help-0

[glass lichen]

12 year old

[round socket]

whats wrong witha. 12 year old

[shrewd oracle]

Also 12 is against discord regulations

[round socket]

wtf is there an age limit

LMFAO

[shrewd oracle]

Yes

[glass lichen]

TOS says 13+

[sly mantle]

this would've been good to know

[round socket]

IM JOKING IM NOT 12

[vast blaze]

iirc 12 y/o are against discord regulations yeah

[shrewd oracle]

13 is the ages for tod

[round socket]

I WILL SHOW PROOF

IM NOT 12

IF NEEDED

[fading citrus]

[alpine sable]

show

[round socket]

i will show to a mod

not yall

everyone targets me for what

[shrewd oracle]

I mean you just admitted you were?

[tepid otter]

https://i.imgur.com/fwA0Idv.png

[round socket]

i was jokin.

[tepid otter]

can someone help me out with this question

[shrewd oracle]

You mean lying

[round socket]

why u so all up in my pussy

omg

ur not even a mod

[glass lichen]

not 100% sure on regression, but you can get 3 equations and 3 unknowns for the co-efficients

[round socket]

let the MODS question me

[shrewd oracle]

Never claimed to be?

[sick kettle]

can someone check this over

[round socket]

let rokeppajanpu question me!

[sly mantle]

@round socket watch your language and your behavior

[round socket]

WHAT

[wary stream]

Is this linear algebra related? Because if it is, do you know how to do least squares?

[round socket]

nah uhhh bc they said some nasty stuff too

[tepid otter]

it's algebra 2

[round socket]

aww naw baby this aint gon work!

[sly mantle]

your chaotic behavior has made it very hard for helpers to interact with you

[glass lichen]

quadratic is defined by 3 points iirc, so the least squares would be the same as the regular solution

[round socket]

I LITERALLY

DID U EVEN SEE HOW THIS STARTED

i @ ed my friend to see if he was available to help me

[alpine sable]

just stfu

[round socket]

and mosh stuck their head all up in my

nvm

and got all up in my business

[glass lichen]

You pinged a random user instead of just posting the question like everyone else

[round socket]

HE WASNT RANDOM THO

AND I WANTED TO KNOW HE WAS AVAIBLE

[glass lichen]

the question did not get posted until long into the tantrum.

[round socket]

bc you came at me for NO reason

going all like "did i ask'

like

hello!

[glass lichen]

hi

[sly mantle]

admittedly some helpers have been a bit cold in their approach in helping you

[shrewd oracle]

Thats not what the question channel is for?

[wary stream]

Well, stop pinging people because people are busy all the time. They don't have to make time in their schedule just to help you

[round socket]

he was gonna help me with my math??

wym this isnt what this channel is for

HE IS MY FRIEND THO LOL

[sly mantle]

but you're not helping at all by blowing up chat with your all caps tantrum

[round socket]

thorry

yall actting like i @ some rando

he is my bestie we go WAY back!

[wary stream]

You pinged specific people. Anyone can help you

[round socket]

i know i just knew he was avaible

wanted to know

if he was

[shrewd oracle]

What i mean is this is for people who want to understand a question off of one of the helpers, if a friend wants to help then dm them

[round socket]

dm ISNT an option for us

[shrewd oracle]

Then go to math general and not clog up the channel for others

[round socket]

but..

[sly mantle]

everyone please drop this convo

[round socket]

i

was

gonna ask for help doe

and i dont wanan ask in general

rokeppa wanna help me with me test 😏

[sly mantle]

sigh

[round socket]

im gonna sneak u into school

we can call

[alpine sable]

weirdo

[round socket]

and ill put u in an airpod

and you can whisper in my ear the answers

[glass lichen]

No

[sly mantle]

you can if you want a ban

[alpine sable]

bruh

[sly mantle]

bye

[shrewd oracle]

He actually did it...

[glass lichen]

#discussion

[alpine sable]

Oh I so sorry yes please I do need help

[mortal harbor]

i have a question which i just thought and i have no idea how to approach it
lets say you have a random point p what is the average distance to the circumference of a circle

[ionic jewel]

im willing to bet its the radius

yeah

it is

[foggy onyx]

(sin(30)(cot(30)(?) = answer should be 1.

anyone can help me with this

[mortal harbor]

you see logically i get that however proof?

[fierce frigate]

can anyone explain linear equations to me?

i havent used them in a long time

and i have a test about it tomorrow

[ionic jewel]

ur missing some parenthesis, but just use a calculator/unit circle

[foggy onyx]

no-

[grand oasis]

What's with that question mark next to cot(30)?

[foggy onyx]

im asking for help im not answering the previous question from the guy-

[ionic jewel]

for each point on the circumference of the circle with distance to the point of d, there is a point directly opposite of it where the distance id the diameter - d

you can do this for points on one half of the circle, to get diameter distance across the circle, and therefore diameter/2 = radius average distance

unless you happen to mean closest point on the circle

[mortal harbor]

nah thats what i meant thank you

know to generalise my question is how do you get the average distance of a point to a function?

like 0,0 and x^2 from like x= 0 to 10

or something

[ionic jewel]

uh

i could probably figure it out

[mortal harbor]

thank you

[ionic jewel]

do you know calculus

[mortal harbor]

i had that thought and its bugging me

somewhat

i know the basics

but im only year 12 in the uk

[shrewd oracle]

You do all diff calc in yr 12 no?

[mortal harbor]

no

[shrewd oracle]

Oh thought ya did

[mortal harbor]

well what does it consist of

like trig and stuff is year 13

and then we do some more in further maths in year 13 also

[shrewd oracle]

Ah

[ionic jewel]

so I suppose we are looking at a sum of the distances from the point to each point on the function

divided by the length of the function

$\int_a^b \sqrt{(x_0-x)^2 + (y_0-f(x))^2} , dx$

[ocean seal]

bunny

$\int_a^b \sqrt{(x_0-x)^2 + (y_0-f(x))^2} \, dx$

[ionic jewel]

seems like this would do what we want?

then divide by arc length

[mortal harbor]

so thats the whole sqrt bit is the distance of a certain point?

