help-39

[hot timber]

wait

[hollow sonnet]

maybe draw a probability tree

[hot timber]

I think you have to make a progression of probabilities

[eager jewel]

yes

^^^

[hot timber]

oh yeah

[eager jewel]

huh so we multiply by (1-1/2)?

[golden bramble]

yeah in addition to this

since if A gets it, there is no next turn

[hot timber]

a first try probability is 1/2. for B it’s 1/6,

[eager jewel]

and we also have to multiply (1/2) for A to get it on the 2nd turn right

[hot timber]

wait I think I blundered

[golden bramble]

yeah

[hollow sonnet]

1/3 for B I think

[hot timber]

if they roll at same time

[hollow sonnet]

C is 1/6

[eager jewel]

ok so
1/2 + (1-1/3)(1-1/6)(1-1/2)(1/2) + ...

[golden bramble]

it is 1/6 for B first try, since (1-1/2)(1/3)

[hollow sonnet]

they say its rolled in order of A,b,c

[golden bramble]

yes, then what is the common ratio?

[eager jewel]

2/3 x 5/6 x 1/2 = 5/18

[clear shore]

You said the answer is 9?

[hot timber]

this gives me motivation to recap my sequence and series chapter

[hot timber]

Allen eh?

[clear shore]

|| I found that P[A] = 1/2 and P[C]=1/2*4/6*1/6 give exactly 9||

[hot timber]

Aight so

[clear shore]

the fact that after each round the probablity reset

[golden bramble]

ok sum it up then. Do you recall the sum of an infinite geometric series?

[eager jewel]

ye ye a/1-r

[hot timber]

for C a = 1/6*2/3

[clear shore]

So we might or might not need to consider doing sequence

[eager jewel]

ok that gives me
P(a) = 9/13

ok so now lets do P(c)

[hot timber]

Do the same for c

[eager jewel]

ye first term is (1-1/2)(1-1/3)(1/6)

[hot timber]

the question is actually very good

[eager jewel]

2nd term is (1-1/2)(1-1/3)(1-1/6)(1-1/2)(1-1/3)(1/6)

[golden bramble]

ok

you should be able to tell what the common ratio is

[eager jewel]

common ratio = (1-1/6)(1-1/2)(1-1/3) again??

[golden bramble]

yep

[iron basin]

bingo

[eager jewel]

OH OK YES

got it

[hot timber]

The r is equal for them both?

[eager jewel]

P(c) = 1/13

[hot timber]

Yessir!

[golden bramble]

yes

[clear shore]

xd

[eager jewel]

this shit only makes sense after the exam bro

[golden bramble]

yep, and that's why Hunter's spoiler works, since the 1-r cancels

[hot timber]

I have one tomorrow 😭

[eager jewel]

all the best

[golden bramble]

but I had them find the actual sum just in case the actual probability is asked

but for the ratio you could shortcut 😛

[eager jewel]

o

[hot timber]

it’s ez tho (binomial and PnC)

[eager jewel]

not easy under exam pressure 😔

[hot timber]

ehh

[eager jewel]

well thank you everyone for helping

.close

[pearl pond]

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[clear shore]

Yk I believe this argument also true, P( A roll a prime number | no one win)P(no one win)=P(A) and P( C roll a 1, A and B didn't win | no one win)P(no one win)=P(C)

[pearl pond]

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[loud viper]

I figured out p=2 so the question boils down to finding lambda such that lambdax²-4x+9 has two integral roots

[eager jewel]

find disciminant

[hot timber]

Discriminant?

[loud viper]

Wait

[hot timber]

Find the minima/maxima of the lambda such that discriminant >= 0

[loud viper]

Discriminant=16-36lambda

What to do further?

[hot timber]

substitute discriminant = 0

find lambda

[loud viper]

Lambda=4/9

[hot timber]

that’s the maximum value

Wait

[eager jewel]

why discriminant = 0

[hot timber]

wait the question asked for the sum

[hot timber]

To find maximum value

[eager jewel]

how do u know roots are equal

[tardy reef]

you want the roots to be integers, discriminant = 0 doesnt help there

[loud viper]

Why do u need to find max value of lambda?

If lambda was also integer then I would have done it

[eager jewel]

@loud viper ok now u have
x= 4+-sqrt(16-36lambda)/2lambda

now its given both roots are positive

[loud viper]

Yes

[eager jewel]

and also discriminant is an integer

[loud viper]

How?

[eager jewel]

cause its given roots are integers

[loud viper]

It could be something like 4/9,16/25

[hot timber]

oh wait I overlooked that

[eager jewel]

will roots be integers then tho

[loud viper]

Maybe

[tardy reef]

One thing I would suggest is, since you know that the roots are positive integers, look for the values of lambda that would allow the sum of roots of the quadratic to be integers and work from there

[brazen hare]

sum = -b/a, multiply = c/a

apply yhat?

[eager jewel]

but it says theres only 1 value of lambda

[hot timber]

We don’t know a

The answer is 4/lambda

eh

[brazen hare]

no see

integral

so 9/lambda should be an integar too

and its only possible if

[hot timber]

wait

[loud viper]

Lambda=p/q p,q coprime and p divides 9

[brazen hare]

sum is 4/lamba

[loud viper]

p divides 9, p divides 4

p=1

Lambda=1/integer

Oh! I got something interesting from that

[brazen hare]

no

apply what he was saying

maximum

[eager jewel]

@loud viper do u have the answer to this q ?

[brazen hare]

.

[loud viper]

12

[brazen hare]

solutions are 3 and 9

[loud viper]

And lambda=1/3

[brazen hare]

roots

[eager jewel]

ohk yeah got the same thing...i took sqrt(D) = 2

[loud viper]

But we have to prove that

[brazen hare]

we have 0 < lamba <= 4/9 right?

[loud viper]

Yes

[loud viper]

1/3,1/4,1/5..... Too many values man

[brazen hare]

dont use hit and trial

try factorization?

[loud viper]

I mean how?

[eager jewel]

bro take sqrt(D) as an integer..it has to be even cause we have 2 in denominator and u cant take 4 or one of the roots will become 0 so it has to be 2

[brazen hare]

lemme write it down

[loud viper]

I didn't get it

[eager jewel]

which part

[brazen hare]

x1 + x2 = 4/lam and x1x2= 9/lam

so this gives us

4x1x2 = 9x1 + 9x2

yes?

