help-13

[safe cloud]

yeah

[green crater]

u can see that the function on the right is even

[tropic oxide]

do you know what happens when you integrate functions of those types over a symmetric interval

i.e. from -a to a

[green crater]

what can u say abt that?

[safe cloud]

yeah

am I supposed to do like f(x+b/2a)

transform onto origin or smth

[green crater]

mhmmm

[safe cloud]

not origin but

[tropic oxide]

you dont need to do any transformations

[safe cloud]

y axis

[tropic oxide]

you dont need to do any transformations

[green crater]

think abt this

[safe cloud]

bc that would then make it even no?

[tropic oxide]

no, you're rushing a bit

first state what happens when you integrate an odd function over [-a,a] and also what happens when doing the same to an even function.

[safe cloud]

yeah integrating an even function between -a and a lets u utilise the symmetry

[safe cloud]

odd is 0

[green crater]

ye

and even?

[safe cloud]

even is 2\int_0^a

[green crater]

yes!

so now what can u say abt the term on the right

[safe cloud]

so on the right its even

[green crater]

mhmm

[safe cloud]

hang on lemme remember texit

[green crater]

u can just go on paint and screen shot xd

as u want

[safe cloud]

$$ \int_{-5}^{5}ax^2+c dx $$

rippp

[green crater]

ahahahah

i get what u mean

[wraith dagger]

AnitaG

[safe cloud]

close enough

[tropic oxide]

$\int_{-5}^5 (ax^2 + c) \dd{x}$

[wraith dagger]

Ann

[safe cloud]

yes thank you

[green crater]

so now u integrate on the same values

can u separate the integral on the left

[safe cloud]

oh hang on is it bc

[green crater]

to make it look like this one

[safe cloud]

bx is odd

so if u write it as the sum of integrals the bx term disappears

[green crater]

yessir

[safe cloud]

righttt eighttt

[green crater]

good job

[safe cloud]

thanks, I feel a bit dim now 😭

shouldve spotted that from a mile away

[green crater]

no worries

[safe cloud]

appreciate it guys

[green crater]

kinda counterintuituve at first

[safe cloud]

yeah a bit

.close

[cedar kiln]

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[limber turtle]

Claim

[limber turtle]

Sending work wait lads

Is this correct.

.

Bro

?

Finally

[buoyant latch]

Ya

[limber turtle]

Poopoo keyboard

[limber turtle]

Slay

Ty

I enjoyed this one

Bit tricky

[cedar kiln]

@limber turtle Has your question been resolved?

[limber turtle]

oh

i forgot

.close

[cedar kiln]

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[cedar kiln]

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[sudden smelt]

What is the proof of a 2nd degree equation of the form $ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0$ representing a pair of straight lines if \begin{vmatrix}
a & h & g\
h & b & f\
g & f & c
\end{vmatrix} = 0

[wraith dagger]

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

[karmic field]

you know how you tell if all that is representing a pair of straight lines?

[sudden smelt]

wdym

[karmic field]

like how the equation needs to be able to be written as for that to represent a pair of straight lines

[sudden smelt]

as a product of two straight lines?

[karmic field]

yea

[sudden smelt]

mhm

[karmic field]

maybe you can get that det inside the equation

[sudden smelt]

yeah thats my doubt. how do we do that

i cant seem to figure it out

[karmic field]

u = x y 1

i think

as a vector

[sudden smelt]

oh man vectors and determinants are further down my course though. i was so sure there is some calculus or algebraic way to solve this

ofc i know to solve basics qs from them but i dont think im well versed enough in the theory to derive this thru vectors or dets

.close

[cedar kiln]

Channel closed

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[karmic field]

might be

but idk try expanding the determinant

see if it helps you factor the expression out to 2 straight lines

[sudden smelt]

i found a way

apparently if u rewrite the 2nd degree equation as a function of x and find its roots, you'll get conjugate pairs and a term under the root. this term is a quadratic in y and since x must be a linear function of y, the term in this root must be zero or a perfect square to be a straight line. and now, to solve for y as a perfect sqare, its discriminant must be zero. upon solving that equation we get this matrix. therefore the condition satisfies

thank you for helping out!

.close

[karmic field]

yea expanding the determinent you can get the comparison

[sudden smelt]

mhm

[cedar kiln]

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[zenith sedge]

is this classified as valid proof?

[night wedge]

You need to show 2x=6 not use it

I think

[cedar kiln]

@zenith sedge Has your question been resolved?

[night wedge]

Ending at 0=0 isn’t how to show a statement as something false can imply something true, but it will be valid if you are able to reverse your steps.

[zenith sedge]

so im confused

what would be the way to approach this

discarding the way i did it

[night wedge]

You want to show the first line of your working right

I’m not sure

It would be nice to know what z is

If we can find it

[night wedge]

I feel like it’s impossible now I’m confused

[dreamy void]

Since you know that there exists some z such that for all y the equation holds, you may try to think about how to choose such y, so that you arrive at what you want to conclude.

[night wedge]

But surely we need z-y+5=0 though

[dreamy void]

Yes, else it doesnt work out

[mental trail]

Adding to what was already said. If you want to prove A => B, ofc you start by supposing A. But then supposing B as you did is completely useless

It's like saying "If A and B are both true, then I couldn't find any contradictions, so it's fine, so A => B"

Both of those "so" are false reasonings

But the one I'm gonna spend most time on is the "if A and B both being true works, then A => B"

You need to check that A forcefully implies B

[mental trail]

So doing this doesn't even show why A being true and B being false wouldn't work

Now, as a hint on how to prove your statement [.... => (2x = 6)]

||Prove the "A" statement is false||

[night wedge]

But we need to show it for all y

[zenith sedge]

okay wait

would this work

[flint cape]

It does

It's a little clunky, simply because of how you've negated it

Are you aware of what contrapositive refers to? @zenith sedge

[zenith sedge]

yeah i do

a implies b is the same as not b implies not a

[flint cape]

Right

We can use that here

2x != 6 implying [some statement] is the same as [the negation of that statement] implying 2x = 6

And that last one is identical to "x = 3"

So if we take the logical negation of this, it then suffices to show that x = 3

[zenith sedge]

so x-7 = (z-y + 5)(z^2+1) - 4 imples that x = 3

[flint cape]

Then this logic still works

[imma be honest I'm slightly getting sleep deprived, something is ringing bells here to me but it could just be the lack of energy ]

[cedar kiln]

@zenith sedge Has your question been resolved?