[ionic jewel]

$\frac{1}{b-a} \int_a^b \sqrt{1+[f'(x)]^2} , dx$

[ocean seal]

bunny

$\frac{1}{b-a} \int_a^b \sqrt{1+[f'(x)]^2} \, dx$

[ionic jewel]

$\frac{\int_a^b \sqrt{(x_0-x)^2 + (y_0-f(x))^2} , dx}{\frac{1}{b-a} \int_a^b \sqrt{1+[f'(x)]^2} , dx}$

[ocean seal]

bunny

[ionic jewel]

you know i cant imagine this doesnt cancel at all

[ionic jewel]

hmm that was the idea yes

[mortal harbor]

you see i dont understand the second bit

[mortal harbor]

this one

[ionic jewel]

second part is the arc length foruma

from a to b of f(x)

[mortal harbor]

ahh i havent learn that

imma research that

thank you so much!

makes sense

[ionic jewel]

wait one second

[mortal harbor]

okay

[ionic jewel]

hmm i found a different formula thats slightly different

oh it might be the same

I feel like the one I wrote makes sense but feel free to read https://math.stackexchange.com/questions/2413594/average-distance-between-a-point-and-a-curve-segment

[alpine sable]

Yo can you help me? With my question?

[ionic jewel]

but they are using parametizations and integrating with respect to curve length which is a little out of what i can think through right now

[mortal harbor]

yeah your one does make sense, i have a hard time following these ngl

[alpine sable]

@alpine sable it's this one

Nvm he isn't here anymore

[ionic jewel]

can only ronald help

[alpine sable]

Anyone else can help?

Nah nah

I just need the initial equation

That's it

[ionic jewel]

2L + W = 60
A = LW

[alpine sable]

Ahhhh thank you so much

It just couldn't wrap my head around it

[ionic jewel]

hopefully what I wrote makes sense

dont just be using it

[alpine sable]

It does

You need 2 sides so 2l +w and area is obv lw

[ionic jewel]

yep

[slow phoenix]

anyone able help me through this?

[ionic jewel]

$\sin\theta = \frac12$

[ocean seal]

bunny

[ionic jewel]

do you know the unit circle?

[slow phoenix]

ahha yeah

i figured it out i think @ionic jewel

[sturdy fog]

for question 5 the answer is 320 and i got 280 from 24.5tan85

why do we have to add 40m?

[slow phoenix]

is that the year 10 textbook

[sturdy fog]

yess

are you australian hahaha

[slow phoenix]

lol i remember using it

yeah lol

[sturdy fog]

could you tell me why we have to add the height of the shorter building?

[misty path]

YALL ROCKIN WITH DA BABY PHONE

[hidden horizon]

bruh this is math server

[misty path]

😂

[lunar vine]

x+2yt3z=12
x-3y+4z=27
-x+y+2=7

how do i solve this

[proud axle]

I would answer but just checking is the 2yt3z 2yt^3z=12?

[alpine sable]

How many baseball teams of 9 members can formed from 14 players?

[proud axle]

.-.

[misty path]

You computer has a virus

[alpine sable]

#2 lmao

[proud axle]

can I just

#rules

[alpine sable]

bc the question is non sense

[karmic pewter]

14C9 breh

It's not non sense 🙂

[alpine sable]

or im just dumb lol

[proud axle]

only 1. So you can't play baseball

[alpine sable]

lol

[proud axle]

in order to play baseball you need two teams of 9

[karmic pewter]

Bruh

The question is asking the arrangement of the team

Permutation and combination

[alpine sable]

ye

[karmic pewter]

14!/5!9!

[alpine sable]

btw how is it 14C9?

[karmic pewter]

^

[alpine sable]

oh

[plain dove]

help please

[scarlet bluff]

Bruh

[jagged imp]

match coefficients on both sides

[scarlet bluff]

You go to the ten thousandths place

Then 5 and above add 1 to the thousandths

4 and below subtract one

[alpine sable]

When 149 is divided by a 2-digit number, it has a remainder of 5. Find all the 2-digit divisors

im confused

[bronze spade]

No

[slow phoenix]

this is the most confusing question, anyone keen to help?

[bronze spade]

If it’s true for all real values of x, then it is also true for x=0 right ? Then k=10a - 2a^4

[jagged imp]

you can get that result by matching coefficients in x

im not trying to be confrontational but I don't understand what's wrong with matching coefficients

[bronze spade]

What do you mean “matching coefficients” ?

It might be me that does not understand what your idea is. Could you explain ? 🙂

The only solution here is to have every single coefficients equal to 0.

[jagged imp]

the coefficient of x^2 on the LHS(a^2-3a-2) must be equal to the coefficient of x^2 on the RHS(0), a^2-a-2=0, and in specific for k=10a-2a^4 you're matching the constant term.

[bronze spade]

Sure, it works but you have to justify why you can use this method

[jagged imp]

"linear algebra basis blah blah(kinda circular ig)", or alternatively, assume some quadratic ax^2+bx+c is equal to another quadratic dx^2+ex+f for all values of x. Then, setting x=0 we have f=c, then write both f and c as one of the two, subtract from both sides. Then you can take the derivative of both and use the same principle

Its just common knowledge that this is true for polynomials in my school curriculum at least

your method is faster and fine too

and same argument for higher degree polynomials

[bronze spade]

Yea, what you’re invoking is the uniqueness of the writing of a polynomial, no need to derivate. Otherwise I also like this argument : you rewrite the equation putting everything in the left hand side. You get an polynomial that is null for every x. It has a infinite numbers of roots. It must be the nul polynomial. Then a^3-3a-2=0, a^2-a-2=0 and 2a^4-10a+k=0.

[jagged imp]

...

you're the one who asked me to justify it

you said not to just invoke it

[bronze spade]

I guess solving those 3 equation might bring up a value of k 🙂

Yea, k = -12

I just wanted to know what was on your mind. The manners considered as common in your school might not be commonly shared by everyone. Thanks for sharing 🙏

[dire nacelle]

Pls help asap

[bronze spade]

What did you try so far ?

[alpine sable]

A factor has a remainder of 2 when it is divided by 3. It has a remainder of 3 when it is divided by 4. Find two such numbers that are smaller than 30

[dire nacelle]

nothing

[jagged imp]

is this a test?

[dire nacelle]

homework

[bronze spade]

Do you have any idea ?

[dire nacelle]

no

[alpine sable]

stop asking people that. why can't people learn math by themselves? Why do you never assume self-taught?

[dire nacelle]

🙇‍♂️ pls help with direct answer

[alpine sable]

@dire nacelle stupid you'll never learn that way

[dire nacelle]

i know

but i'm having lack of time

[jagged imp]

what does this question have to do with how a person learns math

[alpine sable]

choosing beggar

[dire nacelle]

pls help

that would be great

[alpine sable]

u always assume they're "homework" or "test" or "quiz" and never "ohh so you want to learn math on khan academy cuz u wanna git smart??"

