help-13

[noble igloo]

(then multiply by this - but thats trivial since this is well just a constant)

[noble igloo]

<@&286206848099549185>

[cedar kiln]

@noble igloo Has your question been resolved?

[steel karma]

https://www.quora.com/What-is-the-average-distance-between-two-random-points-in-a-circle

top answer does it by integrating just like ur trying to do

[cedar kiln]

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[noble igloo]

Thank you

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[crimson sedge]

is $\frac{2\cos x-1}{\sqrt{1-\cos^2 x}}=\tan(x/2)$

[wraith dagger]

XxmastersmithxX

[south tundra]

No

You could check that in desmos btw

Or any other graphing calculator

[deft forum]

No

$\frac{1- \cos x}{\sqrt{1-\cos^2 x}}=\tan(x/2)$

[wraith dagger]

Deep

[south tundra]

Absolute value tho

[deft forum]

Yup

[crimson sedge]

wym

[south tundra]

There should be | | around the tan

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[bitter salmon]

Can someone help me
I do not know how to solve number 3 and 4

[tropic oxide]

how familiar are you with basic trigonometry?

[south tundra]

Use definition of tan in a right triangle in order to get expressions for tan(25) and tan(a) in 3

[tropic oxide]

like, SOHCAHTOA

[bitter salmon]

yeah i know a bit
sin = opposite/ hypotenos
cos = adjacent / hypotenos
tan = opposite/ adjacent

[bitter salmon]

I still do not understand

[south tundra]

Okay, what would tan(25) be by looking at the triangle on the right?

[bitter salmon]

Ohhhh waiit wait so
tan (25) = h /12??

[south tundra]

Yes, can you solve for h from here?

[bitter salmon]

yep thanks

i got 5.6

Im solving the other triangle now

[cedar kiln]

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[bitter salmon]

Thanks for the help !

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[karmic terrace]

I need help with number 3

[karmic terrace]

I keep getting a surd number but when i look at the answers its just a fraction

[dull anchor]

what have you tried?

[karmic terrace]

Sorry if its a bit messy

I got (-79 ± √ 6241 - 540)/-2

[knotty moat]

we have two equations l + b = 7.9 and lb = 13.5

[violet rapids]

Equations are set up

Correctly

[dull anchor]

ahh you have to divide the equation by -1 first and then use the formular to solve for x

you got -x^2+7.9x-13.5 = 0

[karmic terrace]

Oh ok

I try

I got it

Thanks big

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[dull anchor]

no problem😁

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[wanton glacier]

$r^5+r^6=r^{11}?$

[wraith dagger]

hyperlix26

[violet rapids]

Careful

Does this look like a law of indices?

[wanton glacier]

no

[bright surge]

what's your question?

[violet rapids]

#❓how-to-get-help

[bright surge]

maybe find values of r?

[wanton glacier]

alright, so it stays like that

[violet rapids]

Yel

[dull anchor]

bruh

[violet rapids]

Yep

[wanton glacier]

I just wanted to know if that worked

[violet rapids]

Know the exponent rules!!!

[wanton glacier]

alright

thanks

.close

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[wanton glacier]

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✅

[wanton glacier]

$a+ar+ar^2$ that's 3a right?

[wraith dagger]

hyperlix26

[wanton glacier]

$3ar+r^2?$

[wraith dagger]

hyperlix26

[tropic oxide]

no

$a + ar + ar^2$ is not equal to $3ar + r^2$

[wraith dagger]

Ann

[wanton glacier]

but it's 3a

$3a+r+r^2$

[wraith dagger]

hyperlix26

[wanton glacier]

is that it

[violet rapids]

No

@wanton glacier none of these are like terms!

[jaunty mural]

$$5\times2+5\times2^2+5\times2^3$$
$$= 5\times2+5\times4+5\times8$$
$$= 3\times5\times2+4$$

Is what you're trying to do?

[wraith dagger]

Shuri2060

[jaunty mural]

You seem to have to wrong idea about how to factorise

[cedar kiln]

@wanton glacier Has your question been resolved?

[wanton glacier]

oh wait that's it

you can't separate r and a

so it stays like that

all of the terms have a

so you can factor that out

$a(r^2+r+1)$

[wraith dagger]

hyperlix26

[jaunty mural]

Yes. Better.

[wanton glacier]

$ar^5(r^2+r+1) \implies ar^7+ar^6+ar^5$ right

[wraith dagger]

hyperlix26

[wanton glacier]

<@&286206848099549185>

[jaunty mural]

not implies.

equals.

And you can check equality of algebraic expressions in wolfram.

,w simplify [a r^5(r^2+r+1) - (a r^7+a r^6+a r^5)]

the heck?

when its not being dumb that is

[wraith dagger]

Results provided by WolframAlpha

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

Requested by Shuri2060#9525

[jaunty mural]

[wanton glacier]

hmm

wait

since I have this $ar^7+ar^6+ar^5$

[wraith dagger]

hyperlix26

[wanton glacier]

and this

I can do this right? $r^5(a+ar+ar^2)$

[jaunty mural]

Yes, you can check in wolfram (with spaces or on the actual website)

[wraith dagger]

hyperlix26

[jaunty mural]

if your manipulations are correct or not

see above

[wanton glacier]

,w simplify r^5(a+ar+ar^2)

[wraith dagger]

Results provided by WolframAlpha

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

Requested by hyperlix26#0041

[wanton glacier]

alright

thanks

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[crimson sedge]

Hello

[crimson sedge]

For the derivative of e and log, if we apply the chain rule then why do we need ln e or ln a?