[night wedge]

I don’t see why it would be true for all y though no matter what z we choose

[glad peak]

It’s the other way round. For all integers z, you can choose an integer y such that z-y+5=0. i.e. for any z, let y = z+5

[night wedge]

Not when it’s negated

[glad peak]

In that case it’s fine. Same principle applies. For any integer y, there exists an integer z such that z-y+5=0 and that integer z is simply z=y-5

[night wedge]

You can’t just swap those around though

Z is defined first

[glad peak]

Why would order matter?

[night wedge]

One says it’s true for all y after picking a z, so z cannot depend on y, and we can’t choose y

Other says for all y i can then pick a z

Which is easy just define z so the term is 0

[glad peak]

Neither z nor y are strictly defined prior to the statement. They’re just know to be elements of Z

[cedar kiln]

Channel closed

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[glad peak]

As I see it, it reads as ‘there exists an integer z, for any integer y such that (statement)’. So it is perfectly reasonable to let z depend on y, after all they will both just be integers

.reopen

[cedar kiln]

✅

[night wedge]

I thought the variables can only depend on previous and not future declarations

Statement 1: For all a there exists b such that a=b, true

Statement 2: there exists b such that for all a a=b, false

It’s not commutative, but it is in one direction

In general, if statement 2 is true, it gives statement 1 but not in reverse

<@&286206848099549185> can we get other helpers input on this?

[cedar kiln]

@zenith sedge Has your question been resolved?

[tawny sage]

In general this is false.
"There exists an integer z, such that for all integers y, z+y = 0" is a false statement
"For all integers y, there exists an integer z such that z+y=0" is a true statement

order very much matters and you cant let variables defined earlier in a statement like this rely on later ones

[night wedge]

Wait you can let variables depend on later ones?

[tawny sage]

i missed a "t" oops

[night wedge]

But back to the original problem

I don’t see how the contrapositive is true

We need a z that satisfies it for all y

This isn’t even my own problem but I need help on this

Surely as y varies the term will change because z^2+1 does not equal 0

[tawny sage]

I think the proof would look like this:

For any x, and any z, let y = z+5. Assume that x-7 = (z-y+5)(z^2+1) - 4. This would imply that x-7=0-4 and therefore that x=3, contrary to our assumption.

We don't reorder the quantifiers because we're not negating those

[night wedge]

Don’t we flip quantifiers on not

From forall to exists

And vice versa

[tawny sage]

I don't see any reason to negate the entire statement. If we do, to try to find a contradiction, then we're left with a "for all statement" for y. Those are hard to prove directly

[night wedge]

If we are trying to find the contrapositive we do?

[night wedge]

^

(I didn’t mean it like do you know what a contrapositive is, I meant it like this is what we’re trying to do)

[tawny sage]

if you wanted to do the proof via contrapositive it'd look very similar. We're trying to prove a statement of the form P->Q, by proving that -Q -> -P

starting with -Q:
exists z, for all y, x-7 = (z-y+5)(z^2+1) - 4
exists z, s.t for y=(z+5), x-7 = (z-y+5)(z^2+1) - 4
exists z, s.t x-7 = -4
x=3
2x = 6 (-P)

[night wedge]

Why can you set y=z-5

We’re doing over all y in integers

[tawny sage]

it's an implication, a statement being true for all y implies that it's true for a specific y

[night wedge]

Oh I thought we were trying to show that statement

I see now

Ofc I got stuck when I not it it isn’t true anymore

[tawny sage]

lol

[night wedge]

False implies ? is true

Wait then wouldn’t any x satisfy that

Since we are starting from false we can show anything

[tawny sage]

assuming something is false is very different than assuming a false thing

[night wedge]

Oh I see

[tawny sage]

if you assume 0=1 you can prove anything
if you assume x=1 is false, then what you can prove is very limited

[night wedge]

If you replace x with 0 in that message i think I get it

[tawny sage]

yeah sure that works too

it's probably a better way of explaining it

[night wedge]

Aren’t we assuming a false thing though (in the proof)

[tawny sage]

nope, "0=1 is false" is a true thing

[night wedge]

Like that 2nd line of the message here (sorry this was a bit unclear*)

[tawny sage]

that's not a false statement, it's just a statement that implies that x=3

[night wedge]

There is no z though

If it’s true, then give me a z

[tawny sage]

like this line?

exists z, s.t for y=(z+5), x-7 = (z-y+5)(z^2+1) - 4
if z = 0 and y = 5 then this statement could be either true or false depending on what x is

[night wedge]

No the one above it

[night wedge]

^

[tawny sage]

ah I think I see what you're asking, let me think

[night wedge]

Clearly I found a contradiction within mathematics, I should get a fields medal

I must have make a mistake somewhere in the logic

[tawny sage]

true true

[night wedge]

I might ask tomorrow it’s 2:30am here

[tawny sage]

I'm not sure the exact formal logic way of stating the principle of explosion, so I'm not sure the exact restriction on assuming false things

[night wedge]

Just stayed up for this one confusing question

I think tomorrow morning

[tawny sage]

yeah I gotta go in a minute here anyways

[cedar kiln]

@zenith sedge Has your question been resolved?

[cedar kiln]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[fallen yew]

For linear algebra. not sure where to start, text book says this but ive never done that in class

[hollow trail]

i would recommend forming the matrix with u,v,w as its columns as shown in the picture

[fallen yew]

alright

ok done

[fallen yew]

when it says AX i dont know what that means thats the issue

[hollow trail]

have you worked with linear systems of equations before?

[fallen yew]

yes

[hollow trail]

and have you worked with them in matrix form?

[fallen yew]

yes

just elementary operations so far

[hollow trail]

so if we have a vector x which has the unknowns and b which has constants, and A is a matrix, then Ax = b is a linear system of equations

[fallen yew]

OHHHH WAIT I HAVE DONE THIS

i just dont understand it very well

here let me try

[fallen yew]

am i onto something

[hollow trail]

that's a good start

once you have a system of linear equations, it is a good idea to apply gaussian elimination

[fallen yew]

what is that

i have never heard of that

[hollow trail]

have you not done any row operations?