[jagged imp]

not assuming shit

i asked a question

[dire nacelle]

if you don't want to it's ok no need of judgeing

[jagged imp]

leave this now

no point arguing

[bronze spade]

For the first one, I recommend considering the fact that $x^y = \exp(y \ln(x))$

[ocean seal]

allevmelc

[dire nacelle]

ohh

[bronze spade]

Use the same idea for the right hand side and then differentiate the equality. You’ll be able to determine dy/dx and evaluate it at x=e

For the second one, you could brutally apply Sarrus’ formula

“Rule of Sarrus”

[woven pollen]

m/n = 1/4 ?

[bronze spade]

You’ll be able to write f and derivate it

[dire nacelle]

you sure

[dire nacelle]

ohh

[woven pollen]

nope. I'm doing it again now.

[dire nacelle]

ok

Thanks for helping you guys are like angels to me

[woven pollen]

my best guess so far.

[shy tendon]

Writes insane math stuff
"my best guess so far"

[woven pollen]

😆

[median cedar]

What is the magnification when an object is placed at 2f from the pole of the convex mirror? *
1 point
-1
-1/3
-3/2
-2/ 3

[hidden horizon]

this is a test dont ask answers pls

[north hemlock]

So uh

using volume of cone, $V=\pi r^2 \frac{h}{3}$ we get $V\approx 200$ but if we use the given equation and do $2\sqrt[3]{\frac{3V}{2\pi}}$ we get $V \approx 25$?

[ocean seal]

Jukelyn

[north hemlock]

this is a bit confusing

oh

I see

nvm

lol, I did 2 times that radical for some reason

nvm

[short void]

Help

No calc question.

So.. my classmate signed me up for the Kenyan math Olympiad and I am screwed. This is an example of a past question. My biggest worry is that I have no calc. Apparently, round 1 is the easiest.

[echo spire]

What grade are you in? @short void ?

[short void]

I am in 12th grade (I think).

[jagged imp]

Am i missing something here? Is an equals sign missing here or...

Its not bad cropping i just found the paper myself. question could just be a typo or i could be a dumbfuck somehow

[echo spire]

Well, the only answer that makes sense is B xD

Yeah, I think that I know a way to resolve it let me try

[orchid python]

I don't get it, can any explain, thanks a lot:

e · d ≡ 1 (mod φ(n))

Then e * d becomes:

e · d = k · φ(n) + 1

[ocean seal]

Nam V. Do

[orchid python]

for some integer k

[short void]

Thanks, pls let me know 🙂

[short void]

What question?

[echo spire]

Try to raise the whole equation to the power of 8, it will be (x+1) at the power of 7. And maybe you can work something with it.

[jagged imp]

what equation

this this is just an expression there's no equation

I'm genuinely so confused

[short void]

I thought of putting (x+1)^1/2 and then solving, but still it doesn't work

[orchid python]

I got this now

[echo spire]

I should go back to my trigonometry

[short void]

No, no, no. Help me out.. I will help you with trigs after.

[echo spire]

I got no other idea

[orchid python]

How can I prove

φ(n) ≡ k * φ(n) mod m

[ocean seal]

Nam V. Do

[echo spire]

I got a question if a triangle has consecutive sides will it have consecutive angles as well?

[ionic jewel]

if this means even remotely what I think it does, no

but please clarify

[astral cipher]

Not always

[gray isle]

that statement is so vague

any two sides of a triangle are inherently consecutive

and technically any two angles in a triangle are also consecutive

why does that say test? <@&268886789983436800>

[fading zephyr]

this is not allowed

please remove it

[grand oasis]

Hey, anyone have any idea what ¹𝟺9⠡ means?

I have a problem that require me to translate the hex code of UTF-8 characters to Unicode code points, and I also have to explain the message behind it. They give me a hint: "A well-known sequence", but I have no idea how is it related to this.

The original hex code is c2 b9 f0 9d 9f ba 39 e2 a0 a1.

[rocky dagger]

maybe that's some weird binary..? no idea

[spice cape]

<@&286206848099549185>

[noble sinew]

start with option A - she gets paid a flat 14.90 dollars per phone sold

so find how many she sold and then multiply those 2

[spice cape]

For second question

[noble sinew]

so how much did she earn for option A and B

for b) it doesn't matter on option A how much the phone costs. So just recalculate with option B with the new info

[foggy onyx]

someone help me rn

i don't know how to solve-

[spice cape]

that is not a right angle btw

[upper kayak]

open channel?

ok

ln(x^2-2x-2) <= 0

my answer is -1 <= x <= 3

is that correct?

[gray isle]

no

you seem to be forgetting one of the restrictions of logs

[upper kayak]

hmm

,w ln(0)

[ocean seal]

Results provided by WolframAlpha

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

Requested by gang#5039

[upper kayak]

not undefined so 0 <= x <= 3?

[gray isle]

no

the argument of the log needs to be greater than 0 too
and you should consider another inequality (or compound inequality)

[upper kayak]

i got
-1 <= x <= 3
x <= 1 - sqrt(3)
x >= 1 + sqrt(3)

,w -1 <= x <=3; x <= 1-sqrt(3); x >= 1+sqrt(3)

[ocean seal]

Results provided by WolframAlpha

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

Requested by gang#5039

[upper kayak]

🙁

[rigid smelt]

for the 1 plus minus sqrt(3) it should be a greater/less than sign, not greater\less than or equal to

and combine the solutions

exclude out the intervals doesnt fit into the domain

which is ( (1-sqrt3) , (1+sqrt(3) )

[upper kayak]

why?

[rigid smelt]

because ln(0) is undefined

[upper kayak]

oh

[rigid smelt]

so if you put an equal there, you are basically including an output of 0

[upper kayak]

got it

thanks

[keen zephyr]

Can I get help on 12 please

[fierce frigate]

when u ask about linear equations and get ignored

bye

[stark rampart]

what #

[keen zephyr]

12

I understood up to drawing the diagram

afterwards i don't know what to do with it

[stark rampart]

have u tried doing it algebraically

i think that's the most straightforward way to solve it

[keen zephyr]

Nope I didn't know you could do it algebraically

[stark rampart]

Let $r=|w|$:

[ocean seal]

Armanzel (Ian)

[stark rampart]

coz for a complex number, you need both the magnitude and the argument

u can find w+z in terms of r (since it's the only thing missing from the two numbers)

$w+z=5e^{\frac{\pi i}{10}} + re^{\frac{2\pi i}{5}} \$
$w+z=5(\cos\frac{\pi}{10} + i\sin\frac{\pi}{10}) + r(\cos\frac{2\pi}{5} + i\sin\cos\frac{2\pi}{5})$

[ocean seal]

Armanzel (Ian)

[stark rampart]

now u just group the real parts and the imaginary parts

then remember that

$$\tan(arg(w+z)) = \frac{Im(w+z)}{Re(w+z)}$$

[ocean seal]

Armanzel (Ian)

[stark rampart]

Im is imaginary part and Re is the real part

[keen zephyr]

Ah ok this is making way more sense. Thanks!