[lone rivet]

what do u mean

could u rephrase that

[dense wing]

cause... that's the derivative

[crimson sedge]

If we have function x^2 and we take the derivative

[lone rivet]

its 2x

d/dx

[crimson sedge]

2(x) (1)

[dense wing]

$\dv{x}a^{u(x)}=a^uu'(x)\ln(a)$

[wraith dagger]

Mosh

[crimson sedge]

Yes, why ln(a) is there?

[dense wing]

$\dv{x}a^x=a^x\ln(a)$

[wraith dagger]

Mosh

[lone rivet]

i've never seen this before

actually stumped

[dense wing]

$\dv{x}a^x=\dv{x}e^{\ln(a)x}=\ln(a)e^{\ln(a)x}=\ln(a)a^x$

[wraith dagger]

Mosh

[lone rivet]

the derivative of a^x is xa^(x-1)

oh

[dense wing]

No.

[crimson sedge]

So, here we are applying chain rule thrice. Am I right?

[dense wing]

It isn't.

[dense wing]

no

once

a^x is a function

u(x) is a function

[lone rivet]

OH WAIT THAT IS FOR X TO THE POWER OF SMTH

[dense wing]

you have a single layer of composition

[lone rivet]

THIS IS A TO THE POWER OF X

ok i understand

[dense wing]

the derivation of the a^x rule requires chain rule on e^u(x)

[crimson sedge]

Got it.

Thank you very much @dense wing 😊

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[coarse cloud]

If I were to calculate the log of a scientifically notated number ig 5*10^5 could I simply remove the *10^5, calculate the log of 5 and then add it back in after?

[graceful karma]

$\log(a \cdot b) = \log(a)+\log(b)$

[wraith dagger]

PapaBread

[toxic owl]

beat me to it haha

[graceful karma]

Indeed

[coarse cloud]

Thank you sm

both!

[graceful karma]

Np

[coarse cloud]

Okay just you

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[glacial solstice]

I need help with vectors, the question just confuses me

[jaunty mural]

If you draw a diagram...

it wouldn't be as confusing, I hope

[glacial solstice]

Alright lemme try

oh yeah i did it

For AB i got b

let me try CD

[jaunty mural]

idk how you think you were going to do a vectors question with no diagram

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[mighty quarry]

im lost

[violet rapids]

is this a quiz

or a test

[mighty quarry]

sat prep

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[summer arch]

shouldn't it be g o 1A

and not 1A o g

[cedar kiln]

@summer arch Has your question been resolved?

[spring bolt]

Can we get any context at all?

[summer arch]

sure

sorry about that

It's the proof that functions can have only one inversive function

We assumed that f^-1 and g are the inversive functions of function f

[spring bolt]

what are A and B?

[summer arch]

1A: A -> A

1B: B -> B

f: A -> B

f^-1: B -> A

g: B -> A

[spring bolt]

if you want to flip them, it would need to be $g \circ 1_B$, then

[wraith dagger]

phlp

[spring bolt]

It should work either way. Why don't you try both?

...but the original is correct. g maps B to A, $1_A$ maps A to A, so $1_A \circ g$ maps B to A to A

[wraith dagger]

phlp

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[hardy umbra]

I am having trouble finding the second derivative here. Its a summation of two product rules correct?

[gaunt hamlet]

Yea

[hardy umbra]

I think I figured out my mistake as soon as I posted it. Just messy handwriting...

thanks for confirming though

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[hollow sapphire]

.reopen

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You can't reopen another person's channel

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[cedar kiln]

@wet gorge Has your question been resolved?

[wet gorge]

<@&286206848099549185>

[dire geode]

This doesn't really look like math

[cedar kiln]

@wet gorge Has your question been resolved?

[wet gorge]

is formal language theory not a branch of math?

it's in my computer science program as a math class

it seems very similar to set theory to me but idk

[dire geode]

Regular expression is definitely not a math thing

You're more likely to get help in a CS server

[toxic owl]

The notation might be math related but the theory and specifically what property it is relates more to CS

[cedar kiln]

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[wet gorge]

i cant find any CS servers

[wet gorge]

if someone would like to help with my original question i would appreciate it

[obsidian coral]

There's one in #old-network

[radiant topaz]

.

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[mental torrent]

help?

[tropic oxide]

,rccw

[tropic oxide]

what is giving you trouble here?

[mental torrent]

how do i start?

[tropic oxide]

have you worked with piecewise functions before?

[mental torrent]

nope

[tropic oxide]

so this is your first time seeing a definition like this?

[mental torrent]

yes

[tropic oxide]

okay

are you familiar with programming to any extent

[mental torrent]

a little

[tropic oxide]

ok great

def f(x):
    if x <= -2:
        return -4
    else if -2 < x < 1:
        return x
    else if x >= 1:
        return x^2 - 1

this is that function definition translated into python (with some small syntax non-conformities)

does this help you understand the definition better

[mental torrent]

yup ok

[tropic oxide]

are you able to continue from here

[mental torrent]

do i substitute in -1?