[fallen yew]

yes i have

[hollow trail]

like finding reduced echelon form (ref)

[fallen yew]

yes

[hollow trail]

so gaussian elimination is the process of finding ref

[fallen yew]

ohh ok

have i done ot

[hollow trail]

i would usually recommend leaving it in matrix form while doing row operations but that works

[fallen yew]

ohhh okay

thank you :)

.close

[cedar kiln]

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[cedar kiln]

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• Show your work, and if possible, explain where you are stuck.
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[supple falcon]

I need help solving this for $(x, y) \in \mathbb R$: $\begin{cases} x^2(x - y) + (y - 1)^2 = 0\ 4x^3 - 9x^2 + 7x + 3y^2 - 10y + 5 = 0 \end{cases}$

[wraith dagger]

1 divided by 0 equals Infinity

[supple falcon]

where do i start?

and what is wrong with this latex 😭

i could try to factorize the second expression

but

,w simplify 4x^3 - 9x^2 + 7x + 3y^2 - 10y + 5

[wraith dagger]

Results provided by WolframAlpha

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

Requested by iaminfinityiq

[wicked mantle]

I need help solving this for
\$(x, y) \in \mathbb R$: $\begin{cases} x^2(x - y) + (y - 1)^2 = 0\ 4x^3 - 9x^2 + 7x + 3y^2 - 10y + 5 = 0 \end{cases}$

[wraith dagger]

🌙 ЅκψΑиdΝιɡħτ

[supple falcon]

anyways, where do i start from this?

[tropic oxide]

did yall mean $(x,y)\in\bR^2$ tho

[wraith dagger]

Ann

[supple falcon]

okay okay

you get my point

x, y are real

[tropic oxide]

anyway uhhh this looks mega yucky

[supple falcon]

,w simplify x^2(x - y) + (y - 1)^2

[wraith dagger]

Results provided by WolframAlpha

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

Requested by iaminfinityiq

[supple falcon]

impossible to factorize over R

[stiff brook]

,w factorize x^2(x-y) + (y-1)^2

[wraith dagger]

Results provided by WolframAlpha

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

Requested by leftover5270

[stiff brook]

cool

now both of them can be 0

[tropic oxide]

oh it factorizes does it

[supple falcon]

yo wolfram alpha, what's wrong

[supple falcon]

ye i see

[supple falcon]

but how can i get an idea of this factorization?

[tropic oxide]

divine inspiration

[stiff brook]

u mentioned trying to factorize the 2nd expression

[supple falcon]

im being serious 😭

[stiff brook]

it wasnt divine inspiration, it was 1 divided by 0 equals infinity inspiration

[supple falcon]

IM BEING SERIOUS

😭

[stiff brook]

im being serious too

i wouldnt know how to approach it without wa

[prisma pike]

Only made sense to me after I graphed it on desmos

[opal hinge]

Just try to factorize everything if you lucky you'll get this fast

[digital bane]

this looks like some next-level factoring by grouping after expanding as necessary

[ancient lodge]

tbf the first equation isn't awful to solve as a quadratic in y

the discriminant simplifies quite nicely

[cedar kiln]

@supple falcon Has your question been resolved?

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• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[supple falcon]

New question: I simplified to $4y^3 - 18y^2 + 7y - 15 = 0$ for the first case

[wraith dagger]

1 divided by 0 equals Infinity

[supple falcon]

I found out that one of the roots was $\frac{3 + \sqrt[3]{3}}{2}$

[wraith dagger]

1 divided by 0 equals Infinity

[supple falcon]

How do I work this around algebraically?

[ebon stump]

What do you mean?

[supple falcon]

how do i get from this

[supple falcon]

to this

[ebon stump]

Oh, er... that's complicated

[supple falcon]

by not using the general formula or cardano's method

[supple falcon]

oh wait, it's $4y^3 - 18y^2 + 27y - 15$

[wraith dagger]

1 divided by 0 equals Infinity

[supple falcon]

my bad

[ebon stump]

Oh, you missed a 2 originally did you?

[supple falcon]

yeah

still

how can i work around to get $\frac{3 + \sqrt[3]{3}}{2}$ as a root

[wraith dagger]

1 divided by 0 equals Infinity

[ebon stump]

The simple answer is "solving cubic equations is hard"

[supple falcon]

if you scrolled up, i just used this help channel and there should be a system of equations in there

[stiff brook]

where does this question come from and what tools can u access

[supple falcon]

which question?

[supple falcon]

i have access to a calculator

[stiff brook]

the original one

[supple falcon]

and for now, wolfram alpha

[ebon stump]

If you're just doing it by hand, then even the university course on Galois theory doesn't really let you solve cubic equations like this without putting in significant effort

[stiff brook]

okay u can use that

[supple falcon]

.

[ebon stump]

That said, given the one solution, I can tell you the other 2 lol

(Over the complex numbers)

[supple falcon]

knew it

[stiff brook]

yes if u have one real root, u can do synthetic division to retrieve a quadratic

but anyways, u have a calculator!!

[supple falcon]

yes i have

[supple falcon]

uh

[supple falcon]

well the factor would be something like

$(y - \frac{3 + \sqrt[3]{3}}{2})$

[wraith dagger]

1 divided by 0 equals Infinity

[supple falcon]

💀

[stiff brook]

yes

[supple falcon]

goodness

irrational coefficients

[supple falcon]

how does a calculator do synthetic division 💀

[stiff brook]

if u have a calculator, u can solve for all 3 roots directly.

[ebon stump]

Shortcut using Galois theory if allowed:
You can say that because this equation was a cubic equation, and this root lies in $\mathbb{Q}(\sqrt[3]{3})$ which is a degree 3 extension of the rational numbers, the other roots are the just the Galois conjugates which can be determined by replacing $\sqrt[3]{3}$ with the other complex cube roots of 3.

[wraith dagger]

Haine

[supple falcon]

i disabled complex roots on the calculator

so only 1 root

[stiff brook]

well enable them

[supple falcon]

written as 2.221124785

[supple falcon]

im working with reals

[ebon stump]

There is only 1 real root

[stiff brook]

then u dont need the complex roots

[ebon stump]

You were just wanting to be able to solve it right? But that's like much harder than any of the other steps here.

[supple falcon]

yeah

[ebon stump]

Well, unless you just use a calculator like wolfram alpha or are happy with numerical approximations

[supple falcon]

i was doing the exercise given above (which just got flowed out for some reason)

somehow i got to $4y^3 - 18y^2 + 27y - 15 = 0$

[wraith dagger]

1 divided by 0 equals Infinity

[crimson sedge]

1 divided by 0 is undefined pal

[supple falcon]

get out

[wraith dagger]

Haine

[supple falcon]

original question.

