help-13

[winter mantle]

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[pastel vault]

woops, 0.75 cos(x) + 1.25

[winter mantle]

thanks

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[subtle dirge]

how do i show the proof to this by the inequality? is it alr okay if i solved the discriminant and got the desired inequality?

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[void sand]

you can prove it using the discriminant yes

[sonic helm]

Yep works

[subtle dirge]

okie thank you ❤️

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[slate fog]

Did a problem on counterwights and had to find the mass of a counterwight for an elevator with a mass of 600kg and an acceleration of .5 m/s^2. The picture is the equation my teacher used but I have no idea how he got there. He said use Newtons second law but how do you get to the mass of counterwight equation from F=ma?

[odd verge]

Drawing a diagram could help.

[slate fog]

Is this right?

My partner and I just keep finding the same mass for both objects

[odd verge]

Does it specify the direction the elevator is accelerating?

[slate fog]

We have the first part about the velocity but this is the whole question, so no?

[odd verge]

Okay, let's assume it is moving down.

[slate fog]

As it down is the positive direction?

[odd verge]

So m1(mass of fully loaded elevator)=600kg
m2(mass of counterweight)
a=acceleration

[odd verge]

Wdym?

[slate fog]

Like the downward acceleration uses a positive sign in the equation? Or are we defining up as the positive direction where any acceleration in the equation would be negative

[odd verge]

Downward means a is positive

[slate fog]

Okay

[odd verge]

If it is in equilibrium the upward force= downward force

[slate fog]

Is the equation generally elevator/counterweight? Is the denominator about the counterweight?

Like the ag of the counterweight?

[odd verge]

Wdym?

[slate fog]

I'm just trying to understand the equation. Like how we got m(g-a)/a+g

Like is the mass of the counterweight the elevator's mass times gravity minus acceleration over the counterweights acceleration plus gravity? Either way, I still don't know the algebra used

[odd verge]

oh
g-a here indicates it is accelerating against gravity

[slate fog]

Okay makes sense

So the original equation would be something like m1g-T=m2(a-g)? And we rearrange to solve for m2?

Or no tension because it would be the same on both sides?

[odd verge]

Wait

@slate fog

[slate fog]

Oh lord that makes sense

okay, I understand what to do from there and I understand how you got there.

Thank you very much!!

[odd verge]

Or you can also say that
m1 is accelerating against the then m2 has to do the opposite

[slate fog]

Thank you!

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[odd verge]

Yes

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[cosmic umbra]

i need help

[misty cave]

is it good?

[alpine geode]

try writing out the matrices for f and g and multiplying them

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[plush sinew]

I'm so confused

[plush sinew]

I said 1/8

and it said I was wrong

[gritty viper]

how did you get 1/8?

[viral glade]

Arent there 9 parts in total

[plush sinew]

It's cut into 8 parts and there's 1 of Brocoli

[plush sinew]

Oh right, there is

You were right 😭

mb

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[snow zephyr]

I don’t know how to form it like where the + goes the x y and and and
Hope someone’s German 🙏

[cedar kiln]

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[snow zephyr]

@uncut veldt

[cedar kiln]

@snow zephyr Has your question been resolved?

[snow zephyr]

<@&286206848099549185>

[snow zephyr]

<@&286206848099549185>

[hollow wasp]

for which problem

@snow zephyr Welches Problem

Ich kann 17 probleme nicht beantworten

Im Allgemeinen hat bei diesen Aufgaben ein Objekt den Wert x, ein anderes den Wert y, und beide addieren sich zu einer Summe. Dies wird zweimal gemacht, so dass Sie zwei Gleichungen haben, um eine Variable zu eliminieren.

