help-42

[fast heath]

yes

and the system is in equilibrium

[simple musk]

but

how many newtons left upward force

[fickle hearth]

Make your own channel don't use this

[simple musk]

fr

[fast heath]

what a scale does is it finds the force in the string/spring and divides it with g

so it gives mass directly

[simple musk]

wow

[simple musk]

you mean the tension

[fast heath]

yeah

since the unit is gm

it has already divided by g

if the unit was newton

it is the force

[fast heath]

so whats the net force upwards

[simple musk]

i need to pass grams to kg?

wassup

[fast heath]

since all units are in gms theres no need to do that

[simple musk]

sum of forces in y = 0 btw, equilibrium

[fast heath]

yes correct

[simple musk]

but how many newtons is left force upward

i dont get it

just pass grams to kg and multiply by gravity

[cobalt briar]

yes

[fast heath]

yes

[simple musk]

ok

[fast heath]

correct

now whats the weight

[fast heath]

yes

[simple musk]

is the left upwards force the weight i just calculated

but

[fast heath]

call it force

[simple musk]

why do they give the length of the bat

[fast heath]

for part b ithink

[cobalt briar]

yeah

[simple musk]

center of mass is 110/2 cm

or no?

[fast heath]

no

[simple musk]

xd

[fast heath]

the mass distribution isnt even

the scales tell us that it isnt even

[simple musk]

ohh

let me find the right force then

in upwards

[fast heath]

yes so that first

do*

[simple musk]

how do i find where is center of mass

[fast heath]

do you know the formula for centre of mass

(summation m1x1)/(summation x1)

[simple musk]

is there a formula?

hmm

[fast heath]

yeah i mean

the basic

one

[simple musk]

wdym?

[fast heath]

you should study about it first i guess

[calm coral]

@simple musk Has your question been resolved?

[simple musk]

help bro please

this is sum of momentum?

[simple musk]

.

[fast heath]

no

[fast heath]

i can help but you dont know the formula

let x be the distance of centre of mass from the right hand side

[simple musk]

ok

[simple musk]

so?

let me do the drawing

[fast heath]

according to formula
x = (m1x1 + m2x2)/(x1 + x2)
so m1 = 659gm
x1 = 110cm
m2 = ( the mass you got for right hand side i forgot the value )
x2 = 0 cm

[simple musk]

what

x2 = 0cm

[fast heath]

yeah since we are taking the distances frim the right hand side

[fickle hearth]

How did you get x2 =0?

[fast heath]

since we are taking distance from the left hand side

[simple musk]

weight of the bat is 936g

[simple musk]

oh ok

[fickle hearth]

Oh okay

[fickle hearth]

But the formula is wrong, it should be m1+m2 in the denominator

[fast heath]

yeah sorry

im sleepy

rn

its m1 + m2

[simple musk]

let me fix the drawing

[simple musk]

x2 = d1 = 0

[fast heath]

correct

[simple musk]

wait but why place it to the right

is there any benefit?

or to the left

[fast heath]

yeah so one distance becomes 0

less calculation

[simple musk]

bro i am doing something wrong

just dunno what

[fast heath]

i know

and its my bad

[fast heath]

@simple musk

[simple musk]

so what is the formula

d1 + d2 = (659g + 277g)/(d2 + 0)

[fast heath]

summation(m1d1)/summation(m1)

so here

[fickle hearth]

It's:
x = (m1x1 + m2x2)/(m1 + m2)

[fast heath]

yeah

im so sorry

[simple musk]

wait but

i am doing something wrong?

[fast heath]

why did you change the x to d1 + d2

[simple musk]

if x2 = 0, x1 = 110cm

[fast heath]

yes

[simple musk]

d1 + d2 = 110cm

idk, is confusing

[fast heath]

no but x is the position of centre of mass

keep it as it is

[fast heath]

yeah thats why you should study first im not the best teacher

[simple musk]

what is x1 and x2 btw?

we were saying x2 = 0

[fast heath]

its the distance of masses from our reference point

we are taking reference at m2

[simple musk]

the reference point is at m2, ok

[fast heath]

so d2 = 0

and d1= 110cm

and that will give us Xcm which is of the distance if centre of mass from our reference point

[simple musk]

no, wait a moment

[fast heath]

ok

[simple musk]

x2 = d1

is what makes it confusing

[fast heath]

😭

[fast heath]

lets get rid of x1 and x2

just use d1 and d2

[fast heath]

yes

[simple musk]

yep

[fast heath]

and d2 = 110 cm

[simple musk]

i think the numerator is incorrect

659g . 0cm

[fast heath]

once again we are taking the reference poimt at m2

so distance of m1 would be 110cm

[fast heath]

you didnt multiply with 110

last step

[simple musk]

ohh

[fast heath]

yeah

do you have the answer ?

[simple musk]

but i don't get it, 77cm apart from the right side
?

[fast heath]

yes

[simple musk]

like 77cm starting from m2

[fast heath]

tes

yes

so (110-77)cm from m1

[simple musk]

answers will be uploded in 2 more days lmfao

[simple musk]

👌

[fast heath]

oh alr

[simple musk]

hardest shit i have ever done in life

[fast heath]

its actually very easy trust me

you cam watch some yt vids to learn it

[simple musk]

didn't knew this formula existed, teacher said use point of reference right side mass m2 and equal the sum of momentum to 0

whatever that means

[fast heath]

momentum??

