help-39

[fresh silo]

mhhm there is a possibility to find a and b

but idk how

<@&286206848099549185>

[midnight haven]

Idk I'm stressing over chemistry rn 💀😭

[fresh silo]

im lost

[midnight haven]

Have u yk ask ur teacher abt this

[fresh silo]

i dont have teacher

[midnight haven]

Oh

UMM

[errant osprey]

bro

u gotta move the x^2

to the left

and also the -2x

and then do t9he

thingy

the complete the squarws

so here

after step 1

u got:

y^2-x^2+2x=2

then u gotta

(y)^2 - (x^2-2x) = 2

and you keep oing with that

to get the equation

and the9n

remember

wait jsut ping if u need help after that

[pearl pond]

@fresh silo Has your question been resolved?

[fresh silo]

i need help

after that

[pearl pond]

@fresh silo Has your question been resolved?

[pearl pond]

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[errant osprey]

Yeah

[errant osprey]

mb

for the late

[prime bramble]

do you have another question?

if not, you should close this channel

you've opened it but there's nothing so far

[versed mica]

.close

[pearl pond]

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[gusty prism]

what do they mean

by for every 1 mol h2o

2 mol h

Oh

yep im an idiot

nvm

.close

[pearl pond]

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[tropic pecan]

Help q.5

[silver nest]

trig functions

[latent quail]

wat have u tried?

[tropic pecan]

Idek what to do tbh

Wait

Is it

Sine?

Do I do sine?

To find side b

Hellooo?

Imma js assume it is

[woven garnet]

Not necessary as the side lengths have a known ratio since it is a “45-45-90” triangle

Sides a and a with hypotenuse a sqrt2

[tropic pecan]

I ain't doin alla dat

I already did sine

She ain't checking the hw anyways

[pearl pond]

@tropic pecan Has your question been resolved?

[pearl pond]

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[buoyant garnet]

One message removed from a suspended account.

[buoyant garnet]

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[dusty flame]

im cofnsued abt ur metjod

[buoyant garnet]

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[dusty flame]

${2(x-1) + 2(y-1)\frac{\dd y}{\dd x} = 0}$

[jolly parrot]

k

[buoyant garnet]

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[dusty flame]

shouldnt this be the first step

[buoyant garnet]

One message removed from a suspended account.

[dusty flame]

${ \frac{\dd y}{\dd x} = -\frac{2(x-1)}{2(y-1)} = -\frac{(x-1)}{(y-1)}}$

[buoyant garnet]

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[jolly parrot]

k

[buoyant garnet]

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[echo needle]

help

[echo needle]

system

3x-2y=6,5x-2y=2

[spare lark]

Try to eliminate one of the variables

By adding, substracting lines together

[midnight haven]

*equations

[spare lark]

(Yeah equations, ive done too much matrices with it xd)

[midnight haven]

lol

okay rice how ya doin

[echo needle]

variables is the letters right?

[midnight haven]

yes

[echo needle]

how do i get x/y = ......

so i can replace it in the other equations

[midnight haven]

elimination is easier in this case but substitution works as well

[echo needle]

Eal and Eaz?

[midnight haven]

Eq1 & Eq2

Eq = equation

sorry i was trying to go fast

[echo needle]

is that b in the first equation

[midnight haven]

b?

oh no

6

the number

[echo needle]

you did subtraction here?

i dont think i've seen this method before

how's the elimination ?

[midnight haven]

erm

basically you try to get one variable to cancel so you have 1 equation with 1 variable and can solve for it

[echo needle]

variable is the letter?

[cosmic garnet]

yes

x, y

[echo needle]

i think you try to eliminate the number that's attached to the variable so then you can have something like y or x = ....... which you can replace in the next equation

[midnight haven]

that would be substitution

[echo needle]

yes im doing the subsitition one

hows that work

[midnight haven]

1 sec

[echo needle]

how is x=2+2/3y

[echo needle]

also, the second equation was 5x not 3x

[midnight haven]

it is 5x

[midnight haven]

we solved for x in terms of y

[echo needle]

you solved the first or second?

to replace it

how did you remove the number and got only x

[oak quiver]

You need help in solving for x and y?

[midnight haven]

replaced x with an expression in terms of y

see the arrow?

alr i gtg now tho

someone else will be able to help

cya!!!

[echo needle]

yes

[oak quiver]

Have you learned how to solve such eqs?

[echo needle]

there isn't a variable by itself so how do remove the number next to it so i can make a result out of that variable and replace it

[oak quiver]

That number is called the coefficient

To remove the coefficients, just divide the equation by it

Like 2x + 3y = 4
x + 3/2 y = 2

[echo needle]

so i have 4 choices here? right?

to divide

because 4 of them have variables

[oak quiver]

Can you pls write and show what eqn you are dealing with

[echo needle]

this

[oak quiver]

This?

[echo needle]

yes

[oak quiver]

Oh

Ok now tell me what you are gonna divide these by?