[snow zephyr]

My problem is I don’t know how to do it I mean how to calculate yea but how to structure it don’t know

[hollow wasp]

Ok, try 6. 3 cheeseburgers and 2 fries for 11,40€. Ok so we have $3x+2y=11,40$ because the prices must add up to 11,40. So x represents the price of a burger and y represents the price of fries. Okay. Now the next says 2 burgers and 4 fries is 16,40€. Since we defined our variables we have $2x+4y=16,40$

[wraith dagger]

Surf

[hollow wasp]

we can use these two equations to determine the values of the two variables by eliminating one

if you do not know how to do elimination, i would suggest looking it up

i have to go now though i am quite busy, viel glück

[snow zephyr]

Can we do number 12

[snow zephyr]

Oh okay yeah thank you

[hollow wasp]

yes just one second

[snow zephyr]

Sure

Or

Can I send you what I did and you’re saying if it’s right or not

[hollow wasp]

That works too

[snow zephyr]

I think that’s the easiest way

[hollow wasp]

agreed

[snow zephyr]

I didn’t do every single task but the task i found easy

[hollow wasp]

in these problems you don’t really want to have a variable on the right side

[snow zephyr]

Oh okay

So variable should only be on the left side

[hollow wasp]

for twelve let x = monatlicher Grundgebühr and y = den Kosten pro Kubikmeter Wasser

yes

[snow zephyr]

Yea

[snow zephyr]

Sometimes they could be on the right side right?

[hollow wasp]

so for 12 you have x+17,8y=35,54 (März) and x+23,3y=45,44 (Juli)

[hollow wasp]

sure but to solve the problem you will always bring it to the left side so that you can set the equation equal to constant

[snow zephyr]

Yea ik that but for the equal structure yk

[snow zephyr]

Yk when I see the answer if the task it makes sense to me but anytime I do it by myself it just doesn’t makes any sense

[hollow wasp]

Well, what did you think when you did #12

[snow zephyr]

So I at first I thought x is water and y is monatliche Gebühren

[hollow wasp]

well, it could be either or

the variable names don’t matter

so that part is correct

[snow zephyr]

Oh okay great

[snow zephyr]

Yea

[hollow wasp]

what is the next step ?

[snow zephyr]

For the calculation?

[hollow wasp]

yes

i want to know your thought process so it can be corrected

[snow zephyr]

The i would bring all variable to the left side and the numbers to the right side then divide or do the math depends on how it is set up

[hollow wasp]

well, you have to set the equation up

i’m looking at your work for #12, i don’t understand why you did what you did

[snow zephyr]

I mixed that up like x and 35,45
Y and 45,44

[hollow wasp]

the totals will never have a variable on them

[snow zephyr]

Oh okay yea

And what about task 17 I checked it and the answer for it is
X+6=3y
Y+9=4x

You know what I’ll try my best tomorrow hopefully it’ll be a C at least but thank you for your help I really appreciate it

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[indigo oxide]

why dont u do the -12x * -20

?

[hollow trail]

they're not multiplied

-12x-(9x+20) = -12x + (-1)(-9x+20)

[cedar kiln]

@indigo oxide Has your question been resolved?

[indigo oxide]

wtf

oh

bro man

im so stupid i dont know why

am i consuming too much social media

nvm

but bro

is 0+ and 0-

if its from right or left

or from above or below

[idle tusk]

what?

[indigo oxide]

yk when calculating the limits

[idle tusk]

is that related in any way to the integral

[idle tusk]

yes

[indigo oxide]

no

[idle tusk]

ok i got confused lol

[indigo oxide]

is 0+ describing

like

im rly rly confused

[idle tusk]

0+ is from the right

you can think of it as the the function with positive values getting smaller and smaller

[indigo oxide]

why do some say from above or below

[idle tusk]

like f(1), f(0.1), f(0.01), f(0.001), f(0.000000000001)

[idle tusk]

same thing basically

from above is 0+

from below is 0-

[indigo oxide]

yeah

makes sense

lol

[hollow trail]

depends on how you orient your number line

[indigo oxide]

also,

how do i know if its 0+ or 0-

do i have to pciture the function

in my haed

head

[idle tusk]

fair enough

[idle tusk]

wdym

show an example problem

[indigo oxide]

how do i know if its 0+ or 0-

ok wait

bro ill just do a curve like problem

when i do the limits part

ill come back

[idle tusk]

ok

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[indigo oxide]

@idle tusk here bro

u see this

how do i find out if its 0+ or 0-

[idle tusk]

,rccw

[indigo oxide]

yeah

umm

u there?

