help-13

[reef hare]

Idk 💀

[cerulean sail]

Remember the idea here, if I'm typing this in bed while drinking hot chocolate, then I'm typing this in bed

Why is P(B) being the bigger one a problem?

[reef hare]

so if P(B) and P(A) both happens, then P(B) has to happen

like in a venn diagram B would completely surround A

wait

so then P(AnB) would be P(A)

OHHHHHHHHHHHHHHHHHH

[cerulean sail]

Yep - you can't be more than the smaller one of P(A) and P(B)

[reef hare]

Ohhhhhhh

That makes so much sense

Thank you so much

Omg

Wonderful

.close

[cedar kiln]

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[blissful summit]

how do i integrate this function

[inland sky]

u substitution

[cedar kiln]

@blissful summit Has your question been resolved?

[blissful summit]

this is u sub??

i tried u sub 1/t

idk what to do after

<@&286206848099549185>

[crimson sedge]

What did you get through u substitution

[cedar kiln]

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[crimson sedge]

.reopen

@blissful summit

[blissful summit]

sin^2u

[cedar kiln]

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[blissful summit]

• outside the fraction

[blissful summit]

negative outside the integral

@crimson sedge

[crimson sedge]

okay okay

[crimson sedge]

write in terms of cos

[blissful summit]

so like pythag identity

[crimson sedge]

1-cos2u/2

[blissful summit]

whatttttttttttttttt

how do you that

[crimson sedge]

it is a trigonometric identity

[blissful summit]

(sin u)^2 = 1-(cos^2(u/2))?

tbh i don't think we learned this at all

[crimson sedge]

$sin^2(u) = 1 -(cos2u)/2$

[blissful summit]

but yes i do see the identity online

yeah

[wraith dagger]

Kenshin

[blissful summit]

then its easy

so this is the only way then?

if you know an identity?

[crimson sedge]

It is easy this way

[blissful summit]

right

because we didn't even learn reduction formulas

and i haven't seen this identity before

[crimson sedge]

Otherwise you will have to derive the identity, and then use it anyways

[blissful summit]

right

how can i derive the identity

might come in useful

double angle?

[crimson sedge]

$cos(2u) = cos^2(u) - sin^2(u)$

[wraith dagger]

Kenshin

[blissful summit]

ohhhhhh

ok

thanks so much

definitely doable then

[crimson sedge]

No problem :)

[crimson sedge]

Cool

If you’re done with it, you may close the channel

[blissful summit]

ofc tysm

.close

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[dense locust]

Hello

[dense locust]

I have no clue where to start....
I did this so far for A

[vital wing]

this suspiciously looks like integration

[warped drum]

$f'_2(x)=f_1(x)=x^2$

[wraith dagger]

LF

[warped drum]

what could $f_2(x)$ be?

[wraith dagger]

LF

[warped drum]

keep in mind it has to pass through the origin (so f_2(0)=0)

[dense locust]

Thats correct : D

[dense locust]

Ohh

[vital wing]

welp
I guess you have to integrate now
and then you make sure that f_n(0) = 0

[warped drum]

yes

[vital wing]

finish devising some of the f_n(x) and you will have an easier time doing part b

[dense locust]

So if im getting this correct,
f2(x) +c = f1(x)

[warped drum]

no

[vital wing]

that looks way off

[dense locust]

and to find c, you integrate f1(x) , and then throw it into integrated f1(x) = f2(x)

[vital wing]

that still looks way off

[warped drum]

$f_2'(x)=x^2$

[wraith dagger]

LF

[dense locust]

Yes

[warped drum]

what function has $x^2$ as a derivative?

[wraith dagger]

LF

[dense locust]

whats a derivative... i tend to suck at english side of math

[pine sandal]

differentiation

chapter 8 of the book

[dense locust]

2x

[vital wing]

2x is the derivative of x^2
x^2 is the derivative of [?]

the second line is what you had to do

[dense locust]

1/3 x^3 +C

[vital wing]

ayo

did you forget the constant ("+ C")-

[dense locust]

wait so what does this mean?

[vital wing]

x^2 is the derivative of f_2

now that you have f_2, make it equal to 0 when x is 0

[dense locust]

So that means if u differentiate f2(x) you get f'2(x)

[vital wing]

ye

[dense locust]

ok how do you show something is the integrated form of something

[vital wing]

chapter 8 of what book?

[vital wing]

you just did 🤔

[dense locust]

f2(x) ---> f'2(x)

• F2(x) differentiated = f'2(x)

F'2(x) ---> F2(x)
This gets integrated

right?

[vital wing]

ye!

[pine sandal]

+c

[dense locust]

RIGHT!

Ok, im following

[vital wing]

plus the constant, ye

[dense locust]

Just got confused what everything means

Thats kinda why

Ty guys

[vital wing]

nice catch but you forgot to catch yourself

np!

[dense locust]

: >

[pine sandal]

?

[dense locust]

https://tenor.com/view/the-office-bow-michael-scott-steve-carell-office-gif-14973267

[vital wing]

type .close when you are done!

[dense locust]

.close

[cedar kiln]

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[cedar kiln]

@crimson sedge Has your question been resolved?