[simple musk]

will do brother

[simple musk]

she is probably tripping

[fickle hearth]

Well it might have been mass moment

[fast heath]

yeah probably

theres no velocities here

[simple musk]

wdym?

[fickle hearth]

Mass moment=m1x1+m2x2

Essentially our numerator

[fast heath]

yeah btw that works too

you can try if you want

[simple musk]

what is mass moment btw?

oh maybe is a translation issue from my end

maybe she meant that but in my language

[simple musk]

yep but what about the denominator

[fickle hearth]

Mass moment is... Well I don't know how to exactly define it.

Sorry

Our teacher just told us the centre of mass formula and told us the numerator meant mass moment

[simple musk]

interesting

[fickle hearth]

Do you know what momentum is? Maybe I can explain it in that way

[simple musk]

p = mv

[fickle hearth]

And conservation of it?

[simple musk]

just a measurement of how difficult is to make a particle in motion to stop

[fickle hearth]

Yh and mass movement is the measurement of how hard it is to move an object. Or make it comes in motion

[simple musk]

total linear momentum of a closed system remains constant if no external forces act on it

[simple musk]

oh ok

[fickle hearth]

Yup, in such a way you can conserve mass moment too.
Like if the centre of mass of the object changes or not based on the change in mass or distance of the object.

But tbh you should ask your teacher Abt it for more clarity

Bcuz I might be wrong about some points

[simple musk]

i appreciate it tho

i think i kind of get it

i remember doing this in the past, just i am rusty af

i really appreciate you guys helping me solve this lmao

i will continue, ty

.solved

[calm coral]

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[simple musk]

well T2 = T2x

and T2y = 0

wdym?

[clear raft]

You need to start with a free body diagram drawing, in order to organize your thoughts

[simple musk]

T2y = 0

[simple musk]

T2x = T2

[clear raft]

sure. we can go with that. Now write the global summation of forces. What are all the forces in the x direction? what are all the forces in the y direction?

[simple musk]

ok let me do trig

[clear raft]

as for the sign of T_1 and T_2, we can consider that later

[simple musk]

ok gimme a moment sir

i need to go to the toilet

[calm coral]

@simple musk Has your question been resolved?

[simple musk]

i need one more moment i need to have dinner with parents

[clear raft]

that's going to be a lot more than one moment lol

maybe you should close this channel and take up a different one later

the problem is close to solution. just need to write your global sum of forces and then solve. you already have most of the pieces

[simple musk]

just

i will come back in 30 min

[simple musk]

im not in a hurry, just take your time to reply when i get here

[simple musk]

ok so

[simple musk]

i love your pfp btw

, av @clear raft

[potent lotus]

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[simple musk]

luckily it was easy exercise

. solved

.solved

[calm coral]

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[swift dragon]

,av @simple musk

[potent lotus]

938c2cc0dcc05f2b68c4287040cfcf71's Avatar

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[simple musk]

bro

i moved earth and heaven

contacted gazillion ups argentina mails

they managed to reschedule the delivery by 2 of april

can you believe that

just a lucky day to be alive man

[swift dragon]

so much fucking work

cuz the guy couldnt read instructions

you should definitely leave a bad review

[calm coral]

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[heady walrus]

How do u add the graph

[swift dragon]

you look at every point

and add their values

(f+g)(x)=f(x)+g(x)

[heady walrus]

Only the x?

[swift dragon]

im not sure what youre asking

[heady walrus]

The x values

[swift dragon]

can you show what you mean

[heady walrus]

Can u give me an example

[swift dragon]

look at x=1

what is f(1) and g(1)?

[heady walrus]

1 and -3

[swift dragon]

no

first value is wrong

lemme ask an easier point

x=3

[heady walrus]

2

[swift dragon]

and the second one?

f(3)=2, yes

but g(3)?

[heady walrus]

-3

[swift dragon]

yes

so now what is f(3)+g(3)?

[heady walrus]

-1

[swift dragon]

yup

so (f+g)(3)=-1

thats one point

now find them all

[heady walrus]

Oh I see

Ty

[swift dragon]

and thats your graph

[heady walrus]

.close

[calm coral]

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[shadow harness]

for each diagram the number of squares of area 4cm2 is A the number of dots is D and the number of 2cm lines is L find a connection between A D and L that is true for all diagram

[versed sluice]

,rotate

[versed sluice]

Try writing A, D and L values for each of the cases

[shadow harness]

wdym by that?

like in terms of the diagram number

[versed sluice]

For each diagram, 1 2 3 and 4

Yes

[shadow harness]

for A i got n^2

for D i got (n+1)^2

and for L i got 2(n^2+1)

but im not sure what to do with them

@versed sluice

[versed sluice]

I think that's it?

[shadow harness]

it says find a connection between them

what does that mean?

@versed sluice

[versed sluice]

An equation containing them

Check this: D + L = A + 3

Does this hold true?

Can u verify, I'm a little busy now

Nope, it doesn't

[shadow harness]

ye

not true

[versed sluice]

There's this formula that newton discovered

For a 3d object: V + E = F + 2

V = vertices
E = edges
F = faces

To apply to a 2d object, take the rest of the plane as a face too

[versed sluice]

Isn't that wrong?