[stark rampart]

cheers!

[wet dragon]

109 is 100% if 30 is 27.5% then 3 is %? how do you count it?

[upper kayak]

so the number these numbers are percentages of, is 109

that's because 109 is 100% of that number

you can verify that by checking that 30 is indeed 27.5% of 109

and you want to know what percent of 109 is 3

take it from there

[wet dragon]

yes and i don't know how to count it

i'm dumb

teach me?

please

https://tenor.com/view/nervous-sweating-wet-gif-7398449

[upper kayak]

a percentage is just a fraction with denominator 100

[wet dragon]

wait i just got it

it's around 2.8%

thanks!

[upper kayak]

so i want to know what percentage is 3 of 109

that's just a fraction

3/109

now make the denominator 100

to do that, divide both numerator and denominator by 1.09

2.75/100

rewrite as percentage

2.75%

you can round that to 2.8%

i rewrote them as f(g(x) + h(x)) = f(g(x)) + f(h(x))

no idea what to do next

[candid sluice]

do you think its true

[stark rampart]

hmmmm

$g(x) = x, h(x) = x+2, f(x) = \sqrt{x}$

[ocean seal]

Armanzel (Ian)

[stark rampart]

implies

$\sqrt{x+x+2} = \sqrt{x} + \sqrt{x+2}$

[ocean seal]

Armanzel (Ian)

[stark rampart]

so what do you guys think abt thi

this*

[neat merlin]

I'll try to solve

x = zero?

[stark rampart]

you aren't solving for x

[neat merlin]

a

What do i need todo

[stark rampart]

im giving what's called a coutnerexample

this contradicts f(g+h) = f(g) + f(h)

[neat merlin]

sorry im new at this server

[stark rampart]

oh hi im also new haha

you want to show that this works for all values of x

but since it doesn't, then it isn't tru

[neat merlin]

$\left{\begin{array}{l}1+x y=\frac{x^{2} y^{2}}{2 x-y}+\frac{2 x-y}{x y} \ \frac{2 x-y}{x y} \sqrt{2 x-y}=4-3 x y\end{array}\right.$

[ocean seal]

qcanser

[golden arch]

A 3 x 3 matrix |A| = 3. How much is |5A|? I expected it to be 15 but my correction key says 375

[stark rampart]

125 times 3

coz each term increases by a factor of 5

and to find |5A| you multiply three items at a time

[neat merlin]

Armanzel, where are you from and how old are you?

[stark rampart]

so each term is 5^3 times more than the original

im from the philippines and im 19

[golden arch]

Ah thanks i understand thanks!

[stark rampart]

cheers!

[stark rampart]

find x and y?

[neat merlin]

Yup

[stark rampart]

oh i got a neat solution

[neat merlin]

No there are 2 real solution for x and y

[upper kayak]

no

but how do i prove that

[stark rampart]

Let $a = 2x-y, b = xy$ \
$1+b = \frac{b^2}{a} + \frac{a}{b}$ \
$\Rightarrow b^3-ab^2-ab+a^2 = 0$

u can factor

[ocean seal]

Armanzel (Ian)

[stark rampart]

$(b^2-a)(b-a) = 0$

[ocean seal]

Armanzel (Ian)

[stark rampart]

so two situtations: $b^2 = a, b = a$

[ocean seal]

Armanzel (Ian)

[stark rampart]

this simplifies the problem to some degree

second equation of the system is: \
$ \frac{a\sqrt{a}}{b} = 4-3b$

Letting b = a or b^2 = a arrives at answers for b and a

$ \frac{a\sqrt{a}}{b} = 4-3b$

[neat merlin]

@ocean seal

[stark rampart]

i got $(1,1)$ and $(-\frac{1}{2}, -2)$

[candid sluice]

you can just find a counterexample

[ocean seal]

Armanzel (Ian)

[neat merlin]

Ohhh, perfect

i made this replacement for a fun solution to this equation

[candid sluice]

for a simple one, pick $f(x) = x^2$ and then try to find some h and g such that the statement isnt true

[ocean seal]

Pappa

[upper kayak]

ok

[stark rampart]

i think u have another solution

$(4+\sqrt{19}, \frac{4\sqrt{19} - 16}{3})$

[ocean seal]

Armanzel (Ian)

[stark rampart]

from the b^2 = a case?

there's another one which is just the conjugate of that

[neat merlin]

$6x^5 - 30x^3 + 30x - 13 = 0$

[ocean seal]

qcanser

$6x^5 - 30x^3 + 30x - 13 = 0$

[neat merlin]

How to solve this?, I didn’t 😦

[stark rampart]

did rational roots work?

there's at least one negative root that

that's for sure

[candid sluice]

solving that equation by hand is going to be awful

it doesnt have rational roots

also the only real root is positive

[shrewd sable]

There isn't though

[stark rampart]

oh what

oh did i do rule of signs wrong?

[candid sluice]

the root is $\sqrt[5]{\frac{2}{3}}+\sqrt[5]{\frac{3}{2}}$

[ocean seal]

Pappa

[stark rampart]

damn HAHAHAHA

what's ur method @candid sluice

[candid sluice]

wolfram alpha

[stark rampart]

oh HAHAHAH

[candid sluice]

solving that thing by hand is probably a punishment in a circle of hell

[stark rampart]

hmmm

maybe there is some technique to it

coz the root looks unbelievably neat

[candid sluice]

working backwards maybe doing something like $x = y + \frac{1}{y}$ might help

[stark rampart]

and it's the only root

[ocean seal]

Pappa

[stark rampart]

good idea

[candid sluice]

but there arent any general techniques to solve $ax^5+bx^3+cx+d=0$

[ocean seal]

Pappa

[stark rampart]

yeah

there is no formula

for degree 5 and above

sadly it's been proven hahaha

[hollow owl]

how do I found AB?

[neat merlin]

Genius wolfram...

[neat merlin]

Can anyone solve one from this? I really need help, I haven’t solve of this

[upper kayak]

hahahahah

[neat merlin]

[ocean seal]

qcanser

[neat merlin]

By not using wolfram etc

[stark rampart]

mathway HAHAHHA

[upper kayak]

things like these is the reason wolfram was invented

anyway let me pull up desmos real quick

[neat merlin]

Where are you from and how old are you?

[upper kayak]

why and why?

[candid sluice]

for the first one first factor out x^5 out of the first 4 terms

[upper kayak]

ez

[hollow owl]

and how do I found ab?