[latent totem]

It's simply
One occurs multiply by second doesn't occur +
Second occurs multiply by one doesn't occur

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

@crimson sedge Has your question been resolved?

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[cerulean pine]

does anyone know what t- formulae is

[cerulean pine]

and how to solve question b using it

[cerulean pine]

thank you

[cerulean pine]

how would i apply this to question b though

[warm wraith]

honestly no idea just trying to help

i think theyre asking about question b though

[stuck tinsel]

Oh

My bad

[cedar kiln]

@cerulean pine Has your question been resolved?

[stuck tinsel]

@cerulean pine

[cerulean pine]

oh thank you so much

🤗

i really appreciate it

[cedar kiln]

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[lime wagon]

I'm computing this fourrier series, but im unsure of how to continue from here. Can anyone help please?

[lime wagon]

actually the first big term might simplify to 0.

I still need help 🙏

[dawn junco]

f(x) = sin(x/2) you're really sure about that in the first place?

@lime wagon

[dawn junco]

yeah

absolute value of sin(x/2)

[lime wagon]

yeah i kept it in mind

should have written it I agree

[dawn junco]

doesn't seem like it

ah ok now I see it

[lime wagon]

what do I do with all those cosine terms

[dawn junco]

yeah cos(n*pi + pi/2) is just 0

[lime wagon]

okay yeah thats clear

I have one more question

I am very unsure of the trig identities that let you express cosines and sines in terms of an integer to the power of n or something. Such as cos(npi)=(-1)^n. This one is very obvioous, but theres more complicated ones if whatever is inside the cosine or sine is a sum for example. Is there a table or something such that I can identify the pattern. Or is it just trial and error until you have achieved an expression that yields the same answer as the cosine function evaluated for the same n.

[dawn junco]

it's not a trig identity really

pattern matching and trial and error as you said

[lime wagon]

oh ok

[dawn junco]

they're not gonna ask you batshit crazy identities to derive in an exam setting

[lime wagon]

I wouldnt be surprised

[dawn junco]

most of the time it's cos(n*pi) or something closely related anyway

it's not terribly much harder than that in the context of fourier series

[dawn junco]

I mean they even reminded you of the product to sum identity in the question, that's quite nice of them

[lime wagon]

yeah thanks a lot...

I wish i could get an advantage over other students with a simple identity like this

Okay one more question

cos(pi+npi)=cos(1+n)pi?

why do i see everywhere that they take the x out of the cosine

[dawn junco]

parentheses

[lime wagon]

cos(pi+npi)=cos((1+n)pi)

[dawn junco]

you have an example ?

[lime wagon]

its just weird to me because the calculator doesnt understand it when I dfactor the pi out. even with brackets.

but calculators are not allowed in the exaam anyhow

[dawn junco]

wym they take the x out of the cosine ?

I don't see that anywhere

[lime wagon]

not out sorry

[dawn junco]

or you're just talking about the factoring bit

[lime wagon]

just factor out

they factor starting from here

[dawn junco]

well that's cause they used the product to sum stuff

it's just detail anyway

[lime wagon]

yeah I guess but i never factorred it. maybe it will make my life easier

thanks. Ill keep practicing

.close

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[robust cove]

Hello there, can someone help me with this problem?

[cedar kiln]

@robust cove Has your question been resolved?

[cedar kiln]

@robust cove Has your question been resolved?

[cedar kiln]

@robust cove Has your question been resolved?

[opaque harbor]

consider this triangle, where that third vertex is the centre of the circle

this allows you to write h in terms of r (or r in terms of h) in the equation for volume, and then you can maximise that as a function of r (or h, depending on which way you did it)

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[summer lintel]

where the SLZ thing is made up of matrices that have det(A)=1 for all intergers

I have to that the if A is in Sln(Z), then it is invertible and A^-1 is in SL-N(Z)

I have shown the det(A) is equal to 0 which if one of the conditions

how do i now show that it also only has intergers in A^-1

[crimson delta]

cramers rule

[summer lintel]

we havent read that chapter. is there another way?

[crimson delta]

adjugate matrix?

[summer lintel]

adj/det?

[crimson delta]

well you maybe mean the correct thing

[summer lintel]

I am just not sure how to proceed

[summer lintel]

this is what i mean

[crimson delta]

well the 1/det(A) is just 1

can you show that adj(A) has integer entries?

[summer lintel]

Im trying to, but im not sure how

[crimson delta]

what is the definition of adj(A)

[summer lintel]

C^T, right

[crimson delta]

and what is the definition of C

[summer lintel]

I recall it is, but not sure why it is like so

[crimson delta]

then look it up in your notes

[summer lintel]

Cofactor matrix

[crimson delta]

which is defined how?

[summer lintel]

(-1)^i+j* M_ij

i believe

[crimson delta]

and M_ij is defined how?

[summer lintel]

the minor det of A by deleting the ith row and the jth column

[crimson delta]

is that an integer?

[summer lintel]

yes?

[crimson delta]

why?