Oh, wait nothing

Nvm

No, it is

Take n = 2

[shadow harness]

o yeah

[versed sluice]

We get L = 10, but L = 12

[shadow harness]

let me check it

[versed sluice]

I think there's another + n at the end

[shadow harness]

2n(n+1)

[shadow harness]

wait but how do we approach the question?

[versed sluice]

Square it

L²

L² = 4AD

[shadow harness]

isnt it A+D=L+1

[versed sluice]

How?

Oh yeah it is

[shadow harness]

would that be an answer? im still kinda confused about what the connection means

[shadow harness]

wait how did u get this?

[versed sluice]

L² = 4n²(n+1)²

A is n² and D is (n+1)²

[shadow harness]

yeah but did u have to like guess that L^2 was gonna be a certain amount times of ad

[versed sluice]

No

We can see L = 2n(n+1) and A = n² and D = (n+1)²

That just gives us the hint to square L

[shadow harness]

oh yeah

[shadow harness]

wait so what should i put as my answer?

[versed sluice]

What does the teacher want?

Addition or multiplication

It depends honestly

Is it possible to put both?

[shadow harness]

question doesnt really specify

it only asks for a connection

[versed sluice]

??

[shadow harness]

i mean i could

[versed sluice]

Then why not

[shadow harness]

js wondering rq for another question if it asks for 12% of 180 people should i round it to 22 180x0.12=21.6

[versed sluice]

Ig u should

Considering u wouldn't want to give someone 0.6 of a person

[calm coral]

@shadow harness Has your question been resolved?

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[cunning reef]

Hai! i went to a math competition around last year, and one of the problems was a sequence question which simplified to x_n = sin(something) and y_n = cos(something)
x_(n+1) = (some terms of y_n, y_(n-1) and x_n,x_n-1)
y_(n+1) being similar
i remember each coefficient was real numbers, and involved sqrt(3) . Is there a sequence representation of trig that i dont know about?

[velvet osprey]

do you remember specifically what the recurrence was

[cunning reef]

i do not sadly💔

but it looked really cool, but i couldnt do the question and most of the people around me said it simplified to cos and sin after the competition was over

[unkempt drift]

wait for the solutions to be published

[cunning reef]

the solutions never get published

neither do the questions

[unkempt drift]

what kind of ass competition did you do

[bronze adder]

I recall a similar question I once did that involved something like this

But I've lost the source to it

I think there was AM-GM involved

[cunning reef]

it was a competition made by senior highschoolers like imo participants for middleschoolers

from a school, and the competition isnt that serious. i still dont know why the questions arent posted tho

[cunning reef]

oh what

[velvet osprey]

one whose organizers are very anal about cheating ig

[cunning reef]

well the organizers are the highschoolers😭

[velvet osprey]

💀

[calm coral]

@cunning reef Has your question been resolved?

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[deft arrow]

should I use integration by parts?

[vagrant oak]

That could work

After IBP, you should be able to apply u sub

[deft arrow]

0 to (\pi/2) integration ( \ln(\sin x) \sin x ) so here i will use IBP
$$
I = \left[ -\ln(\sin x) \cos x \right]_0^{\pi/2} + \int_0^{\pi/2} \frac{\cos^2 x}{\sin x} , dx
$$

pls check

[potent lotus]

Tom

[deft arrow]

-ln(0)?

[vagrant oak]

I expected the opposite but this could also work

[vagrant oak]

Okay no

Its improper integral

[vagrant oak]

This results in -inf + inf or sth like that

Try it the other way around

Like sinx is derivative of -cosx

[calm coral]

@deft arrow Has your question been resolved?

[calm coral]

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[calm coral]

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[grave jacinth]

Hi can someone help me with this problem please? I'm like lost from the start. So no progress so far

Write the polynomial g(x) in standard form with the integer coefficients given the following roots:
(3 + root2), (- 1 - i)

[velvet osprey]

what is "V2" supposed to mean

[grave jacinth]

Root 2

Sorry copy paste error

[velvet osprey]

ok

the idea will be to assemble g(x) via its factors

@grave jacinth if you were told that, say, 7 is a root
can you come up with the factor that would give this?

[grave jacinth]

I don't really understand what you mean by this I'm sorry

but I'm trying to set the problem up as the first step

with the conjugated

Conjugates

Currently

[versed sluice]

See,

When u have integer coefficients,

The roots r always in conjugates

Right?

[grave jacinth]

Um

Yes

[versed sluice]

(if they r irrational or non-real)

[grave jacinth]

They always have conjugates

[versed sluice]

Yes

So

[velvet osprey]

that's going to come later

[versed sluice]

If 3 + √2 is a root, 3 - √2 is also one

[velvet osprey]

you have to understand the relationship between roots and factors

[velvet osprey]

,rccw

[velvet osprey]

x-(3-sqrt(2))

[grave jacinth]

Do you mind if you explain it to me? I have a bad habit of just doing the math without understanding the concept

[velvet osprey]

not x-(-3-sqrt(2))

[grave jacinth]

Fixed, thank you

[velvet osprey]

in general, before any complications like "integer coefficients", to get a number r as a root, you must have (x-r) as a factor.