[tropic oxide]

i mean...

yes, if you so insist,

but you have to identify where to substitute it into

based on those case conditions in the function defn

[mental torrent]

oh

i’m kinda slow haha i don’t get it

[tropic oxide]

okay, so: which of the following three inequalities is satisfied by x = -1?

x <= -2
-2 < x < 1
x >= 1

[mental torrent]

the second

[tropic oxide]

okay, and what's the formula next to it in the definition?

[mental torrent]

x

[tropic oxide]

so what's f(-1)?

[mental torrent]

-1?

yeah all cool me too

i think they are separate equations

i just checked my answer sheet and it says -4 lol what

[tropic oxide]

yes

[tropic oxide]

hold on what?

[mental torrent]

oh then the answers are probably wrong right?

[tropic oxide]

are you sure you're looking in the right place for the answer key

[mental torrent]

yeah idk it says -4

i’m pretty sure it’s -1 tho

thanks for helping me out!

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[wide furnace]

The supposed answer to this question is 0.65

But to my understanding it should be 0.8

Can anyone explain why?

[tropic oxide]

you're confusing 'independent' with 'mutually exclusive'

[wide furnace]

So how do you work it out?

[faint viper]

P(A)+P(B)-P(A and B)

[tropic oxide]

recall what the term 'independent events' actually means

and do that ^

[wide furnace]

Oh my

A mental atrophy on my part then

Thanks

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[fluid swift]

Show that angles MNR and RAN are equal.

[cedar kiln]

@fluid swift Has your question been resolved?

[crimson sedge]

MAR or MNR

[fluid swift]

MNR

[crimson sedge]

Is RA a bisector?

[fluid swift]

yes of course

[crimson sedge]

DR too?

[fluid swift]

yes

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.reopen

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✅

[cedar kiln]

@fluid swift Has your question been resolved?

[fallen heath]

the motive is to prove that the angle bisector of A is concurrent with the incenter of triangle DMN yes !? (interesting 1.1)

[fluid swift]

Actually it's to prove AMRN is cyclic. (I just don't know how else to do that.)

[bright surge]

<@&286206848099549185>

[cedar kiln]

@fluid swift Has your question been resolved?

[solid quarry]

@fluid swift show that points M A R N lie on a circle

[steel karma]

im curious

i dont think they are necessarily cyclic

ping me if u have proof

or at least not enough info from the question to show they are cyclic

@fluid swift post the full info of the question if u have it

[fallen heath]

(Just fyi they are indeed cyclic.)

[crimson sedge]

I am clueless.
I still can't prove it

I am bad at geometry :(

Is this an Olympiad question?

[fallen heath]

if I'm not wrong about this, it's probably an entrance level oly question (@_@;)

[fluid swift]

that's, like, the most useless advice ever

[solid quarry]

i know srry

[fallen heath]

Eden anything you tried?

For all I know, I can see anti-parallel lines and recall my oly. camp which I flawlessly bunked [and now rethink my life choices]

[fluid swift]

I messed around with the circle D for a bit. Found nothing.

[crimson sedge]

Where did you get this problem

[fluid swift]

IMO 2004 p1 - Maybe this is an xyproblem moment :\

[fallen heath]

hence proved lmao

[fallen heath]

not really!! It tends to get like that at times when one's unable (@_@;)

[crimson sedge]

I need to start studying geometry from basic again

Ping me if u get sol

[steel karma]

u have a proof?

im imagining ok suppose it is cyclic, then what if I take the centre of the circle to be A'

A' is an angle bisector too, just like A

What's the difference between A and A' then in the construction?

[fallen heath]

Apart from it being an IMO problem and the GeoGebra calculator confirming the facts, I'd need some more time if you want more proof lol

[steel karma]

hold on

i see now

A'RD is collinear

while ARD is not in the diagram

[fallen heath]

the construction is as follows:

Consider triangle ABC, draw a circle with BC as diameter, let the circle intersect sides AC and AB at points M and N respectively. Let the midpoint of side BC be D. Let the angle bisector of MDN and angle bisector of angle BAC intersect at R. Prove that AMRN is cyclic.

[fallen heath]

extend NM to meet BC at G. Maybe you can use menelaus to somehow manage to get the similarity in triangles AFM and NFR or something

@fluid swift If you're not exclusively looking for a geometric proof, maybe this could satisfy you atm:

[fluid swift]

i dont even know menelaus....

[fallen heath]

Do you know sine law?

[fluid swift]

i actually dont want to use trig

so

well

[fallen heath]

Ah'

[fluid swift]

yes i can prove using sine law

[fallen heath]

smh, so bad :c I figured the trig solution

[fluid swift]

but i dont ws=ant to do that

[fallen heath]

but just can't find a geometric one

however there being a trig solution implies similarity is due

[crimson sedge]

Oh then replace sine with ratio and prove

Uh I mean don't use sine cos
But instead use ratio side

Lik sin =ab/BC like that

[fallen heath]

there you go.. just focus on proving this is cyclic lol

[fluid swift]

i literally have no idea what this is

[crimson sedge]

Nvm

[steel karma]

did you just go straight to international math olympiad and dunk it in help

[fluid swift]

no

[steel karma]

lol

[fluid swift]

this is practice

[steel karma]

without ever doing IMO

[fallen heath]

it's not that tough bm, you could do it with trig easy, but uhh

mate's looking for a geometry soln.