[summer lintel]

pretty sure it just says it in the book

i dont rly know why is it that tho

[crimson delta]

it is the determinant of an integer matrix

to compute the determinant, you only need to add/subtract/multiply the entries

[summer lintel]

oh

[crimson delta]

so it is also an integer

[summer lintel]

that is the definition of M_ij

cool

so i and j are intergers

so adj(A) = (-1)^(i + j)*M__ij?

[crimson delta]

no that makes no sense

on the left is a matrix, on the right is a number

[summer lintel]

ok my bad

but the adj is made up of smaller matrcies who only have intergers in them

[crimson delta]

it is not made up of them

it is made up from their determinants

[summer lintel]

but the determinants integers will continue to be intergers?

[crimson delta]

the determinants are again integers, yes

[summer lintel]

mh

ok

@crimson delta ,right?

[crimson delta]

could be more precise, but yes

[cedar kiln]

@summer lintel Has your question been resolved?

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[past narwhal]

i just need an explanation distinguishing one tailed t tests and two tailed t tests

[past narwhal]

for example one exercise asked to state whether its a one tail or 2 tail test and the prompt was that the 6th grade boys weigh less than the girls

so i had to create a H0 and H1 hypothesis ; the boys weigh less than the girls or the boys do not weigh less than the girls

but idk how to tell if its 1 tailed or 2 tailed t test

<@&286206848099549185>

im thinking its a 1 tailed

someone help this isnt that hard of a question 🙏🏻

[hollow trail]

2-tailed is good for if you want to find if something is not equal to your null hypothesis (you want to find if it's different but don't care if it's more or less)
1-tailed is good for if you want to find if something is less than or greater than your null hypothesis, in one direction only (e.g. if you want to find if it's greater but don't care if it's less)

[past narwhal]

got it

thx

.close

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[shut ember]

x^2*y''' = 2y' (Integrate the Euler equation).
Can't do the ones containing x

[bold hinge]

Hint

Assume y = x^n

[hollow minnow]

Wherea are you stuck?

[bold hinge]

Then you should be able to solve easily!

After multiplying through by x

[shut ember]

found that n1 = 0 and n2 = 3. What next?

[bold hinge]

Check your math again

You should be solving x^3 y"' = 2xy'

Yep so you have n1 = 0 with multiplicity 2, n2 = 3

Mb, forgot the 2

[shut ember]

so we get n1= 0 and n2= 3?

[bold hinge]

Yes, but remember multiplicity

[shut ember]

i did like this

y = x^n
y' = n (x^n-1)
y''' = n(n-1)(n-2)(x^n-3)

x^2n(n-1)(n-2)(x^n-3) = 2n(x^n-1)

we remove x

n(n-1)(n-2) = 2n

n^2 - 3n^2 = 0

n^2 (n-3) = 0

so like n1=n2=0, n3= 3?

[bold hinge]

Yes

[shut ember]

so y1=y2=1 and y3=x^3?

[bold hinge]

Well not quite

You must have two different solutions for n1 and n2

1, log(x), x^3

[shut ember]

how if n2=n1 tho

[shut ember]

did you find how?

[shut ember]

can you take a look?

[hollow minnow]

So you got 0,0,3

[shut ember]

yep

[hollow minnow]

Well then because the multiplicity is 2 it gives you y1 = c1,y2 = c2log(x) and the other root gives c3x^3

[shut ember]

and then the final answer is c1+c2log(x)+c3x^3?

[hollow minnow]

Yeah

[shut ember]

well thank you then!

[hollow minnow]

[shut ember]

thanks to you too @bold hinge

.close

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[vestal fern]

how to simplify tan(2x)=1

[idle tusk]

do you know how to solve tan(x)=1

[vestal fern]

yea

special triangle

[idle tusk]

what is x equal to, then?

[vestal fern]

is it jsut double the angle

ok thanks

pi/4 *2

pi/2

[idle tusk]

no

[vestal fern]

what

[idle tusk]

2x=pi/4

[vestal fern]

oh

[idle tusk]

but also don't forget +2pi k

[vestal fern]

wait

why is it not 2*pi/4

[idle tusk]

$\tan(🐟)=1$

[wraith dagger]

artemetra

[idle tusk]

solve for 🐟

🐟=pi/4 + 2*pi*k

but also 🐟 is 2x

so 2x=pi/4 + 2*pi*k

[vestal fern]

x=pi/4

tan(2x)

2*pi/4

[idle tusk]

no

i mean

[vestal fern]

yea ik waht u mean

imo its a bit weird

[idle tusk]

tan(pi/2) ≠ 1 then

[idle tusk]

it really isn't

[vestal fern]

im thining backwards

u know x why not just plug x back into tan(2x)

[idle tusk]

x=pi/8

[vestal fern]

but it wouldnt equal one

[idle tusk]

2x=pi/4

[vestal fern]

yea ik

[idle tusk]

tan(2x) = 1

ta-da

[idle tusk]

if it helps, make a substitution k=2x and solve tan(k)=1, then substitute 2x back and solve for x

idk

[vestal fern]

OH

I get what u mean

[idle tusk]

ye

[vestal fern]

thanks

[idle tusk]

no problem

if you are done type ".close"

[vestal fern]

.close

[cedar kiln]

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[brave wedge]

j wanna check my answer - i got (4t+8t^3)/(2t^2+1)

[flint plinth]

a_T is what, tangential acceleration?