<@&286206848099549185>

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[sharp vortex]

calculate biggest and smallest point, im stuck here

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[magic spear]

Hi

[magic spear]

How do i answer this? Im rly confused

[digital cliff]

(x-a)^2+(y-b)^2=r^2
is for a circle with center (a,b) and radius r

[magic spear]

thank you so much

[magic spear]

wait

could i

write

y^2+x^2=6^2?

[digital cliff]

yeah

[magic spear]

thanks

have a good day

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[dire geode]

do you just enter that

[proper niche]

Is that arc notation meant to signify those angles are the same? If so, you have congruent triangles so solve x+26 = 3x ...

[static idol]

I got 13

x = 13

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[forest olive]

Hi

[forest olive]

Ehm

[mighty shuttle]

Well, you can start off by proving that traingles YJM and YLK are congruent

[dull oxide]

It looks about half as long

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[dense gyro]

hellopp

[cedar kiln]

Please don't occupy multiple help channels.

[dense gyro]

I need help with finding a good website that gives good algebra 1 practice questions

I neeed word problems for algebra 1 domain and range

[cedar kiln]

@dense gyro Has your question been resolved?

[woeful isle]

Read the last 2 points

• This section is meant for math questions only. If your question is off-topic, try #discussion or #chill.

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[dense gyro]

hello

[dense gyro]

I need a good website to get questions for algebra 1

[dusk goblet]

good old sal khan

[dense gyro]

I cant seem to find word problems for domain and range there

[dusk goblet]

word problems?

[dense gyro]

yup

I have a few examples I can give you

[dusk goblet]

have you tried asking your teacher

or

looking up algebra 1 word problems

[dense gyro]

I have

but they seem to be what I am not looking for

I remember my uncle showed me this website it was good but I cant remember the name

it started with a "k"

[dusk goblet]

khan academy

maybe try algebros

they had good stuff for calculus

idk about algebra

but you can try

flippedmath.com

[dense gyro]

oh kk

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[tulip steeple]

How do i do b

[tulip steeple]

So what i was doing

This is what I got for the basis of the image F

But what exactly do I do for the kernel

Do i just solve the system of equations after the row echelon

<@&286206848099549185>

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[tulip steeple]

<@&286206848099549185> 😭

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[ornate rune]

Is fibonacci series and fibonacci sequence the same?

[pastel vault]

when people say Fibonacci series, they really mean Fibonacci sequence

[fair geyser]

"series" means the total sum of everything converges

as far as i understand

so yeah there's no fibonacci series

[pastel vault]

could be a divergent series

no one said it had to be convergent

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[dire thunder]

hey guys i am dumb as all hell farmboy what would i need to focus on to pass my asvap army test math in general is my problem

[cedar kiln]

@dire thunder Has your question been resolved?

[bleak viper]

go through the syllabus

overview all the topics and see which one are the easiest for you

then watch some video for them and do questions

after completing the ones which are easier for you, do the harder ones

[dire thunder]

ok man thanks

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[arctic hare]

can someone help with this

our school is starting circled after Christmas but I wanted to start a little bit early because I love math

[vestal bluff]

do you know the general form of the equation of a circle

[crimson sedge]

Hello, i need help with heron's formula

[vestal bluff]

wowie thats cool

[vestal bluff]

#❓how-to-get-help

[arctic hare]

x^2 + y^2 = r^2

[vestal bluff]

close, its $(x-h)^2 + (y-k)^2 = r^2$, where $(h,k)$ is the center of the circle

[wraith dagger]

ashy!