[fluid swift]

nvm i guess i can stick with trig.

.close

[cedar kiln]

Channel closed

Closed by @fluid swift

Use .reopen if this was a mistake.

[fallen heath]

wait Eden lol

[steel karma]

what next

[crimson sedge]

Oh

[fallen heath]

wow (@_@;)

[steel karma]

i claim this

[crimson sedge]

Wait how do I solve it help

[steel karma]

.open

[fallen heath]

no clue what you did >_<

.claim

I claim thee

[crimson sedge]

.claim 😎

[fallen heath]

*this

[crimson sedge]

Wut

Shit

[cedar kiln]

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[fallen heath]

claim

[fallen heath]

UwU

[knotty moat]

lol

[steel karma]

i just put the circumcentre of the MRN on the diagram

that's all i labelled

[fallen heath]

to prove ANRM is cyclic

[crimson sedge]

Here

[steel karma]

pls

my diagram

[fallen heath]

no mine wth,, my channell

[knotty moat]

..

[steel karma]

yours is ambiguous

if MRA = ARN then they dont need to be concyclic

[knotty moat]

its looks cyclic already so it is cyclic

[fallen heath]

except given MRA not ARN UwU

[steel karma]

that was what tripped me up earlier

i did exactly what u did

i reduced it to ur diagram

and then was like hey taht doesnt need to be concylic

lol

[fallen heath]

with trig it's easy to see: sin(AMR) = AR (sin(MAR) / MR) = AR(sin(NAR) / NR) = sin(ANR)

[steel karma]

what is S

[knotty moat]

nvm I step out of trigo

[gentle lintel]

AMR must be 90° no?

[fallen heath]

👀

[crimson sedge]

Yes

Yes

[fallen heath]

wha-

[crimson sedge]

See

[fallen heath]

why-

[steel karma]

what are the implications of this

[crimson sedge]

Look at original pic

[fluid swift]

AMR+ANR=180deg

[fallen heath]

implication: AMR + ANR = 180 or AMR = ANR ; but AMR = ANR is ruled out atq

[steel karma]

i see

[fallen heath]

so we done UwU

but

the thing is

[steel karma]

nice>

[fallen heath]

how to do geometrically T_T

[gentle lintel]

(ansh come help me later ;-;)

[steel karma]

prove the sine rule geometrically

and then just use it

or cosine rule

cosine rule works the same here

[crimson sedge]

Oh wut this was so easy there was just a trick to it

[fallen heath]

original picture ? what?

[steel karma]

clusterfuck

[fallen heath]

How tf is AMR = ANR = 90°?

[knotty moat]

I was trying to prove it congruent lol

[crimson sedge]

This

[steel karma]

of unnecessary information

in the original pic

[fallen heath]

Yeah lmao

[crimson sedge]

TriMCB is on radius of circle

[fallen heath]

reducing it is the best lol

[knotty moat]

nothing else is given in question?

except the diagram

[fallen heath]

bruh ... you've taken too much wine

#chill

[knotty moat]

ok

[crimson sedge]

Yeah

Uh NVM I wil go sleep

[steel karma]

tbh i think its fine

u use sin rule to prove it

since sin rule is proved geometrically also

[fallen heath]

Ikr

but then one must be able to see it without trig as well

[steel karma]

its an olympiad problem

whatever it takes lol

[knotty moat]

we can join points M and N?

lol

and

it is a isosceles triangle

angle NMR = angle RNM

what is its use lol

nvm

[fluid swift]

nvm i'll just use trig

[steel karma]

i like combinatorics olympiad the most

those are the most fun to me

even tho i suck at those

[fallen heath]

a question for you as well then

[steel karma]

no like geometry

[fallen heath]

Help me out with this one:

[steel karma]

post

[knotty moat]

I will just store questions to solve later lol

[fallen heath]

Number of n lettered words made out of n A's and n B's such that the number of A's from the left is at all times not less than the number of B's from the left.

[steel karma]

it's literally from international math olympiad

[knotty moat]

IMO ?

[steel karma]

yes

this question that was posted

[knotty moat]

cool

lol

[steel karma]

i dont understand the q

[knotty moat]

..

[steel karma]

if you make it out of n As and nBs isnt it a 2n lettered word

"A's from the left is at all times greater than the number of B's from the left."

what does this mean?

[knotty moat]

AAABB

[steel karma]

A's from the left?

[knotty moat]

type series

[steel karma]

B's from the left?

[knotty moat]

ig

oh nvm

[fallen heath]

yeah

Like.. AABABAA is valid

because 2A before first B and 3 A before 2nd B

but AABBB isn't valid

[steel karma]

ok so n lettered words

[fallen heath]

because there's 2 A before 3 Bs

[steel karma]

it's a n lettered word

made up of As and Bs

[fallen heath]

yeah

[steel karma]

not n As and n Bs

or is it a 2n lettered word

with equally many As and Bs

[fallen heath]

n lettered word

(@_@;)

[steel karma]

ok

[knotty moat]

no value of n?

just a variable?