[grave jacinth]

Oh yeah

I knew that

[velvet osprey]

which you already seem to have done

[grave jacinth]

yes

[velvet osprey]

but you could not verbalize it

[grave jacinth]

yeah 😭

I assume I would just go about solving it now?

[versed sluice]

Do u know this thing where in a polynomial $ax^n+bx^{n-1}+...$, sum of roots = $-\frac{b}{a}$ and so on with other things

[grave jacinth]

I have no idea what that is

[potent lotus]

Suika

[grave jacinth]

Never learnt it

[versed sluice]

If you're given ax² + bx + c and roots as x1 and x2

They didn't teach u that x1 + x2 = -b/a?

[grave jacinth]

no

[versed sluice]

x1 x2 = c/a

Ok

[grave jacinth]

is that like factoring

[versed sluice]

Yes

So, basically

[grave jacinth]

Don't think we have it

[versed sluice]

Okay

[velvet osprey]

unnecessary

[versed sluice]

Not needed?

[grave jacinth]

thank the lord I had no idea what that was!

🌞

[versed sluice]

Okay

[velvet osprey]

ok so this is good ish so far but you still need to fix the same sign error with the complex roots

so you will have your factorized thing as

(x - 3 - sqrt(2))(x - 3 + sqrt(2))(x + 1 - i)(x + 1 + i)

[grave jacinth]

Only the second term changed

changes

In the

parenthesis

[velvet osprey]

multiply the first and second factor together using the difference of squares

[velvet osprey]

yes you don't just negate the entire thing to get the conjugate

only the "complicated" part (radical or i)

[grave jacinth]

Wait but my original thing was (-1 -i)

[grave jacinth]

How come the 1 is positive here

[velvet osprey]

that's the root you want

x-(-1-i) = x+1+i tho

[grave jacinth]

Oh

Yes that's true

Okay I've fixed both sides

[velvet osprey]

.

[grave jacinth]

A^2-B^2

Equals (a+b)(a-b)

[velvet osprey]

don't mix capitals and lowercase

anyway yes apply that identity to the product of just the first two factors

[grave jacinth]

Sorry

[grave jacinth]

I think I did the first side

Here's my second side

[velvet osprey]

ok now what is i^2

[grave jacinth]

I think -i^2 is

Uhh

[grave jacinth]

-1

[velvet osprey]

right

[grave jacinth]

because root one times root one

[velvet osprey]

... no

[grave jacinth]

Oh

[velvet osprey]

i^2=-1 is just the defining property of the imaginary unit i

[grave jacinth]

Oh

Okay so

Just the defining thing

And the negative makes it positive

therefor x^2+2x+2

[velvet osprey]

yes

so your polynomial is now

(x^2-6x+7)(x^2+2x+2)

[grave jacinth]

Awesome

[velvet osprey]

now you have to expand this and you are done

[grave jacinth]

okay let me figure out how to do that

And this can be further grouped

Is this correct

[velvet osprey]

,w expand (x^2-6x+7)(x^2+2x+2)

[potent lotus]

Results provided by WolframAlpha

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

Requested by ann_dec

[velvet osprey]

ok yeah you got it right excellent

[calm coral]

@grave jacinth Has your question been resolved?

[calm coral]

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[blazing coyote]

If we know the directions, i'd say probably

[swift dragon]

what if u=v?

wait

scratch that

😅

[marble sun]

Since there's two non parallel vectors and the function is in 2 variables, it's probably going to allow us to find the grad uniquely

[potent lotus]

neon

[bronze adder]

two equations and two unknowns 🔥

[blazing coyote]

so $\left( \pdv{f}{x} , \pdv{f}{y} \right) \cdot \mathbf{v} = a; \left( \pdv{f}{x},\pdv{f}{y} \right) \cdot \mathbf{u} = b
\
a,b \in \R$
\
two LEs in two vars

[potent lotus]

What a wonderful world !

[blazing coyote]

And we're done

Cool

thanks

.close

[calm coral]

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[calm coral]

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[grizzled coral]

convert to polar form

i got

$v = [\sqrt{41}, 53.13^o , -38.66^o ]$

[potent lotus]

Yousif

[grizzled coral]

but answer shows

$v = [\sqrt{41}, 53.13^o , 38.66^o ]$

[potent lotus]

Yousif

[grizzled coral]

can someone solve it and lmk whos right

[calm coral]

@grizzled coral Has your question been resolved?

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[remote mural]

i would love help in solving this 5 variable simultaneous equation :

(25548331/15720)(0.0024m+1.8848)(dr+0.5r+0.05d+1.025)=l/0.0408 , (141151/78600)(2894.61+905a)(0.0024)(dr+0.5r+0.05d+1.025)=l/16.32, (141151/78600)(2894.61+905a)(0.0024m+1.8848)(d+0.5)=l/0.0272, (141151/78600)(2894.61+905a)(0.0024m+1.8848)(r+0.05)=l/0.0544, (d/0.0544)+(r/0.0272)+(m/16.32)+(a/0.0408) = 16883/204

im calculating something but i meet problems when isolating variables

[vagrant oak]