[brave wedge]

yeah

r' dot r''/magnitude r'

a(t) dot unit tangent

acceleration in direction of tangent vector

[flint plinth]

ok, can you show your intermediate results? i don't really want to calculate it myself to check your answer because i'm lazy 😁

like what did you get for r' and r''

[brave wedge]

ye np

one sec

r'=1,2t,2t^2

r''=0,2,4t

[flint plinth]

yep correct

[brave wedge]

magnitude of r'= sqrt(1+4t^2+4t^4)

[flint plinth]

yes

[brave wedge]

which is j (2t^2+1)^2

so bottom is 2t^2+1

r' dot r''=4t +8t^3

[flint plinth]

yep all correct

[brave wedge]

bet

ok

so

while i have u here

https://cdn.discordapp.com/attachments/1024102668391682078/1186036136150323200/Screenshot_2023-12-17_150352.png?ex=6591c93a&is=657f543a&hm=9c2b555ac65d06da80e755ee499583fb99f1e7477b0ca88aaf30c56fc5d9ae6e&

is the only way to do this bash

the algebra is very bad

[flint plinth]

remind me the formula for curvature

i can't recall it offhand

[brave wedge]

r' x r'' / r'^2

r'^3*

[flint plinth]

ok

[brave wedge]

top is magnitude

bottom is magnitude as well

[flint plinth]

yea that's gonna be a lot of fun

algebraic mess

[brave wedge]

yeah like

thats really ass

[flint plinth]

at least you can plug in t=0 after you find the derivatives

that'll clean things up

[brave wedge]

?

before i cross

i can plug in 0?

[flint plinth]

"curvature at time t=0"

[brave wedge]

yeye

but does it still work if i plug in pre cross

[flint plinth]

yes, find the derivatives then you can plug in t=0 before proceeding

[brave wedge]

holy shit

[flint plinth]

yep it'll work

[brave wedge]

sorry

that actually helps a lot lol

[flint plinth]

yea

otherwise it's gonna be a horrible mess

[brave wedge]

thank god ik that i defo wouldve j skipped during test lmao

yeah

@flint plinth LOL

LOOK

i got sqrt(42/27)

not sure if its right

[flint plinth]

if you show your work i can check it, but i'm definitely not gonna compute it myself haha

[brave wedge]

yeah

one sec

@flint plinth

[flint plinth]

yikes, kinda hard to read, can you type it in?

[brave wedge]

Lol sorry

one sec

Can you read this better

[mortal hemlock]

try using latex

$\frac{1}{1+x}$ just an example

[wraith dagger]

Flappie

[brave wedge]

@flint plinth

[cedar kiln]

@brave wedge Has your question been resolved?

[subtle harbor]

,rccw

[subtle harbor]

it's right, verified with calculation :)

desmos 3d can handle this computations :)

[cedar kiln]

@brave wedge Has your question been resolved?

[brave wedge]

LES GO

ty

[subtle harbor]

np :)

[brave wedge]

.close

[cedar kiln]

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[scenic canopy]

How do I do question 9

[scenic canopy]

I know how to do it with two values but not with infinity

[cerulean sail]

Do you know how to find the sum to infinity for a general geometric sequence (such that the sum exists)?

[scenic canopy]

What like using the equation with the first term and the common ratio

[cerulean sail]

Yea, that would do

[scenic canopy]

Yes then

[cerulean sail]

Could you state it then please?

[scenic canopy]

a/1-r

[cerulean sail]

Yep, that'll do, so if you have the geometric sequence having the first term $a$ and common ratio $r$, you know that $\abs{r} < 1$ (because the sum to infinity exists!) and that $\frac{a}{1 - r} = 19683$

[wraith dagger]

@cerulean sail

[scenic canopy]

Ok

[cerulean sail]

Can you make a similar statement for the sum of the first 8 terms?

[scenic canopy]

a(1-r^8)/r-1 = 19680 r<1

[cerulean sail]

Careful - make sure the numerator and denominator are "the same way around", so like either
[
\frac{a(1 - r^8)}{1 - r} = 19680
]
or
[
\frac{a(r^8 - 1)}{r - 1} = 19680
]

[wraith dagger]

@cerulean sail

[cerulean sail]

But otherwise yea

[scenic canopy]

Yeh that was a mistake my bad

[cerulean sail]

Anyways, you now have both
[
{\color{green} \frac{a}{1 - r} = 19683 } \text{ and } \frac{{\color{green} a}(1 - r^8)}{{\color{green}1 - r}} = 19680
]
happy with solving those?

[scenic canopy]

Yeh this was the bit I got stuck on what do I do now

Because there’s two unknowns I didn’t know what to do

[wraith dagger]

@cerulean sail

[cerulean sail]

Do the colours give you an idea?