[fallen heath]

nu 0_0

[knotty moat]

lol

[steel karma]

my first instinct is to write a recursion

[gentle lintel]

hmm so

[steel karma]

so we split it into the difference of total As and Bs

[gentle lintel]

the first letter must be A

[steel karma]

or for some string that is valid let f(string) by the difference in the number of As and Bs

[gentle lintel]

big hmm

[steel karma]

and we can sum across all strings of length n

i should write it down

[knotty moat]

I got nothing but I will watch

[gentle lintel]

so suppose we have n A’s and n B’s in two bags

how many ways can we form a sequence

such that there are always less A’s than B’s?

[knotty moat]

@fallen heath you have algebra question? I want to see if I can solve lol

or maybe later

[crimson sedge]

Is the solution algebraic or numeric?

[gentle lintel]

we’ve got two counters that can count from 0 to n, how can we always have the left counter displaying a bigger/equal number

maybe this is easier

[steel karma]

this is idea

[fallen heath]

lmao

[knotty moat]

lol

[crimson sedge]

So if there are p A's ther can only be p-1 B's or less than that such that it starts with 2 A's

[fallen heath]

Wait lemme figure one.

[knotty moat]

ok

[steel karma]

cause if you have a valid string

you can either append an A or B to the end of it

to get a string of length n+1

if the string ends with the same number of As and Bs you cant append a B

[crimson sedge]

Shouldn't the string be of n

[steel karma]

that's where this chart be comes from

[steel karma]

if oyu have a string of length n*

you can append an A or B at the end

to get a string of length n+1

and if the string ends with the same numbers of As and Bs you cant append a B

[fallen heath]

there you go (@_@;) but I already know the ans. so it's just for gift sake-

[knotty moat]

lol

..

let me try lol

[steel karma]

that's a recursive way to find the number of ways

quickly

but idk how it relates to a simple formula

[fallen heath]

(@_@;) if it's a recursion.. maybe I can follow from there

Do you know one that helps here?

[steel karma]

nope

im looking at the first row

1 / 0 / 1 / 0 / 2 / 0 /5 / 0 / 14 / 0 / 42

what are these numbers

it's like a pascals triangle

but cut

[fallen heath]

welp, think about it and lemme knowww (ヘ･_･)ヘ┳━┳
I think I need to revise how sine law was derived in the first place smh

[gentle lintel]

big thonk

i arrived at y² - 22y - 3 = 8√y

[fallen heath]

XD question's free for anyone..

[gentle lintel]

is there an elegant way to solve this

[steel karma]

actually hold on

These are catalan numbers

1,1,2,5,14,42

[fallen heath]

wha-

howw

[steel karma]

🤷‍♂️

[gentle lintel]

how do ;-;

[fallen heath]

n = 3 is already out of your sequence lol

[gentle lintel]

me dumb

[fallen heath]

n =3 has AAA, AAB, ABA

[fallen heath]

.

[fallen heath]

use bwane (@_@;)

[steel karma]

no

i was looking at the top column

[fallen heath]

it has a very exact solution that gives a very exact answer

an irrational root

[gentle lintel]

rational root theorem tells me there are no integer roots lol

[steel karma]

the recursive formula is:

For n odd:
f(n+1) = 2f(n)

For n even
f(n+1) = 2f(n) - C(n/2) where C(x) is the xth catalan number

[gentle lintel]

so by common sense i do (x² + ax + b) (x² + px + q) expansion

[steel karma]

i was looking at what we are subtracting

[gentle lintel]

then do annoying as fuck system

[steel karma]

and noticed it is the catalan numbers

[gentle lintel]

👏

there should be elegancy somewhere

[fallen heath]

XD

[steel karma]

i found the name of ur problem

"dyck words"

[fallen heath]

there is, but you'll blow your brains when you realise it

shut up bm lmfao

[gentle lintel]

sure

sad jj

[fallen heath]

*when you realize the solution yourself (@_@;)

[gentle lintel]

i dont care tell me :((

[fallen heath]

I read on dyck words too.. not the same thing tho

one hint and it'd be out there in the open smh

(the solution)

[gentle lintel]

purchase a shotgun?

[steel karma]

im just reading on catalan numbers

wow this is a rly big field of maths i had no clue about ;')

or idk how big it is

but a lot of interesting things to learn it seems

[gentle lintel]

no idea 😦

[fallen heath]

maybe tmrw :p

.close

[cedar kiln]

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[fallen heath]

lemme know if you figure plz ;-;

I did something similar previously in hs but uh (ヘ･_･)ヘ┳━┳ I forgot how

[cedar kiln]

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[scenic cargo]

What does the number outside of a square root mean? Example:

[scenic cargo]

Does it mean 2 times the square root?

[stable solstice]

yea

[scenic cargo]

Ahh, i see, thanks man

.close

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[cedar kiln]

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[azure canyon]

can someone help me find a

y = a(x - h)^2 + k

yea so i got everything but a, i tried puting in the given point but got it wrong

yes

ohhh

i got it now! thanks!!!!!