[vestal bluff]

you have the center

and the radius

so plug those in

[arctic hare]

it says that circle with centre (0,0) has equation x^2 + y^2 = r^2 here

[vestal bluff]

well with center 0, 0 this is correct

yeah okay mb

[arctic hare]

so how do I

[vestal bluff]

just plug the radius in

r is the radius

[arctic hare]

r=3?

is that it or am I tripping

[vestal bluff]

r is sqrt 3

[round geode]

$r=\sqrt3 \ x^2 +y^2= (\sqrt{3})^2$

[vestal bluff]

$r=\sqrt3 \n x^2 +y^2= (\sqrt{3}^2$

eh

[wraith dagger]

Nyxzore

[round geode]

Jeez I've never suffered with latex like that before

[arctic hare]

but wouldn’t the square root and square cancel to make r=3

[vestal bluff]

r^2 is 3

[round geode]

You said r=3
We are saying r^2=3

[arctic hare]

fair enough

is that it

[round geode]

Yes

[arctic hare]

just x^2 +y^2 = 3

[vestal bluff]

yup

[arctic hare]

Ty ty

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[mint raft]

Hi, I need help with finding the function of this parabola but I don't know how to do it.

I know that it should be: y=ax^2+bx+c

[normal cipher]

Its not symmetric tho

Are you sure its $ax^2 + bx + c$

[wraith dagger]

Carbonite

[mint raft]

Now that you say it

It's true

No I'm not sure, I just know a parabola should be x^2

[mighty mango]

which is what you want, a function or an equation?

[mint raft]

A function

[mighty mango]

if it is a parabola, it can't be written as a function

it's lean

[normal cipher]

Yea

[mint raft]

Ohhh okay

And how do I find the function of a lean?

Wait I think I know it

Nvm, thanks!

.solved

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[strange arrow]

If you role a normal 6 face die with numbers 1 to 6, 4 times and find product of it. What are the chances that product is divisible by 4?

[strange arrow]

I solved this in 2 way and got 2 different answers and LLM cant really agree on which one is correct

In order to be product of 4 numbers to be divisible by 4 it has to have at least two factors of 2 (a 4 or combination of 2 twos or 2 and a 6)
1 - 0 factor
2 - 1 factor
3 - 0 factor
4 - 2 factor
5 - 0 factor
6 - 1 factor

[normal cipher]

This is an AMC question

[strange arrow]

so if i find probability of product having factor of 0 and 1 i can substract that number from 1

[strange arrow]

wdym

[normal cipher]

nvm

We know that total number of rolls is 6^4

[strange arrow]

yes we do

and total number of rolls that have 0 factor is 3^4

[normal cipher]

Yes

[strange arrow]

so probability of 0 factor is 1 /16

[normal cipher]

Now only factor of 2 need consider

[strange arrow]

why? we only need to find factor of 0 and 1

now heres the problem i do not understand

[normal cipher]

Wait hear me out

[strange arrow]

okay

[normal cipher]

If there rolls 2 2s or 2 6s, condition is satisfied

Has to do with power of two in the product

Then we can compute for sequences with only one 2 or only one 6, and the rest are odd

Using similar method @strange arrow

[strange arrow]

urs is way better

[normal cipher]

Yay

[strange arrow]

and also none of 2 6 4

[normal cipher]

Yes only one 2 or 6

[mighty drift]

it's an LLM, you shouldn't trust it anyway

[strange arrow]

desperate times call for desperate measures

[normal cipher]

Tf

[mighty drift]

<@&268886789983436800>

[strange arrow]

is server getting ddosed

[normal cipher]

Beat me to it

[strange arrow]

nah only in this chanel

[hollow kiln]

do you know what that means

[livid hound]

confurm

[strange arrow]

yes i used incorrectly here

i half understand it

[normal cipher]

