help-26

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[tardy lotus]

can someone check my model? my leslie matrix gives me complex eigenvalues so i think im doing something wrong but i cant tell what...

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@tardy lotus Has your question been resolved?

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@tardy lotus Has your question been resolved?

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[dusk raft]

poor evil bean

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@tardy lotus Has your question been resolved?

[graceful minnow]

guys I need help finishing the question it's doing my head in it is about material science is there anyone who can help me

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[spare niche]

can someone please help me

[whole geode]

!status

[topaz sinew]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
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5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

@spare niche Has your question been resolved?

[spare niche]

1

[whole geode]

Begin using a graphing calculator to graph the function as instructed

You can also use an online tool such as desmos

[spare niche]

okay done

[whole geode]

Now using that tool find the area between the curve and the x axis

Again, as instructed

[spare niche]

but how do i do that using desmos?

@whole geode like this?

[whole geode]

That's the integral proper, but yeah.

[spare niche]

but i should put 1 below instead of 0 right?

[whole geode]

Notice that you had to negative the integral (reversing the bounds) because the integral is the signed area

You can also integrate from -2 to 1 of abs(f(x))

[whole geode]

Alternatively, you need to integrate these two parts separately

[spare niche]

oh okay

[whole geode]

Because integrals are signed areas, not total areas

[spare niche]

like this

and then for second part i do, -Area 1 + Area 2 = 2.25

and for first part it is 2.66666+0.416666 = 3.0826

right? @whole geode

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[spare niche]

.reopen

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✅

[whole geode]

https://www.desmos.com/calculator/i466k0yer2

I would do it like this

[whole geode]

But other than slightly off math, yes that is workable as well

[spare niche]

what does | | mean?

[whole geode]

Absolute value

[spare niche]

oh okay

thanks!

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[oak cairn]

Can someone please please please help me with my data management project!??

[oak cairn]

I need to calculate the probability of getting $1 and $2

So for $1 (if someone can make sure im correct)

I got: Winning $1
Scoop 1:
(1/3)(1/2)(4/5)(3/4) = 0.1

Or

2 scoops:
(1/3)(1/2) [ (4/5) (3/4) ] (3/4) = 0.075

Possibility of getting $1 = 0.175

For the two scoops, is it correct that I the icecream part right because I DONT want GOLD since it leads you to $2, so I avoided it and instead out 5 i wrote 4/5 and then 3/4

[topaz sinew]

@oak cairn Has your question been resolved?

[oak cairn]

<@&286206848099549185>

[scarlet nimbus]

Yes bro

What help?

[oak cairn]

everything is in the chat :)

[scarlet nimbus]

I think it’s correct

I have to leave

Bye

[oak cairn]

bruh

<@&286206848099549185>

[vestal sigil]

its correct, but just one thing

for 2 scoops, you multiplied by (3/4) twice

theres only 1

[topaz sinew]

@oak cairn Has your question been resolved?

[oak cairn]

but it should be twice

bc first u pick 1/3 for the cone, then u pick 1/2 to get 2 scoops, and then ur suppose to get 2 scoops that arent gold so dont u first pick 4/5 and ur second scoop is 3/4

then u pick 3/4 for the toppingss

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[strong sable]

a curve ( y = f(x) ) located in the first quadrant. The curve passes through the point ((\frac{\sqrt{2}}{2}, \frac{1}{2})). For any point ( P(x, y) ) on the curve, the normal line to the curve at point ( P ) intersects the ( y )-axis at point ( Q ). Additionally, the segment ( PQ ) is bisected by the ( x )-axis.\

My doubt is : Since PQ is bisected by the x-axis, shouldn't point P be in the fourth quadrant? However, P is on the curve f(x), and this function is in the first quadrant. Isn't this a contradiction?

[thorny flame]

Scoria

[loud oasis]

why does PQ being bisected by the x-axis imply P is in quadrant IV?

[strong sable]

I think like this

[night imp]

P lies on the curve so p will be on quadrant 1

Your P is going 4th quadrant

Why

Ohhh I get it, I missed to read the last line mb

[loud oasis]

note it says Q is on the y-axis, not necessarily the positive y-axis

[strong sable]

aah !

tysm !

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[neon iron]

the answer is incorrect right?

[neon iron]

inner area in sintheta excluding costheta from pi/4 to pi/2

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[fresh adder]

y= log ( sin e^{logcos^2x})

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Please don't occupy multiple help channels.

[little pine]

is the question to find the derivative of that?

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[silk pumice]

i guess it is asking for the integral 😂

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[marsh surge]

Considering A is infinite, how can I prove that A x A is also infinite?

[marsh surge]

A is an infinite set*

[odd pagoda]

find an infinite subset

[marsh surge]

Oooo so {(x, x) | for all x from A} being infinite and also a subset of A x A proves it?

because x contains all of the elements of A?

[neon iron]

Depending on how rigorous you need to be, you might need to prove that the set you just defined is infinite

Which should be trivial, you just need to prove it's the same size as A

Do you know how to do that?

[marsh surge]

Uhhh not really on paper

this is not a homework btw, I'm trying to self-study

[neon iron]

Right

[marsh surge]

But how would I be proving that?