[verbal whale]

Notice that you can isolate 2y from both, which I believe is easier

[oak quiver]

||let him do by his method||

[echo needle]

im gonna remove the coefficient one of the first equation, so i'll divide the whole equation by 3 so

[verbal whale]

Unless you've only learnt substition so far

[oak quiver]

Sure

But it might be easier to substitute y instead of x

[echo needle]

3x-2y=6 / 2 = 1/5x-y=3

i divided by 2

[oak quiver]

1.5

Not 1/5

[echo needle]

3/2 is 1.5

yes but you cant write with the . and ,

[oak quiver]

kk np

[oak quiver]

Yes so now what

[echo needle]

now its y=6-1.5

[oak quiver]

Forgot x

And it should be 3

[echo needle]

y=3-1.5

3x

y=3x-1.5

wait no

wait yes

y=3x-1.5

[oak quiver]

3-1.5x

[echo needle]

yes

[oak quiver]

No wait

1.5x - 3

mb

[echo needle]

no its

y=3-1.5x

[verbal whale]

Nope

[oak quiver]

Bro you can takeover

[verbal whale]

You started from 3x - 2y = 6, right?

[echo needle]

yes

divided all by 2

to get a solo y

[verbal whale]

Yeah

So you get 1.5x - y = 3

[echo needle]

and i got 1.5x-y=3

[verbal whale]

Exactly

[echo needle]

so then its y=3-1.5

[verbal whale]

No...

[verbal whale]

Move y to the right hand side

So you get 1.5x = y + 3

And now you subtract 3 from both sides

So you obtain 1.5x - 3 = y

[echo needle]

:(

wait

im finding value of y not -y so

i multiply it by 1

and then get y

and get y=3+1.5

[verbal whale]

Multiplying by 1 means doing nothing

[verbal whale]

I don't get how you arrived here

[echo needle]

someone said do that when you have a negative variable

[echo needle]

im doing substitutional method

so i can replace that on the other equation

[verbal whale]

Yeah but still I don't get what you're doing

[toxic fractal]

multiplying by 1 is not the same as multiplying by -1

[oak quiver]

Can you write on paper and show

[toxic fractal]

where's the original problem? there's a lot to scroll here

[oak quiver]

How 1.5x - y = 3
Becomes y = 3 + 1.5x

[oak quiver]

.

[toxic fractal]

okay, remember you can only do two things on an equation.
adding the same to both sides, or multiplying both sides by the same

so using the first eq:
3x-2y=6
you divide by 2
1.5x-y=3
you add y
1.5x=3+y
you substract 3
1.5x-3=y

[toxic fractal]

3 divided by 2 is not 1/5

and the 2nd implication is also not true

[toxic fractal]

you're not dividing the rhs

[toxic fractal]

.

[toxic fractal]

anything you do in one side, you need to do on the other

[oak quiver]

Correct

Now show what operation you make to get y = 1.5x + 3

[oak quiver]

Show step by step

Like adding 2 and all

[echo needle]

y=3-3/2

right?

[oak quiver]

No

[echo needle]

y=3-3/2x

[verbal whale]

No...

You are always getting the signs flipped, I don't know how

[verbal whale]

Which one of these are you using now?

[echo needle]

y=3+3/2x

[verbal whale]

Don't tell me the results, just answer this

[verbal whale]

.

[verbal whale]

Btw still wrong

[echo needle]

im working the first one

first equation

[verbal whale]

Yeah...

[verbal whale]

But which one of these two laws are you using?

[echo needle]

substitutional method

[verbal whale]

Forget that for now

You don't know how to solve a first degree equation, not a system!

Substitution method is for systems, but here you're only solving one equation

[verbal whale]

These two laws should be well known to you, since you're studying systems. But it seems not at all
That's why we're asking you to show the steps you're doing

[echo needle]

what you do to one side you do to the other

[verbal whale]

Can you answer this or not?

I don't know how to help you if you don't collaborate 😬

[echo needle]

i can choose ?

[honest spear]

do you know what substitution or elimanation is?

[echo needle]

subsitutional method yes

[honest spear]

Okay so I have 2 equations with 2 variables. I want an one equation with one variable so that that variable can be solved

[echo needle]

2 equations with 4 variables

[honest spear]

no

x=x yes?

the x in one equation is the same value as the other

[echo needle]

first eq x y and second eq x y

4 letters

[honest spear]

but 2 variables

[echo needle]

ok 2

[honest spear]

so if you solve for y in the first equation you get

...

[echo needle]

whats the calculation ?

[honest spear]

i'm asking you

[echo needle]

we using the sub method yes?

[honest spear]

let me explain the process to solve for y. We want y by itself so i would have to minus 3x from both sides then divide both sides by -2 to get just y; yes?

[honest spear]

yes

[echo needle]

they said if i want y by itself i need to divide all by 2 which is y's coefficient

[honest spear]

-2 is the coefficient

because i want y = ... not -y =...

[echo needle]

yes so i divide the equation both sides by -2?

and the get a solo y

[honest spear]

yes

try it and show me what you get

[honest spear]

now to get y = ... you'd have to ...

[echo needle]

i have the y

+y is y

[honest spear]

yes but you want y = something not -1.5x + y = constant

when you graph an equation its generally in the form y = something in terms of x

[echo needle]

-2y divided by 2y is +y cus -x- is +

[honest spear]

yes

[echo needle]

oh ok wait

[honest spear]

if i asked you what does y equal

you'd have to answer y is equal to ...