Sorry i gtg, hope the method helps @strange arrow

[strange arrow]

thank u no problem

[hollow kiln]

brute forcable

[strange arrow]

how to post latex formulas

[strange arrow]

any probability problem is if u have enough time

[hollow kiln]

$\frac{3}{2}$

[wraith dagger]

haygiya

[hollow kiln]

i know this much

[hollow kiln]

the distribution would be normal, if you could calc the first half youd get the 2nd half

[strange arrow]

{only \ 1 \ twos \ or \ sixes:}
\
\frac{1}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} = \frac{5^3}{6^4} = \frac{125}{1296}

\
{Only odd:}
\
\left( \frac{1}{2} \right)^4 = \frac{1}{16}

\

1 - \left( \frac{1}{16} + 2 \times \frac{5^3}{6^4} \right)

my bad

[hollow kiln]

dollar signs at both ends

[strange arrow]

$

{only \ 1 \ twos \ or \ sixes:}
\
\frac{1}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} = \frac{5^3}{6^4} = \frac{125}{1296}

\
{Only odd:}
\
\left( \frac{1}{2} \right)^4 = \frac{1}{16}

\

1 - \left( \frac{1}{16} + 2 \times \frac{5^3}{6^4} \right)

$

[hollow kiln]

hmm

[strange arrow]

${only \ 1 \ twos \ or \ sixes:}
\
\frac{1}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} = \frac{5^3}{6^4} = \frac{125}{1296}
\
{Only odd:}
\
\left( \frac{1}{2} \right)^4 = \frac{1}{16}
\
1 - \left( \frac{1}{16} + 2 \times \frac{5^3}{6^4} \right)$

[wraith dagger]

whatihavedone

[hollow kiln]

i’d go calculate following
P(X=4)=(1/6)^4
P(X=8)=…
P(X=12)=…
P(X=16)=…
P(X=20)=
P(X=24)=…

[strange arrow]

approximetly 74.4%

[hollow kiln]

could only take so much time

[strange arrow]

u gotta be trolling

[hollow kiln]

why

[strange arrow]

there gotta be way more optimal solution

[hollow kiln]

i mean

computers exist for a reason

idk you’re probably right

[strange arrow]

wait i have an idea

just roll 4 dice 1000 times

and count it

u just gave me idea to how to check every probability problem answer

[hollow kiln]

😃

[mighty drift]

you can easily find the probability of rolling at least one 4
after that you can find P(no 4) P(divisible by 4 | no 4) using a binomial distribution

[strange arrow]

thats initially what i did

[cedar kiln]

@strange arrow Has your question been resolved?

[strange arrow]

answer is around 77.2%

<@&286206848099549185>

If you role a normal 6 face die with numbers 1 to 6, 4 times and find product of it. What are the chances that product is divisible by 4?

[desert nova]

you can think of the fact that if an even number multiplies an odd number at least one time, then the product is always even

[strange arrow]

$1 - \frac{5^4}{6^4} - \frac{2^3 * 4}{3^4}$

[strange arrow]

yes

[desert nova]

is the second fraction for removing the P of getting 1 * 1 * 1 * 1 ?

[strange arrow]

what do u mean by that

sorry this is wrong let me fix it real quick

[desert nova]

it's okayy

[strange arrow]

$(1 - \frac{5^4}{6^4}) + (1 - \frac{2^3 * 4}{3^4})$

wait its over 1

what

my bad

found another issue with the first part

$(1 - \frac{1^4}{2^4}) + (1 - \frac{2^3 * 4}{3^4})$

[wraith dagger]

whatihavedone

[strange arrow]

im now very confused and i have 4 different answers

$(1 - \frac{1^4}{2^4} + \frac{2^3 * 4}{3^4})$

$1 - (\frac{1^4}{2^4} + \frac{2^3 * 4}{3^4})$

[wraith dagger]

whatihavedone

[strange arrow]

1 - (no even numbers + only 1 of 2 or 6)

yep its definitely wrong

[strange arrow]

nvm i have done it by myself

.close

[cedar kiln]

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[distant gale]

Can someone explain how to get the limit of this function? I would say its 0, but the correct result should be infinity.