[ebon stump]

Oops, did it backwards

Deleted source instead of image

[crimson sedge]

#include <iostream>
#include <string>

int main(){
std::cout << "anything divided by 0 is undefined buddy" << std::endl

return 0;

}

[prisma pike]

You're not tough man

[ebon stump]

0/0 = 0 in the 0 ring 😉

[supple falcon]

#include <iostream>

int main() {
    std::cout << "stop raiding my channel and get out\n";
    std::cout << "why the hell do you need the <string> header LMAO\n";
    return 0;
}

[prisma pike]

Bro forgot the semicolon too🙏 🙏

[supple falcon]

#include <iostream>

int main() {
    int x;
    std::cout << "Which roast do you want? ";
    std::cin >> x;

    std::cout << "Don't care, who needs your opinion\n";
    return 0;
}

[opal hinge]

What're we doing here guys

[uneven sage]

lmaoo

[crimson sedge]

#include <iostream>
#include <string> 

int main () {

std::cout << "you technically cant output strings without the string class youre outputting chars rn buddy " << std::endl;




return 345;

}

[ebon stump]

Getting distracted by the looks of it

[uneven sage]

is this c++

ah

[fossil dawn]

folks, let's not get out of hand

gdi

!redit

ah fuck it, for regular discussions, please move to #discussion or #chill

[opal hinge]

!redir

[cedar kiln]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[uneven sage]

[supple falcon]

anyways i need my question done

[wicked mantle]

wait what? It’s not done yet?? 😂

[uneven sage]

what's your question again? i can't find it..

[opal hinge]

-# What're you cooking, LNRD

[fossil dawn]

check pins for question

[supple falcon]

im at the cubic

[uneven sage]

okie

[crimson sedge]

void iamInfinityCannotDoMath(int x, int y){

try {

if ( y = 0 ) {

std::throw runtime_error("Anything divided by 0 is undefined buddy ");

}

int main(){

int x = 1;

int y = 0

 iamInfinityCannotDoMath(x,y);


return 9999;

}

[supple falcon]

YO

💀

[crimson sedge]

[fossil dawn]

this is getting unproductive. please leave this channel be if you're not helping the helpee.

[wispy marten]

Yo!

[wicked mantle]

@crimson sedge Brother, we need you to stop

[crimson sedge]

i gotchu

[crimson sedge]

mb just thought of having some fun

@supple falcon goodluck with your math

[supple falcon]

lol

[supple falcon]

ain't runnin

[crimson sedge]

i need to include libraries

[tropic oxide]

indentation 💀

or rather lack thereof

[supple falcon]

fr 💀

[supple falcon]

anyways, how can i work this around?

im talking about real roots here

not the cardano's method

[crimson sedge]

is this ODEs

[supple falcon]

not using a formula to find the roots

[supple falcon]

wolfram alpha's doing

[crimson sedge]

whats the topic i mean

[supple falcon]

try-catch gone wrong 💀

[slender atlas]

didn't someone mention that there was only one real root?

[crimson sedge]

its been a while since i coded anything

sign that i need to practice

[supple falcon]

ik

but how can i work around to get that root?

[slender atlas]

so you know what the root is, and you want to actually prove by hand that it is a root. am I understanding this?

[crimson sedge]

what class is that

[supple falcon]

kinda like that

[supple falcon]

yes, you are understanding it

[supple falcon]

algebra

[slender atlas]

I see. but earlier I think you mentioned you are allowed a calculator, correct? is there a reason you would like to prove this by hand?

[crimson sedge]

whats your question again

can you ping it

or smth

[supple falcon]

because i have to write that on paper

i can't just say that this has a root of $\frac{3 + \sqrt[3]{3}}{2}$

[wraith dagger]

1 divided by 0 equals Infinity

[wicked mantle]

@crimson sedge and again, you should stop disturbing him further

[supple falcon]

directly

[crimson sedge]

im trying to help

[slender atlas]

because given that the root is irrational, and such a complicated one at that, not only does the RRT fail here, but trial division would also suck given that a cube root is involved here.

[supple falcon]

ik

[crimson sedge]

yea..

[slender atlas]

oh but, you can try to depress the cubic here by setting y = (x - b/3a), if you've heard of that.

[supple falcon]

yea i've heard of that

how do you solve a depressed cubic?

it's in the form of $ax^3 + bx + c$ if im not wrong?

[wraith dagger]

1 divided by 0 equals Infinity

[slender atlas]

yes, and this one, after expansion, might cancel out nicely.

[upper ruin]

You can replace every instance of x with this and then check that you get polynomial = 0

[supple falcon]

that's obvious

[upper ruin]

And so???

[supple falcon]

that's plug and checking

[upper ruin]

You asked to prove it's a root

[upper ruin]

Of course! It's just the definition of root

[supple falcon]

this channel got 2 people ain't serious

well

when you are trying to solve $x + 23923048910483138904 = 0$

do you say

[wraith dagger]

1 divided by 0 equals Infinity

[upper ruin]

Why are you calling me not serious??

[supple falcon]

we have $-23923048910483138904 + 23923048910483138904 = 0$

[wraith dagger]

1 divided by 0 equals Infinity

[upper ruin]

Ok you Re trolling then

[supple falcon]

so $-23923048910483138904$ is the solution to the equation

[wraith dagger]

1 divided by 0 equals Infinity

[supple falcon]

do you do that?

[upper ruin]

<@&268886789983436800>

[slender atlas]

either way, if you want to start, try depressing your cubic first.

[supple falcon]

ANSWER ME

[frank minnow]

What’s going on

[upper ruin]

I'VE ALREADY ANSWERED YOU!

[supple falcon]

this guy is not being serious on helping me with my problem

[wicked mantle]

Bruh

[upper ruin]

Me? Maybe you mean LNRD

[supple falcon]

and you

[stiff brook]

we have told you countless times to use a calculator. i dont understand why u insist on solving it by hand

u can go use ur general cubic formula if u want.

[frank minnow]

Can someone give me a summary of what's happened

[upper ruin]

Lol then you don't deserve to stay in this server, let me tell you

[upper ruin]

#help-13 message

[wicked mantle]

I legit have no idea why Alberto is considered not serious

[crimson sedge]

👋

[wicked mantle]

@supple falcon is there any reason that you insists on solving it by hand?

[supple falcon]

i can't just write that on paper that:
$4y^3 - 18y^2 + 27y - 15 = 0$
This equation has a root of $y = \frac{3 + \sqrt[3]{3}}{2}$

[wraith dagger]

1 divided by 0 equals Infinity

[supple falcon]

I'm not allowed to do that on paper

[upper ruin]

But can't this be also done by hand? @wicked mantle

[supple falcon]

and especially on my test

[slender atlas]

actually, one question.