[loud viper]

Yes

[loud viper]

Further?

[brazen hare]

my method is kinda hit and trial after this

lemme think of another

[eager jewel]

if ur going to do hit and trial anyway then why not do it in the start

[loud viper]

Sorry guys I ve to go

Some very important work

[eager jewel]

ok

yes pls save the world

[brazen hare]

makes it easier

to find x1x2

since it becomes x1 = 9x2/4x2 - 9

and its integer

[hot timber]

?

[brazen hare]

only integer value comes when x2 = 9

[hot timber]

Didn’t quite understand that result

I’m actually unwell rn so I’m not thinking straight

[brazen hare]

x1 is integer

[hot timber]

And so is x2

[brazen hare]

so for it to divide fully

it should first be multiple of 9

and greater too

[hot timber]

Oh yeah

[brazen hare]

-9 confirms it for us

in denominator

[hot timber]

mhm

[pearl pond]

@loud viper Has your question been resolved?

[pearl pond]

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[midnight haven]

how should i approach this?

[toxic lichen]

what are $\wedge$ and $\vee$ supposed to mean here

[jolly parrot]

Ann

[hot timber]

what is that

[toxic lichen]

!xy

[pearl pond]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[midnight haven]

oh my bad!

[brazen hare]

and / or?

[toxic lichen]

if they are "and" and "or" then why do we have something that looks like a real-number algebraic expression on the right lol

again, @midnight haven please just post the original question

in full

no matter what language it's in

[brazen hare]

lhs bits?

[midnight haven]

V is the LCM
and upsidedown V is GCD

The Least Common Multiple and The Greatest Common Divisor

[hot timber]

ok thanks

aha

[midnight haven]

thxmods

[hot timber]

the hack gambit

[brazen hare]

huh what happened

[pastel umbra]

scam message; got deleted

But yes; the notation appears to be specifically French

[midnight haven]

yeah it is i had to search the english terms

[pastel umbra]

Soit g := gcd(x, y)
-# Let g := gcd(x,y)

On peut dire quoi avec x, y alors ?
-# What can we say about x, y then?

[midnight haven]

if g=gcd(x,y)

x=x'g y=y'g such that gcd(x',y')=1

[pastel umbra]

ye

je continue en anglais ?
-# Shall I continue in English?

[midnight haven]

what ever you're comfortable in

i don't mind

[pastel umbra]

I'll stick with English then

[pastel umbra]

How then can we describe the LCM, given this?

[midnight haven]

x'y'g?

[pastel umbra]

ye

Can you rephrase this, and simplify?

[midnight haven]

so you get basically hm nice

$x'y' +1 =2x' +y'$

[jolly parrot]

Drk

[pastel umbra]

ye

Now this has just become a Diophantine equation

Solve however you like 🙃

[midnight haven]

okayyy we only did a general method for $Ax+By=C$

[jolly parrot]

Drk

[midnight haven]

but i think this is doable

thanks a bunch !

.close

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[torpid wolf]

cant find the stationary points

[pearl pond]

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[toxic lichen]

!msgdel

[pearl pond]

The original post of this help channel has been deleted, and it will abruptly close and possibly lock. (This is irreversible.) Please claim a new channel, and don't delete the first message of any future channel you claim.

[torpid wolf]

i deleted the old one because i realised i did something stupid with step 3

should be fixed now

[steady vapor]

I think you should try and differentiate the function again

[torpid wolf]

wdym

did i get the derivative wrong

[steady vapor]

Yes

[torpid wolf]

ah

ok i will try that

[steady vapor]

Do you always use the negative exponents instead of actually writing a numerator and denomenator?

[torpid wolf]

yeah

because this way i can use the product rule instead of the quotient rule

and uh i cant memorise the quotient rule easily

and the quotient rule generally feels more tedious

[steady vapor]

Okay well whatever feels easier for you

But yeah try to differentiate it again and see if you get something else

[torpid wolf]

ok

.solved

[hot timber]

.close is the one

[robust lynx]

It’s closed.

[robust lynx]

[pearl pond]

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[kindred forge]

5a

[cobalt hinge]

Ooh, did this one a while back
Hint: consider f(lambda), the characteristic polynomial, and that f(lambda)=det(A-lambda*I). How could you get the constant term from this form?

[kindred forge]

I understand setting lambda to 0 will give it to you, but why does that work for matrices who do not have an eigenvalue of 0?

[cobalt hinge]

Uhh

Do you remember the structure of a polynomial?

[cobalt hinge]

f(0) is the char poly evaluated at lambda=0, it never says that f(lambda)=0

Does that make sense?

[kindred forge]

ok, so we can evaluate at 0 even if 0 is not necessarily an eigenvalue?

Let me put this into writing and then you can assess the logic to make sure I’m not tripping over anything

[cobalt hinge]

Because eigenvalue means det(A-lambda*I)=0. here lambda=0 is not necessarily an eigenvalue since f(lambda)=0 is not given

[barren path]

just writing the det(A - lambda*I) and looking at the constant wouldnt give the answer?

like with an example

thinking of an 2x2 matrix

[a b
c d]

[kindred forge]

I can definitely show examples, I just need to show general proof here

[nocturne plover]

The reason it works is that the characteristic polynomial of $A$ is $\det(A - \lambda I)$

[jolly parrot]

Katharine

[nocturne plover]

And if you set $\lambda = 0$ then you get $\det(A)$

[jolly parrot]

Katharine

[nocturne plover]

but setting $\lambda = 0$ is exactl the same as looking at the constant term in the characteristic polynomial

[jolly parrot]

Katharine

[nocturne plover]

meaning they are one and the same

[kindred forge]

Thank you

.close

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[pearl pond]

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[shut bane]

I think this problem is incorrect it never mentions F. In the triangle 𝑨𝑩𝑪,𝑫,𝑬 are on 𝑩𝑪̅̅,𝑨𝑪̅̅ respectively, such that 𝑩𝑫/𝑫𝑪=𝟐/𝟑; 𝑨𝑬/𝑬𝑪=𝟑/𝟒. Find the value of 𝑨𝑭/𝑭𝑫∙𝑩𝑭/𝑭𝑬.