-1/9

no

tf

why you need a girl?

i see you i see you bro

LMAO

we all need some, we kinda down bad huh

send me the problem bro

im two time back to back gold champ

[obsidian coral]

But why? That's weird

[dire geode]

@brisk forge stop being weird

[azure canyon]

LMAO

MANS DOWN BAD

[cedar kiln]

@azure canyon Has your question been resolved?

[cedar kiln]

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[stark creek]

yo anyone know y we do L = 50 - W?

[obsidian coral]

,rotate

[stark creek]

ah ye but y do we do L = 50 - W

y not W = 50 - L

[obsidian coral]

You can, if you want

[stark creek]

wait but would it give the same results?

[obsidian coral]

It should

[stark creek]

so are they both the right answer?

[obsidian coral]

Yes

[stark creek]

oh wait let me ask another anser

*answer

[cedar kiln]

@stark creek Has your question been resolved?

[cedar kiln]

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[cedar kiln]

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[copper peak]

does anyone know how to enter a matrix in mymathlab by code, instead of manually selecting from the menu

[copper peak]

when i paste this @MATX{{-3};{1};{1}}, it works perfectly

when i type it out it doesnt work

[cedar kiln]

@copper peak Has your question been resolved?

[exotic elk]

u can do it in wolfram alpha like {{1, 2, 3}, {3, 2, 1}, {1, 2, 3}} + {{4, 5, 6}, {6, 5, 4}, {4, 6, 5}} for example

https://www.wolframalpha.com/input?i={{1%2C+2%2C+3}%2C+{3%2C+2%2C+1}%2C+{1%2C+2%2C+3}}+%2B+{{4%2C+5%2C+6}%2C+{6%2C+5%2C+4}%2C+{4%2C+6%2C+5}}&assumption="ClashPrefs"+->+{"Math"}

or symbolab with

\begin{pmatrix}2&3&4&5\\ 6&7&8&9\end{pmatrix}```

[wraith dagger]

MattDog_222
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

BritS

$\begin{pmatrix}6&9&6&9\\ 6&9&6&9\end{pmatrix}$

[crimson sedge]

There you go

[copper peak]

yeah but these dont work in mymathlab which is my homework website

and entering it manually from the menu is very tedious

[crimson sedge]

Excuse me did you say mymathlab?

Nvm

[copper peak]

mylab math

[crimson sedge]

https://youtu.be/onC-pL6mnVg

[exotic elk]

that doesn't show the code

[copper peak]

yeah it doesn't

[exotic elk]

try

\begin{bmatrix}3&3\\ 23&46\end{bmatrix}```

bmatrix might work

[crimson sedge]

||Why did I see these numbers?|| <-- ignore this

[copper peak]

nope

[copper peak]

this works when i copy paste it

[wraith dagger]

MattDog_222
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[exotic elk]

its more subtle now

[copper peak]

so maybe i have to add something before it?

[exotic elk]

could try \[\begin{bmatrix}3&3\\ 23&46\end{bmatrix}\]

[wraith dagger]

MattDog_222

[exotic elk]

or dollar signs $\begin{bmatrix}3&3\\ 23&46\end{bmatrix}$

[wraith dagger]

MattDog_222

[crimson sedge]

Try writing [[6 9][6 9]]

[copper peak]

doesnt work

i tried it

there must be a way

[exotic elk]

idk im just fishin, [1,2;3,4]

[copper peak]

no lmao

okay i give up

thanks for helping tho lol :)

.close

[cedar kiln]

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[cedar kiln]

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[sterile maple]

Hi,
I have no idea how to solve this sum. I tried making a triangle with an angle 30 at the corner but it hasn't worked. Could someone please help? Thank you

[dusty hazel]

Right, so um the triangle ABC is equilateral correct?

[sterile maple]

yes

[dusty hazel]

What does that tell you?

About the angles of the triangle.

[sterile maple]

all are 60

[dusty hazel]

Correct.

Since the triangle def is formed by lines parallel to the original triangle, they both should be similar
True?

[sterile maple]

ya

[dusty hazel]

So each angle of def is?

[sterile maple]

all are 60 even the inner ones

[dusty hazel]

Correct.

So the inner triangle is equilateral.

[sterile maple]

yes.

[dusty hazel]

Since the original triangle had sides equal to 10cm, and new one has 1cm less in all directions.
Shouldn't each side be 2cm less than 10?
It's not I think, still.

[sterile maple]

it's not that simple unfortunately

[dusty hazel]

I thought that was the case this time.

Haha, my bad.

I'll calculate

[sterile maple]

So I thought of making a smaller triangle

like this

[dusty hazel]

Where the perpendicular is 1cm.

Label the points it'd be easier.

So we get the base √3 in this smaller triangle, no?

Same for all other sides?

[sterile maple]

Yes

[dusty hazel]

Since equilateral triangle is a regular figure.

So yeah new sides of this triangle,
Shouldn't they be like 10-(2√3)

In cm.

[sterile maple]

Oh so

[dusty hazel]

so perimeter would be three times this.

Unless we did something wrong.

[sterile maple]

30 - 6root3 then?

[dusty hazel]

I believe.

[sterile maple]

Perfect This is the answer in my tb

Thanks man!