[scenic canopy]

Oh I see

So you use the 19683 as a/1-r on the other equation

[cerulean sail]

Would be a good idea to

[scenic canopy]

So it’s just 19683(1-r^8) = 19680 and then I just solve it

[cerulean sail]

Yep, that gets you r, then you can then find a

[scenic canopy]

Alright thank you

I got 0.33… as r is that correct?

[cerulean sail]

Yep, r = 1/3 - note they also tell you r is positive too

[scenic canopy]

So to find a do I just sub in r into the sum to infinity equation

[cerulean sail]

Yep, just put the r you found in one of the other equations, the infinity one is the easier one

[scenic canopy]

Could you help me with another question

[cerulean sail]

Sure, which one?

[scenic canopy]

Number 7 on the photo I sent

[cerulean sail]

Cool cool

Assuming this is A level maths right?

[scenic canopy]

Yeh

[cerulean sail]

Thought so cool, so we have

[
\int \sec(x) + \tan^2(x) \dd x
]

[wraith dagger]

@cerulean sail

[cerulean sail]

Do you have an idea of how to deal with it?

[scenic canopy]

Um I know all the techniques like substitution and by parts but I never know which one to use

[cerulean sail]

This one is slightly easier and doesn't require either: you can split it up into
[
\int \sec(x) \dd x +\int \tan^2(x) \dd x
]
do you know how to do either of those?

[wraith dagger]

@cerulean sail

[scenic canopy]

Oh ok thats easier than I thought

[cerulean sail]

Yep hopefully it should be! one of them you don't even need to really "do" (cause it's a formula booklet given )

[scenic canopy]

Is it ln(secx + tanx) + tanx - x + c

[cerulean sail]

Looks good to me

[scenic canopy]

Do you mind helping me with one more😂

[cerulean sail]

sure thing, which one?

[scenic canopy]

Number 6

[cerulean sail]

Cool cool, they gave you a tough one there

[
y = \ln( \sin(x^2 - 3) )
]

[wraith dagger]

@cerulean sail

[cerulean sail]

And they want the gradient of the normal when x=2

Have an idea of how to deal with it?

[scenic canopy]

By parts?

[cerulean sail]

Well, they're asking for the normal, and by parts is for integration

[scenic canopy]

Well you have to differentiate and then sub in 2

But where to start with the differentiation I don’t know

[cerulean sail]

Maybe it might help to label things in bits

So like
[
y = \ln(u) \text{ and } u = \sin(x^2 - 3)
]
of course it should hopefully come as no surprise that you need the chain rule for it(!)

[wraith dagger]

@cerulean sail

[scenic canopy]

Ok

[cerulean sail]

hopefully that should make applying chain rule a tiny bit easier - let me know how it goes!

[scenic canopy]

Is it 1/sin(x^2-3) * cos(x^2-3) * 2x

[cerulean sail]

Yep, looking good, though one more thing you can do with that to make it a tiny bit nicer!

[scenic canopy]

Sin(x^2-3)^-1 * cos2x(x^2-3)???

[cerulean sail]

Not quite what I had in mind, though true

You have
[
\dv{y}{x} = \frac{2x \cos(x^2 - 3)}{\sin(x^2 - 3)}
]

[scenic canopy]

Oh put the cos part over the sin part

[wraith dagger]

@cerulean sail

[cerulean sail]

Yep haha

Just realised though this is more about finding the normal rather than just what dy/dx is I guess good practice either way hehe

[scenic canopy]

So I just sub in 2 to find the gradient right

[cerulean sail]

Yep, that would be the tangent gradient, then you know what to do with that

Alternatively, you may find it easier to take dy/dx and then do -1/(dy/dx), which might make working stuff out a tiny bit easier, but either way it's minor

[scenic canopy]

I got an answer of 229.16 for the tangent gradient which doesn’t look right

[cerulean sail]

Hmmm

Check the calculator mode - remember for calc, all trig functions should be in radians

[scenic canopy]

2.57 in radians

[cerulean sail]

Yea I get that too

Then of course, you want the normal

[scenic canopy]

Alright thank you for all the help

[cerulean sail]

A pleasure! have a good one

[scenic canopy]

And you

[cedar kiln]

@scenic canopy Has your question been resolved?

[cedar kiln]

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[rapid pond]

how tf do you solve this equation

[rapid pond]

like is there a good way

[muted bear]

Im assuming you have a triangle and youre solving for the sides

!original please

[cedar kiln]

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.

[rapid pond]

A rectangular box has width $12$ inches, length $16$ inches, and height $\frac{m}{n}$ inches, where $m$ and $n$ are relatively prime positive integers. Three faces of the box meet at a corner of the box. The center points of those three faces are the vertices of a triangle with an area of $30$ square inches. Find $m+n$.

[wraith dagger]

Aurora

[rapid pond]

@muted bear

aime problem

the question i asked, was how to solve that equation which is part of the solution

[muted bear]

Ah

[rapid pond]

that was my first instinct when solving this problem, then i got stuck on that step

the solution only says "solving, we get 41"

so how do you solve that equation

[muted bear]

Can you send a link to the sol please

[rapid pond]

https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_7

im looking at solution 3

[gusty forum]

holy sqrt bash

multiply every bracket by 2 and simplifying gets you 4 brackets with 3 similar ish looking terms each which should be bashable for aime kids

ok it simplifies to 2 differences of squares

ye that’s pretty fast as far as bash goes

[rapid pond]

........