[frank minnow]

Okay so based on my quick skim of chat
@supple falcon wants to know how to solve this cubic by hand
@upper ruin is going ahead and saying that its not really possible to solve this by hand and you should just use another source to find this is a root and plug it in
And you're both getting mad at eachother about that

[supple falcon]

you have to do it step by step

[wicked mantle]

It can, but why complicate our lives

[supple falcon]

go ahead

[slender atlas]

you're allowed to use a calculator. may I know what expectations for steps there is to solve this by hand if you're allowed a calculator? especially on a step as tedious as this?

[frank minnow]

Is this correct?

[supple falcon]

well for me it's correct

[upper ruin]

Because he wants to 🙈

[frank minnow]

Is my summary correct from your end?

[supple falcon]

if i didn't learn the quadratic formula

for example: for $x^2 - 3x + 2 = 0$, i would actually have to factor out if i didn't learn the formula

[wraith dagger]

1 divided by 0 equals Infinity

[supple falcon]

my calculator is just a tool to help me determine the roots

[supple falcon]

well it's how it works in my country, i don't know why

[stiff brook]

.

[slender atlas]

if that's really the case, your only hope for this question is to pray that the depressed cubic will save you (and hint: it does).

[supple falcon]

alr thanks

[slender atlas]

that's all I can say, since finding the depressed cubic is just mechanical steps.

[frank minnow]

Okay so I don't find that @supple falcon is acting in bad faith here, so I won't be timing them out or anything. As much as their question kinda doesn't have a better answer than "Use cubic formula", asking questions that are kinda not the right question to ask isn't against the rules.

@upper ruin I'd recommend if a helpee is irritating you this much in future to just ghost them or take a step back from the conversation as getting irritated with them doesn't help anyone.

@crimson sedge if you want to pick fights with someone bc of their name feel free to do that in the many social channels we have. Otherwise don't engage if you're not contributing anything useful

[supple falcon]

thanks for resolving this out

[crimson sedge]

dude i wasnt tryna fight anyone, we were just having fun

[frank minnow]

You can have fun in other channels, the help channels are expected to be somewhat formal and academic

[wicked mantle]

No way you say this in front of a mod 💀

[supple falcon]

fr 💀

[frank minnow]

If you're caught "just having fun" again in here I will be getting

[crimson sedge]

i gotchu

[stiff brook]

has the question been resolved

[crimson sedge]

seems like it

[supple falcon]

yes

i'll close now :)

[crimson sedge]

(or not, its a cubic equation) (you can solve it but not by hand)

[supple falcon]

thanks guys

.close

[cedar kiln]

Channel closed

Closed by @supple falcon

Use .reopen if this was a mistake.

[upper ruin]

Quite much, yes

[cedar kiln]

Available help channel!

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[vast scaffold]

.reopen

[silent stratus]

Have I done it correct??

[opal hinge]

,rccw

[rotund sapphire]

you did one thing wrong

[opal hinge]

Can you send the problem? It would help

[gray hamlet]

look at the 4th line closely

[cedar kiln]

@silent stratus Has your question been resolved?

[cedar kiln]

Channel closed

Closed due to timeout

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Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[lost meteor]

im struggling with this. im soposed to determine if this has a limit, is konvergent and / or monoton. but im just confused on where to even start here. i managed to do the previous ones but this one seems complicated.

[lost meteor]

für = for

its in german, sorry

[dreamy void]

Die Folge erinnert ein bisschen an Fibonacci

[lost meteor]

du kannst deutsch

was eine erleichterung

aber wie ist das Fibonacci

das checke ich nciht

[dreamy void]

Also ich meine von der Struktur nur

Fibonacci war ja f_n = f_(n-1)+f_(n-2)

[lost meteor]

aber wenn man ein paar glieder berechnet kommen werte wie 1/2 dann 3/4 dann 5/8 dann 11/16

scheint schon sehr random zu sein

[dreamy void]

Die scheint zu steigen

aber immer unter 1 zu bleiben

[lost meteor]

steigen? 0,5 , 0,75 , 0,625 , 0,6875

das ist so ein bisschen hin und her. es sei denn ich habe es falsch gerechnet

[dreamy void]

Ah ne hast Recht

[lost meteor]

was jetzt?

[crimson delta]

na eine frage hast du damit ja beantwortet

[lost meteor]

das sie nicht monoton ist

oder was

[gray hamlet]

und dass sie eine limit hast

[lost meteor]

ja, aber wie kann man diese Limit berechnen?

[crimson delta]

ne das noch nicht

[gray hamlet]

doch

es ist konvergent somit es eine limit hat

[crimson delta]

was passiert wenn du jede zweite zahl anguckst

[crimson delta]

das wurde noch nicht gezeigt

[lost meteor]

ich bin verwirrt

ich dachte man kann damit nur sagen (bis jetzt) das es nicht monoton ist

[crimson delta]

ja

[lost meteor]

und dann?

[crimson delta]

ich glaub das funktioniert

nicht sicher

man muss halt abundzu ein bisschen rumspielen

[lost meteor]

ich checke nicht was du damit meinst

[gray hamlet]

$a_{n}=a_{n-1}+(-\frac{1}{2})^{n-2}$

[wraith dagger]

Roy

[dreamy void]

Ich denke, du sollst die Folge in zwei Teilfolgen trennen, jeweils mit geraden und ungeraden Gliedern

[lost meteor]

wie geht das

[dreamy void]

Ich denke wenn beide gegen den selben GW konvergieren, dann gilt das auch für die ganze Folge

[lost meteor]

bist du sicher?