[toxic lichen]

yeah that does look suspicious

does it come with a diagram or is it text-only?

[shut bane]

Text-only

Without diagram

So is incorrect?

[toxic lichen]

yeah, i'd return it with a comment like "point F not mentioned until the goal; problem inherently unsolvable unless and until information is given about point F"

[shut bane]

Ok thanks

.solved

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[shut bane]

.closed

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[hollow sonnet]

how was $\frac {\mathrm{d}}{\mathrm{d}x}(\frac{u}{v})=\frac {u'v+v'u}{v^2}$ derived

[jolly parrot]

victorian aristocrat

[toxic lichen]

that should be a minus up there not a plus

[hollow sonnet]

oh

mb

[toxic lichen]

aside from this

product and chain rules

also btw: $\dv{x} \left( \frac{u}{v} \right)$

[jolly parrot]

Ann

[hollow sonnet]

never knew thats a possible format

thanks

[toxic lichen]

did i answer your question to your satisfaction

[hollow sonnet]

yes

will do the rest myself

.close

[pearl pond]

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[woven river]

hi, I cant find the correct answer for this one

[woven river]

here is my work, anything wrong?

[pearl pond]

@woven river Has your question been resolved?

[woven river]

help 🙏

<@&286206848099549185>

[storm gull]

Did u try chatbots

Maybe deepseek

Sure humans are better

Still

[woven river]

they suck at solving these

[storm gull]

So u tried em?

[woven river]

yea

[storm gull]

Deepseek?

[woven river]

i dont even understand how they did it

[storm gull]

So you want a good explanation and not just the solution?

[woven river]

basically im asking this

what is wrong with my work

please ping me when answering, thanks

[pearl pond]

@woven river Has your question been resolved?

[woven river]

HOLY I SOLVED IT

BROOO my upper limit was wrong ALL ALONG

FFFFFFFFFFFFFFFFFFFFFFFF

ok thanks guys

😭

.solved ....

[pearl pond]

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[meager scroll]

a)if all the elements of each row of a square matrix add to 0, solve Ax=0 to prove det(A)=0
b) if they add to 1 show that det(A-I)=0. Does this mean det(A)=1?

[meager scroll]

i solved the first 2 by finding that 0 and 1 are obvious solutions so since they have 2 solutions they have infinite so det=0

but idk how to answer the last part of if det(A)=1

[meager scroll]

i think this is correct but lmk

[pearl pond]

@meager scroll Has your question been resolved?

[pearl pond]

@meager scroll Has your question been resolved?

[pearl pond]

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[dusty jungle]

1 != 2 2!= 1

[cursive wraith]

R6 o R6 is the relation {all (a,c) such that a != b and b != c for some b}

well, every R^2 couple is in that relation

because if a = c, then take idk b = a + 1

a != b, and b != c

so a (R6 o R6) c

and if a != c

take b = (a+c)/2

so a (R6 o R6) c as well

so R6 o R6 = R^2

This is the definition of R6 o R6, but the objective is to find what it is

if a= b and b = c, you get a = c

so equality o equality = equality

and that's it

the point is to write it in the most simplified form

saying equality o equality = equality is already good

if you know equality = {(a,a), a in R}

then it's good to write it too

no unfortunately

which one is in simplest form? {all (a,c) such that a != b and b != c for some b} or R^2

you won't get many points if you just write out the definition of composition

you have to build up your intuition for that, if all you had to do to answer "Write ... in simplest form" is write the definition

then you probably didn't find the simplest form

and?

are you confusing with b)?

tbh this might be enough

you could maybe express it mod 12

but it's a little far fetched

no like, find the residues mod 12 that a-b can be

so we're dealing with a single mod

the intersection case is gonna give you a single residue mod 12

the union is gonna give you multiple

because 0 mod 4 is the same as 0,4 or 8 mod 12

and 0 mod 3 is the same as .... mod 12

idk what you mean by that

but you could always reason mod 12

if multiples of 3 are 0,3,6 or 9 mod 12

and multiples of 4 are 0,4 or 8 mod 12

multiples of 3 that aren't multiples of 4 are .... mod 12

if you want to write it differently than with just the definition

you need to go mod 12 I think

[pearl pond]

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[scarlet aurora]

Jo ich bin 12 klasse und ich und ein homie schreiben morgen mathe Schulaufgabe, wenn jemand uns im Call helfen könnte werd ihr wirklich die Goats.

[rough forge]

Viel Glück ig

[tropic saddle]

also ihr müsst nochmal lernen wie man um hilfe fragt. das ist hier nen englischer server also in der falschen sprache anzukommen ist schonmal schlecht. und dann solltet ihr wenigstens noch sagen um welches thema es überhaupt geht

[pearl pond]

@scarlet aurora Has your question been resolved?

[scarlet aurora]

was bringt mir wenn mir ein englischer typ schreibt

und ich erwarte das hier auf dem server einfach leute sind die alles können

deswegen thema egal

[tropic saddle]

nicht jeder hat bock in nen verkackten call mit euch hirnis zu gehen ohne überhaupt zu wissen worum es geht. aber naja good luck

[scarlet aurora]

hahaha du musst wirklich so ein opfer sein das dich sowas ooc traurig macht

[rough forge]

Schon mieses Gefühl wenn man die einzige (deutschsprachige) person beleidigt, die einen den hintern morgen retten könnte

[scarlet aurora]

bro obv hilft er nicht

außerdem nicht so deep es wäre einfach nochmal gut gewesen hätte jemand erklärt

[rough forge]

ja klar, das hört sich auch nach einer roblox session an

[scarlet aurora]

ihr seid nicht meine helden die mich noch retten können sorry wenn ihr dachtet dass ihr wenigstens in mathe besonders seid

[rough forge]

ich dachte ich wäre gut in mathe, bis ich erfuhr dass N und Q die gleich kardinalität haben

was macht man noch so für mathezeugs in der 12.

[scarlet aurora]

https://tenor.com/view/erm-aksuali-veli-vel-nerd-nerd-emoji-gif-11908435316515627035

[scarlet aurora]

ist dir langweilig?

[rough forge]

ja

[pearl pond]

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[desert plaza]

can someone check these plz?