"bashable for aime kids"

nah cuz what tf????

this is too hard

[gusty forum]

i mean if you bash that it should be enough for aime time allotment lol

but simplifying to 2 differences of squares is faster

[rapid pond]

oh true

wait i dont see the difference of squares

brb imma eat dinner

[gusty forum]

average aime experience: find something that looks solvable, spend 25 minutes bashing, learn later you’re 1 transformation way from something that maybe takes 5 minutes to solve

[gusty forum]

$RHS = \frac{1}{32}(a+b+c)(-a+b+c)(a-b+c)(a+b-c)$

[wraith dagger]

IV

[gusty forum]

where a=10 and b and c are the stuff that will take me years typing on a phone

[cedar kiln]

@rapid pond Has your question been resolved?

[rapid pond]

oh thats true

wait what could i difference of squares though

i dont see anything helpful when it comes to difference of squares rn

[gusty forum]

makes the bashing a bit easier no?
also im pretty sure the ss on aops is wrong lmao

they have an extra 1/2 i think

[rapid pond]

oh ig true

there was an extra 1/2 in the front using herons

[gusty forum]

isn’t that already in the first bracket

[rapid pond]

anyways i think its just a bash from here

like literal bash

[gusty forum]

ye

[rapid pond]

ty ig althoughit looks like theres no good way

.close

[cedar kiln]

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[prisma haven]

<@&286206848099549185> I'm stuck on part a

[gusty forum]

!15mins

[cedar kiln]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[gusty forum]

also use ε as an intermediate transformation

[prisma haven]

I made an augmented matrix and got the B matrix into rref

but my answer doesn't look right

<@&286206848099549185>

[cursive nacelle]

.show you work

[gentle pecan]

Hello @prisma haven I am going to start working on your problem. If you have Twitch feel free to swing by and help me out!

[cedar kiln]

@prisma haven Has your question been resolved?

[cedar kiln]

@prisma haven Has your question been resolved?

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[prisma haven]

.reopen

[cedar kiln]

✅

[cedar kiln]

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[brave wedge]

was wondering if i could get a hint on this

[brave wedge]

kind of lost

i know i have to maximize lives

L=1000w+3000x

[dire geode]

Maximize this L subject to the given equation in the question

[gusty forum]

i mean this is lagrange multipliers but idk why the wording makes me feel it’s not

[brave wedge]

like use it as another variable?

this is lagrange?

ohh

i thought its j regular optimization

i forgot how to do lagrange ngl

[gusty forum]

half of my brain was like this is instant from lagrange and the other half says this is precalc

but lagrange is pretty quick ye

[brave wedge]

alr

[dire geode]

You could also just solve for w or x in your given equation, then plug it into L and get L in terms of just one variable

So yea regular optimization works too

[brave wedge]

yeah i j thought that like

if doing that

then we dont maximize right

like how would that be maximizing

also does it matter which equation we use the lagrange operator on

it shouldnt write....?

right*

[cedar kiln]

@brave wedge Has your question been resolved?

[brave wedge]

<@&286206848099549185>

[dire geode]

Does what matter

Try anything

[dire geode]

No. Of course you're still maximizing

Why do you think you wouldn't

[brave wedge]

like

[brave wedge]

how does taking tangent of j the resources

and solving for w and x

give us

anything

like if we solve for w and x for live

lives

right x=3w

and plug

how do i know thats maximizing

idrk what that means is what im saying

like how come that is maximizing

[dire geode]

I thought you knew how to do optimization

[dire geode]

Read this and do a couple examples
https://tutorial.math.lamar.edu/classes/calci/optimization.aspx

[brave wedge]

alright thanks

[brave wedge]

yeah i j thought that like

usually we minimize or maximize something right

how does minimizing or maximizing resources

have anything to do with lives

like how does maximizing resources solve for lives

[granite knoll]

you aren't maximizing resources, you're constrained by that equation

[dire geode]

That's not what I said at all

[dire geode]

Maximize L means maximize what

[brave wedge]

lives yeah

my b im being a little silly

ty

how would u solve for a single variable tho

in that equation(constraint)

[brave wedge]

like we have to use lagrange no?

[dire geode]

.

[dire geode]

No

You're asking a lot of questions that can be answered if you just tried something

[dire geode]

Read

[brave wedge]

got it

i will try doing now

i got the answer using lagrange

im j trying to see if the other will work so ima try that rq

w=12 x=8 from lagrange

holy

ok so

what i did was

[brave wedge]

for regular optimization i j did x(3x-2w)+w^2=144

set x=2w/3 to simplify

got w=12

then x=8

[dire geode]

Where did this come from

[brave wedge]

from the x(3x-2w)

i j wanted to get w by itself

but ig im supposed to solve for x in terms or w or w terms of x right

[dire geode]

No idea what justification you're making

[brave wedge]

yeah

im having a little trouble seperating a variable into terms of the other

like in his examples right

pauls

it usually is easy to seperate one variable

put it in terms of the other

but here its a little more difficult bc there w^2 -2xw and 2x^2

unless i j treat it as a constant

one sec

nah idt that works either bc we have a constant

hmmm

[brave wedge]

how would u solve for one variable in terms of another

idt u can

using constraint

[dire geode]

,tex .quadratic formula

[wraith dagger]

riemann

[brave wedge]

yeah

w as constant

but v messy no?