[honest spear]

perfect

now you can SUBSTITUTE the value of y into the second equation

because now you know what y is equal to

[echo needle]

but shouldnt 1.5x be in fraction 3/2 instead

[honest spear]

genrally thats the more preferred way to write it as its easier to read

but they are the same value so its just preferance

[echo needle]

yes but is teacher going to mark as error ?

i dont think i ever write it with dots and ,'s

always transform to fractions

[honest spear]

wouldnt make sense to; I'd ask your teacher - since its a format preference

but you should get used to $\frac{3}{2}$

[jolly parrot]

Nyxzore

[honest spear]

since its the standard way of writing it

[echo needle]

you mean i should do fractions instead?

[honest spear]

yes

you should

[echo needle]

ok lets proceed

[honest spear]

.

[honest spear]

proceed

[honest spear]

just send me the end result

[honest spear]

no

whats 5-3

not -2

[echo needle]

its 2

[honest spear]

solve for x

[honest spear]

so whats next?

[honest spear]

good

[honest spear]

...

continue

[honest spear]

what?

$y = -3-3 = -6$

[echo needle]

what is $

[jolly parrot]

Nyxzore

[honest spear]

its the discord prompt to use latex

[echo needle]

latex?

[honest spear]

this bot allows you to type in latex which is a library made for representing maths

[honest spear]

what's next?

[echo needle]

next step?

[honest spear]

yeah

[echo needle]

nothing

[honest spear]

correct

[echo needle]

x=-2

y=-6

:D

[honest spear]

congrats

[echo needle]

can we another

[honest spear]

practice more

[echo needle]

please

you best helper

[honest spear]

yes but lets rather do it in dm so that we aren't hogging help channels

@echo needle .close please

[echo needle]

.close

[pearl pond]

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[topaz mauve]

How Hard would it be to model the surface area of a Car since it has a very irregular form.

[oak quiver]

Depends on the car

[night beacon]

yeah what car

[topaz mauve]

Like a toy car

something like this for example

[pearl pond]

@topaz mauve Has your question been resolved?

[inland lantern]

this is surely not a math question. look at 3d modelling techniques, and decide what LOD (level of detail) you need

if you mean model it in some way that doesn't use a computer the answer is probably it's basically impossible

[pearl pond]

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[eager summit]

I'm making some physics calculations for my game regarding object's velocity, we want to create drag by multiplying velocity by some 0-1 value each ingame frame

velocity = velocity * 0.9

however I need velocity to be framerate independent, meaning that no matter how many times this calculation is executed per second, after one second has passed the value should be the same

For this example it is achieved by adding

velocity = velocity * 0.9^(DeltaTime / ExpectedDeltaTime)

Where Delta Time is time elsapsed since the last frame (for 1 frame it equals to 1, for 2 frames it's 0.5 etc.)
This formula works correctly, unless I start adding to that value beforehand

velocity = velocity + 5 * DeltaTime
velocity = velocity * 0.9^(DeltaTime / ExpectedDeltaTime)

In this case value does not end up being the same. Any ideas (is it even possible) on how to keep the addition in formula while making the velocity reach the same value after the same time elapsed?

[fluid axle]

value does not end up being the same.
the same as what ?

you're comparing the results of these two different codes right ?
velocity = velocity * 0.9^(DeltaTime / ExpectedDeltaTime)

velocity = velocity + 5 * DeltaTime velocity = velocity * 0.9^(DeltaTime / ExpectedDeltaTime)

[eager summit]

It's different when the Delta Time is different.
To be more clear

1. We run this calculation FRAMES times
2. DeltaTime for each calculation equals to 1 / FRAMES
3. After such process with any given positive "FRAMES" value the velocity should be the same

ExpectedDeltaTime is just constant value, let's say 0.02

[eager summit]

I'm only interested in this one

velocity = velocity + 5 * DeltaTime
velocity = velocity * 0.9^(DeltaTime / ExpectedDeltaTime)

idk how to put it more math-friendly, like
v = (v + 5 * dt) * (0.9^(dt / 0.02))
repeat this X amount of times
where
v = 100 (initially)
dt = 1 / X

for X = 50 and X = 10 this should yield the same result

[pearl pond]

@eager summit Has your question been resolved?

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[midnight haven]

by considering the partial sums prove that the sum 1/k(k+p) k>=1 converges. what is the sum as k=1...inf?

[midnight haven]

i did partial fractions and got to s_n = 1/p(sum 1/k - sum 1/(k+p)) but i dont know how to show it converges

i forgot calc 2👍

it looks like this

so i cant cancel out the terms

[merry carbon]

Are you sure you can’t?

[midnight haven]

ummm..

i dont think so😭

unless i do crazy stuff like 1/p(inf - inf) = 0🥰

[merry carbon]

Maybe try writing out what happens if n was 2p

[midnight haven]

then we would have
1/p(1+1/2+...+1/p- 1/(2p+1) - 1/(2p+2) - ... - 1/(3p))

[merry carbon]

I mean, sure, do you see where the cancellation happens at least?