[desert plaza]

I posted on other channel but it keeps closing before someone gets to me

[shrewd mist]

(a) looks good

b looks good, at least the logic. There might be a math error but it doesn't look like it

Where did the 4 come from in (c)?

Area of a circle is pi*r^2

[pearl pond]

@desert plaza Has your question been resolved?

[pearl pond]

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[desert plaza]

Oh I thought it’s the SA of the sphere

just as a general note

for the ratio

do I just take the total height and radius

and make a ratio with it?

[pearl pond]

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[viscid shale]

la / al costado de R^2 es diferencia de conjuntos of es un "tal que"?

[stoic imp]

talcual gordis

[stoic imp]

todo bien? como va?

[viscid shale]

bien

[shrewd mist]

yeah making a ratio with the height and radius (via similar triangles) will let you eliminate one of h and r. For c you don't need to carry that info forward, you can just get r from h (similar triangles) and then forget about h for the rest of the problem. If you have any more follow ups open another channel since this one has expired.

[pearl pond]

@stoic imp Has your question been resolved?

[stoic imp]

can u help me

[velvet ruin]

Can I get a translation for this 😭?

[viscid shale]

well, thinking about it the intersection cap is probably not the correct symbol, anyways

[velvet ruin]

What exactly are you confused about on this?

If it's the general way on how to solve it

• find critical points inside the region
• analyze the boundary then compare values

[viscid shale]

@stoic imp

[stoic imp]

according to who?

[velvet ruin]

wym?

[stoic imp]

whats the logic behind what you are saying

care to elaborate?

[velvet ruin]

Well, you're looking for the absolute max and min, right? It'll only make sense you find the critical points

[stoic imp]

care to elaborate on the general strategy you are proposing here?

is not as simple as just finding the critical points of f and seeing if they fall inside the region if that's what you are saying

[velvet ruin]

Partially differentiate them first

Then set it to 0

[stoic imp]

we are missing the extreme values happening at the border

of the region

@velvet ruin

[velvet ruin]

wym?

[stoic imp]

you don't know what lagrange multipliers are correct?

[stoic imp]

explain your idea first

how do you plan on solving this exercise from start to finish in words

@velvet ruin

[velvet ruin]

Doesn't that make it more complex?

[stoic imp]

not necessarily

explain your idea first from start to end

in words

@velvet ruin

[velvet ruin]

The idea is basically

• partial differentiation then equate to 0
• substitute the x you got to the fx then you should find your critical point
• you could then analyze the boundary, it has 3 parts. Find the derivative and such
• then you can compare the values you found

I'm sorry for I reply in a sluggish manner, it's currently 5am here and I have not yet slept.

[stoic imp]

care to elaborate on the third step? what does "and such" mean in math words?

[velvet ruin]

You'll be analyzing the boundary with certain values of x and y, basically just find the critical points

That should be the whole idea of it

[stoic imp]

im pretty sure step 2 is handwavy as fuck and wrong to say the least? you equate the gradient to 0, that is you set fx = fy = 0, for this you get a couple solutions in (x,y) form and those are your critical points, you prolly meant this but without all the dyslexia just to start with, second of all, after you analyze the three part boundary that's basically finding the pairwise intersections between the constraints of the region which should give you a couple points. (x,y) values but we are still not done yet, because the extremas might still lie in the boundaries but not necessarily in those pairwise intersections this is where lagrange multipliers come in

we don't necessarily need lagrange multipliers for this, I am just saying how I would do it

@velvet ruin

[stoic imp]

you are being handwavy again, you need to elaborate more otherwise we can't understand your idea

[velvet ruin]

Well, I only followed EVT for this

And this is clearly a closed and bounded region

I'm only stating what I think could be a possible solution

[stoic imp]

the region is compact, yes AFAIK

[stoic imp]

extreme value theorem says that if your region is compact and f is continous then the absolute minimum and maximum of f are guaranteed to be in the compact region, it never mentions critical points

[stoic imp]

idk what theorem are you reading but im pretty sure that's not EVT, care to elaborate on that?

[velvet ruin]

Its more of a connections. EVT is one, another could be Fermat's

[stoic imp]

I see, now you are being a little more clear communicating your ideas, but can you elaborate more on whats the proposed solution you are saying for this?

[velvet ruin]

Mb really, I'm really running low. As for the elaboration, give me a sec

The first step is verification of the conditions, making use of EVT. Then comes the analysis of the interior which makes use of Fermat's, searching for critical points inside the region where the gradient is zero. Then comes the third step, which is the analysis of the boundary, (single-variable EVT), breaking down the complex boundary into 3 more simpler ones (1D and used single-variable calc). So with those 3 segments, we find the max/min of each. This should be able to cover all the potential extrema on the boundary segments and at the possible/identified intersections.

With all that done, you should now be able to compare and create your conclusion. This should be guaranteed by evt

[stoic imp]

how do you plan on analyzing the boundary of the region?

[velvet ruin]

Wdym? It should be divided into 3 parts then finding the min/max.

[velvet ruin]

It can be split into three.
X = 0
Y = 0
Y = 5-x²

[stoic imp]

what about x= √5?

are we ignoring that one or what

[velvet ruin]

It belongs to a specific corner point though? When y=0 x =√5

The point is basically intersection between the boundaries

So, you could say y=0 covers it

[stoic imp]

this never talks how to deal with the intersection between the boundaries how are you handling that

or maybe it does mention it at the end but I never understood what you mean by this

[stoic imp]

the intersection between boundaries

analyzing each boundary itself is different than analyzing the pairwise intersections

[velvet ruin]

This is still under evt (single variable)

[stoic imp]

again, evt guarantees the extreme values exist between or in the intersection of the boundaries but doesnt give us the critical points nor the maximum or minimum occurring at the pairwise intersections of the boundaries of the constraints

we can find the pairwise intersections of this boundaries and then label them as candidates for extremas, and the world is not falling apart I say

[stoic imp]

ah I see whats your plan now, you are planning on parametrizing this f(x,y) to make it a single variable function

using the boundaries of the region

basically converting this multivariate problem into a single variable question

@velvet ruin

[velvet ruin]

Yea

This would allow for the use of standard single variable calculus techniques

Then the resulting values would be compared to find the overall max/min

[stoic imp]

I don't like it

[velvet ruin]

Well, it's on you if you want to go for it or not.