,w (25548331/15720)(0.0024m+1.8848)(dr+0.5r+0.05d+1.025)=l/0.0408 , (141151/78600)(2894.61+905a)(0.0024)(dr+0.5r+0.05d+1.025)=l/16.32, (141151/78600)(2894.61+905a)(0.0024m+1.8848)(d+0.5)=l/0.0272, (141151/78600)(2894.61+905a)(0.0024m+1.8848)(r+0.05)=l/0.0544, (d/0.0544)+(r/0.0272)+(m/16.32)+(a/0.0408) = 16883/204

[potent lotus]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by methisalwaysright

[vagrant oak]

dang

[remote mural]

TAT

[vagrant oak]

,w solve (25548331/15720)(0.0024m+1.8848)(dr+0.5r+0.05d+1.025)=l/0.0408 , (141151/78600)(2894.61+905a)(0.0024)(dr+0.5r+0.05d+1.025)=l/16.32, (141151/78600)(2894.61+905a)(0.0024m+1.8848)(d+0.5)=l/0.0272, (141151/78600)(2894.61+905a)(0.0024m+1.8848)(r+0.05)=l/0.0544, (d/0.0544)+(r/0.0272)+(m/16.32)+(a/0.0408) = 16883/204

[potent lotus]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by methisalwaysright

[remote mural]

is my question too hard

TAT

[vagrant oak]

no, not really

i dont get why wolfram cant do it

I suppose that this is not a homework question, right?

Or anything like that

do you actually wanna solve it yourself or do you only want the answer?

[remote mural]

nah i graduated, its like a project thing

[vagrant oak]

yeah, i see

[remote mural]

i would like to solve it myself

but it has been 3 days

anyways here is the original problem

[vagrant oak]

oh it's not linear

damn, that complicates it

(25548331/15720)(0.0024m+1.8848)(dr+0.5r+0.05d+1.025)=l/0.0408

(141151/78600)(2894.61+905a)(0.0024)(dr+0.5r+0.05d+1.025)=l/16.32

dividing these 2 equations, you should get an equation with only m and a

(141151/78600)(2894.61+905a)(0.0024m+1.8848)(d+0.5)=l/0.0272

this one just adds d

the original problem would be helpful probably, or at least the original equations

it seems like you already expanded sth

[remote mural]

in genshin, damage is calculated as
dmg = talent% x scaling x (1+DMGBonus) × EnemyDefMult×EnemyResMult
where in my case EnemyDefMult and EnemyResMult is constant of 190/393 and 1-(-0.3/2) respectively

in my case talent% is a constant of323% or 3.23
DMGBonus has a constant and a variable lets call it m, DMGBonus = 0.16+[(0.002+0.0004)m+302]/100

wait im not done

i want to calculate the maximum average damage
lets call crit rate r and crit damage d
AvgDmg = dmg(1+d)(r)+dmg(1-r)

the scaling of this character is attack where i call a
scaling = 905a + 2894.61

to find maximum damage, i will have to use the lagrange multiplier on the function AvgDmg(a,m,r,d)

the constraint of the system is: (d/0.0544)+(r/0.0272)+(m/16.32)+(a/0.0408) = 16883/204

[remote mural]

yea

[vagrant oak]

Yeah, dont do that

keep it in the factored form

i just noticed something

your factors are e.g. like
(d - 0.5)

[remote mural]

e = 12.5122+10.1303a is the equation of e and a i got

[vagrant oak]

and the -0.5 is same for all equations

[vagrant oak]

you can got more

[remote mural]

icic

[vagrant oak]

by dividing the factored forms you can get many equations with just 2 variable

[remote mural]

well i got e[(r/300)+1/6000] = 2r^2 - (603r/450) + 6247/9000

[vagrant oak]

#(m + #)(d + #)(r + #) = l
#(a + #)(d + #)(r + #) = l
#(a + #)(m + #)(d + #) = l
#(a + #)(m + #)(r + #) = l

[remote mural]

and d = 2r-0.4

[vagrant oak]

this is how first of your few equations look like

# is constant

[remote mural]

and im stuck

[vagrant oak]

you wont need that for now

[remote mural]

icic

[vagrant oak]

divide the first 2, you get equation with m and a

[remote mural]

wdym

[vagrant oak]

divide 1 and 3, you get a and r

[remote mural]

wait devide how

[vagrant oak]

i mean that your factors are exactly the same in all equation

[remote mural]

i mean some have the same

[vagrant oak]

LHS1 = RHS1
LHS2 = RHS2

LHS1 / LHS2 = RHS1 / RHS2

[vagrant oak]

except the last one, all of them have

[remote mural]

like in those 4 equations, except the constraint, each variable has 3 equations with the same factors

[remote mural]

so when i reach the point where i cannot use the equation with the same factors

[remote mural]

i meet things like this

[vagrant oak]

just do all the pairs, that equations is overcomplication

[remote mural]

ok let me try

[vagrant oak]

#(m + #)(d + #)(r + #) = l
#(a + #)(d + #)(r + #) = l
#(a + #)(m + #)(d + #) = l
#(a + #)(m + #)(r + #) = l

These are your 4 equations, the 5th one will be ignored for now

divide the first 2, you get linear equation in m and a

divide 1 and 3, you get eq in a and r

divide 1 and 4, you get eq in a and d

divide 2 and 3 you get eq in m and r

divide 2 and 4 you get eq in d and m

divide 3 and 4 you get eq in d and r

using only the last 3 I mentioned you should be able to find m, d, r

all of them shall be linear btw

[remote mural]

thats where i got well i got e[(r/300)+1/6000] = 2r^2 - (603r/450) + 6247/9000

[vagrant oak]

there should be no square in r

[remote mural]