[midnight haven]

kind of

[merry carbon]

It’s one of those “special” telescoping series where you have cancellation happening “weirdly”

[midnight haven]

hmm

so as n gets larger

the first sum stays as is but the 2nd sum gets smaller and smaller

then shouldnt it diverge as the 2nd sum is going to 0

[merry carbon]

Second sum?

[midnight haven]

• 1/(2p+1) - 1/(2p+2) - ... - 1/(3p) this part

[merry carbon]

Everything there vanishes so it converges

You’re only left with the first few terms as you’ve noticed, the 1 + 1/2 + … + 1/p

[midnight haven]

ohhh right its the first few terms not the entire harmonic series

so for sufficiently large n there wont be any terms left

[merry carbon]

Yea, just a finite amount of terms (similar to “usual” telescoping sums of vanishing sequences, where you have only the first term remaining after telescoping)

[midnight haven]

okkkk

so the limit will be 1/p

[merry carbon]

The limit of the telescoping series we have?

[midnight haven]

yep

of s_n = 1/p(sum 1/k - sum 1/(k+p))

oh wait i tried for n=3p we are still left with sum 1/k (k=1..p)

so the sum is whatever sum 1/k (k=1..p) is (finite) divided by p

thank u

.close

[pearl pond]

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[limber oasis]

what is your doubt?

[sharp smelt]

Need my proof verified

[limber oasis]

share proof?

[sharp smelt]

Typing it out rn

We first prove if $7 \mid a$ and $2 \mid a$, then $14 \mid a$.
\
$a=7k_1$; a=2k_2$
\
As both are true simultaneously , we have
\
$a^2=14(k_1)(k_2)$
\
Therefore $14 \mid a^2$.

[jolly parrot]

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

[cursive wraith]

how does 14 dividing a^2 prove it divides a?

[sharp smelt]

It doesn't

[cursive wraith]

you're supposed to use gauss theorem or lcm properties

[sharp smelt]

I think I'm expected to do this using first principles

wait

what am I doing

[limber oasis]

I was hoping you'd know

[sharp smelt]

if $7 \mid a$ and $2\mid a$ , then $ lcm(7,2) \mid a$

[jolly parrot]

A dense set

[limber oasis]

yes

[sharp smelt]

lcm(7,2)=14

thus $14 \mid a$

[jolly parrot]

A dense set

[cursive wraith]

Does "first principles" include the fact that the lcm is a divisor of any common multiple?

was that at least proved in what you're supposed to be able to use?

[sharp smelt]

That's part of my definition of the lcm

[cursive wraith]

so your definition of the lcm is a theorem?

because to show that there exists a common multiple of a and b that divides all others is not trivial

[sharp smelt]

I mean that's trivial using WOP

[cursive wraith]

well ordering principle shows existence of a smallest positive common multiple in terms of natural order <=. and now?

[sharp smelt]

consider the set of common multiple of $a$ and $b$. It's easy to prove that $ab$ is one of them. thus the set is non-empty. We can thus conclude that the set has a least element that acts as the lcm

[jolly parrot]

A dense set

[cursive wraith]

the set of positive common multiples* but ok

there is a least element for <=

why does it divide all others?

[sharp smelt]

By definition

[cursive wraith]

no?

the definition you gave me, shows that it's smaller than all other positive common multiples

but why does it divide them?

[sharp smelt]

My definition the lcm is the least natural number that is a mutiple of both a and b

[cursive wraith]

ok

3<4, does 3|4?

why would a "least" element divide the others, you're missing this argument

[sharp smelt]

no

The set is the set of multiple of both a and b

[sharp smelt]

ah

I'm not sure how I'd prove this TBH

[cursive wraith]

alright

well there are several ways to prove this

you can first show that gcd(a,b) = d is divisible by any divisor of a and b (this can be done using WOP and euclidian division)

and that a = da', b = db' with gcd(a',b') = 1

afterwards just define lcm = |ab|/d = d|a'b'|, which is a common multiple of a and b (lcm = |ab'| = |a'b|)

show that it does divide any common multiple of a and b

[sharp smelt]

hmm, okay

[cursive wraith]

and you're done

[sharp smelt]

I see

thanks

I think I'll prove that $14 \mid a \implies 7 \mid a \land 2 \mid a$ first

[jolly parrot]

A dense set

[sharp smelt]

$a=2(7)k_1 \implies \frac{a}{2}=7k_1$ $a=2(7)k_1 \implies \frac{a}{2} = 7k_2$. As the integers are closed under division, this implies that $7\mid a \land 2 \mid a$

[jolly parrot]

A dense set

[sharp smelt]

Ok, have to go for a club meeting now

thanks for the help

[cursive wraith]

the integers are closed under division now?

[cursive wraith]

alr

you only had to write a = 7(2k1) and a = 2(7k1)

[pearl pond]

@sharp smelt Has your question been resolved?

[pearl pond]

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[midnight flicker]

Determine the value of the parameter m at which the mass equation
four solutions:

[nimble locust]

Question @midnight flicker

if i post a question how soon will it be answerd, and how many can i post at once? (max)

[ionic harness]

Replace t=x+1. It does not change the number of solutions. Then consider some cases, like t<0 and t>=0. Do the same with the outer abs. Draw the graph of that. Then the answer is kind of obvious.