[upper kayak]

OA = 4
MB = 1
for the triangle formed by the red line and OM,
the short leg is OA - MB = 3
using pythagorean theorem, AB^2 = 4^2 - 3^2

[hollow owl]

OK OK thanks

[harsh acorn]

$$x=2 (mod5)$$
$$x=1 (mod8)$$
$$x=3 (mod14)$$

[ocean seal]

Hyl1s

[harsh acorn]

asks for x's the smallest value

[stark rampart]

chinese remainder theorem

hahahaha

there is a straightforward way to solve this

since x = 2(mod 5), you can let x = 5a+2, where a is an integer

then subsitute in the next equation

$5a+2 \equiv 1(mod 8)$

[ocean seal]

Armanzel (Ian)

[stark rampart]

where u get $a\equiv3(mod 8)$

[ocean seal]

Armanzel (Ian)

[stark rampart]

then let a = 8b + 3 for some integer b

and just continue

17 is the answer

i thinks o

[stark rampart]

but u gotta let x be a positive number, or else you can have an immensely negative x

[harsh acorn]

@stark rampart Can you show me how to solve it with chinese theorem?

I tried but I stuck in
$$6x_2=1(mod 8)$$

[stark rampart]

okay so

[harsh acorn]

can't find x

[ocean seal]

Hyl1s

[stark rampart]

wait i think this is wrong

coz u divide an even number by an even number and u get an odd remainder

maybe backtrack

so the answer is generally in the form

[harsh acorn]

$$N_2*x_2 = 1(mod 8)$$
$$70x_2 = 1(mod 8)$$
$$6x_2 = 1 (mod 8)$$

[stark rampart]

560n + smth

[ocean seal]

Hyl1s

[harsh acorn]

Isn't it like that

[stark rampart]

what does N2 stand for

oh u did the first and the third equation first?

[harsh acorn]

I get x_1= 6 in first one

[stark rampart]

oh wait

u always get the answers in terms of variables

[harsh acorn]

and as I know $$x= N_1X_1B_1+N_2X_2B_2...$$

[stark rampart]

the best way is to start with the one with the biggest divisor (which is 14)

ohhhhhh

oh i see

[ocean seal]

Hyl1s

[stark rampart]

so u do the generalized one

ok ok

[harsh acorn]

yea with that it looks easier for me :p

[stark rampart]

so like $N_1(5)(8) + N_2(8)(14) + N_3(14)(5)$

[ocean seal]

Armanzel (Ian)

[stark rampart]

this is good because

when we take mod 5 for example, the first and third term of this vanishes

since they are both divisible by 5

[harsh acorn]

Isn't N1= 5*8?

oh

my bad

I understand it now

[stark rampart]

ok ok

i think

oh waittt

u can't use chinese remainder

[harsh acorn]

bcs 8 and 14

2

[stark rampart]

u can only use it if all divisors are coprime

[harsh acorn]

oh

[stark rampart]

yeah yeah

oh i got lucky with the 17 HAHAHHAH

that's whyyy i couldnt find solutions too

[harsh acorn]

so how can I get rid of 8 and 14?

[stark rampart]

u can split mod 14 into mod 7 since 3 mod 14 is same as 3 mod 7

obviously x is odd so no need to worry

$$x=2 (mod5)$$
$$x=1 (mod8)$$
$$x=3 (mod7)$$

maybe this one works

[ocean seal]

Armanzel (Ian)

[harsh acorn]

Let me try it now

[stark rampart]

this one's doable

[burnt pagoda]

guys i need help with an equation

u guys want me to show u

[stark rampart]

go ahead

[scenic flame]

What's the reason there's no third derivative?

can anyone help

[stark rampart]

there's no third derivative?

i have a secret technique for this one hehehehe

$\tan(\frac{\pi}{4}-\frac{x}{2}) = \frac{1-\sin x}{\cos x}$

[ocean seal]

Armanzel (Ian)

[stark rampart]

this is a cool identity

the difficulty in differentiating becomes enormously easier

this can be derived from

$\tan{\frac{x}{2}} = \frac{1-\cos x}{\sin x}$

[ocean seal]

Armanzel (Ian)

[stark rampart]

then replace x with

$x \Rightarrow \frac{\pi}{2} - x$

[ocean seal]

Armanzel (Ian)

[stark rampart]

the tangent identity will help find the derivative easier :3

[harsh acorn]

@stark rampart I get 569 but it is false

[stark rampart]

ohhhh

[harsh acorn]

,w 569 = 2 (mod 5)

[ocean seal]

Results provided by WolframAlpha

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

Requested by Hyl1s#3863

[stark rampart]

what values of n1 n2 n3 did u get

[harsh acorn]

n1= 56 x1=2
n2= 35 x2=3
n3=40 x3=2

[stark rampart]

oh wait what is x1 x2 x3

[harsh acorn]

$$N_1X_1= 1(mod 3)$$
$$N_1= 56 , x_1=2$$

[ocean seal]

Hyl1s

[harsh acorn]

,w 56*2= 2 (mod 5)

[ocean seal]

Results provided by WolframAlpha

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

Requested by Hyl1s#3863

[harsh acorn]

56 = 8*7

product of other modulus

[stark rampart]

i think the n1 x1 method is harder

there should only be 3 missing variables

[scenic flame]

thanks again @stark rampart

[stark rampart]

cheers!

[alpine sable]

How do I solve this?

[minor thorn]

set them equal to each other and solve for x

[smoky snow]

Hey guys, my question is easy to understand, but (imo) hard to solve. Here it is:

Imagine you have an N-sided regular polygon. Let's call the distance from the center of the polygon to one of the vertices A. At each vertex of that polygon, you center a square of side length B. Now, given a value for N and A, what is the largest B that does not let the squares overlap?

Note: all the squares are oriented in the same way.

Here's a drawing to better show what I mean

[stark wadi]

Mike is running towards the last pie for sale at the school bake sale. Jack see Mike running and Jack decides to sprint for the pie. Mike runs at full speed, 5.3 meters/second, while Jack speeds up at a rate modeled by the function 1.5x^2 + 0.3x. The pie is 25 meters away from Jack when he begins to run. Mike is already 4 meters ahead. Who reaches the pie first? Explain how rounding an exact answer could affect the results.

Mike: y = 5.3x + 4
Jack: y = 1.5x^2 + 0.3x

Does anyone know how to determine when they reach the pie?

[harsh acorn]

Can I solve this question without hard force?

[upper kayak]

mike's distance from the pie is 21 meters

set y = 21 and solve

that's how much time it takes mike to reach the pie

then in jack's equation, set y to 25

that's the time it takes him to reach the pie

[supple egret]

Hey guys, can someone help me on how to compure a Fourier Series representation of f(x) = ch(x) on (-pi, pi) domain

I m struggling with calculating An

chx is an even function so that helps in a sense that i don't have to calculate Bn

I m not sure how to use $ syntax to present where i stopped

If i try to write it as text it will be too messy and unreadable probably.