[crimson delta]

ja das gilt

[lost meteor]

und wie teile ich es auf

[dreamy void]

Ich denke es gibt auch einen Ansatz für lineare Rekursionsgleichungen, aber sei das mal hingestellt

[dreamy void]

Du betrachtest a_(2n) und a_(2n-1)

[lost meteor]

ok

wie soll ich daraus 2 folgen aufstellen

[crimson delta]

ja man kann auch direkt ne geschlossene form berechnen aber darauf zu kommen ist glaube ich hier nicht erwartet

[dreamy void]

Ja denke ich auch

[crimson delta]

du hast die folge x_n=a_(2n)

und die folge y_n=a_(2n+1)

[lost meteor]

sowas habe ich noch nie gesehen

[crimson delta]

berechne doch erst mal die ersten paar mitglieder von beiden folgen

[lost meteor]

was ist dieses x_n und y_n

[dreamy void]

einfach namen für die teilfolgen

sei x_n die folge mit geraden gliedern und y_n die folge mit ungeraden

[crimson delta]

du nimmst a0, a2, a4, a6, a8, a10, ...

und packst die in eine folge

und dann a1,a3,a5,a7,a9,... und packst die in ne andere

[lost meteor]

ich verstehe nicht wie ich die in eine FOlge packen sollte. DIese Glieder kann ich doch nur mit der originalen FOlge berechnen

[crimson delta]

naja du hast ja noch die gesamte folge

bloß halt aufgeteilt

ich behaupte nicht dass du die x_n berechnen kannst ohne die y_n zu kennen

[crimson delta]

ich möchte immer noch dass du das tust

[lost meteor]

habe ich

[crimson delta]

mehr

[lost meteor]

ok

a7 = 21/32 und a8= 43/ 64

ich darf eigentlich kein taschenrechner nutzen

d.h. das in dezimalzahlen angeben wäre nicht erlaubt. es sei denn ich mache schriftliche division

[crimson delta]

naja haben alles 2er potenzen unten

kann man also leicht vergleichen wenn man will

man muss nicht immer mit dezimalzahlen rechnen

[lost meteor]

wie muss ich also weiter

[crimson delta]

was passiert wenn du dir jetzt die zahlen x_n und y_n anguckst

fällt dir etwas auf?

[lost meteor]

was ist überhaupt x_n und y_n

sorry

[crimson delta]

x_n

[crimson delta]

y_n

[dreamy void]

[
\begin{array}{cccccc}
(x_n) = (\textcolor{red}{a_0}, & \phantom{a_1}& \textcolor{red}{a_2}, & \phantom{a_3} & \dots ) \
(y_n) = ( \phantom{a_2} & \textcolor{blue}{a_1}, & \phantom{a_2} & \textcolor{blue}{a_3}, & \dots )
\end{array}
]

[crimson delta]

(ich hab x_n die geraden indizes)

[lost meteor]

der nenner ist bei y_n immer das 4x

als das davor

aber bei x_n auch

ansonsten fällt mir nicht mehr auf

[cedar kiln]

@lost meteor Has your question been resolved?

[crimson delta]

was fällt dir auf bzgl monotonie und so zeug

[cedar kiln]

Channel closed

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[vagrant yoke]

Hello , I want to learn calculas so can any suggest me from where can I learn it ? 🙂

[night wedge]

Might want to talk in #precalculus

[opal hinge]

books and youtube

[vagrant yoke]

ok

but

which books I have to prefer for learn about calculas

on online 🙂

[night wedge]

Help is usually for specific problems but sometimes we do talk more generally here, but I feel it’s more suited for the discussions channels like #precalculus

As for the question, you will need to know about limits

[vagrant yoke]

for integration ? or in discord

i think for integration

i do not prefer rectangle method for calculas because it give some more area of curve right ?

[cedar kiln]

@vagrant yoke Has your question been resolved?

[night wedge]

I’m not sure what you’re saying here, the rectangle method does give the correct area under the curve with small enough rectangles so it’s not incorrect, but the trapezium method just approaches the answer faster

[gray hamlet]

best is khan academy

they have pretty good lessons

[cedar kiln]

Channel closed

Closed due to timeout

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[cedar kiln]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[supple falcon]

Back to this question again

[supple falcon]

I'm in a progress of this question, AGAIN

so according to the help i got so far

i managed to factorize one equation to $(x - y + 1)(x^2 - x - y + 1) = 0$

[wraith dagger]

1 divided by 0 equals Infinity

[supple falcon]

i handled one of the cases, which is $x - y + 1 = 0$

[wraith dagger]

1 divided by 0 equals Infinity

[supple falcon]

which is the polynomial i talked also in here

[supple falcon]

how do i handle the other case, which is $x^2 - x - y + 1 = 0$?

[wraith dagger]

1 divided by 0 equals Infinity

[wicked mantle]

Question👇

[supple falcon]

yeah yeah

[supple falcon]

thanks

[supple falcon]

that turned into

$\begin{cases} (x - y + 1)(x^2 - x - y + 1) = 0 \ 4x^3 - 9x^2 + 7x + 3y^2 - 10y + 5 = 0 \end{cases}$

[wraith dagger]

1 divided by 0 equals Infinity

[supple falcon]

I handled the case that $x - y + 1 = 0$

[wraith dagger]

1 divided by 0 equals Infinity

[supple falcon]

How do I handle $x^2 - x - y + 1 = 0$

[wraith dagger]

1 divided by 0 equals Infinity

[marsh mesa]

Good morning. @supple falcon

[supple falcon]

good evening

[supple falcon]

what do you want

<@&286206848099549185>

[frail citrus]

cree-

[supple falcon]

anyways can you help me on this?

[native heath]

why

[supple falcon]

solve for reals

[supple falcon]

so for this system

[supple falcon]

i managed to turn into this one

[marsh mesa]

Your username is technically wrong. Division by zero is undefined. If you wanted to be precise, it should read ‘1 divided by 0 equals undefined.’ Less catchy, but correct.

[supple falcon]

and i handled this case

[supple falcon]

but how can i handle this?

if you want to mess with me, read this

@marsh mesa

[marsh mesa]

You might want to repeat elementary algebra before tackling these exercises. Just a suggestion.

[supple falcon]

im in elementary algebra

[supple falcon]

secondly, if you can't help then please don't raid

[slender atlas]

I'm not sure if there's anything to gain from the first case to use in the second case, because I can't help but notice that the last three terms of the second case are the same as the first one.

[supple falcon]

i don't want to handle this kind of trolling for the 2nd time of the day

and the 3rd time for the last 2 days

[supple falcon]

i mean, i see it's kinda the same

but unfortunately, the term was $-x$ compared to $x$

[wraith dagger]

1 divided by 0 equals Infinity

[supple falcon]

this one, i solved for $x$ through $y$ and substituted into the second equation

[wraith dagger]

1 divided by 0 equals Infinity

[supple falcon]

but if i try to solve for $x$, then i would have $\pm$

[wraith dagger]

1 divided by 0 equals Infinity

[supple falcon]

oh...

[slender atlas]

I'm tempted to get y as a quadratic in x, substitute into eq. 2, and then factor by grouping or RRT reduction.