[midnight flicker]

ohh

[midnight flicker]

let’s try

[midnight flicker]

what

i don’t get it

[nimble locust]

Is it ok if i post a few questions on one of the channels or am i only allowed to post one

[midnight flicker]

its okay

do whatever you want

[novel wedge]

Please post a question on an unoccupied channel and you are allowed to post multiple on one UNOCCUPIED channel

[midnight flicker]

thank you!

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[alpine portal]

alright so we got a,b,c odd integers
they said prove that ax^2+bx+c doesnt have integer roots

[alpine portal]

root btw is a solution of an equation such as A(x) = 0

[pseudo oxide]

quadratic formula

[jolly parrot]

ren

[pseudo oxide]

use that

[alpine portal]

@pseudo oxide i substitued the coefficents using the odd number algebraic form

2k+1

[pseudo oxide]

ok

[alpine portal]

then i found out it was a square root of 21

it was like this

.close

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[midnight haven]

if 0<x<1, then will 1 + x^2 + x^4 + x^6 + x^8............. be infinite?

[harsh gulch]

no

[midnight haven]

ok so it will limit to a finite value?

[harsh gulch]

yeah

[midnight haven]

ok

close

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[eager moss]

where am i going wrong

[pearl pond]

@eager moss Has your question been resolved?

[velvet tiger]

the last step?

[pearl pond]

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[midnight haven]

im not sure how to do b

for a, i got 25.6

but i dont know what angle i need to find..

[pearl pond]

@midnight haven Has your question been resolved?

[midnight haven]

<@&286206848099549185>

[pearl pond]

@midnight haven Has your question been resolved?

[midnight haven]

<@&286206848099549185>

[broken fossil]

bro why was no one here for da pro

i cant really answer this tho

[pearl pond]

@midnight haven Has your question been resolved?

[midnight haven]

Idk..

[pearl pond]

@midnight haven Has your question been resolved?

[noble nova]

idk a lot about this, i'm just a HS student, but what comes to my mind is that you could use a vector whose components are in the proportion mentioned in the question "...the ball has velocity (12i + 15j) m/s..." so if i is horizontal and j is vertical you would have a sort of triangle ABC, right in B, with AB = 12x and BC = 15x (note that the ball is travelling horizontally so your angle will be the angle made by the vector itself and its horizontal component which you could get with arctan), idk if that works out, I don't really understand this kind of questions very well cause there is a lot I don't know yet, if I'm wrong I'd love to be corrected.

[pearl pond]

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@jovial imp Has your question been resolved?

[pearl pond]

@jovial imp Has your question been resolved?

[pearl pond]

@jovial imp Has your question been resolved?

[midnight haven]

Thanks

[pearl pond]

@jovial imp Has your question been resolved?

[pearl pond]

@jovial imp Has your question been resolved?

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[inner pier]

Could someone explain these please? I don’t quite get It

[west sapphire]

which one?

i suppose you're ok with the last one

[regal herald]

what is this meant to be showing, im a lil curious

[inner pier]

Indeed, I am, and well, I mean all of them

[leaden wadi]

The last line should have been x^(1/2 + 1/2) = x^1. That would have made it more clear.

[inner pier]

Oh, it’s Professor Dave’s math video explaining square roots, I don’t really understand this though

[inner pier]

Ah I get it now, makes sense

[pearl pond]

@inner pier Has your question been resolved?

[inner pier]

😭

[broken fossil]

power rules pal

[inner pier]

Uh, yeah I figured as much, I just don’t know the rules

Why are the rules like that?

What do they mean

Why?

[pearl pond]

@inner pier Has your question been resolved?

[sacred crest]

sqrt is equivalent to ^(1/2) precisely because it is the only exponent for which the quake root property holds (as shown in the picture)

[pearl pond]

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[hearty roost]

How can I prove the series converges or diverges?

[craggy tendon]

write the general term of the series

[hearty roost]

Ok

What do I do next?

[craggy tendon]

what is it

[hearty roost]

Summation from n=1 to infinity of (-1)^n+1 times A(n) where A(n) is a sequence defined as 1/n if n is odd and 1/n^2 if n is even

[craggy tendon]

(2n)^2 - (2n-1)/(2n-1)(2n^2)

you should get that

try ratio test and root test

if that doesnt work

lmk

[hearty roost]

Wait im sorry

Im confused as to how that’s the general term of the series

[craggy tendon]

magic

[hearty roost]

If n=1, then dont you get 3/4?

[craggy tendon]

yes

[hearty roost]

So are you grouping two terms togehter

[craggy tendon]

yes

[hearty roost]

Oh

How did you know to do that?

[craggy tendon]

because like

the negative

thing

is annoying

[hearty roost]

Oh

Well thabks

[pearl pond]

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[flint musk]

Can someone help me figure out what went wrong while trying to integrate this?

[flint musk]

I tried using the power reducing trig identity to make it easier to integrate

[dusty flame]

-sin(u)

no?

$\frac{\dd}{\dd x} -\sin x = -\cos x$

[jolly parrot]

k

[flint musk]

oh my god thank you

Oh yeah, I accidentally took the derivative of cosine instead of anti derivative

thank you again

.close

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[patent spindle]

help regarding differential equations

[pearl pond]

@patent spindle Has your question been resolved?