[smoky snow]

any <@&286206848099549185> available for this? I'm still not sure how to solve it

[rugged ermine]

um first off dont ping em they will come to you

[ivory grove]

no his questions been there for 15 mins

[rugged ermine]

oh

welp

oke

[rich niche]

some way to simplify this

j is the variable here btw

[pliant estuary]

Hey guys, can someone please show me, how can I simplify this equation?
$$U_2=U_1\frac{\frac{R\frac{1}{j\omega C}}{R+\frac{1}{j\omega C}}}{R+\frac{1}{j\omega C}+\frac{R\frac{1}{j\omega C}}{R+\frac{1}{j\omega C}}}$$
I´ve had some attempts but Im getting non senses. For context, $U_n$ are voltages, $j$ is imaginary unit, $\omega$ is angular frequency, $R$ is ohmic resistance and $C$ is capacitance. Any help would be appreciated (btw. this is a equation for output voltage of Wiens harmonic oscilator). Ive asked this question yesterday but Im dumb and made a typo in the equation.

[ocean seal]

marejak023

[summer quiver]

Didn't I help you simplify this already?

[alpine sable]

anyone to help in another channel?

[pliant estuary]

Yes, but today Ive noticed that Ive unfortunately made a typo 😬

ive typed jwC instead of 1/jwC

[summer quiver]

But I thought you wrote out the steps as we did them?

[alpine sable]

.

[final crag]

Could someone help me with this?

[summer quiver]

So basically, for a function to have an inverse, it needs to be injective (or one-to-one), which means that for every argument x, f(x) has a unique output

For example, consider $f(x) = \ln x$ . This function is strictly increasing, so there are no two numbers $a,b,a \neq b$ such that $\ln(a) = \ln(b)$. Therefore $f(x) = \ln(x)$ is invertible, and its inverse is $f^{-1}(x) = e^x$.

[ocean seal]

iateawalrus

[final crag]

so the first one has an inverse?

[alpine sable]

How does this equal to 1/2

[summer quiver]

Try l'Hopital's rule?

[alpine sable]

Is it the only option?

[summer quiver]

No idea

[alpine sable]

I hate l'Hopital for trigonometry

But thanks

[summer quiver]

Looks like it. Try creating the inverse

[summer quiver]

Sorry man :/

[alpine sable]

Np thanks

[final crag]

can you give example of what inverse is

like inverse of abc is cba?

just work from the back of the string?

[alpine sable]

multiply top and bottom by the conjugate @alpine sable

[final crag]

write it backwards

[summer quiver]

Well, suppose we had a function f(abc) = cab

[final crag]

then it would be bac?

[summer quiver]

Then we could have an inverse function g(abc) = bca

Let's try f(g(abc))

[final crag]

wait what

[summer quiver]

?

[final crag]

how is it bca

[summer quiver]

What did f do to abc?

[final crag]

it grabs one from the back and then one from the front

[alpine sable]

Thanks it worked

[summer quiver]

Np

[final crag]

not sure im confused

[summer quiver]

Alright

Let's try your functino bac

[final crag]

i have to give a function

[summer quiver]

A property of an inverse function is that if f and g are inverses, then f(g(x)) = g(f(x)) = x

Let's try f(g(abc)) if g(abc) = bac

Then f(g(abc)) = f(bac)

What f does is, it rearranges the string in the order 3, 1, 2

[final crag]

f(cdeabfg) = abcdefg?

[summer quiver]

so that f(abc) = cab

So f(bac) = cba

cba is not the same as abc, so f(g(abc)) =/= abc, so g(abc) = bac is not an inverse of f(abc) = cab

[alpine sable]

(5b + 3) + 4 (3b - a)

how do i solve this

[icy trail]

distribute the 4 into the second bracket and collect like terms

[gray oxide]

An earthquake happens underneath the sea which creates a big wave. The water depth* d* (m) in a port that gets reached by the wave gets the formula

.....

$$d(t) =11-12sin((2pi*t) /15)$$

0<t<T

t= time in min

T = period? ( when a whole rotation happens?)

Between which timestamps is the port dry?

I've translated this so if this sounds strange I'll try using another word if u just ask me

The period is 15min

Beyond that I have no idea what to do

[ocean seal]

!callidus

[gray oxide]

<@&286206848099549185>

[kindred tartan]

Let the coefficient of correlation of two variables X and Y be larger than zero. What will be the effect on it, if the data of X are multiplied by the factor of 2?
a) The effect depends on the data of X.
b) It depends on Y.
c) The coefficient will be doubled.
d) It will be increased fourfold.

[tepid otter]

can anyone help me on this question

[candid torrent]

@tepid otter do you know completing the square method?

[tepid otter]

uhh no

i tried looking up but i found nothing

[candid torrent]

ok so basically it works by converting the function into vertex form by adding a term and subtracting the same term

[wary stream]

There's tons of resources on completing the square

[wary stream]

Like this

[lofty siren]

Hey everyone, does anybody know the answer to this one? Solve the following 10 powers In this question, print all determinable decimals.

[zenith pagoda]

use prime factors

15 = 3 × 5
10 = 2 × 5

thus, 2×15 = 10×3

then 3 × 10 × 10^(-9) = 3 × 10^(-8)

the result of this is 0.000 000 03

[wary stream]

You realize you're not suppose to give full solutions, for your suppose to assist at the parts the people gets stuck, according to the #❓how-to-get-help rules

[zenith pagoda]

it doesn't say I'm not supposed to give full solutions

[wary stream]

"Try your best to explain and not give answers"

[zenith pagoda]

yes

[wary stream]

You gave the answer here

[zenith pagoda]

I explained and elaborated.

It doesn't say "do not give away the answer"

get a mod to arbitrate or leave me alone

[wary stream]

And fully showed how to get to the solution

[lofty siren]

he was just trying to help why are u being so mean? and im still looking into it trying to learn

so technically i am still learning

cuz im trying to get to the conclusion

[zenith pagoda]

<@&268886789983436800> is backseat moderation against the rules? because that's what's happening right now

[wary stream]

I wasn't being mean. They showed the entire process on how to achieve the solution. We're here to assist on the parts you get stuck, not give full work

[glass lichen]

I mean, you did just post the solution

instead of asking what they were stuck on and explaining

[gray oxide]

So do I get to retag helpers after 50min or?.....