If you want to take a different approach, you can freely do so.

[stoic imp]

where you from?

[stoic imp]

so three different parametrisations you say?

[velvet ruin]

Yes

[stoic imp]

how to do it

@velvet ruin

f(x,y) = (x-1)(x-y)

f(0,y) = y-x

f(x,0) = x^2 - x

f(x, 5-x^2) = (x-1)(x -5 + x^2)

@velvet ruin

this is not looking so great

f(x, 5-x^2) = x^2 -5x + x^3 - x + 5 - x^2

f(x, 5-x^2) = -6x + x^3 + 5

[stoic imp]

are u here?

[naive marlin]

tu pablo inglese ?

[stoic imp]

ye

[stoic imp]

@naive marlin

[pearl pond]

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[lyric karma]

Idk how to start

I made the diagram

Ping when someone comes ty

[humble root]

Hi! Here's a hint: How would you find the base of the triangle?

[spiral coyote]

Make equations using tangents of x nd y

[lyric karma]

Using tan

[humble root]

Good, so - show me how you would do it :)

[lyric karma]

Let x = distance of tower from point z to y, and B distance from z to tower

Tan z = P/B
Tan y = P/B-x

What next

[lyric karma]

I have no clue 😭

[humble root]

Good! Here's a hint:
We want to solve for P, which is the height. We know that:
tan(z) = P/B
What happens if you make B the subject?

[lyric karma]

What does making B the subject mean >.<

B = P/tan(z) ?

[humble root]

Yep!

That's what I mean :)

[humble root]

Good so, what next?

[lyric karma]

B-x = P/tan(y)

And then I try to substitute or eliminate and find values from the 2 equations?

[humble root]

Perfect, yep!

That's exactly it.

[spiral coyote]

The variable for "x" in thr ques is d

But yeah just eliminate B now and just some working and you've got ur answer

[pearl pond]

@lyric karma Has your question been resolved?

[pearl pond]

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[exotic gale]

George has a right rectangular prism with three distinct integer edge lengths, and its volume is 2026 cm^3. In square cm, what is the surface area of George's prism?

[exotic gale]

i just know xyz=2026

and 2(xy+yz+xz)=S

S being surface area and x,y,z being the edges

i don't know what to do next im stuck

[plush bramble]

Do you know how to factor 2026

[exotic gale]

yeah

[chrome plank]

2026 doesn't factor into three integers

[exotic gale]

1013 times 2 times 1?

[chrome plank]

oops, you're right

[boreal raptor]

forgor about 1 😭

i wouldve too 😔

[exotic gale]

.close

[pearl pond]

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[pearl pond]

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[cloud shuttle]

In the law of cosines, why do we need to square both sides if we can get the third side simply by a = b-c

[shrewd mist]

What do you mean? a = b - c is not a true statement

[cloud shuttle]

suppose two vectors b and c, how would you find the line joining their heads?

i'm a beginner!

[west bluff]

Isn't that just the line joining the points with the same coordinates as the vectors?

[tulip ore]

cosine rule requires magnitudes of a vector which is calculated as sqrt(x^2 + y^2)
the ^2 is unavoidable on the inside

[west bluff]

Like, you can always take a vector to the origin and from there, the point with the coordinates of the vector will be "at the head of the arrow"

[tulip ore]

also as a reminder, ||a - b|| is not the same thing as ||a|| - ||b||
for example you can consider a = (1, 0) and b = (0, 1)

[cloud shuttle]

on coordinate planes i think we say those 'lines' are vectors?

[west bluff]

No, the lines are sets of points

A vector and a point determine a single line though

[cloud shuttle]

processing....

https://classroom-images.cdn.askfilo.com/classroom/1697932607435_lrtqzsjj_1661626.jpg

why we are squaring |a|

[viscid shale]

You probably already know pythagoras

Which is just a specific case of Cosine Rule

[cloud shuttle]

yes, a bit

[cloud shuttle]

yes

[viscid shale]

a^2 = b^2 + c^2

you cant just drop all the squares

We say that a^2 = b^2 + c^2 doesnt imply that a = b+c

More over, the only case in which that two equations satisfy at the same time is when b or c is equal to 0

which obviously doesnt constitute an actual triangle.

[cloud shuttle]

I was trying to derive this using vector method

[viscid shale]

I can see it, but the reason of not using a = b + c - 2bc cos(A)

is precisely the same as i just explained

[cloud shuttle]

okay, they are different. but, then, how said to inventor "now square it"? I mean, you can say our problem was just to find the third side of a triangle and we could solve it by just finding the vector that is joining the heads of the two given vectors.

And the image i sent is doing about same thing but then both sides got squared. Why?

[viscid shale]

We are trying to prove cosine rule

using vectors, that is.

[cloud shuttle]

No, I'm trying to understand why it is as it is to solve a problem that could be solved simply.

I just want the reason of squaring, thats it!

[viscid shale]

squaring it makes it easier to work with it, basically

there is another option:

which is using the euclidian distance sqrt(x^2+y^2)

this is how it works by doing that:

$\sqrt{x_0^2+y_0^2}=-\left(\sqrt{x_1^2+y_1^2}+\sqrt{x_2^2+y_2^2}\right)$

[cloud shuttle]

how?

[viscid shale]

but is so happens to be that when using cosine rule for triangles (not vectors)

you dont have access to the xy coordinates

only the lengths and angles

therefore, this whole thing is basically useless for actual triangles

[cloud shuttle]

Now, it makes sense!

[viscid shale]

in vector algebra you could access the xy of a, b and c, but for triangles you dont, so you have to work their modules*, and that requires doing a few more operations

[cloud shuttle]

wasn't it (x_2 - x_1)^2 + ( y_2 - y_1 )^2

[viscid shale]

nope, the module of a 2d vector is just pythagoras

[cloud shuttle]

okay, thanks!

[viscid shale]

anyways, hope that helps, is just vector algebra operations that you do to find the values you want only using modules

[cloud shuttle]

.close

[pearl pond]

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[pearl pond]

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[empty rover]

I need help with this

[velvet ruin]

What do you need help with for this?