[neon iron]

First let me teach you what "same size" means exactly in set theory

Do you know what a bijection is?

[marsh surge]

bijection as in a function that is bijected?

like |X| = |Y| (X and Y being the domain and codomain)

being surjective AND injective?

[neon iron]

I don't know what "bijected" would mean honestly

[neon iron]

yes

a 1-to-1 mapping

[marsh surge]

yeah right

[neon iron]

If there exists a bijection from a set to another then these sets have the same size

That's how we define "same size" in set theory

[marsh surge]

oooo

[neon iron]

By the way, the technical term for "size" when talking about sets is "cardinality"

Now all you need to do is find a bijection from {(x, x) | for all x from A} to A (or vice-versa) to prove they are the same cardinality

Then since A x A is a superset of that you'll have basically proven |AxA| >= |A|

[marsh surge]

oh wow

[marsh surge]

Can I just do that by
x being any element from A, we have a bijection: x -> (x, x)
As simply as that?

[neon iron]

The phrasing looks a bit off but you perfectly understood the idea

Okay technically you have to prove it's a bijection but come on

this'll do

[marsh surge]

aaaa righttt

[neon iron]

I think that actually for any infinite set A, |AxA|=|A|, but I'm not sure and I don't see how to prove it

[marsh surge]

Hmmm

I've always had problems like

writing proper proofs

like it might be an English problem but I don't know the wording

do you recommend any like

resources?

[neon iron]

this reminds me

[marsh surge]

Like for example for this, I am not sure how I would go about proving it in a way that is accepted mathematically. Sure, this:

x being any element from A, we have a bijection: x -> (x, x)
works but as you said it's a little... meh.

[marsh surge]

Ah lol so many people have the same problem

[neon iron]

Define a function f: A -> A x A such that for all x in A we have f(x)=(x,x)

Then to prove it's bijective you'd just have to prove it's both surjective and injective

Injectivity is just: Take a in A and b in A such that a =/= b. We have: f(a)=(a,a), f(b)=(b,b), and (a,a) =/= (b,b) so f(a) =/= f(b)

We have a =/= b => f(a) =/= f(b), which is equivalent to f(a)=f(b) => a=b, which is the definition of injectivity

Oh wait

I wasn't supposed to say "in A", but "in <the set you just defined>"

[marsh surge]

yeahhh

but I understand it now

damn

so it all boils down to the definition of the basic things like bijection. I just substitude my set in it and it would be properly written I assume?

[neon iron]

Then you do surjectivity, so just prove that for every element in the set you defined you can find an a such that f(a)=<the element in question>

[neon iron]

Yeah basically

Just like

Avoid literally writing like I just did

I'm not being very rigorous either

and certainly not eloquent

well mostly just not eloquent enough

I'm speaking like I'm on Discord

uh

Welp this lasted for longer than it should've... Anyways, unless you have any more questions, you can now close this channel by typing .close

[marsh surge]

Ah, but wouldn't it be enough that like, since they are bijective, |A| = |B| => proving B is infinite, and since B is a subset of A x A, A x A is also infinite

[marsh surge]

Yeah I understand 🙏

[neon iron]

"since they are bijective"?

Like, the sets?

[marsh surge]

yes yes

[neon iron]

That doesn't mean anything

[marsh surge]

(x, x) and the other set

Oh wait

[neon iron]

There exists a bijection between them, yes

[marsh surge]

Ahhhh yeah

[neon iron]

The function is bijective

Not the sets

Any more questions?

[marsh surge]

I understand now

thank you very much

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[sinful jackal]

4e^(8t)-18=15e^(-t) solve for t

[fair thorn]

you mean

??

[thorny flame]

ren

[sinful jackal]

Well

The original question is this

(4e^(10t))/(3e^(2t))-6=5e^(-t)

This is what i simped it to

[night imp]

That's okay but you can write
e^-a = 1/e^a

That's what ren did with the right hand side

[sinful jackal]

Oh okay

[night imp]

Now you can do cross multiplication but idk what's next, I am bad with e questions

[sinful jackal]

Me too

[fair thorn]

probably just cross multiply, substitute e^t = x and pray

[chilly walrus]

precisely lol

[sinful jackal]

Im sorry for the delay

Why would i replace e^t for x

[torpid matrix]

I haven't read the problem but it's surely to make it look less scary

[sinful jackal]

Yeah but like

I need to understand the reasoning

[sinful jackal]

Forgive me, but is there a law of mathematics where e^x equals the same number as the base x?

[torpid matrix]

they are suggesting replacing the entire entity of e^t with x I believe

which changes nothing mathematically but makes it look nicer

[sinful jackal]

Oh

[torpid matrix]

maybe I should read the problem before commenting

[sinful jackal]

Dang so its not just me

This question got hands

Atleast im not alone

[ashen trout]

Send the actual qn

[sinful jackal]

This is the og @ashen trout

[ashen trout]

👌

[ashen trout]

is the RHS meant to be 5e - t or 5e^-t

[sinful jackal]

5e^-t

Im sorry

[ashen trout]

ok

alg

[ashen trout]

Can you send a photo of the question

or like a screenshot

of where you got it from

[sinful jackal]

Well

[sinful jackal]

This was copied from the whiteboard at school

[ashen trout]

oh ok

[ashen trout]

Do you have the answer?

