}{\theta}) = -3$?

[flat frigate]

Ann

[stray ridge]

Aah

G log (1/8)

= -3

[quasi bison]

that is an eight???

[stray ridge]

Yea

[quasi bison]

does not look like an eight at all

but whatever

i am assuming that g is supposed to be the base?

[stray ridge]

Ues

G os base

[quasi bison]

so you wrote $g^{-3} = \frac{1}{8}$

[flat frigate]

Ann

[quasi bison]

which is correct based on what you've said so far

[stray ridge]

Yes

[quasi bison]

and what is troubling you from here?

[stray ridge]

How to calculate g

[quasi bison]

if it was raised to a positive power would you know what to do?

[stray ridge]

Remove the negative and the 1?

[quasi bison]

no, not what i'm asking

let me give you a similar but different equation:

$z^5 = 92$

do you know how to solve this equation?

[flat frigate]

Ann

[stray ridge]

No

I only know how to calculate the exponent

Not the base

[quasi bison]

so you have not encountered things such as roots?

or fractional powers?

[stray ridge]

I have

I just haven’t had to calculate the base before

[quasi bison]

...

???

so you've never solved any equation like z^5 = 92

[stray ridge]

No

[quasi bison]

that's strange

because it contradicts your claim that you know what roots are

[stray ridge]

I’ve started logarithm like 3 days ago

[quasi bison]

otherwise you would've had no trouble going from $z^5 = 92$ to $z = 92^{1/5}$

[flat frigate]

Ann

[stray ridge]

Aah

I know that root of x is x^1/2

Or 1/2*x is squareroot of z

x

Cube root is 1/3

Oh

So

,rotate

[stray ridge]

So g is 2?

2^3 is 8

2^-3 = 1/8

Is this correct?

,rotate

[quasi bison]

yes

[stray ridge]

Thank you 🙏

[safe radish]

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[light bridge]

Just got another quick question, Im trying to calculate the adjacency matrix of a 3x3 matrix that is

[light bridge]

[0 1 0]

[0 0 1]

[0 1 0]

but like is the reachability of this matrix calculatable?

as M2 is just the same thing?

nvm

.close

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[safe radish]

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[spiral crescent]

I dont understand 😞

[rocky gazelle]

,rccw

[spiral crescent]

Hello do you know?

<@&286206848099549185>

[safe radish]

@spiral crescent Has your question been resolved?

[safe radish]

@spiral crescent Has your question been resolved?

[final loom]

Is this resolved?

Given: $\sqrt{-(x - 6)(x + 4)} \geq \sqrt{-x(x - 8)}$

[flat frigate]

Arya

[final loom]

So first of all you're supposed to make the terms subject to the "square roots" into non-negative (as you're clearly not dealing with sqrt of negative terms)

[spiral crescent]

No both are not done

Im asking where is the mistake in my process

[final loom]

,rotate

[final loom]

oh

that

that's because you got a bit hasty

6x - 24 ≤ 0 and not ≥ 0

@spiral crescent

[spiral crescent]

Where is this on the solution, i dont remember exactly ☠️

Wait let me open the copy

[final loom]

hahahaha

that's the third line lol

[spiral crescent]

Ah i divided by

6

☠️

[final loom]

hmm? yep you still have the inequality sign wrong

[spiral crescent]

Ohhhh

Wtf

[final loom]

lol

Gg, you probably took care of the rest ig

[spiral crescent]

And the other one?

[final loom]

hmm lemme see

[spiral crescent]

Yes i got the correct ans for the 1st one now

[final loom]

_< I love it when beginners make these cute unnecessary steps.. I have an advice if you're willing to listen :p