[empty rover]

I dont know how to begin

[velvet ruin]

for a, you can use Pythagorean identity

b is quotient

for c, it'll be straight forward since you'll know what tan x and sec x atp

[empty rover]

but dont I need to identify which quadrant they belong to?

[warped violet]

Yes

That is the first thing you need

Finding in which quadrant is the angle x

[warm current]

Not really. As eclipse mentioned, you can use pythag theorem.

[warped violet]

Well yeah okay cos is positive so it doesn't matter

But it still doesn't harm

Useful to know if the solutions you get are what they should be

[pearl pond]

@empty rover Has your question been resolved?

[empty rover]

how can we use pythagoras theorem tho?

[velvet ruin]

The Pythagorean identity specifically the

sin²x + cos²x = 1

Use that to find cos x

[pearl pond]

@empty rover Has your question been resolved?

[dull totem]

or just

the theorem

sine is opp/hyp

cosine is adj/hyp

find the adjacent side using the pythagorean theorem

just draw a triangle

[cloud shuttle]

that's the correct answer

[pearl pond]

@empty rover Has your question been resolved?

[pearl pond]

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[hearty fiber]

Isnt this solution incomplete as circles can also be in the 2nd and 4th quadrant, i think?

[compact ridge]

but then such a circle wouldn't be tangent to the axes

[hearty fiber]

How can we say that with certainty ?

[compact ridge]

I think the diagram is pretty clear already

if you think about the circles that are below the line, the circle can't cross the axes or else it wouldn't be tangent

same with the case above the line

there's one case for each (above and below the line)

[hearty fiber]

But can we say that with certainty ?

Like any sort of proof would really help, please

[wild fable]

south w the clutch

[compact ridge]

you don't understand what 'tangent' means....

[hearty fiber]

I do know what it is, its a line that just touches a circle at a single point without ever getting inside it

[compact ridge]

okay, and if you increase the radius beyond the case where the circle is tangent

the circle will intersect the line twice right?

[hearty fiber]

Cant something like this exist ?

[compact ridge]

I recommend you try drawing this out yourself

ah okay let me think

[iron basin]

yes
but that seems most intuitive and harder to miss maybe

[compact ridge]

I think it just follows from the algebra, no?

they missed a case yeah

2 solutions for the first equation and 2 solutions for the other

so 4 in total

[hearty fiber]

Thank youu very much

[compact ridge]

sorry about that I didn't realise also the book was wrong

wait it's not this but I bet there's another case

[hearty fiber]

My wording was kinda ambiguous too it's fine

[hearty fiber]

Also how do we use desmos like that ?

[iron basin]

uh... type?

[hearty fiber]

I just know the very basics of it

[iron basin]

^ for power is the one extra thing

you could type the other symbols using your normal intuition

[hearty fiber]

How did you input the command to find all the possible circles which are tangent to the axis and the line too

[iron basin]

no i uh calculated it

it cant do that

[hearty fiber]

Ohh 😅 nvm then

[pearl pond]

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[sharp smelt]

so this just the pdf / the marginal at y=1/4?

[naive marlin]

bonjour

[sharp smelt]

which would be $\frac{24xy}{ \int_{0}^{y-1} 24xy dx}$

[jolly parrot]

wai

[sharp smelt]

I then plug in y=1/4 in the denom

[sharp smelt]

bonsoir

[naive marlin]

there is the formula right

[sharp smelt]

yea, just wanna make sure I'm not messing up

[sonic patrol]

hm i don't get this

[sharp smelt]

this is P(X|Y=y)

[sonic patrol]

it should be $\frac{\int_{- \infty}^{\frac 12} f(x, \frac 14) dx }{\int_{- \infty}^\infty f(x, \frac 14) dx}$

[jolly parrot]

Pseudo (Cat theory #1 Fan)

[sharp smelt]

huh?

[sonic patrol]

is that not the conditional probability approach...?

[sharp smelt]

that is but why is what I did wrong

[sonic patrol]

this doesn't even seem to be a number

[sharp smelt]

okay, lemme re-write it

[sharp smelt]

what do you think the equation should be

[sonic patrol]

what i wrote

[sharp smelt]

sure, but the bounds are different

0 to 1-y, right

[sonic patrol]

i don't get where this y - 1 comes from

[sharp smelt]

the region given

[sonic patrol]

right..

[sharp smelt]

oh, I think I get it now

[sonic patrol]

again the answer to this will be a function of x and y

but the answer you should get is a number

[sharp smelt]

so $\frac{\int_{0}^{\frac{1}{2}} 6xdx}{ \int_{0}^{y-1} 24xy dx}$

[jolly parrot]

wai

[sharp smelt]

I'm finding the marginal

[sharp smelt]

and then I evaluvate this at 1/4

[sonic patrol]

what does that even mean

[sharp smelt]

I plug in y=1/4 after evaluvating this

[sonic patrol]

doesn't that give you a function of x

[sharp smelt]

why do I need a function of x...

the marginal is wrt Y

[sonic patrol]

if you take this expression

and plug in

y = 1/4

your answer is not a number

your answer still depends on x

[sharp smelt]

you plug in x=1/2 too

[sonic patrol]

well you should've mentioned that, for one

for two i'm not convinced this gets you the correct answer

[sharp smelt]

you think I should fix y before integrating?

[sonic patrol]

to be honest i have no idea what you're trying to do or where you're getting your expressions from

[sharp smelt]

in essence I'm asking if yout think this is what I should have done

Thanks!

.close

[pearl pond]

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[proper nova]

How do I solve this???
The question states as follows: $(a, b, c) \in \mathbb R^3$, $\sum ab = 1$. Find $\min P = 3a^2 + 3b^2 + c$

[jolly parrot]

1 divided by 0 equals Infinity

[pearl pond]

@proper nova Has your question been resolved?

[proper nova]

<@&286206848099549185>

[blissful cloak]

with (\sum ab=1) that is to say (ab+bc+ac=1)?