[dusty hazel]

You're welcome!

[sterile maple]

.close

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[dusty hazel]

I figured some stuff on my own too.

First of all we have the general form
ax²+bx+c=0
Where a,b,c are odd and integers.

a≠0

Also if the equation has unreal roots that proves the thing easily, so we will talk about real roots only.

So ∆≥0

Also, for rational roots, ∆ must be a perfect square of a rational number.

So we have to contradict this.

also ∆=b²-4ac
b² is odd, 4ac is even, ∆ is definitely odd.

Also, b a and c are integers, therefore ∆ is an integer.

Let ∆ be square of some integer $\sigma$

[wraith dagger]

Sakata Yaksha

[dusty hazel]

Sigma is odd as ∆ is odd.

therefore ∆=b²-4ac
And $∆ =\sigma^2$

[wraith dagger]

Sakata Yaksha

[dusty hazel]

$\sigma^2=b²-4ac$

[wraith dagger]

Sakata Yaksha

[dusty hazel]

$4ac= b² - \sigma^2$

[wraith dagger]

Sakata Yaksha

[dusty hazel]

b and sigma both are odd therefore their difference is even. This is what bugs Me, we had to contradict this from happening.

[cedar kiln]

@dusty hazel Has your question been resolved?

[dusty hazel]

<@&286206848099549185>

[cedar kiln]

@dusty hazel Has your question been resolved?

[jaunty mural]

I suspect if you take this mod 4, the result might come out

[dusty hazel]

Oh thanks, finally someone saw my question.

[jaunty mural]

or some other mod if not thay

[dusty hazel]

Mod 4 is simply four though, and that's what I wrote.

[jaunty mural]

what?

[dusty hazel]

Elaborate what you meant please, I didn't follow.

[jaunty mural]

consider cases in modular arithmetic

It doesnt work for now, actually.

But thats the first thing that came to mind

Ah ok, I can see a factorisation

[dusty hazel]

Right.

[jaunty mural]

Youre not stuck here.

[dusty hazel]

Am I not?

I can factorise that surely.

Also b and $\sigma$ are odd.

let $b =2k+1 and \sigma =2h+1$

[wraith dagger]

Sakata Yaksha

Sakata Yaksha

[jaunty mural]

what?

[dusty hazel]

I mean

[jaunty mural]

https://media.discordapp.net/attachments/903430334681608232/942395923345047563/748029603087384657.png

[dusty hazel]

I got till here.

[jaunty mural]

You literally wrote it in this form

There is an obvious factorisation to do here, surely.

[dusty hazel]

Ik

I'm doing that obviously.

$(b+\sigma)(b-\sigma)$

[wraith dagger]

Sakata Yaksha

[jaunty mural]

right.

[dusty hazel]

I also concluded

[jaunty mural]

Then there should be cases to consider in modular arithmetic or similar

[dusty hazel]

b and $\sigma$ are both odd.

[wraith dagger]

Sakata Yaksha

[limpid plume]

Both b - sigma and b + sigma should be even

[dusty hazel]

I have to prove this using basic stuff nothing that fancy.

[dusty hazel]

Correct, we are contradicting though. Please read all my previous messages if you can, I'd appreciate it. I concluded lots of stuff, and it'd make us all understand what I am approaching and what I should have done if that's now what I should have done.

[limpid plume]

So, the statement you have started with - does that have any counter examples?

If you're arriving at a contradiction then the statement might jot actually be true

[dusty hazel]

I mean I have to prove it does not have rational roots, so I assumed it has rational roots. Now we're contradicting this.

[limpid plume]

Let me try something and then come back if I find something

[dusty hazel]

Sure, thanks!

[limpid plume]

[jaunty mural]

do we not get a contradiction mod 8?

[dusty hazel]

What's that?

[jaunty mural]

have you seen modular arithmetic?

[dusty hazel]

Like I told, that's not what I can use in this question.

[jaunty mural]

I'm asking, I haven't tried btw

[dusty hazel]

Also, I have not.

[jaunty mural]

are you sure you can't use it?

[dusty hazel]

I can obviously.

I'm not allowed to.

That's what I meant.

[jaunty mural]

why

[dusty hazel]

My bad.

[jaunty mural]

modular arithmetic is no different from essentially considering remainders

you can literally do this with just the numbers and it makes no difference in logic

[dusty hazel]

Oh the name seemed fancy lmao.

So, can you give an example real quick?

[jaunty mural]

you should lookit up

modular arithmetic

quicker

than me saying

Look on brilliant or similar

Numbers mod 4 are 0, 1, 2 or 3 for example...

[dusty hazel]

mod 4
I thought that was |4|

Absolute value function.

I'll google it real quick then.

[limpid plume]

@dusty hazel Can we prove the discriminant of the new equation with k as variable is less than zero?

And even if it is not less than zero, it is not a perfect square?

[limpid plume]

They are talking about congruence

Like

8 = 0 mod 4

13 = 1 mod 4

[dusty hazel]

Yes, I got it.

We use mod only and only for modulus generally.

So I didn't know earlier.

[dusty hazel]

Yes it's not necessarily a perfect square, true.

But if it is a perfect square, the roots are automatically rational.