[supple falcon]

i forgot i can substitute $y = x^2 - x + 1$

[wraith dagger]

1 divided by 0 equals Infinity

[supple falcon]

oh nvm

but if i needed more help

i'll just post here

[stiff brook]

this is roughly what happened but since the first eq was factorable, it split into nice cases

[slender atlas]

I'm not sure exactly what cases you meant, but I was thinking of this.

[supple falcon]

oh god

[supple falcon]

this case

[slender atlas]

substituting this into eq. 2 gives us a quartic, but through RRT you can reduce the quartic into a quadratic.

[supple falcon]

,w verify 4x^3 - 9x^2 + 7x + 3(x^2 - x + 1)^2 - 10(x^2 - x + 1) + 5 = 3x^4 - 2x^3 - 10x^2 + 11x - 4

[wraith dagger]

Results provided by WolframAlpha

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

Requested by iaminfinityiq

[slender atlas]

(if I did not screw up, which I'm willing to very much bet I did.)

[supple falcon]

,w simplify 4x^3 - 9x^2 + 7x + 3(x^2 - x + 1)^2 - 10(x^2 - x + 1) + 5

[wraith dagger]

Results provided by WolframAlpha

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

Requested by iaminfinityiq

[supple falcon]

i think my calculations are wrong lol

[stiff brook]

yes rrt does work then

looks like it factors over Q

[supple falcon]

can you try to check for me where did my calculation gone wrong?

$4x^3 - 9x^2 + 7x + 3(x^2 - x + 1)^2 - 10(x^2 - x + 1) + 5 = 0$

[wraith dagger]

1 divided by 0 equals Infinity

[supple falcon]

$\Leftrightarrow 4x^3 - 9x^2 + 7x + 3x^4 + 3x^2 + 1 - 6x^3 - 6x + 6x^2 - 10x^2 + 10x - 10 + 5 = 0$

[wraith dagger]

1 divided by 0 equals Infinity

[supple falcon]

$\Leftrightarrow 3x^4 - 2x^3 - 10x^2 + 11x - 4 = 0$

[wraith dagger]

1 divided by 0 equals Infinity

[slender atlas]

my constant term is -2.

[supple falcon]

oh i saw where im wrong

okay i see

i think i can resolve the rest of this

thanks guys!

.close

[cedar kiln]

Channel closed

Closed by @supple falcon

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[sonic fossil]

!status

[cedar kiln]

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.
4. I got an answer and would like my work checked.
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

[sonic fossil]

1. i don't know where to begin

i have tried on this question 3 to 4 times before but i wasn't successful to find answer. Any helper take me to final answer

.close

[cedar kiln]

Channel closed

Closed by @sonic fossil

Use .reopen if this was a mistake.

[cedar kiln]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[plain arrow]

help i just have a small question

[plain arrow]

if i divide 10.02 cm by two, is the answer 5.01cm or 5.010cm

[slender atlas]

is this physics?

[plain arrow]

just solid mensuration

[slender atlas]

and if it is, how did you obtain 10.02cm - through measurement, or given?

[plain arrow]

it was given by a caliper

so like measured, but given

if that makes sense

[native heath]

depends on how many sigfigs that is

conventionally thats the former

[slender atlas]

I would say 5.010cm, because the original measurement has 4sd.

[gray hamlet]

5.01

[plain arrow]

i know it is four and therefore 5.010cm but when comparing to the source material it says 5.01

the problem was worded exactly as "A caliper was used to measure the cross-section of a cylinder. It gave a reading of 10.02cm. Calculate the area of the cross-section."

so like i really dont know as this is just a practice problem and i dont know which sigfig to use whenever it comes in the real exam

[native heath]

by rights it should be 5.010

[plain arrow]

so i add a trailing zero?

[neon crystal]

yes

[plain arrow]

ahh

i should contact my lecturer with this but it's very deep into the night here right now

[neon crystal]

to be honest

you dont know the measurement tool

uncertainty

[plain arrow]

a venier caliper

Vernier

10.0 from the graduation, .02 from the alignment

[native heath]

if it was more than pm 0.01 they wouldnt report the last sigfig now would they

[native heath]

oh

ok then reporting the last sigfig is weird to begin with

[slender atlas]

in physics, that would normally not count as a significant digit (in chemistry it does though, for some reason).

[night wedge]

Shouldn’t it be 10.00?

If we’re dealing with 0.01

[native heath]

it should be 10.0

[native heath]

what

[plain arrow]

the caliper has graduations of .1 centimeters

it is 10.0, but with extra

[slender atlas]

anyway, since you know this, there seems to be only 3 significant digits, not 4. the last digit is not reliable because it's eyeballed.

[plain arrow]

the top and bottom scales are aligned to yield .02

i think

[night wedge]

Shouldn’t the error be 0.1 or 0.05 then?

[plain arrow]

it is not eyeballed i think

[native heath]

whats your uncertainity

im so confused

what

[plain arrow]

https://www.reddit.com/r/mechanical_gifs/comments/6i0iia/how_to_use_a_vernier_caliper/#lightbox

[native heath]

whats the least reading of your caliper

[night wedge]

This is why I sucked at uncertainty in physics

[plain arrow]

the least reading is 0.1 centimeter

[nova snow]

perky

[native heath]

ok then your measurement is 10.0 not 10.02

cuz you cant measure that 0.02

also lmfao vernier caliper with least reading of 1 mm?

trash caliper what

[plain arrow]

wait what is the least reading

[native heath]

least counts are 0.1 mm typically

[plain arrow]

the alignment between the top and bottom scales?

or the graduation of the top scale

the graduation of the top scale is 1mm, but the alignment is 0.1 mm

i am so sorry for the confusion

[native heath]

its the difference in the least count of the main scale and the vernier scale

[plain arrow]

ah

the least count is 0.1mm then

the main scale is 0.1cm

[native heath]

alignment doesnt matter

its

sigh

[plain arrow]

i am sorry

[native heath]

what are the graduations on the vernier scale

its like 0.9 mm typically

[plain arrow]

yes it is 0.9mm

[native heath]

ok so LC of 0.1

0.1 mm

[plain arrow]

i see

[native heath]

so 0.1 mm of uncertainity

[plain arrow]

does this mean that the 0.2mm is insignificant

i mean the .02 cm part

[native heath]

what does that mean

its just whatever you have while you measured?

[plain arrow]

so it is significant

is the diameter therefore 10.02cm ?