[sinful jackal]

I do not

[ashen trout]

bruh

what topic is this under

@sinful jackal

[sinful jackal]

Well im not exactly sure, the teacher gave us a bunch a questions she said was modeled after the last three units. Which were logs, limits and derivatives

[ashen trout]

not possible

[sinful jackal]

NOT POSSIBLE

Okay maybe its the scope of the question then

She did say solve for x when there was no x

[sinful jackal]

Like she said solve for x and the question had no x. So im assuming she meant solve for t

But maybe theres something that has to do with x?

Atleast tell me why so I can understand?

[ashen trout]

you’re probably missing part of the qn

that’s my only guess

i can’t be 100% sure but without answers to sub into your equation we can’t know

[sinful jackal]

That sucks

Hmm

[topaz sinew]

@sinful jackal Has your question been resolved?

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[pearl fog]

im 99% sure its C and im guessing ACD is congruent to CBE

but i only found 2 (AC=BC and BAC=ACB)

not sure how to use the info that bfc=120

[mellow arrow]

congruency?

triangles ADC and CEB seem to be congruent

[ruby tree]

BFC = 120 means CBF + BCF = 60

And you have other 60º angles there

[pearl fog]

oh

ok im stupid makes sense

ty

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[pearl fog]

honestly didnt know why i didnt think cbe and acd is the same

[mortal steeple]

https://tenor.com/view/skill-issue-gif-24176277

lmao

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[deep bay]

I’ve been trying to do this question for the last 30 minutes. I’m sure 1b is correct but i’m not sure how to do 1a

I see, where have i went wrong?

I’ve done the question using u = cosx but the question is asking me to do u = sinx

Okay, thank you 🙏

Ah

Agreed it is

Thank you for your effort tho

How’d you do it?

Oh wow okay

Thank you so much

You’re a legend

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[little iron]

why are there two minus in (x--3)square?

[little iron]

why is the 3 negative?

ohh I think I got it

its because this 3 is positive, so I have to multiply 2 negatives to get the positive back. right?

or am I cooked?

@rustic pecan

wdym "with signs"?

then why not just put a positive 3

ight this made it worse

how is -3 & k = -4?

so the top one and the bottom ones are the same?

ight so how is -(-4) = -4?

oh ok

so (x+3)square and (x-(-3))square the same?

yes I think so

[little iron]

why am I wrong here tho?

yes

but these two are the same because the second one's -3 is multiplied by another negative, so its the same right?

if Im right here I 100% understand

ight tysm bro

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[edgy kelp]

Find the value of k given that there are equal roots. the answer is meant to be k=-3/2, k=3 but I cannot seem to get this answer can anyone help

[edgy kelp]

yeah i get that part its a sheet on using the discriminant to find k so i used b^2-4ac=0 but when I put my values in i keep getting the wrong answer and i just dont know if its my method or my substitution or my factorising that has gone wrong i tried to use chat gpt but that was useless

[native slate]

What are your values of a b and c

[edgy kelp]

a= (k+1) b= -2(k+3) c= 3k

[whole geode]

That is correct, what about your descriminant?

[edgy kelp]

b^2-4ac=0 so (-2(k+3))^2 - (4)(k+1)(3k)=0

[whole geode]

ok that is correct as well. Simplifying this down you get what quadratic?

[edgy kelp]

I ended with -14k^2-24k-18=0

[whole geode]

That is not correct.

Can you show your work?

[edgy kelp]

so i started by expanding (-2(k+3))^2 and got (4k^2+24k+36) and then i done the other side and got -12k^2-12k

[whole geode]

ok

so how did you get -14k^2 from 4 - 12? and so on?

[whole geode]

anyway, this is correct

[edgy kelp]

ahh i see where i went wrong with that im now getting -8k^2+12k+36

[whole geode]

That is correct, you can simplify this by dividing through by -4

to get an easier to manage quadratic

[edgy kelp]

thankyou so much

[whole geode]

you're very welcome

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[thorn surge]

i need help with this

[thorn surge]

idk how to start 😭

[plush tulip]

if you swap 2 rows of ur matrix, how does the determinant change?

[thorn surge]

swap row 2 from the orignal matrix?

or from A?

[plush tulip]

any matrix

[thorn surge]

isnt determinant multipled by negative 1?

[light mesa]

yes, swapping rows changes the sign

[thorn surge]

only if its swapped twice?

like 2 rows?

[plush tulip]

what

[light mesa]

no if you swap twice, then you get the same value as you started with, since -1*-1=1

[thorn surge]

is this only on nxn matrix?

[thorn surge]

ah

[plush tulip]

basically u wanna use these to relate ur matrix to the original one

[light mesa]

This is all you need to know to solve these 👆

[thorn surge]

ok

[plush tulip]

btw who's in your pfp

[thorn surge]

oh lol its my lil sister. im on her laptop rn

[plush tulip]

okay

is this her account

[thorn surge]

yup

[plush tulip]

i was thinking it was someone from kpop

[thorn surge]

no lol

[thorn surge]

im still confused

idk how to use this to answer the question

[plush tulip]

do you know row operations

[thorn surge]

ye

swap rows, multiply by scalars, add subtract rows

[plush tulip]

these describe the effect of row operations on the determinant

[plush tulip]

yes

[whole geode]

It might also help to specify that the transpose also doesn't change the determinant

[plush tulip]

also that's it's multilinear in each row/column but maybe that's more advanced

[thorn surge]

you mentioned effect of row opperations on determinant

but why would the effect on a determinant be helpful?