[fading zephyr]

this isn't a big deal, you can just drop it. mosh and dldh06 are not wrong in that a more guided approach is preferred

mixolydian did at least provide the intermediate steps, which is nice, but it is indeed preferred to not just give out answers.

[lofty siren]

its not a biggie tho

[fading zephyr]

agreed

[lofty siren]

👍🏼

[wary stream]

It's not, I was just saying, that they should teach/explain the parts that you got stuck on until you understand what the concept is, instead of showing the full work and solution

[fading zephyr]

everyone take it easy 😛 this just escalated a little due to the "tone" in which things were said

and yes, @gray oxide , you can first post your question again once this channel (or another one) is free, and if another 15 mins go by, you can ping again

[gray oxide]

If someone asks a question on the 15th minute do I wait or ping?

[fading zephyr]

you can ping, and/or tell them the channel is taken

[gray oxide]

Ok I see thank you

[sly halo]

I'm playing this video game, and you can upgrade items. There is a .1% chance for an upgrade to be successful, 1 in 1000. However, on the 1000th try, it is a guaranteed upgrade. How do I find the average number of tries it will take to upgrade?

[zenith pagoda]

does it scale up?

or are all probabilties the same, except the 1000th

[sly halo]

all the tries have a .1% chance of success, but if you failed 999 times, the 1000th is guaranteed

so normally the avg would be 1000 times, but this guaranteed upgrade thing must make it lower, Im just not sure by how much

[wicked axle]

can someone show how to do 4B?

[gaunt coyote]

hello
anyone here is good at excel and math?
i have a very complicated question
tryd now like 4 ppl here they tryd it but no succes so far
best would be when u are a trader

[glass lichen]

draw it

then use cosine law/sine law

or convert it into cartesian vectors and add them

[wicked axle]

this was my attempt

but im 99pct sure its wrong

[wary stream]

Find the x and y components of each vector, sum up the like components, then use pythagorean theorem for the resultant

[wary stream]

Where did the angle 70 come from?

[wicked axle]

yeah i think thats my biggest mistake

[wary stream]

Draw your x and y axes and plot on that

[barren torrent]

https://cdn.discordapp.com/attachments/238956921581862913/848644781499482132/unknown.png\

https://cdn.discordapp.com/attachments/238956921581862913/848644781499482132/unknown.png

@wary stream

can you help me?

[wary stream]

Create proportions

[wicked axle]

so i tried again

but i still think that angle is 70

i feel like im missing something

180-50-60

[wary stream]

The angle between the two vectors is definitely not 70

[wicked axle]

not that angle

the one i drew in the image

[wary stream]

You drew the angle between the two vectors

And labeled as 70

[wicked axle]

no the blue arrows are the two vectors

and that forms and angle of 110

the 70deg angle is the angle thats formed when drawing the parallelogram @wary stream

[zenith pagoda]

so the expected value is E = 1*p + 2(1-p)p + 3(1-p)²p + etc.... because you are basically expecting to fail n-1 times before getting it on the nth try
I will try to shortcut this https://www.geeksforgeeks.org/expected-number-of-trials-before-success/
Normally your answer would result in E = 1/p but we have to replace the 1000th value (subtract the usual and add the 100%) as such:
E = 1/p - (p)1000(1-p)^(999) + 1000

now, just plug in the 0.1% as p = 0.001

[sly halo]

@zenith pagoda hmm that gave me 1999

[zenith pagoda]

oh wait

I messed up

lets see

[sly halo]

just thinking about it, my guess was it would be 750

[wary stream]

So I don't do parallelogram method, instead I break up the vectors into x and y components because then you can sum them up component wise and then use pythagorean theorem to get the resultant and then use arctan for the angle. I solved it myself, got the same r as you(which I didn't think I would) but the angle I got was different

[sly halo]

either that or 500

[wicked axle]

ah okay

i'm not far into this unit yet so we didn't learn about that

are you familiar at all with the parallelogram method? because that's how we're expected to do it for these questions

at least for the time being i guess

[wary stream]

It's just using sines and cosines

[zenith pagoda]

I'm gonna get a piece of paper to work this out, I made a mistake bc I forgot that you basically have to ignore all values after 1000

[dusk coral]

some someone please explain how they got the 0 and undefined im so confused

<@&286206848099549185>

[wary stream]

Don't ping the helpers right away, you have to wait at least 15 minutes before pinging

[dusk coral]

ohhh

my bad

[zenith pagoda]

https://www.wolframalpha.com/input/?i=sum+np(1-p)^(n-1)+from+1+to+999+given+p%3D0.001

there was no simple solution I could find

[sly halo]

264?

that cant be right

[zenith pagoda]

hmm

[sly halo]

I feel like it has to be between 500 and 1000

[zenith pagoda]

yes thats for sure

[young arch]

@dusk coral This is what the original tanx trig function looks like, hold on theres more to the explanation

[zenith pagoda]

ah I found my mistake, I think. So I completely left out 1000 but the correct thing to do was to add on the final term as being 1000(1-p)^999

note that you do not multiply by p, because the probability is 1

[young arch]

so the amplitude is 4 for your graph y = 4tan2t and the period is going twice as fast (the 2t) so that means your function will go between the interval pi/4 to 3pi/4 as represented in your graph

if you plug in the initial point and the terminal points of the function y = 4tan2t then it would be 4tan2pi/4 = 4tanpi/2 and 4tan6pi/4 or 4tan3pi/2

so take a look at the unit circle and see where the values are, so for 4tanpi/2 focus on pi/2 on the unit circle and see the sine and cosine values, tangent = sin/cos so here it is (0,1) for the values of sine and cosine

so if tan = sin/cos then tan = 1/0 = undefined

[zenith pagoda]

so add 368.0635 to the result, 264.241 + 368.0635 = 632.3045

[zenith pagoda]

this is it. the true final result

[young arch]

so 3pi/2 for tan would be -1 / 0 so undefined as well, not zero

[sly halo]

@zenith pagoda hmm, ty 🙂

[kindred tartan]

<@&286206848099549185> .

[sly halo]

@zenith pagoda this is what I was trying to do in excel

this is assuming a 10% chance of success

and hopefully my normal distribution is right

[zenith pagoda]

hmm

I'm not entirely sure how to read it

[sly halo]

so the first line is the different possible results, a result being "it took me 7 tries to get a successful upgrade"

the second line is me attempting a normal distribution. so for each possible result, how many times did it land on that many tries for a successful upgrade. you can see that 10 tries was the most common and 1 / 19 tries are the least.