[livid hound]

result should be 0

[strange arrow]

try to find denominator of fraction

after taking l'hopital 2nd time

[distant gale]

Thank you for your help! :) Shouldn't the numerator tend to 3 and the denominator tend to infinity as x approaches infinity? Thus, it would be 3 divided by infinity = 0?

[sand cradle]

Yes

,w lim x^3 e^(-x^2) for x to infty

[wraith dagger]

Results provided by WolframAlpha

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

Requested by kepe_5235

[distant gale]

Thank you ^^

[cedar kiln]

@distant gale Has your question been resolved?

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[crimson sedge]

Whyyy

[crimson sedge]

Why is the e^-a getting affected by an Abs. t?

[idle tusk]

-a * t = -at, -a * (-t) = +at

it's how absolute value works

[idle tusk]

|t| = +t when t >= 0 so -a*|t| = -at when t >= 0

[crimson sedge]

This makes sense

But why the second part of the equation is: +at when |t| < 0 ?

[crimson sedge]

-t would not exist since it's an absolute

It would only exist within the absolute markers | | and never outside of it

[idle tusk]

not when |t| < 0, but when t < 0

[idle tusk]

no

if t = -1, then |t| = 1

-t = 1

-(-1) = 1

[crimson sedge]

Right, so the -a part would not be affected, no?

[crimson sedge]

My bad, yeah

[idle tusk]

why not

it's multiplied with |t|

[astral bay]

if t = -1 then -a|t| = +at

because both of those are just -a

[crimson sedge]

but |t| = 1 and -a * 1 = -a

I feel like I'm missing something here

[astral bay]

yes, and +at is -a

because t = -1 and a * -1 = -a

[crimson sedge]

Wait... so he's removing the absolute by reversing the sign???

[idle tusk]

yep

[crimson sedge]

Damn, it finally makes sense 😭

Thanks a bunch you two

What is the command to say this problem is done?

oh found it

.close

[cedar kiln]

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[cedar kiln]

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[charred badge]

Given is a grashopper population of 250 grashoppers.
It triples with perfect conditions every 4 days.
The growth can be described as an exponantial growth.
Determine the fitting growth factor w with the initial stock and growth factor.
(Sorry for the horrible english, just ask if you dont understand something)

So the right solution is: f(w) = 250 * (1,333...) to the power of w
My solution is: f(x) = 250 * 3 to the power of x/4

w is in both examples in days which means if w is 4 the result is 750.

[charred badge]

Is my solution any right?
Because it has the exact same outcomes for every number you put into w

[cedar kiln]

@charred badge Has your question been resolved?

[charred badge]

<@&286206848099549185>

[cedar kiln]

@charred badge Has your question been resolved?

[sand osprey]

yes, it's right

[cedar kiln]

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[hazy quartz]

This is what I found

Any help

[cedar kiln]

@hazy quartz Has your question been resolved?

[icy zenith]

@hazy quartz

lovely discrete math.

if you want solution, you will have to wait. agree?

surjective function: for every $h$ in $F$ must be such $e$ in $E$ such that $f(e)=h$

[wraith dagger]

mathell

[icy zenith]

when working with reversed functions, we always must keep in mind that domain and codomain will change their places. For instance, $Y$ codomain, and $X$ domain of function $f$ which maps X into Y. for $f^1$ we will have $X$ as CODOMAIN, and $Y$ as DOMAIN.