[pearl pond]

@patent spindle Has your question been resolved?

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[slow pivot]

Hi, I need a desmos graph and don't really understand how to get it. Could someone please make one? All of the math is done, I only need the one thing. Thank you!

[slow pivot]

Nevermind, I figured it out. Bot, please .close

.close

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[versed mica]

@rare edge

@hard kite

what’s good

tough

[rare edge]

Yo

[rare edge]

Idk what to do

I know the vertical asyomptotes are -2 and 4

Y intercept 2 z intercept 1

X*

<@&286206848099549185>

<@&286206848099549185>

[untold raven]

hi

[rare edge]

I need help

[untold raven]

7 or 8

[rare edge]

Let’s start with 7

[untold raven]

it is grph of some inverse fn shifted by some theta

[rare edge]

Fn?

Like a Yn?

[untold raven]

fn= function

[rare edge]

Ohhh

Got it

[untold raven]

like tan inverse

[rare edge]

^

I said these

[untold raven]

ye but tht wud help

[rare edge]

Okay so now what

[hard kite]

i think its a rational function isnt it

[untold raven]

oh

[rare edge]

Sooo

I got 33 minutes to do these 2

[untold raven]

which exam

[rare edge]

It’s homework

[untold raven]

find the x corrdinte pls

[rare edge]

There’s only 1?

[untold raven]

i done

x-16/(x+2)(x-4)

@rare edge

sry

[rare edge]

What’s this

[untold raven]

wAit

[rare edge]

This is how they formatted the last one

[untold raven]

wht is AAA

in their answer

[rare edge]

That’s not the answer

The answer key is

Under it

For another problem

They want it written like that

[untold raven]

ok ans is 16*(x-1)/(x+2)(x-4)

[rare edge]

Nice I got that too

What did you get for 8

@untold raven

@untold raven

[untold raven]

x*(x-3)/(x-4)(x+4

[rare edge]

Nice I got that too what about 9

I think I got that one wrong

[rare edge]

And the last too are multipled btw

[rare edge]

This was wrong?

[untold raven]

u r submitting in 8 tht is 9

[rare edge]

That was answer for 8

[rare edge]

What

Dm me @untold raven

[pearl pond]

@rare edge Has your question been resolved?

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[tough trench]

help idk what its asking for

[compact ridge]

do you not understand the example, 236?

[tough trench]

wait its just 2x3x6 i has to equal the last 2 values

like 36

[compact ridge]

yeah

[tough trench]

do i just guess n check

loll

[compact ridge]

so using the hint, a * b * c = 10b + c

so the left hand side must be a two digit number ofc

[compact ridge]

there are only so many possibilities, correct

[tough trench]

wait i odnt get the hint lol

[compact ridge]

so if I have say 21, that's just 2 * 10 + 1

or 87 is 8 * 10 + 7

[tough trench]

wait but what does it help wth

[compact ridge]

I'm not sure actually

it seems though that a can only be 1, 2, 5 though

[tough trench]

why and how

[compact ridge]

we want to use Simon's favourite factoring trick I guess

say, note that (b - 1)(c - 10) = bc - 10b - c + 10

so when a = 1 we must have b * c = 10b + c

[compact ridge]

so from this, what must (b - 1)(c - 10) equal?

[tough trench]

im fconfused

[compact ridge]

can you make one side of the equation 0?

[compact ridge]

ping

[pearl pond]

@tough trench Has your question been resolved?

[tough trench]

lowk i think i guessed n check

i got 128, 612, 324, 236

[compact ridge]

oh okay, yeah I was going to teach you a more systematic method

but seems like you didn't need it

[tough trench]

wiat im hesistant of 128

does it work

[compact ridge]

ah yeah cause 1 * 2 * 8 = 16

it's not 28

[compact ridge]

so this equation shows that a = 1 is impossble

[tough trench]

ohh

[compact ridge]

though yeah I should have checked a = 3 as well

[tough trench]

thanks for helping me tho!

[compact ridge]

no worries!

[compact ridge]

the other number is 315

ah I know how to solve it systematically now, turns out we needed a different technique

[tough trench]

wait so does 128 count as a number

[compact ridge]

no

[tough trench]

okay knew it i was like wait it dont wokr

thanks!

[pearl pond]

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[brittle onyx]

how do i do this qu?

[worn bear]

I feel like ull be using the trig identities for angle sums

Eww

[brittle onyx]

ah ok

[brittle onyx]

how am i meant to work out the sum of all them though?

[worn bear]

https://mathcs.clarku.edu/~djoyce/trig/identities.html

Found a link to the udebtities

[brittle onyx]

ok

[worn bear]

Wdym work out the sum

[brittle onyx]

like the multiple choice options give 0, pi, 2pi, and 4pi

[worn bear]

Hmm this is tricky cause of the factors

Huh, i think the answer is zero

Cause the question already is assuming that WHATEVER the variables are

It is given to be true that

acos(x)+bsin(x)=Rsin(x+α)=Rsin(x-β)=Rcos(x+γ)=Rcos(x-δ)

Do u agree with that premise?