[jolly parrot]

ΠαϳαμαΜαμαΛλαμα

[proper nova]

yes

i only learned AM-GM

[blissful cloak]

oh wait yeah it's right there in the question lmao

[proper nova]

and with $(a, b, c) \in \mathbb R^3$, i don't really know what to do

[jolly parrot]

1 divided by 0 equals Infinity

[proper nova]

🙏

[blissful cloak]

I mean you have something you want to minimize and a constraint, just use lagrangian multipliers? Might not be the most efficient or elegant answer but it'll get you there

[proper nova]

wth

[blissful cloak]

(\mathcal{L}=3a^2+3b^2+c+\lambda(ab+bc+ac-1)) then just solve the system for each (\frac{d\mathcal{L}}{da}=0,\frac{d\mathcal{L}}{db}=0,\frac{d\mathcal{L}}{dc}=0,\frac{d\mathcal{L}}{d\lambda}=0)

[dusty jungle]

they said that they only know am gm

[proper nova]

I DIDN'T EVEN REACH CALCULUS YET 😭

[blissful cloak]

oh

[proper nova]

i don't understand what is ts going on 💀

calculus guy cooking in algebra

[blissful cloak]

dw that's something for calc iii, hmmm

[proper nova]

oh nvm

WHAT THE 💀

[proper nova]

please i need AM-GM 😭

[blissful cloak]

idk how you could use AM-GM here... that's an inequality, here you're trying to minimize something...

[proper nova]

isn't minimizing something requires you to make an inequality?

💀

why don't they let me ping helper twice

:(

[warm dagger]

I’ll ping the helpers for you

[proper nova]

can you help?

[dusty jungle]

you need am gm to get ab bc and ca right

how would you do that using three seperate am gms

[warm dagger]

Uhm no 💀

I mean no to helping :))

Not to using am gm

[dusty jungle]

yeah i got that

[proper nova]

i wanted to try using the UCT method, specifically for $3a^2$ in a seperate AM-GM, and same for $3b^2$ and $c^2$

[jolly parrot]

1 divided by 0 equals Infinity

[dusty jungle]

uct?

hm havent heard of it

but normally

[proper nova]

im not really used to terms in english so i basically translated or smth

[dusty jungle]

if i told you ab = 1 and i told you to find the minimum of a^2 +b^2 what would you do

[dusty jungle]

no im sure its a thing but i just dont know it

[proper nova]

hmm

[dusty jungle]

plain am gm

[proper nova]

ah

$a^2 + b^2 \geq 2\sqrt{ab} = 2$

[jolly parrot]

1 divided by 0 equals Infinity

[dusty jungle]

what if i told you to find the minimum of 2a^2 + b^2/2

[proper nova]

uhhhhhhh

[dusty jungle]

(ab=1 again)

[proper nova]

$\frac{2a^2 + b^2}{2}$?

[jolly parrot]

1 divided by 0 equals Infinity

[dusty jungle]

mb its $2a^2 + \frac{b^2}{2}$

[jolly parrot]

CherryMan

[proper nova]

in my draft

i'd do

let the landing point at $a = x$, $b = y$

[jolly parrot]

1 divided by 0 equals Infinity

[proper nova]

so i know that $xy = 1$

[jolly parrot]

1 divided by 0 equals Infinity

[proper nova]

using that, i can do $2a^2 + 2b^2x^2 \geq 4abx$

[jolly parrot]

1 divided by 0 equals Infinity

[proper nova]

similarly, $\frac{b^2}{2} + 8a^2y^2 \geq 4aby$

[jolly parrot]

1 divided by 0 equals Infinity

[proper nova]

so i need $4x = 4y$ or $x = y$ aka the landing point is at $x = y = 1$

[jolly parrot]

1 divided by 0 equals Infinity

[dusty jungle]

yes

so it was really plain am gm again

[proper nova]

but in my work

i can do $2a^2 + 2b^2 \geq 4ab$

[jolly parrot]

1 divided by 0 equals Infinity

[proper nova]

$\frac{b^2}{2} + 8a^2 \geq 4ab$

[jolly parrot]

1 divided by 0 equals Infinity

[proper nova]

so $2a^2 + \frac{b^2}{2} + 2b^2 + 8a^2 \geq 8$

[jolly parrot]

1 divided by 0 equals Infinity

[proper nova]

i tried to do the same thing in the question

but the only thing left to handle is $2b^2 + 8a^2$

[jolly parrot]

1 divided by 0 equals Infinity

[dusty jungle]

ok

[proper nova]

AM-GM again?

[dusty jungle]

$\frac{2a^2 + \frac{b^2}{2}}{2} \geq \sqrt{ab}$

[proper nova]

\geq

[jolly parrot]

CherryMan

[dusty jungle]

ya

so i dont know about uct

but you can split the 3a^2 + 3b^2 + c^2 similarly

into three am gms

[proper nova]

what i did above is basically uct if i translated correctly

[dusty jungle]

which give ab + bc + ca with some coefficient

look, we need ac and bc

so we need two terms with c^2

the only way to split it is c^2/2

now we can either to (a^2 + c^2/2) + (b^2 + c^2/2) + (2a^2 + 2b^2)

or

(2a^2 + c^2/2) + (2b^2 + c^2/2) + (a^2+b^2)

which one makes the numerical coefficient of every term same after applying am gm

[proper nova]

so basically it's just how you distribute your terms

[dusty jungle]

yeah

i think uct can work i wish i knew the method better so that i could try it

[proper nova]

this thing is basically what i meant

[dusty jungle]

oh i think i get it now

[proper nova]

i got smth like $3a^2 + b^2 \geq 2ab\sqrt{3}$

[jolly parrot]

1 divided by 0 equals Infinity

[proper nova]

$3b^2 + c^2 \geq 2bc\sqrt{3}$

[jolly parrot]

1 divided by 0 equals Infinity

[proper nova]

$c^2 + 3a^2 \geq 2ca\sqrt{3}$

[jolly parrot]

1 divided by 0 equals Infinity

[proper nova]

so altogether

$3a^2 + 3b^2 + c^2 + 3a^2 + b^2 + c^2 \geq 2\sqrt{3}$

[jolly parrot]

1 divided by 0 equals Infinity

[dusty jungle]

yeah but thatss a bound for 2a^2 + 2b^2 +c^2

[proper nova]

that's my original approach

[proper nova]

why should i split it like that?