[plush tulip]

the vertical bars mean determinant

you could start by transposing A maybe

to make it look like the matrix ur given

[thorn surge]

im gonna try that

[plush tulip]

then divide the last column by 3 so you get c1 c2 c3

wait actually it's probably better to start with the original matrix

and do operations on that

[thorn surge]

so do i work with orignal matrix and make it look like the one on A?

i tried and i cant get to the answer 😭

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[neon iron]

suppose I have a vector

[neon iron]

Two vectors at an angle of THETA

AB and AC

BAC = theta

now

force along AB = Tcos(theta)

THEN what is the component of AB along AC

it should be Tcos^2(theta) right?

Or simply T?

[mortal steeple]

diagram?

[mortal steeple]

if this is the case

[neon iron]

What the heck

[mortal steeple]

AD = F

[neon iron]

no

see my diagram

[mortal steeple]

BD = F cos(theta)

[neon iron]

where is D?

[mortal steeple]

CD = F cos^2(theta)

the point on the cube

[neon iron]

no i am not talkign about this case

[neon iron]

I am talking about this >:

[mortal steeple]

actually we cant say

like im gonna assume F cosx is a component of F

then you have to be careful while taking a component of a component

[neon iron]

So it depends on the situatio n

[mortal steeple]

yup

[neon iron]

suppose AB and AC are two vecotrs

will the tip of the resultant AD lie on BC?

[mortal steeple]

nope

[neon iron]

just curious

ok

[mortal steeple]

according to paralleogram law of vector addition

it wont

[neon iron]

ok bye

[mortal steeple]

you can search about it

[neon iron]

my parents have come

thanks

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[sweet shard]

!15m

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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.

[crude anchor]

Womp Womp

We don’t answer for you

You shouldn’t use this as a last resort for answers

Use it as a learning resource

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[crude anchor]

that’s crazy

<@&268886789983436800>

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[honest river]

Ive figured out that AG is 6 sqrt 5

[honest river]

What next??

[plush tulip]

use similar triangles

agb and bce

let gf and cd intersect at m

then ig bec and gdf

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@honest river Has your question been resolved?

[dusky zephyr]

@honest river

I’ve got an idea

Because you have 3 sides of a triangle ABG

you can use the cosine rule to find some angles

Use cosine rule to find angle ABG and then find angle GBC

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[grizzled crystal]

This works as a proof for (ii)?

[grizzled crystal]

Our teacher used mutatis mutandis and then I got lost so I did this instead which is more intuitive for me

[honest river]

I havent studied trig yet

can you do it pls

[dusky zephyr]

What level question is this?

Go to dms instead, someone else has occupied this chat

[honest river]

okl

[grizzled crystal]

(Discrete math question above)

[dusky zephyr]

Is this undergrad?

[grizzled crystal]

Yes

[dusky zephyr]

#discrete-math

I’m still in high school but I’m interested

[dusky zephyr]

For early uni

[rocky obsidian]

don't quite understand (ii) though. What is the x and y after the "if and only if".

it seems like if forall x, y in A, x ~ y <=> forall x, [x] = A.

I think this proof works. But since this is a "if and only if", should you prove if [x] = [y], you have x ~ y?

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[jolly patio]

MeV/G to Gy? How do i convert

[light mesa]

Gy is J/kg, so convert MeV to J and g to kg

Actually, they gave you the conversion factor on the slide already 😅

[jolly patio]

So would i multiply Mev/cm^2 by 1.602 *10^-13 or no?

[light mesa]

No, It's 1.602 *10^-10 for J/kg = Gy

It's 1.602 *10^-13 for J/g

[jolly patio]

Oh okay, i got it!

ty

\close

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[wraith acorn]

i have been trying to factor this thing for omega

and honestly i am fucking stumped

tried rational roots, no dice

[sweet shard]

Can you type your equation

[acoustic pecan]

i cant read it well

[wraith acorn]

sure

2w^4 - 5w^2 * g/R - 2g^2/R^2 = 0

is that more helpful

[keen raptor]

substitute u=w^2

[wraith acorn]

ok let me give that a shit

shot(

im starting to think ive messed up making my actual equation somewhere

substisting u made it a bit easier on the eyes but so far the solution still eludes me

i haven't been ablel to get it into a novel form

[sweet shard]

Show the original question and the entire work

[wraith acorn]

the original question involves the equations of motion for a a bead on a ring that's rotating about a point on its im, but i will attempt to whittle the context down to where I think I made an error (the point up to which I'm absolutely sure I've done everything correct)

Right now what I'm trying to do is solve the characteristic equation det(K - w^2m) = 0 to solve for w, which in this context is a normal frequency

the K matrix I have is:

along with the M matrix

if I have made an error up until this point, it is somehwere in the prefactors

again, entirely possible but i really don't think i did

wait

i mixed those suckers up

K is supposed to be M vice versa

in the picture i mean

but i evaluated it correctly im pretty sure

what it should be

ive trolled this entire thing

i griefed hard

i gotta go back to drawing board

thanks for input

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[keen raptor]

Quadratic formula

[wraith acorn]

im a doofus. thank you

i did end up solving it by factoring it but thank you again

thats certainly better than "if u simply look at the expression the solution falls out"

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[sweet trout]

<@&286206848099549185> may i ask the 2.5 times part i just dont understand that part

[whole geode]

@sweet trout you'll need to reopen the thread

[sweet trout]

how?