[limpid plume]

If it's less than zero we have a very solid proof

[dusty hazel]

Yes indeed.

[limpid plume]

And I meant the discriminant of the newer equation

[dusty hazel]

Which one, can you mark that ?

Or reply to it or something.

[limpid plume]

The last discriminant in this paper

[dusty hazel]

I'm extremely sorry, I did not see this at all. I'll check.

[limpid plume]

Oooooooh wait wait

My god it should be positive instead of negative, my baf

[dusty hazel]

Yes, I was about to say.

[limpid plume]

Let me do it from scratch

[dusty hazel]

Hey

[limpid plume]

The first one was correct

[dusty hazel]

k²+k+4ac...

[limpid plume]

I'm sorry for confusing you

[limpid plume]

Exactly

[dusty hazel]

Yeah lol, also this is exactly what my proof looks like.

[dusty hazel]

Here.

Just we use 2k+1 and stuff for odd integers now.

[limpid plume]

Yes, I just assumed your a as 2a + 1 so I can make it necessarily odd from the start 👍

[jaunty mural]

I'm not saying modular arithmetic necessarily helps btw. But it is something worth trying

[dusty hazel]

I just saw what it was, although I'm not aware of it's applications.

You told to use mod 8 though, correct?

[jaunty mural]

squares cannot be 3 mod 4 for example

didnt work though

yes mod 8

To make some parity argument

[dusty hazel]

Ah, alright I will keep that 8 in mind.

So

$4ac= b² - \sigma^2$

[wraith dagger]

Sakata Yaksha

[dusty hazel]

$4ac=(b+\sigma)(b-\sigma)$

[wraith dagger]

Sakata Yaksha

[dusty hazel]

Ohh lol I figured it out maybe.

If only this is correct, I'll be done for the day and sleep nicely.

Yes, I'm very positive I did it.

!!!!!!

@jaunty mural thanks, 8 really is a good number.

@limpid plume thanks, for your help.

So we prove this like this

The difference of squares of two odd integers is a multiple of 8!

While on LHS we have 4ac, where a and c are odd.

Which contradicts.

[dusty hazel]

I just proved this too.

[limpid plume]

That's exactly what I found just now, it's in a Number Theory course I did a few months back

Also

Alternative proof I justade

Made*

[dusty hazel]

Yes, sure I'd like to have a look.

$4ac= 8n$ as $b²-\sigma^2$ is always a multiple of 8.

[wraith dagger]

Sakata Yaksha

[dusty hazel]

We get, $ac=2n$ which is never possible as a and c are odd.

[wraith dagger]

Sakata Yaksha

[jaunty mural]

👌

[limpid plume]

I apologise for bad handwriting

[dusty hazel]

Thanks, I'll also look everything about modular arithmetic today.

[dusty hazel]

Bad handwriting, that's the reason I use the bot XD. I'm no good.

[limpid plume]

Competitive exams have done so much harm to me it's not even funny anymore

Not only it gave me depression at one point, it also made my handwriting bad an my ability to write subjective answers bad

MCQs are just not the way to live 😭

[dusty hazel]

Hmmmm, you're Indian I assume? (A very random guess, no offense)

[limpid plume]

Yes

And none taken

[dusty hazel]

I see, I'm checking your proof, we'll have our talk then.

Yes, your proof was all worth it.

Thanks mate.

Now, let's take this to dms and close the channel as we're done here.

[limpid plume]

I wrote some letters wrong in the last I think, but you got the idea right?

[limpid plume]

Yes.

[dusty hazel]

Yes, pqrs all odd. So yeah I got that.

.close

[cedar kiln]

Channel closed

Closed by @dusty hazel

Use .reopen if this was a mistake.

Available help channel!

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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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[gray vine]

I did something wrong here, but I don't know what:
sin(2x) = 4sin(x)
sin(2x) - 4sin(x) = 0
sin(2x) = sinxcosx + cosxsinx = 2sinxcosx
2sinxcosx - 4sinx = 0
sinxcosx - 2sinx = 0
cosx - 2 = 0

[dense wing]

You divided by sin(x)

[gray vine]

Is that not permissible?

[dense wing]

it is not

cause it clearly loses solutions

[trail jackal]

you only can divide by something nonzero. So if you divide by sinx, you also need to deal with the case that sinx=0

[gray vine]

you also need to deal with the case that sinx=0
Why's that?

[dense wing]

cause sin(x) is a factor of the LHS...

apply zero product property.

ab=0 iff a=0 or b=0

[trail jackal]

you get sinx(cosx-2)=0 so either sinx=0 or cosx-2=0

[gray vine]

ah that's true

well cosx-2 = 0 can't be correct because cosine can't be greater than 1

[dense wing]

yep.

[gray vine]

so we're left with sin(x) = 0, which equals πn/2 + 2πn

oops

i mean π + πn

[dense wing]

Yes, sin(x)=0 when x=pi*n, n in Z

[gray vine]

alright thanks 🙂

.close

[cedar kiln]

Channel closed

Closed by @gray vine

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>.
• 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.

[crimson sedge]

guys can somebody explain to me how this was simplified?

[pallid arrow]

Change tan(x) to sin(x) / cos(x)

[crimson sedge]

then ?

im just really lost on how was this simplified