[dusty jungle]

ok wait

that is one of two options

that is the incorrect one

but for the second option

$2a^2 + \frac{c^2}{2} \geq 2ac\ 2b^2 + \frac{c^2}{2} \geq 2bc \ a^2 + b^2 \geq 2ab$

[jolly parrot]

CherryMan

[dusty jungle]

so $3a^2 + 3b^2 + c^2 \geq 2(ab+bc+ca) = 2$

[jolly parrot]

CherryMan

[proper nova]

ooh

[proper nova]

.close

[pearl pond]

Channel closed

Closed by @proper nova

Use .reopen if this was a mistake.

[pearl pond]

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[midnight haven]

Not sure what to do, pls help

[compact ridge]

for the area of the regular pentagon

[midnight haven]

to find the radius of the smaller circle

[compact ridge]

you should draw the lines OA, OB, OC, OD, OE on the diagram

[midnight haven]

kk

[compact ridge]

I know, there's a lot of steps to this

[compact ridge]

so the pentagon is made up of 5 congruent isosceles triangles

each triangle has area 1/2 ab sin C

[midnight haven]

kk

[midnight haven]

yea

[midnight haven]

then do i like get the area of one?

[compact ridge]

we want to find the radius first though

since we don't know that

[midnight haven]

kk

[compact ridge]

it's right-angled trig

[midnight haven]

how do yk its right angled

oh wait nvm

so i split it down the middle

[compact ridge]

you split this triangle into two equal halves

[midnight haven]

so AOE

kk

[compact ridge]

half of 180 degrees is 90

[midnight haven]

yea

[compact ridge]

yeah, so if M is the midpoint of AE

what's angle AOE?

[midnight haven]

90?

[midnight haven]

cus 180 - (45*2)

[compact ridge]

no

[midnight haven]

oh

[compact ridge]

okay, what's angle AOE first?

[midnight haven]

AOE? uh 45?

because isoceles

[compact ridge]

no, there's no 45-90-45 triangle

okay you know that AOB + BOC + COD + DOE + AOE = 360 right?

[midnight haven]

ye

[compact ridge]

but then all of those angles are the same

[midnight haven]

o

so 360/5

[compact ridge]

yep!

[midnight haven]

so 72 then

[compact ridge]

okay so AOE = 360/5 = 72

[midnight haven]

kk

[compact ridge]

and then how about now?

[midnight haven]

then i can use tan()

on AOM

[compact ridge]

what's the angle in one of the right triangles

[compact ridge]

it's not tan actually

[midnight haven]

we have opposite and we want the adjacent?

oh wait is the radius not the straight line OM

[midnight haven]

sin() if OM isnt radius then

[compact ridge]

no, it's the hypotenuse

yes so you use sin

[midnight haven]

oh shi

oHHH CUS OM ISNT GOING TO THE CIRCLE

kk so it would be 4/sin(72)

so r = 4.20....

[compact ridge]

no, 72 is the wrong angle

[midnight haven]

oh

[midnight haven]

wait how?

[compact ridge]

we split the 72 here

[midnight haven]

ohhhh kk

so 4/sin36

r = 6.80...

so then area of big circle = pi * (6.8...)^2

[compact ridge]

right, and then finally you can do area of triangle = 1/2 ab sin C

5 of those triangles makes up the pentagon

[midnight haven]

so how does the area of the big circle relate to the triangle area

[compact ridge]

the key is white area = 1/2 of big circle

so pentagon - pi (small radius)^2 = 1/2 * pi * (6.8...)^2

[midnight haven]

so i got area to be 25.88.. and i multiplied it by 5 it gave 129.44... not 145.489

wait no thats the area of pentagon right?

[midnight haven]

ohhh

[compact ridge]

,w 5 * 1/2 * (4/sin(36 deg)) * (4/sin(36 deg)) * sin(72 deg)

[jolly parrot]

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Requested by sooouth.

[midnight haven]

o

[compact ridge]

the angle in between OA and OB for example is still 72

[midnight haven]

yea

i did Aoed = (6.8...)(8)(sin72)

,w (4/sin(36 deg)) * 8 * sin(72 deg)

[jolly parrot]

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Requested by xii_w

[midnight haven]

huh

,w (1/2)(4/sin(36 deg)) * 8 * sin(72 deg)

[jolly parrot]

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Requested by xii_w

[midnight haven]

ok thats what i got

so A pent =

,w 5 * ((1/2)(4/sin(36 deg)) * 8 * sin(72 deg))

[jolly parrot]

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Requested by xii_w

[midnight haven]

ok so thats what i have

now A pent - pi(r^2) = 1/2(pi * (6.8...)^2

[compact ridge]

one sec

[midnight haven]

so ,w 5 * ((1/2)(4/sin(36 deg)) * 8 * sin(72 deg)) /(1/2)(pi * 4/sin(36 deg))

,w 5 * ((1/2)(4/sin(36 deg)) * 8 * sin(72 deg)) /(1/2)(pi * 4/sin(36 deg))

[jolly parrot]

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Requested by xii_w

[midnight haven]

so if we divide this by pi

,w (5 * ((1/2)(4/sin(36 deg)) * 8 * sin(72 deg)) /(1/2)(pi * 4/sin(36 deg)))/pi

[jolly parrot]

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Requested by xii_w

[midnight haven]

yeah this deffo aint the radius

my logic was

Apent/(1/2 pi * (6.8...)^2) = pi(smallr^2)

so then to do LHS and divide by

[compact ridge]

okay go back to this

[midnight haven]

WAIT I DIDNT ROOT

nvm thats still big

[midnight haven]

kk

[compact ridge]

here

[midnight haven]

wym by p = (5)(1/2)r^2(sin(72)

[compact ridge]

that's the area of the pentagon

p is the pentagon

[midnight haven]

ohh kk

for area of pent i have129.44

[compact ridge]

yeah you were using the wrong value then

[midnight haven]

i did 1/2(6.8...)(8)(sin72)

,w 1/2(4/sin(32 deg)) * 8 * sin(72 deg)

[jolly parrot]

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Requested by xii_w

[compact ridge]

ah no, cause these triangles are all isoceles

[compact ridge]

OA is the radius
OB is also the radius

for triangle AOB

[midnight haven]

wym..

[midnight haven]

yea arent they the same

so IA abd IV are 6.8...

OA and OB

[compact ridge]

yeah so you should be using 1/2 (6.8....)(6.8....) sin(72)

8 should not appear

[midnight haven]

oohhhh

oh yea i get 27.094..