[brave wedge]

wouldnt it be easier to do it with lagrange

[dire geode]

Easier is relative

Do it whatever way is easier for you

[brave wedge]

yeah i see

[brave wedge]

thank you a lot

i app it sincerely

i understand i was being stupid

at some points

sorry

.close

[cedar kiln]

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[onyx rivet]

what does this question mean?

[onyx rivet]

i dont know where to start

[rare vault]

do you know how to solve simultanous equations?

does "substitution" ring a bell

[onyx rivet]

yes

i know that

but is asking for k

[rare vault]

have you tried it?

[onyx rivet]

idk what k is supposed to be

[onyx rivet]

no i didnt do that yet

[rare vault]

k is a constant, you have to figure out which one

[onyx rivet]

how tho

[rare vault]

well you know how to find the solutions to simultaneous equations

so you should try that to find that

[onyx rivet]

x^2-8x+15 = -6x+k

x^2-2x+15 = k

[rare vault]

ah perhaps a property that you might not know that will help here

[onyx rivet]

i got x^2-2x+15 = k now what do i do

[rare vault]

$\sqrt{b^2-4ac} = 0$ implies there is exactly one solution for quadratic $ax^2 + bx + c = 0$

[rare vault]

move k back to the other side to get a quadratic

[onyx rivet]

x^2-2x+15-k = 0 ?

[rare vault]

yes

now how do you find solutions to that

[onyx rivet]

isnt x^2-2x+15 a quadratic already

[rare vault]

k is a constant

so (15+k) is also a constant

[onyx rivet]

(x^2)(-2x)(+15-k)

so its like that

[rare vault]

???

[onyx rivet]

idk bro

[rare vault]

do you know what a quadratic is

[onyx rivet]

yes

they look like x^2-2x+15

[rare vault]

$ax^2 + bx + c = 0$

right?

[onyx rivet]

yea

what do i do with the k that we added

[rare vault]

x^2-2x+15-k = 0

what are a, b and c?

[rare vault]

its a constant!

its part of the constant

[onyx rivet]

what constant

whats a constant again

[rare vault]

a number

[onyx rivet]

a = 1
b = -2
c = 15

[rare vault]

that doesnt change

[rare vault]

a and b are right

c is 15-k

[onyx rivet]

okay

so

then what

how do i do $\sqrt{b^2-4ac} = 0$ with the k

[wraith dagger]

Rahm Bow

[rare vault]

exactly how you would expect to

knowing what a b and c are

[onyx rivet]

$(02)^2-4(1)(15-k)$

[wraith dagger]

Rahm Bow
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[onyx rivet]

$\sqrt{b^2-4ac} = 0$

[wraith dagger]

Rahm Bow
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

Rahm Bow

[onyx rivet]

wtf

wait

its just

sqrt(-2)^2-4(1)(15-k)

@rare vault right?

[rare vault]

yes

= 0

specifically

[onyx rivet]

ok so what do i do after that

[rare vault]

$\sqrt{(-2)^2-4(1)(15-k)} = 0$

solve for k!

[onyx rivet]

but i thought we are square rooting it

[rare vault]

ya doesnt matter

square both sides then you get this

[rare vault]

there changed it, its the same answer though

[onyx rivet]

the answer is 0

0(15-k) = 0

so is that the answer

[rare vault]

the answer is not 0

can you show your algebra work?

[onyx rivet]

there are 0 solution

[rare vault]

there is a solution

[onyx rivet]

(-2)^2-4(1)(15-k)

[rare vault]

=0

[onyx rivet]

0(15-k)=0

0*15 = 0

0 * k = 0

0 = 0

right?

[rare vault]

no

[rare vault]

what are you doing going from this step to the next one?

im not sure what you are doing, but its something illegal

[onyx rivet]

oh wait i forgot bedmas

i was going from left to right

k = 14

[rare vault]

yes

thats right

[onyx rivet]

what about this question

same thing?

[rare vault]

okay, so ill give you the little trick

so the determinant of a quadratic is $b^2-4ac$. If this is negative, there are no solutions. If this is 0, there is one solution. If this is positive, there are two solutions

[cedar kiln]

@onyx rivet Has your question been resolved?

[cedar kiln]

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[cedar kiln]

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[lavish jolt]

the integral was from -infinity to +infinity

how did it change

[cedar kiln]

@lavish jolt Has your question been resolved?

[short blade]

what is Hn and what is Rodrigues' formula

@lavish jolt

[lavish jolt]

polynomial

i will send the formulas

[cedar kiln]

@lavish jolt Has your question been resolved?

[cedar kiln]

@lavish jolt Has your question been resolved?