That this long equation is given to be true?

(Btw the reason y i ask if u agree with what i said is so we can both agree on each step in the logic that will give us the answer)

Lmk

Ahh basically we know
Rsin(x+α)=Rsin(x-β)
Which is only possible is α=-β

And so we know that α+β must be zero then

And we also know that Rcos(x+γ)=Rcos(x-δ)

Whoch is only true if γ=-δ

Meaning that γ+δ is also zero

Meaning that α+β+γ+δ = 0+0=0

Lmk if that makes sense

No identities needes

[pearl pond]

@brittle onyx Has your question been resolved?

[brittle onyx]

Ahh yea I get it now

Thanks for the explanation

[worn bear]

Np

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[sacred hollow]

Is there a better way the bulking it and counting it my own?

[cinder flower]

dynamic programming by hand would be pretty fast

[sacred hollow]

What is tat?

[blissful salmon]

its like the number of ways of reaching a point on the grid in terms of adjacent points, then you keep going down and right until you find the number of ways for B

[sacred hollow]

Huh?

[cinder flower]

wasn't too bad at all

hopefully no arithmetic errors

i can explain though

[sacred hollow]

So no other way than bulking it?

[blissful salmon]

or you could consider what would happen if the grid was filled

[cinder flower]

what's bulking it?

[sacred hollow]

Tis ok, I just wanted an ez way to calculate it lol

[sacred hollow]

Oh um.. like doing all the arithmetic for each edge I mean like caling the previous ways together

[sacred hollow]

Like?

[blissful salmon]

like you should count the number of restricted paths

[blissful salmon]

then subtract that from total number of paths if the grid was filled

[cinder flower]

if the grid didn't have that hole in the middle, the answer would be the number of permutations of rrrrrrddddd

[blissful salmon]

also i confirmed this a quicker way, and your answer is correct

[cinder flower]

so you can try to adjust it from there but it'll be a little tricky

[cinder flower]

yay

[sacred hollow]

Ook

[blissful salmon]

||it would be C(11,6) - C(6,3)C(5,3) = 262||

[sacred hollow]

Okk

[sacred hollow]

Understood

Thanks guys! Have a great day!

.close

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[foggy mason]

what have you tried ?

[brittle onyx]

hope the quality's not too bad

[foggy mason]

hmm, what i thought about was doing this :

x = (x.a) a + (x.b) b = ka + pb

if we are doing a projection on the vectors a and b, that means x can be written as a linear combination of a and b

[brittle onyx]

How did u get that

[foggy mason]

from what the exercise said

but i think you are correct

[brittle onyx]

Oh sweet do u reckon u can help me continue from what’s I’ve done

[foggy mason]

the whole point is to find x right
you found the projection on a and b of x

[brittle onyx]

uh yeah

[foggy mason]

you can write x in terms of a and b

[brittle onyx]

Uhh

I don’t understand

[foggy mason]

x = something a + something b

and you have the projections
use that

[pearl pond]

@brittle onyx Has your question been resolved?

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[dense iris]

will a simply differentiate to 1?

-_-

[pearl pond]

@dense iris Has your question been resolved?

[cosmic raptor]

it’ll differentiate to 0, since it’s a constant

[dense iris]

ah yeah

thanks

.close

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[hard locust]

why is e u here? isnt it supposed to be dv since its an exponential function? Liate

[light helm]

liate is a guideline
both linear trig and exponential are relatively easy to differentiate/integrate,
so doesn't really matter which way you choose here

[hard locust]

ah

,close

,close

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[shrewd summit]

why is there another (4x^2-3)^4??

[sudden heath]

Because in the product rule, you keep 1 factor and differentiate the other

[shrewd summit]

Ohh so this is the product rule. I'm still in chain rule, that's why I got confused.

thank youuuu 👊🏻

[pearl pond]

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[shrewd summit]

why is 5 on the outside?

is that the rule?

[dusty flame]

${\sin^n(x) = (\sin x)^n}$ if ${n \in \mathbb{Z}^+}$

[jolly parrot]

k

[shrewd summit]

what does the last one mean?

[dusty flame]

if n is 1,2,3,4,...

i havent seen anyone use 1

so i assume it might be just 2,3,4,5...

idk

but i think u get the point

[shrewd summit]

Yeah I do

thanks ☺️

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[supple knoll]

it is same thing written in different form

n could be anything

[shrewd summit]

is this correct?

[shrewd summit]

oh ok, thanks.

[supple knoll]

Is the last thing minus sin x3 or time negative sin x3

Ok I see

Ur right

[shrewd summit]

Yey ☺️

thanks

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[tulip jolt]

how do i prove (combinatorial proof) that there doesnt exist a polyhedron with an odd amount of faces and each face having an odd amount of edges (without using euler law for planar graphs preferably)

[plain rivet]

hmm

[plain rivet]

To prove that there does not exist a polyhedron with an odd number of faces where each face has an odd number of edges, let's use a combinatorial argument based on counting edges and faces.

Step 1: Set up your counts

• Let ( F ) be the number of faces of the polyhedron.
• Let ( E ) be the number of edges.
• Let ( V ) be the number of vertices.