1 and 3 i get d

[remote mural]

1 and 4 i get r

but yeah let me try them

[vagrant oak]

,w factor (dr+0.5r+0.05d+1.025)

[potent lotus]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by methisalwaysright

[vagrant oak]

oh wait

[remote mural]

prb because i substitute d in terms of r

[vagrant oak]

(dr+0.5r+0.05d+1.025)
Are you sure this factor is correct?

shouldnt it be (d+0.5)(r+0.05)

oh

[remote mural]

i dont have my phone w me rn but i substitute cd in terms of r into 3 because i would get a equation with m d and r if i dont

[vagrant oak]

is it (d+0.5)(r+0.05) + 1

[remote mural]

its not a factor

that equation comes from

(1.5+d)(r+0.05)+(0.95-r)

i mean yea kinda a factor

but there was something extra

[vagrant oak]

hm okay

[remote mural]

wait so is there any way to factorise that

even if it would look uglyu

cus it might save my life

[vagrant oak]

it wouldnt help

[remote mural]

oh

[vagrant oak]

what would help is if all the factors were same

(0.0024m+1.8848)

e.g. m appears everywhere only in this form

you could actually make a sunstitution to make it more appealing

[remote mural]

in 2 it appears as 0.0024 only (constant)

[vagrant oak]

Yeah, because in 2 there is no m at all

[remote mural]

i didnt multiply it with the other constant cus it will be messy like 1

with a scary fraction

[vagrant oak]

but whenever m appears it appears liek this (0.0024m+1.8848)

[remote mural]

yes

[remote mural]

yes

[vagrant oak]

and same for a

[remote mural]

yes

[vagrant oak]

the problem is with r and d

[remote mural]

yes hmm

because the partial derivative removes them

[vagrant oak]

(d+0.5)(r+0.05) + 1 = (dr+0.5r+0.05d+1.025)

this is quite interesting though

(d+0.5) is the other way d can appear

and (r+ 0.05) is the other way r can appear

so lets just make all the substitutions to make life less painful

[remote mural]

hahha ok

[vagrant oak]

it suffices to solve this

if you can solve this, you won

[remote mural]

wait can i replace my ugly part with this

[vagrant oak]

solve this and you won

e.g. M = (0.0024m+1.8848)

oh and c's are just constants

solution can be in terms of them

[remote mural]

ohh

[vagrant oak]

ill have to go for few mins now

[remote mural]

thank your for simplifying it XD now imma try

[vagrant oak]

you can try your luck with dividing, ill come back later

[remote mural]

okie

[remote mural]

okie

tq

[calm coral]

@remote mural Has your question been resolved?

[vagrant oak]

managed to get here till now

the DR + 1 part is mildly annoying

[remote mural]

true

[remote mural]

wait how did you get c4c2/c1 a r = l

I will try more till I get something I can work with

Wait no im stucj

JAHAHAH

Should I substitute d or r as itself cus d and r is kinda annoying

[vagrant oak]

I don't remember 💀

it was simple tho

just some subs

[vagrant oak]

Could work i guess

[remote mural]

but it gives me d or r squared and i cant escape that

or it gives me d +1/d or r + 1/r

it makes it harder atleast i think it does

wait it does not i just need complete this

[vagrant oak]

i think i got it

2 eq 2 unknowns

solvable

hopefully

unless i messed up somewhere

ill try to retrace my steps

[remote mural]

yes pls tq im very confused TAT

[vagrant oak]

Yeah that will work

Let's use these 2

they are simple

now we do all the substitutions needed

[remote mural]

btw c4 = 2c3

[vagrant oak]

start with these 2

[remote mural]

could make things easier

[vagrant oak]

Let's not worry about that for now, they're just constants after all

[remote mural]

ok

[vagrant oak]

it makes me type 1 letter less but thats pretty much all

our goal is having those capital letters expressed purely via c's

c's are not variables, we know them, I just dont write them because im lazy af

anyway, let's do the subs

now both left hand sides are in terms of A, R

oh waitttt

$\frac{c_{4}c_{2}}{c_{1}}\cdot A\cdot R=L$

[potent lotus]

MathIsAlwaysRight

[vagrant oak]

was gonna use this for L

but i just realized

that it might just lead us back

but let's try it anyway

oh wait

it uncovers A

take the second eq and divide by A * R

$c_{4}\cdot A\ \cdot\left(\frac{c_{2}}{c_{1}}\right)=\frac{c_{4}c_{2}}{c_{1}}$

[potent lotus]

MathIsAlwaysRight

[vagrant oak]

if this is correct

then it's done

[remote mural]

omg

idk leme test

[vagrant oak]