[native heath]

so yeah your reading is 10.02 cm

unless you account for human errors or smth and consider your uncertainity greater than pm 0.01

half that should be 5.010

[plain arrow]

i see

i was worried because adding a trailing zero isnt intuitive to me

[night wedge]

If we can have 0.01cm before dividing by 2 then when dividing by 2 we will have 0.005

[plain arrow]

for future reference;
if i have an answer whose number of sigfigs is less than that of the input (given value) with the least sigfigs, should i add trailing zeroes

[night wedge]

If the smallest of whatever you’re using is 0.02cm then it should be 0.01cm?

[plain arrow]

the smallest is 0.01cm

[native heath]

what

[night wedge]

I might be wrong but I don’t get significant figures because it can add or remove precision

[native heath]

honestly

read the wikipedia page

https://en.wikipedia.org/wiki/Significant_figures

[plain arrow]

say if i divide 104cm by 2, is my answer 52cm or 52.0cm

[native heath]

rare example of wikipedia being good

[plain arrow]

i assume 52.0

[plain arrow]

thank you

[night wedge]

Like if your initial precision is 0.015 then halving makes it 0.0075 which would appear to imply 0.0001 precision

[plain arrow]

that's my problem

since the number of decimal places go beyond what the measuring device can record

[night wedge]

How precise can precision be

Might be a good question for the physics discord

[plain arrow]

ahh

i am finished with the question

how do i end

!end

[night wedge]

https://discord.gg/physics

!done

[cedar kiln]

If you are done with this channel, please mark your problem as solved by typing .close

[plain arrow]

.close

[cedar kiln]

Channel closed

Closed by @plain arrow

Use .reopen if this was a mistake.

[native heath]

For instance, if a length measurement yields 114.8 mm, using a ruler with the smallest interval between marks at 1 mm, the first three digits (1, 1, and 4, representing 114 mm) are certain and constitute significant figures. Further, digits that are uncertain yet meaningful are also included in the significant figures. In this example, the last digit (8, contributing 0.8 mm) is likewise considered significant despite its uncertainty.[1] Therefore, this measurement contains four significant figures.

specifically, 'Further, digits that are uncertain yet meaningful are also included in the significant figures.'

[native heath]

you def have pm 1 cm of resolution right

which is like pm 0.5 mm of error when halved

52.0 pm 0.5 works

[plain arrow]

pm = plus minus?

ahh

i thought pm meant picometer

[native heath]

yes

$\pm$

[wraith dagger]

Περσυ

[native heath]

so, theres uncertainity in that .0 but its still significant

[plain arrow]

土

i see

.close

how do i reclose

[night wedge]

It’s already closed

[plain arrow]

ah

[night wedge]

Quick, leave before we reopen it

[native heath]

.reopn

[night wedge]

Don’t you dare

[cedar kiln]

Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
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[night wedge]

Try drawing line CE

[wispy wadi]

ok

[wispy wadi]

GFE is congruent to CDE

idk how to use though

[night wedge]

The angles in the triangles will be the same

Can you find angle CEK?

[wispy wadi]

10

thats as far as i got

[night wedge]

Do you agree CE is the same length as EG?

[wispy wadi]

mhmm

[night wedge]

Therefore, CE=EK

As EG=EK

Therefore, CEK is isosceles, and you know one angle

[wispy wadi]

so (180-10)/2= 85

[wispy wadi]

10?

[night wedge]

I think that’s the answer

!done?

[cedar kiln]

If you are done with this channel, please mark your problem as solved by typing .close

[wispy wadi]

i have another

[night wedge]

X here is a length?

Start by labelling the points and try to solve it

[wispy wadi]

ok

[wispy wadi]

i need help

[night wedge]

I don’t know how to solve this one geometrically, sorry

[wispy wadi]

ye

<@&286206848099549185>

[honest field]

oh wow

@wispy wadi gimme a moment

[wispy wadi]

alr

[wispy wadi]

oh

[honest field]

my solution is a bit complicated

[wispy wadi]

its alr

[honest field]

it has a lot of tedious steps

[wispy wadi]

sure

i've seen worse

[honest field]

quite technical

[wispy wadi]

ok

[wispy wadi]

right tri

[honest field]

1st step is to find the length of the blue side and pink side

[wispy wadi]

ok

how though

[honest field]

this is a regular hexagon

[wispy wadi]

so the height is root(12)?

[honest field]

which height you mean?

[wispy wadi]

blue

[honest field]

it's not √12

[wispy wadi]

oh

[honest field]

you can use this blue isoceles triangle to help

[wispy wadi]

ooh

[honest field]

the hexagon's side length is 4

[wispy wadi]

4root(3)

[honest field]

yep

and you should get the length of the pink side

[wispy wadi]

i multiplies root(3*4) on accident

[wispy wadi]

ye

pink = 7

[honest field]

great

now 2nd part

[wispy wadi]

0k

[wispy wadi]

whoa

[honest field]

you see that those 2 perpendicular lines are basically the heights of the pink and blue triangles

let's call those heights a and b respectively

[wispy wadi]

okay..

[honest field]

so the area of the pink one is 7a/2 and the blue one is 7b/2

their total area is 7(a+b)/2

our goal here is to find a+b

[wispy wadi]

what do we set it equal to

[honest field]

now going back to this

we're going to take the area of the big hexagon and subtract the colored areas

[wispy wadi]

okay

[honest field]

you can go ahead and do this

[wispy wadi]

ok

area = 24root(3)

[honest field]

which area you mean?

[wispy wadi]

hex

agon

[honest field]

alr keep going

[wispy wadi]

then tri areas are

3root(3)

1 root(3) and 4root(3)

[wispy wadi]

i got em

[honest field]

what's their total

[wispy wadi]

8root(3)

[honest field]

great

[wispy wadi]

16root(3) = 7(a+b)/2

i think

[honest field]

yep

so what is a+b

[wispy wadi]

4root(2)/7

[honest field]

hold up that's wrong

[wispy wadi]

hmm

32

[honest field]

recalculate again

[wispy wadi]

32root(3)/7

[honest field]

ok nice

onto the last part

[wispy wadi]

kklk

[wispy wadi]

hmm

[honest field]

so we see that the total length of those two perpendiculars is this

now we create a rectangle

[wispy wadi]

ok

[honest field]

you see that rectangle?

[wispy wadi]

kinda

[honest field]

sorry for my ugly drawing

[wispy wadi]

dw

[honest field]

anyways the blue side is just the long diagonal of the hexagon