[cedar kiln]

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[crimson sedge]

its chain rule

[pale lake]

so i get the left bit

idk how they got to the right bit

[pale lake]

how do they get the right bit

[mortal hemlock]

$(y^2)'=2yy'$, but look at what is given as y'

[wraith dagger]

Flappie

[pale lake]

so how does u' = -u?

if we let u = y^2

[mortal hemlock]

thats whats given

[pale lake]

but who said y = u

we said u = y^2

[mortal hemlock]

you have the system given right?

[pale lake]

yes the top bit

[mortal hemlock]

its given that y'=y?

[pale lake]

but we let u = y^2

[mortal hemlock]

where do we let u=y^2?

this definition is after the proof

[pale lake]

ok thats fine

kinda makes sense

bit weird

[mortal hemlock]

first we prove 2yy'=-2y^2 => (y^2)'=-2(y^2)

after that we define u=y^2

if we didnt prove the statment first

we couldnt ahve said u'=-u

[pale lake]

yh

so shouldnt

u'

=

-2u

??

(y^2)'=-2(y^2) then sub in u

[mortal hemlock]

no

well

[pale lake]

that makes no sense tho

[mortal hemlock]

perhaps

probably, yes

there might be a mistake

but im not completely sure

[pale lake]

ok cool ill ask my teacher

btw have u done hyperbolic equalibria before?

@mortal hemlock

[mortal hemlock]

im not sure what that means

but maybe

:)

[pale lake]

hyperbolic equalibira means for the real part of any eigenvalue to be 0

im basically struggling to understand why what theyve shown shows us that the real part of eigenvalues wont be 0

[mortal hemlock]

no clue

sorry

[pale lake]

npl

thanks for the help anyway

.close

[cedar kiln]

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[cedar kiln]

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[thin matrix]

How would i isolate/solve for theta?

[zealous compass]

cry

jokes

only way i can see is to auxillery

[thin matrix]

what is that?

[zealous compass]

google auxillery angle

[thin matrix]

that looks like it could be helpful, i’ll check it out

[zealous compass]

basically instead of a sin and cos term with theta in it you just have one term

so it should get you somewhere with inverse trig shenanigans

[cedar kiln]

@thin matrix Has your question been resolved?

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[subtle quartz]

may I get some help on this? I don’t know where to start. Thanks!

well the image isn’t wanting to load so here’s the question: Write the equation of a line in slope intercept form for a line that has a slope of m=10 and passes through (4, -5).

(no graphing, just a written answer).

[crimson sedge]

Hi stella, you know the intercept point

you can use the equation y = mx + c

$r: \ \ y = mx + c$

[wraith dagger]

Rize

[subtle quartz]

ohhhhh, okay! thank you very much!

I’ll try that!

[crimson sedge]

no problem 🙂

[subtle quartz]

well, if I use that equation, I would be trying to find “c”, correct?

[crimson sedge]

yep

[subtle quartz]

ohhh okay! thank you. I appreciate the help

[crimson sedge]

your welcome stella

[subtle quartz]

:))

[minor crystal]

or you can use the other one

[wraith dagger]

kanna

kanna

[minor crystal]

[subtle quartz]

I am going to write those down, thank you as well Kanna!

[hollow minnow]

In this question you’re given the slope m and a point, it’d make more sense to use the usual y = mx+c formula.

[crimson sedge]

is more direct

[hollow minnow]

Well, actually nvm my nonsense. Pretty much the same.

[minor crystal]

[crimson sedge]

did you find something @subtle quartz ?

[subtle quartz]

somewhat

i don’t know the next step or even if I have it typed in correctly

well, written it in correctly

[crimson sedge]

show your steps if you want

[subtle quartz]

we use “b” instead of “c”, so I just switched it

it’s sending, hold up.

[crimson sedge]

for now is okay

[subtle quartz]

ohhh, alright, that’s reassuring

[crimson sedge]

Remember the point $(4,-5)$

[wraith dagger]

Rize

[subtle quartz]

so would I change it to

-5 = 10(4) + b?

[crimson sedge]

yep

[subtle quartz]

ohhhh, okay!!

[crimson sedge]

next step is plot for verify

[subtle quartz]

could you explain that?

[crimson sedge]

draw the line

in a cartesian plane

[subtle quartz]

on a graph?

[crimson sedge]

graph the line

try with this: https://www.geogebra.org/calculator

[subtle quartz]

okay okay

so do I write the whole equation out on the app?

[crimson sedge]

yes

put y = mx + b

with m = 10 and b that you find

and with that, add (4,-5)

[subtle quartz]

does this look right?

[crimson sedge]

no

[subtle quartz]

wait oops

[crimson sedge]

nope

do you see your equation -5 = 10(4) + b?

solve for b

in other terms

$find \ b \ such \ that -5=10(4) + b$

[wraith dagger]

Rize

[crimson sedge]

when you find b, you can plot on that app the equation of the line

[subtle quartz]

okay so I got

y = 10x - 45

[crimson sedge]

ok

now

plot

[crimson sedge]

$y = 10x - 45$

[wraith dagger]

Rize

[crimson sedge]

and (4,-5)

for write a point just copy and paste

(4,-5)

[subtle quartz]

okay I believe I got it