[whole geode]

This

[sweet trout]

do i send again into another help channel

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[vernal matrix]

.Yes, this one may close/lock at any moment!

.(and just as I say it )

[sweet trout]

damn 💀

[sweet trout]

<@&286206848099549185> may i ask the 2.5 times part i just dont understand that part

[merry hill]

translate

[vernal matrix]

You're good now, just don't delete this message no matter what you do!

[sweet trout]

theres english my guy

[merry hill]

holy shit im blind

[sweet trout]

its normal
i dont like that they didnt put english on the top

[rich cosmos]

Hello group, I hope you are doing great. Could someone help me to solve this exercise, I am a little confuse and lost with it. Thank you so much

[topaz sinew]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[sweet trout]

i mean progression pov

[merry hill]

distance of point P from R is 2.5 times the distance of point P from y-axis

[sweet trout]

then how do i form my equation

oh wait

🗿

is it just 2.5x if x is the distance of point P to R

[merry hill]

yes

wait no

x is distance of point P from y-axis

[sweet trout]

wdym

💀

[sweet trout]

hmm

i kinda get it

[merry hill]

"distance of point P from R is 2.5 times the distance of point P from y-axis"

distance(P, R) is 2.5 times of distance(P, y-axis)

so if distance(P, y-axis) is x, distance(P, R) is 2.5x

[sweet trout]

i see

[sweet trout]

do u mind i ask one more

its not me not knowing is just my stupid ass cant interpret what the question wants 💀

[merry hill]

ask away

[sweet trout]

do i sub MT 2 or MU as 2

i mean

idk how to say it 🗿

[merry hill]

none

its saying ratio of MT and MU is 2:1

so MT/MU = 2/1

[sweet trout]

so MT/MU =2
and then MT= 2MU>

[sweet trout]

.

[merry hill]

yes

[sweet trout]

Is this correct

[thick lily]

,rotate

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[timid goblet]

,rccw

[timid goblet]

note: I don't want to do by graphical method

What did:
square both sides and got (x^2-x-4)(x^2+X+4)<0

then I got 4 roots but idk how to arrange them

[timid goblet]

.

?

oh

[acoustic tangle]

You can compare the two positive ones and the order will be apparent

[timid goblet]

2 positive and 2 negative I need to compare
is it always true in power 4 eqn?

[acoustic tangle]

Not necessarily

A polynomial of degree 4 can have any number of real roots upto 4 (and even if it has 4, it's not necessary that only two of them are positive)

[timid goblet]

if it has 4 roots then relate positive to positive and negative to negative

[timid goblet]

then how do I know which to compare

[acoustic tangle]

In this case you should have roots r1, r2, -r1, -r2, take any pair of these roots besides the pairs {r1, -r1} and {r2, -r2}, and compare them to zero

Then you will know which ones are positive and negative

[acoustic tangle]

(In this case this is true only because the polynomial has no odd degrees in it)

[timid goblet]

compare to zero?

[acoustic tangle]

Yes, do you need help with that?

[timid goblet]

wdym??

[acoustic tangle]

I will give you an example

[timid goblet]

ok

[acoustic tangle]

Say you want to compare (2 - sqrt(5))/11 to zero

Put (2 - sqrt(5))/11 and 0 on the same line and apply the same operations to both of the sides to simplify the inequality (keep track of which steps would flip the sign of inequality)

(2 - sqrt(5))/11 v 0
(2 - sqrt(5)) v 0*11
2 - sqrt(5) v 0
2 v sqrt(5)

Now you need to compare 2 and sqrt(5). They are both positive, so it is safe to square both of them

4 v 5

4 < 5 and none of the steps above flip the sign of inequality, so you can replace v with < on each line

I.e., (2 - sqrt(5))/11 < 0

[timid goblet]

bro I can use calculator decimal

[acoustic tangle]

If you can use a calculator, simply enter the roots and you will be able to arrange them directly

[timid goblet]

but I don't get marks

If I do directly

[acoustic tangle]

Then don't use a calculator

[timid goblet]

I can only check

and use some shortcuts

[acoustic tangle]

Once you've determined which roots are positive, you can compare them the same way

[timid goblet]

oh

but what if it was abs(X)=abs(x^2+x-4)

what should I do in that case

[acoustic tangle]

Depends on the polynomial, if there's some symmetry between the roots, you could use that. If that's not the case, the worst case scenario would be to compare every root with one another

[timid goblet]

u mean positive positive and negative negative?