A = 1

doesnt look correct

[remote mural]

oh

HAHHAH

yea that dosent

[vagrant oak]

yeah so now i need to find up where i messed up

[vagrant oak]

must be this

[remote mural]

im still trying to comprehend your working give me a min

AHHAHA

[vagrant oak]

ill probably be doing that for the next min as well 💀

[remote mural]

i just comprehended it

[remote mural]

idk where this comes from so i can say everything else is correct but this

[vagrant oak]

yeah, i just realized ive made a mistake

i forgot one A there

which makes it equivalent to what ive written before

hmm

actually we have 4 equations 5 unknowns

so it wont have 1 sol

we will later have to use the 5th equation for that

for now we should parametrize it in terms of L

that seems to be the most convenient parameter to choose

[remote mural]

oh its great i expected that

just hope not too much solutions

im not interested to test alot

[vagrant oak]

yeah, by not 1 I meant infinitely many

[remote mural]

hmm

[vagrant oak]

4 equations 5 unknowns is infinitely many

the 5th equation will narrow it down

[remote mural]

oh damn

[remote mural]

hopefully

[vagrant oak]

so for now we should hope to express the cap letters in terms of c's and L

then using the 5th equation we will be able to determine L

[remote mural]

yes

[vagrant oak]

these 2 equations remain true

[remote mural]

but all equations have L in them so its abit hard

[remote mural]

ohh

[vagrant oak]

using these we should in theory be able to do the job of expressing A and R

[remote mural]

yes

[vagrant oak]

so yeah, let's see

[remote mural]

ok

[remote mural]

wait looking at equation 2 made me realise something

[vagrant oak]

ohh i see something as well

nice subtituion

[remote mural]

equation 2 in the image is the outcome when substituion the value of D in terms of R in to the 1st equation

but we could

[vagrant oak]

Yep

thats exactly how i made it

$\frac{c_{2}\cdot c_{4}}{c_{3}}A\left(\frac{c_{1}}{c_{2}c_{4}}\frac{L}{A^{2}}\right)^{2}+c_{2}A=L$

[potent lotus]

MathIsAlwaysRight

[vagrant oak]

well i got the answer..... you wont like it though

this might be algebraically unsolvable

[remote mural]

equate eq 1 and eq3 to get A in terms of R and D, and then i forgot

too messy

[remote mural]

what is it

[remote mural]

ohno TAT

[vagrant oak]

So using first equation of this, we can express R

like this

then substitute

$\frac{c_{2}\cdot c_{4}}{c_{3}}A\left(\frac{c_{1}}{c_{2}c_{4}}\frac{L}{A^{2}}\right)^{2}+c_{2}A=L$

[potent lotus]

MathIsAlwaysRight

[vagrant oak]

terrible

ik

but it will be even terribler

$\frac{c_{1}^{2}}{c_{3}c_{2}c_{4}}\frac{L^{2}}{A^{3}}+c_{2}A=L$

[potent lotus]

MathIsAlwaysRight

[vagrant oak]

some minor simplifications

now multiply both sides by A^3

and this is where the terrible part comes in

oh wait a moment

[remote mural]

idk where did this come from im comprehending AHHAHA

[vagrant oak]

just substituting for R

[remote mural]

ohh

[vagrant oak]

it's the second equation from here with R substituted

notice that it has only L and A

so yeah, simplify and multiply by A^3

$\frac{c_{1}^{2}}{c_{3}c_{2}c_{4}}L^{2}+c_{2}A^{4}=LA^{3}$

[potent lotus]

MathIsAlwaysRight

[vagrant oak]

and finally, reorder

[remote mural]

looks promising

[vagrant oak]

Yeah, looks

[vagrant oak]

now this is a quartic equation in A

aA^4 - bA^3 + c = 0

now would you like to hear the bad news or the good news first?

[remote mural]

good news

[vagrant oak]

We have a formula to solve quartic equations

now comes the bad news

[remote mural]

i dont think its quadratic

4 and 3

[vagrant oak]

Quartic = degree 4

not quadratic, quartic

[remote mural]

ohhh

[vagrant oak]

Yeah, so the bad news is the formula looks like this

[remote mural]

i didnt know its called that but yeah

[remote mural]

holy wshot

wait let me

[vagrant oak]

numerical approximation is probably the best you can do

[remote mural]

fix my newly diagnosed asthma

im dying

help

whats this monster

[vagrant oak]

Quartic formula

the last one to exist

[remote mural]

i didnt even know it existed

[vagrant oak]

any polynomial of higher degree doesnt even have a formula

[remote mural]

hmm

[vagrant oak]

not because it wasnt discovered, but because it cant exist

the question is how to get a numerical approximation for that

[remote mural]

no +- sign in teh equation?

i expect it to have maybe 2 of them

[vagrant oak]

that equation is just for the first root

there are 3 more

[remote mural]

considering 4 variables is 4 answers

[remote mural]

oh damn

AHHAHA

dw im jobless

i got time

i will try

HAHAH

WILL BE A PAIN THO HAHAH

[vagrant oak]

yeah

if you are actuallly going to try

then i'd instead solve for L

that's quadratic

[remote mural]

oh

then lets do that instead

[vagrant oak]

but the degree will rise later on

[remote mural]

oh no hmm

[vagrant oak]

because once you do that, you'll actually have to substittue the expression for L into another equation

[remote mural]

wait i realise solving for A means i will see L alot of times in the answers

[vagrant oak]

mhm

[remote mural]

i wanty to cry

[vagrant oak]

im quite confident that the whole system wont have a nice solution

[remote mural]