Step 2: Properties of faces and edges

If every face has an odd number of edges, we can denote the number of edges in each face as ( e_1, e_2, \ldots, e_F ), where each ( e_i ) is odd.

Since each edge is shared between two faces, the total sum of edges counted across all the faces is:

[
\text{Total edges counted} = e_1 + e_2 + \ldots + e_F = \sum_{i=1}^{F} e_i
]

Step 3: Counting edges in terms of faces

Since each edge contributes to two faces, we can relate this total to the actual number of edges ( E ):

[
\sum_{i=1}^{F} e_i = 2E
]

Step 4: Analyze the parity

Since each ( e_i ) is odd, the sum ( \sum_{i=1}^{F} e_i ) is the sum of ( F ) odd numbers. The parity (odd or even nature) of the sum of odd numbers is as follows:

• If ( F ) is odd (which we are considering in our proof), the sum of ( F ) odd numbers will be odd.
• If ( F ) is even, the sum of ( F ) odd numbers will be even.

But since ( \sum_{i=1}^{F} e_i = 2E ), this sum must also be even (since ( 2E ) is always even).

Step 5: Conclusion from the parity argument

We have two conclusions from our analysis:

1. If ( F ) is odd, then ( \sum_{i=1}^F e_i ) is odd.
2. However, ( \sum_{i=1}^F e_i = 2E ) is even.

This leads to a contradiction because an odd number cannot equal an even number.

Final Statement

Therefore, no polyhedron can have an odd number of faces with each face having an odd number of edges. Hence, we conclude:

There does not exist a polyhedron with an odd number of faces where each face has an odd number of edges.

[jolly parrot]

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

[thin sigil]

This is either an AI answer or there's people out here who have these things ready to copy paste

[tulip jolt]

oh wow i couldve used chatgpt myself

oh yeah but this is trivial

i shouldve been able to come up with that

fuck

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[jolly parrot]

Nyxzore

[plain rivet]

not gpt

but similar

[broken fossil]

aint no party like a gpt party

[thin sigil]

cause a gpt party can't count

(the letters in a word)

[plain rivet]

h-h-hell naw

[pearl pond]

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[tawny pewter]

does wanyone one know why this is wrong

[foggy mason]

-4 instead of -3

[north talon]

-4 not -3

[tawny pewter]

ohh

thankss

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[steady steppe]

can someone help me to prove 8.1 and 8.2?

X,Y,Z are random variables and a,b are real constants

[steady steppe]

maybe 8.2 may be proven by the linearity of expecation

but probably thats not the right way

[pearl pond]

@steady steppe Has your question been resolved?

[pearl pond]

@steady steppe Has your question been resolved?

[brittle tinsel]

what have you tried?

[tulip jolt]

most likely uses chatgpt anyways

[steady steppe]

I didn't make any progress. I just thought about using the definition of conditional expectation but the problem statement does not mention whether the variables are discrete or continuous

[brittle tinsel]

it is discrete since the only value your random variable takes on is a

at least for 8.1

for 8.2 you would need to prove it for both

[steady steppe]

Uff

[steady steppe]

And how to prove 8.1 because on this case I do not have an idea

[brittle tinsel]

E(a | Y) is the same as calculating

E(X|Y), where X is a discrete random variable and P(X = a) = 1

when you take the expected value of anything, it must be a random variable

[steady steppe]

Hmm

[steady steppe]

them basically this is the result? a * P(X=a|Y) = a*1 = a

[brittle tinsel]

correct

[steady steppe]

is this correct?

that double integrals seem strange to me

[steady steppe]

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[midnight haven]

For w(x) idk how to word this but where it says -2 < x < 2 does it mean i need a number that's less than 2 but also not past -2?

[cyan lily]

you can think of it as an x that satisfies both -2 < x in addition to x < 2

-2 < x < 2 is just for convenience

[midnight haven]

Oh so I can just kind of break it apart -2 less than x but x is less than 2

[cyan lily]

yea

[midnight haven]

Okk and for the -4, x is less than or equal to -2

How do i

Do the input thinf with the -4 if there's no x

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[trim mantle]

translation : given the boolean fucntion that is almost KNF , continue

[trim mantle]

.

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@trim mantle Has your question been resolved?

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@trim mantle Has your question been resolved?

[tulip ore]

if you dont have any other ideas, you can always go through all 8 possibilities then do KNF on the Ts you see

[pearl pond]

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[dusk otter]

If I have 3 flags, red, blue and green each being 2 feet tall. How can I find a recurrence relation to represent the number of ways ot place flags along an n foot pole?

[dusk otter]

Would it just be 3a_{n-2}

?

@vestal tapir So is my original answer correct?

[vestal tapir]

i don't know, i was sure it's wrong

it looks insane

so simple

but now i maybe agree

is that assuming as many flags placed as you can?

[dusk otter]

The pole is n feet tall

[tardy reef]

Am I wrong in understanding that you have only 3 flags, with n positions to place them?

[dusk otter]

No, that's correct

[tardy reef]

Then you can place the first flag at n-2 different locations, the second at n-4 and the last at n-6

[dusk otter]

I probably should've shown this sooner, but the question has these options

[vestal tapir]

it's right if you are not allowed to leave empty space