[acoustic tangle]

Not just that, here you had roots of the form r1, r2, -r1, -r2. Say r1 and r2 where the positive ones and r1 < r2, then you would immediately know that -r2 < -r1 < 0 < r1 < r2.
But 2 of the roots being positive and 2 of the roots being negative does not mean they have that form, that's guaranteed when the polynomial has no odd degrees in it

[timid goblet]

what if it had odd degrees

[acoustic tangle]

.

[timid goblet]

all neg with odd then?

[acoustic tangle]

What do you mean?

[timid goblet]

if all roots negative with odd power

[acoustic tangle]

Then you will most likely have to compare each pair of roots until you obtain the arrangement

[timid goblet]

which arrangement?

[acoustic tangle]

Of the roots

[timid goblet]

ok ty

.close

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[mortal heart]

hi guys i need help in 10 ii

[mortal heart]

i solved 10(i) already, the answer is 60 degree and 300 degree

[fair thorn]

x = 1/2 theta

[mortal heart]

OHHH

okk i'll try

[mortal heart]

thank you so much!

btw how about this guys?

i found 12a already, it is x = π/6 and x = 5/6 π

[fair thorn]

i guess just distribute and simplify?

[mortal heart]

is it related with 12a?

i don't understand how to use the answer on a for letter b

should i substitute the x (answer from 12a) to the x in 12b?

[timid goblet]

inequality should flip right?

this solution gives wrong answer

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[mortal heart]

.close

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[timid goblet]

sum of any two angles of a triangle is greater than third angle?

[icy sky]

no

obtuse

[timid goblet]

Yes, the sum of any two angles in a triangle is always greater than the third angle. This property comes from the basic fact that the sum of all three angles in a triangle is always .

Why is This True?

For a triangle with angles , , and :

1. The sum of all angles is:

A + B + C = 180^\circ

A + B = 180^\circ - C

A + B > C

A + C > B

B + C > A

Conclusion

The sum of any two angles in a triangle is always strictly greater than the third angle, not just equal. This is a fundamental property of triangles in Euclidean geometry.

[timid goblet]

I asked 3 math bots all say it's true

[icy sky]

thats why you dont ask math bots

[timid goblet]

chatgpt

didn't expect this from gpt

[neon iron]

gpt sucks at math sometimes

just imagine a triangle with 20° , 20° and 140°

[timid goblet]

got it

.close

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[pearl fog]

maybe you meant sum of 2 sides in a triangle is always greater than the third side

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[indigo estuary]

Need help

,rccw

[indigo estuary]

Idk how to approach this qu

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@indigo estuary Has your question been resolved?

[indigo estuary]

<@&286206848099549185>

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[soft arch]

anyone have tips for getting good at first year linear algebra level proofs? like i have a lot of questions with vectors, matrices (inverses, transpose), and spanning sets where they say "if s = {} or if A = .... such that (some condition), prove that (some fact) is true for all cases of A|b or any other value"

this is kind of vague, but i guess just general problem solving/proving advice would help

[soft arch]

for example there are questions where its like "suppose this set has vectors v1...vk such that none of them are zero or something, prove that this is linearly independant all the time if this other set of some vectors is linearly dependant" or something

[odd pagoda]

generally at that level for most proofs its enough to write down the definitions of what you know and what you want to show and connecting them to each other

and practice

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[short trail]

This might be a hit or miss depending how it is

[short trail]

Prove or disprove for arbitrary sets 𝐴,𝐵,𝐶:

[pseudo sonnet]

what have you tried?

this should be a double inclusion proof

[short trail]

Oh I tried this already and I know it's true

I just don't know how to prove

When A = {1,2,3,4,5}
and B = {3,4,5,6,7}
while C = {5,6,7}

.

[pseudo sonnet]

write down the sets explicitly

what is B \ C?

[short trail]

Then for the left I'd be B \ C = {3,4} and union with A it's {1,2,3,4}

[pseudo sonnet]

[short trail]

For the right, the union of A and B is {1,2,3,4,5,6,7}

Without C it's {1,2,3,4}

So it's true both are equal

Yet I don't know how to prove it now

[pseudo sonnet]

hold on, I think there's been a mistake

[short trail]

No I was using /, because \ didn't work for me as I had the one for commands

[pseudo sonnet]

A u (B \ C) = {1, 2, 3, 4, 5} u {3, 4} = {1, 2, 3, 4, 5}

not {1, 2, 3, 4}

and A u B = {1, 2, 3, 4, 5, 6, 7}, 8 isn't an element of any set here

but removing C from A u B does get us {1, 2, 3, 4}, which you got too

{1, 2, 3, 4, 5} is not equal to {1, 2, 3, 4} though

[short trail]

If A would be {1,2,3,4}, then it would tho

[pseudo sonnet]

sure

but for a statement to be true for all sets A, B, and C, it must be true every single time

and you just found an example where it isn't

so the statement cannot be true in general

it may be true sometimes, but not always

[short trail]

Yeah, it wouldn't always be

[pseudo sonnet]

mhm, so the statement is not true

if it were true however, you'd have to provide a proof, probably by double inclusion

[short trail]

Yeah I just checked

=

[short trail]

I probably have to do double inclusion

[pseudo sonnet]

but the statement isn't true

you can't prove something that is false

[short trail]

Wouldn't I have to prove it's false?