[noble sinew]

This is correct

[sly halo]

then the next two lines are the previous 2 lines multiplied together. one of them is simply multiplication, and the other has the values clamped at 10

given the scenario that after 9 upgrades, it is guaranteed

and then the green cells are the total number of tries / Total successful upgrades, the answer

the non clamped answer makes perfect sense

so I would think the clamped answer is right as well

[noble sinew]

it isn't - here is proof using simulation if you don't trust his valid calculations

drawing 10^7 from negative binomial (counts amounts of fail until 1st success with prob 0.001) - so we +1 to that since we want number of fails + the success one counted.

Then I show the mean and then replace all values where x>1000 with 1000 and take the mean

[sly halo]

ty 🙂 Its not that I didn't trust his work, just curious why mine was wrong

[zenith pagoda]

thanks

[noble sinew]

why would it be a normal distribution (first of all it isn't continious)

[sly halo]

ya, I was just thinking that

[zenith pagoda]

is that R

[noble sinew]

si

[sly halo]

I cant use the same distribution for both experiments

[zenith pagoda]

I gotta learn R

[alpine sable]

a/b and b/a are called convalent?

does someone know analitic geometry? i need help

[noble sinew]

ye R is nice

[alpine sable]

<@&286206848099549185>

[noble sinew]

read rules @alpine sable

but pretty sure no - sounds like a chemistry term

[sand mountain]

lmfao i like them chem questions in math

[icy peak]

is this the right way to right domain and range with the less than symbol?:

{x ∈ 𝑅|0<x}
{y ∈ 𝑅|0<y}

[zenith pagoda]

b/a is the multiplicative inverse of a/b

[zenith pagoda]

sure, but x>0 is more conventional

[celest ledge]

anyone want to take a crack at this? https://math.stackexchange.com/questions/4155551/sum-of-angles-extension-to-crossed-polygons

[icy peak]

bet but thats not incorrect right?

[celest ledge]

there is also this link which might be helpfull. I just cant get the java applets to work http://dynamicmathematicslearning.com/star_pentagon.html

[zenith pagoda]

they are both correct

[icy peak]

do you think professors would dock marks for that?

[zenith pagoda]

what does replace do exactly

does it erase all values after 1000

[noble sinew]

replace values greater than 1000 with 1000

[zenith pagoda]

uhh

do you mind explaining a bit further? how or what does it replace

[noble sinew]

x is a vector that has 10^7 different values

depending on how long it took for each individual simulation for it to get the success

if we got a value above 1000 in x we then replace it with 1000 since that was the requirement we had that we always get a success at 1000

[zenith pagoda]

hmm

pretty impressive

so x is a kind of lazily evaluated thing or something?

is that how you can just add 1 to it

[noble sinew]

it looks like this

[zenith pagoda]

oh

right, so it's an actual vector

[terse merlin]

Hi, is this where I get help?

[zenith pagoda]

of course

[terse merlin]

Do you know who could help me?

[sterile beacon]

Can you draw the energy schedule for the following cases:
a Fredrick takes the elevator up in a house.
b A car increases speed.

[terse merlin]

I'm trying to relearn calc, so i have to do algebra

and for some reason numbers arent adding up and i would like help since i dont have any1 else atm

[zenith pagoda]

is it a specific problem? post it in here (or in one of the other questions channels) and someone will eventually help

[terse merlin]

okay

[sterile beacon]

You could help me in the meantime xD

[terse merlin]

I have to rewrite this in Standard Form. with the points being (0,-2) and the slope being (2/3).

(I know its a very basic problem but im having trouble with the 2/3 part because it always ends up as a fraction for me.

[sterile beacon]

~0.67?

[zenith pagoda]

y = ax

[terse merlin]

stand. requries it to be an integer

[sterile beacon]

Could you multiply everything with 3?

[zenith pagoda]

yeah basically thats the answer

[terse merlin]

i tried that but then it would be 2x

no?

so itd be:

[zenith pagoda]

well lets see

[sterile beacon]

x/3 * 3 = x

?

[terse merlin]

3y+6=2x

how do i get rid of the 2?

[sterile beacon]

/2?

[terse merlin]

on both sides

aka 3/2y

then im back at square 1

[sterile beacon]

What

[zenith pagoda]

ehh I'll just try to work it out right now

[sterile beacon]

oh

y=(2x-6)/3

Generally the y is supposed to become alone

[zenith pagoda]

-2 = 2/3×0 + b therefore b = -2

y = 2/3x - 2

[sterile beacon]

y = 2x/x - 2

[zenith pagoda]

ehm, standard form does not require that a is an integer..?

[sterile beacon]

yes this is the f(x) function

not if it's an f(x) function

[terse merlin]

all numbers have to be integers

[sterile beacon]

the slope is 2/3

[terse merlin]

A is the only one that needs to be a positive

i get that the slope is 2/3 but what does taht do for me if i multipl it by 3

would it be easier if i showed me work?

[sterile beacon]

I didn't understand your question first

[zenith pagoda]

I'm not sure why the integer requirement

[terse merlin]

ok ill post my work

[zenith pagoda]

is it part of the question

[terse merlin]

cuz its standard form

[sterile beacon]

Yeah why is integer required?

[terse merlin]

or i have to convert it anyways

[sterile beacon]

What's that about

[zenith pagoda]

but standard form decidedly does not require that, right?

ok so

I just looked up what standard form is

[terse merlin]

"must be integers and A must be postive"

[zenith pagoda]

it's Ax + By = C

[terse merlin]

mhm

im stuck at this part

[zenith pagoda]

y = 2/3x - 2
2 = 2/3x - y
6 = 2x - 3y

[terse merlin]

wait can i divide 2x entirely?

or wait

no its not division

is subtraction

🤦‍♂️

[zenith pagoda]

this is the answer, A is positive

[terse merlin]

2x-3y=6 is the answer

[zenith pagoda]

yes

just subtract 3y from both sides from your final expression

[terse merlin]

why would i subtract 3y if its suppose to be standard form?

isnt it fine as it is?

[zenith pagoda]

no I meant from wht you had on paper

[terse merlin]

oh yeah mb

thank youu

[zenith pagoda]

whats the interface youre using

[noble sinew]

RStudio

[livid lintel]

does this server help with Stoichiometry

[alpine sable]

Just to make sure, vectors and matrix are pre calculus content, right?

[zenith pagoda]

yeah I thinj so

[alpine sable]

alr

[normal lava]

Hello everyone, can anyone explain on how to answer this question?

Ping me if you have a solution, thank you!

<@&286206848099549185> ^^^

[bronze spade]

Have you tried something so far ?

[fresh parcel]

use the binomial theorem

oh wait

[normal lava]

well Idk what to do to the x+2, but I do know how to expand the denominator

[bronze spade]

Wait, do you know how to expand 1/(2-x)^2 ? If that’s the case you’ve done the biggest part