HAHAHA

[vagrant oak]

and there is a solid chance the solution wont even be algebraically expressible

[remote mural]

its ok i dont need it to be nice

[remote mural]

HAHAHAHA

ohno TAT

[remote mural]

should i attempt this monster

[vagrant oak]

,w solve M(DR+1) = 1,
A(DR + 1) = 1,
AMD = 1,
AMR = 1

[remote mural]

4 times

[potent lotus]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by methisalwaysright

[vagrant oak]

ill try asking WA this

[remote mural]

💀💀💀

[remote mural]

ok

[vagrant oak]

yeah, idk whats wrong but WA still doesnt get it

,w x^2y = 1, x(y^2+1)=3

[potent lotus]

Results provided by WolframAlpha

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

Requested by methisalwaysright

[vagrant oak]

Yep

expectable

not nice

[vagrant oak]

this is how they reduce

[vagrant oak]

so yeah, WA approximated them

which means that the degree must be at least 4

and even if you were lucky and it actually was 4 and thus solvable, nobody will like your solution

numerical solution is better than too ugly algebraic solution

and 4th degree is quite ugly

but im solidly confident now that the degree will rise above that, making it completely unsolvable

i gtg now tho, bye

[remote mural]

omg hmm

[remote mural]

hmm

[remote mural]

okiee thanks for your help anyway

i will try to solve it

now

...

[remote mural]

ok whoever sees this and replies and i dont reply, im probably asleep

[calm coral]

@remote mural Has your question been resolved?

[calm coral]

@remote mural Has your question been resolved?

[calm coral]

Channel closed

Closed due to timeout

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• 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.

[cyan mango]

find absolute extreme of a function first or local values first which is easier?

[sacred zephyr]

local values

[cyan mango]

thank you!

.close

[calm coral]

Channel closed

Closed by @cyan mango

Use .reopen if this was a mistake.

[calm coral]

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.

[icy crystal]

need help

[icy crystal]

need help with equation first level

[remote mural]

what do u need

[glad parrot]

Whats the question

[remote mural]

if u wanna help DM to me

i can give u advice for math problem

[limpid lagoon]

!original

[calm coral]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[icy crystal]

4x+2=5x-1

[limpid lagoon]

start by isolating x on one side and your constants (just plain numbers with no variable) on the other side

[icy crystal]

can i change to 5x-1=4x+2

flip it

[limpid lagoon]

thats the same thing yeah but that doesn't really get you anywhere

how would you get rid of the 4x on the right side?

[icy crystal]

-4x on both sides

[limpid lagoon]

good

what do you have after you do that?

[icy crystal]

x-1=2

[limpid lagoon]

good

[icy crystal]

+1 both?

[limpid lagoon]

yes exactly

[icy crystal]

x=3

?

[limpid lagoon]

yep

[icy crystal]

nice

can u help 1 more

[limpid lagoon]

sure

[icy crystal]

3x-2=-6

[limpid lagoon]

what do you think you should do first

[icy crystal]

can i flip -2 to right side and make it +?

[limpid lagoon]

yes

[icy crystal]

3x=4

here i got stuck

[limpid lagoon]

not 4, whats -6+2?

[icy crystal]

-4

[limpid lagoon]

yes

3x=-4

remember 3x is the same as saying 3 times x

whats the opposite of multiplication?

[icy crystal]

:

[limpid lagoon]

addition is to subtraction as multiplication is to ___?

[icy crystal]

im not sure

bad english

[limpid lagoon]

do you understand that the opposite of adding is subtracting?

in other words, opposite of + is -?

[icy crystal]

yes

[limpid lagoon]

okay, now what is the opposite of multiplying?

[icy crystal]

dividing?

[limpid lagoon]

yes

what can you divide 3x by to get just x?

[icy crystal]

3

[limpid lagoon]

yes

so remember how you added 2 on both sides?

here, you divide 3 on both sides.

[icy crystal]

so x=3 divided by -4?

[limpid lagoon]

other way around

-4/3

[icy crystal]

why other way

[limpid lagoon]

because you are dividing by 3

what you did is 3 / 4

[icy crystal]

oh yes

answer is x=3/4?

[limpid lagoon]

other way!

[icy crystal]

4/3

[limpid lagoon]

and dont forget the negative

[icy crystal]

-4/3

[limpid lagoon]

yes. now you can leave it like that or convert it into a mixed fraction, depending on how your teacher wants the answer

[icy crystal]

okay ty

[limpid lagoon]

np

[icy crystal]

i try to learn these first before convert

[calm coral]

@icy crystal Has your question been resolved?

[calm coral]

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[calm coral]

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[wind bane]

guys what is angular velocity

[sweet meadow]

change in angle / change in time

[glad parrot]

Time rate at which an object rotates around an axis

[wind bane]

will it then be fair to say that the instantaneous angular velocity is d(theta)/dt

[rigid scaffold]

yes

its usually represented by the lowercase omega

[wind bane]

but it will be completely correcrt to write it as d(theta)/dt

[rigid scaffold]

yes

cuz it is a rate of change

e.g. radians per second

thats like v=ds/dt

[wind bane]

but theta/t is not the same as d(theta)/dt. since theta/t is angular velocity. while d(theta)/dt is the angular velocity at a particular moment in time

is this correct?