Or at least explain why

[pseudo sonnet]

a single counterexample suffices to disprove the statement

the statement reads "for all A, B, C, we have the following equality of sets", so if it's false, you just need to prove its negation statement "there exists sets A, B, C, such that the following equality does not hold"

the easiest way to prove existence is to give an explicit example of sets A, B, and C that don't satisfy the equality, which you've already done

[short trail]

It feels somewhat fishy tho, since I am usually a zero in math

[pseudo sonnet]

nothing fishy about it

this is the standard way to disprove a "for all" statement

you find a single counterexample

[short trail]

Well, that's definitely something new

Usually I spend 2 hours here trying to understand how it works

[pseudo sonnet]

that's a good thing

[short trail]

So it's somewhat shocking this went smoothly

[pseudo sonnet]

usually, you need to have some level of intuition for where the statement may go wrong, in order to find a counterexample

that intuition can only be obtained if you spend a lot of time understanding what's going on

so pat yourself on the back

well done!

[short trail]

Well okay, it was more accidental greatness, since I did it wrong and you pointed it out

So yeah, thanks ^^"

[pseudo sonnet]

if you're done, you may close the channel

[short trail]

Alright then, thanks again

.close

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[short trail]

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[royal ermine]

How do I solve the question ( 2√5 + 4√3 as numerator and denominator 2√3 + 8√5 ) times (2√3 - 8√5 as numerator and denominator 2√3 - 8√5)

[thorny flame]

k

[royal ermine]

this isnt foil?

[golden blade]

k

[finite storm]

like

[mortal steeple]

it is

[hallow depot]

what is foil ?

[finite storm]

2√3 + 8√5 and 2√3 - 8√5

?

[finite storm]

just expanding

[mortal steeple]

idk some american thing

they call opening brackets as foil

[finite storm]

yeah

its the methdo of expanidng

[royal ermine]

im so confused 😦

[keen raptor]

As stated, isn't the second fraction just 1?

[mortal steeple]

i mean they need an acronym for this smh

[finite storm]

hypertoast is trying to conjugate

[golden blade]

sorry you the smartest math guy

[royal ermine]

no because of distrubutive proprety

[finite storm]

a = 2 root 3
b = 8 root 5

[finite storm]

it is just 1

what u're doing

[mortal steeple]

thank you for the recognition 🤓

[finite storm]

is called conjugation in math

[hallow depot]

thanks guys I have found it

[finite storm]

its a way to remove radicals from the denominator

[golden blade]

it was literally here haha

[hallow depot]

ahah lol I didn't notice it

[royal ermine]

this is what im trying to do

[finite storm]

yes

its called conjugation

[mortal steeple]

rationalisation too

[finite storm]

u can use

[royal ermine]

but you just said that it would be equal to 1

[finite storm]

conjugation is just multiplying by 1

cuz sqrt{3}/sqrt{3} is also 1

thats why the conversion is applicable in the first place

anyway

[royal ermine]

but here the root 3 is multiplying the 5

[finite storm]

try multiplying the denominators and expand using
${a^2 - b^2 = (a-b)(a+b)}$

[thorny flame]

k

[keen raptor]

Here I got AI to latex it lol

$\frac{2\sqrt{5} + 4\sqrt{3}}{2\sqrt{3} + 8\sqrt{5}} \cdot \frac{2\sqrt{3} - 8\sqrt{5}}{2\sqrt{3} - 8\sqrt{5}}$

[thorny flame]

mommymorphism aficionado

[royal ermine]

YES thats my question

[mortal steeple]

ai latex?

[royal ermine]

ty

[mortal steeple]

bro send me the link

[keen raptor]

I just asked gemini lol

[finite storm]

$\frac{5}{\sqrt{3}} = \frac{5}{\sqrt{3}} \times 1 = \frac{5}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{5 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}}$

[royal ermine]

im so confused but how do i rationalize the denominator from here?

[thorny flame]

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

[finite storm]

@royal ermine notice how u multiply the denominators together?

[royal ermine]

ya

[finite storm]

so

${(2\sqrt{3} + 8\sqrt{5})(2\sqrt{3} - 8\sqrt{5})}$

[thorny flame]

k

[finite storm]

should be the denominator, no?

[royal ermine]

ya

and thats equal to -308

on my scientific calculator

but

how about the numerator?

[finite storm]

,calc 12 - 64(5)

[thorny flame]

Result:

-308

[finite storm]

u expand it using foil

[royal ermine]

but there not in brackets?

[finite storm]

there's no easy way out for the top

the top is

[thorny flame]

k

[royal ermine]

and thats how it ends?

[finite storm]

yes for the numerator

[royal ermine]

ok

but does it have to be in brackets?

[finite storm]

u expand the brackets first

,w (2\sqrt{5} + 4\sqrt{3})(2\sqrt{3} - 8\sqrt{5})

[thorny flame]

Results provided by WolframAlpha

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Requested by nuclearinteraction

[royal ermine]

um?

but that would be irrrational

[finite storm]

so answer is

${\frac{-56 - 28\sqrt{15}}{-308} = \frac{56 + 28\sqrt{15}}{308}}$

[thorny flame]

k

[royal ermine]

why do we move to positive?

[finite storm]

negative cancel negative

[royal ermine]

but theres 3 negatives?

[night imp]

-a-b can be written as -(a+b)

-a-b/-c = -(a+b)/-c
Now we have 2 negatives, can we cancel them?

[royal ermine]

ya which makes positive

ohhhh

thanks!

.close

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[mint adder]

Hi
My question is that Is it possible to understand theorems such as triangle congruent theorem?
My maths teacher says that just rot learn it

[dry hinge]

It is definitely possible to understand. Probably also recommended. Has your teacher shown a proof? If not, you can look up the proofs online: https://www.cuemath.com/geometry/triangle-congruence-theorem/. Once you’ve looked at the proofs, you can ask questions about them here

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[jolly agate]

Hello! How do i even start this integral

For the first integral

Should I substitute ln(y) = u ?

[sweet shard]

Yea try that

After splitting the integral of course

[jolly agate]

yes

So I will have like this

u / (u^2) + 1

so we will have 1 / (u+1) which the integral of this is ln(u+1)

?

correct to think like that ?

[sweet shard]

Almost. Missing factor of 1/2

[jolly agate]

hmmm

[sweet shard]

Use v=u^2 to be thorough

[jolly agate]

I have not learned v ...

[sweet shard]

It's just repeated u sub

But v is the next letter in the alphabet

[jolly agate]

oh

But whee does it come from ?

the 1/2

[sweet shard]

.

[jolly agate]

I cant really see it

[sweet shard]

Try it and find out

[jolly agate]

oh wait

This is like u * 1/(u^2+1)

And i have missed the derivative of u right

First u

fuuuuk

I dont understand man

so u / v +1 ?

how does it help me ? :/

How did she get a 2z up there

[frank tide]

multiply by 2 on the inside divide by two on the outside

[jolly agate]

and why that ?

[frank tide]

because d/dz (z^2) = 2z

its for the u-sub

[jolly agate]

Ohh

wait but

In this case I have that

u / u^2 +1

so she subbed again

Om here

u / (u^2)+1

So i sub again

V= (u^2)+1?

And dv/dy is 2u

And dy = du /2u

ohhh i get it

thank you

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[inland fjord]

can someone explain how to do g) please

the answer key says the asymptotes are x = -2 and x = 4/3

oH wait

nevermind i copied it wrong

its minus 8

nevermind, quadratic formula works

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[broken kelp]

Can someone explain where the integral with 4 and 1 comes from

I understand the computing of the formula, but not why everything is being calculated the way it is

[acoustic pecan]

u=2x-3

2(7/2)-3=4
2(2)-3=1
its the same integral

just a change of variable

x to u

its mainly to deal with the (2x-3)^1/2 which while not difficult really would still be a bit of a pain

u^1/2 distributed is a lot easier

[broken kelp]

I understand that, I meant the integral changing from 7/2 and 2 to 4 and 1

[acoustic pecan]

i showed you

it was the first thing i demonstrated

[broken kelp]

Why does that change the integral?

[acoustic pecan]

if you change a variable

then naturally the limits change to be for that variable

[broken kelp]

I see thanks

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[paper olive]

I find it difficult to do laplace inverse transforms of functions with s^2 in their denominator, does anyone know any formulas that deal with it?

[vernal matrix]

Probably divide it until it becomes a “proper fraction”?

You could e.g. write s^2 = (s + 4)^2 -8s - 16, then you can cancel stuff out, with what results hopefully being easier to deal with

[neon iron]

yeah partial fractions and then look at a table

[paper olive]

wait where do you get the 8s-16 from

ooh I see you expanded it

[vernal matrix]

I wrote s^2 as ((s + 4) - 4)^2 and expanded it (too lazy to try and actually do the polynomial division haha)

Probably might have been more useful as (s + 4)^2 -8(s + 4) + 16 but I never remember the pairs

[neon iron]

hold up, McMeny, were you ever in the IB discord?

Aspect is trolling

your name sounds familiar

[paper olive]

yes..

[neon iron]

ah yes I know you

M23 right

[paper olive]

I am an IB graduate

yessir

[neon iron]

nice

anyway

yeah I suggest using partial fraction decomposition or the method charbit suggested

[paper olive]

oh yeah I forgot about that

[neon iron]

and then you'll either remember or look up the inverse laplace transform of something that'll have c/(s+a) and d/(s+a)^2 where c, d \in R or look it up

here's the table

just split into fractions

try not to give out solutions directly

we want them to figure it out by themselves while providing guidance

[paper olive]

its not really possible to take the inverse laplace of someting like -8s/(s+4)^2 right?

because s is in the denominator

I'd have to use partial fractions for this

[neon iron]

numerator but yeah I don't think we have something for that; partial fractions is your best bet

[paper olive]

hmm interesting; in my professors slides he put the inverse laplace of 1/s as 1, instead of u(t)

[neon iron]

yeah that's right, I think this table is wrong, here's a better one:

[paper olive]

but the delta function's laplace transform is confimed to be just 1 right?

[neon iron]

the laplace transform of the Dirac delta function is 1 yes, if I recall correctly

it's been a while since I did DE so I am a bit rusty, I just remember the general ideas