[wraith dagger]

mathell

[icy zenith]

in our function, $F$ is codomain. $Y$ is the subset of $F$. $f^{-1}(Y)$ will give us some set, let's call it $R$ and it is a subset of $E$. here is the tricky part. R and Y are the subsets of E and F, repsectively, which means that we can apply here our lovely function. $f: R \to Y$. now, $f^{-1}(Y) = R$, thus, $$f(R) = Y$$. we have proved third statements, which is pure definiton of surjection: for every $y$ in $Y$, we have at least one $r$ from $R$ such that: $$f(r)=y$$. we again showed that f is SURJECTIVE

[wraith dagger]

mathell

[icy zenith]

@pastel vault how you doin? can you check my thoughts?

do second one on your own, because it is similar with third one. Now, i will focus on fourth one

[wraith dagger]

mathell

[icy zenith]

if Y is the subset of F, let's define such set, call it $R$ which will be the subset of $E$. again, reversed shit-function. $Y$ became domain, which was codomain in vanilla function, and $R$ became codomain, which was domain in vanilla function. if after $f^{-1}(Y)$ we got empty set, it suggest us that $R$ was empty set. now, let's return back to our lovely vanilla function. $f: R \to Y$. what we can acquire from empty set? yeah, nothing. It suggests that $Y$ is empty set.

now, the tricky part comes. what is the relation of this function to surjection?

let's recall the fact that:

1. empty set can be the subset of every set.

fact one suggest us that empty set is the subset of empty set. funny, right?

recall the definition of surjective function: there is at least one $r$ from $R$ such that $f(r) = y$ where $y$ is from Y.

if Y is just empty set, how can we check the condition of surjection? We are unable.

thus, then the condition for surjection is trivially satisfied because there are no elements in $Y$ to check.

sorry for my broken english. I hope you got the main idea

[wraith dagger]

mathell

[icy zenith]

@hazy quartz

[cedar kiln]

@hazy quartz Has your question been resolved?

[cedar kiln]

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[fallow grotto]

If you have this green circle in the middle of the Y and X axis, if the green circle oscilates on the Y axis and X axis every 0.1 second, would every point (demonstrated by arrows) be touched at some point

[fallen trail]

huh?

Oscillate as in?

[fallow grotto]

up n down

ledt n right

[fallen trail]

It depends at which rate it oscillates

[fallow grotto]

0.1 second

per full 'wavenlength'

[fallen trail]

What's the wavelength?

[fallow grotto]

it goes up and back down to orginal spot it takes 0.1 seconds

[fallen trail]

we need to know the distance it travels

[fallow grotto]

lets say around 1cm

both directions

[fallen trail]

the sine function is used to model oscillations

and one oscillation is one complete curve from 0 to 2pi

[fallow grotto]

yes

hold on

[fallen trail]

I'm not sure if this is correct

But assuming that both rates (of x and y) are constant, would it just move along the line y=x

[fallow grotto]

im putting it to practise now

[azure swift]

Does the circle oscillate along the y-axis and x-axis in the same rate/same phase?

[fallow grotto]

Yes

[azure swift]

Are you aware of Desmos? If you have trouble visualizing this you can simply graph this behavior.

Namely (x-sin t)^2 + (y-sin t)^2 = 1, where t varies continuously

[fallow grotto]

i do know desmos

ill do this now

So by the looks of it, it only moves diagnoly. If I changed it so instead of them being stimusanously at 0.1 seconds, what if I made one 0,1 second and the other 0.05

[azure swift]

One wavelength so to speak in this case takes t = pi amount of time units to traverse, whatever that means, do you care about then exact perceived time unit difference or are you fine by simply getting a difference?

[fallow grotto]

Fine getting a difference

[twilit bison]

if you want it to move in a circle, use the same rate but use cos for one and sin for the other

[fallow grotto]

This is what im trying to achieve

[azure swift]

That would correspond to shifting the phase of sin on one of them by pi/2

[twilit bison]

https://www.desmos.com/calculator/knnzj8oocr

[azure swift]

So sin(t-pi/2)

But if you want to change the frequency, you multiply t by a number

[fallow grotto]

Hold on let me record what I mean to maybe simplify things

https://gyazo.com/c812c7b859e0373f740b91e3537afb20

This is the end goal

[twilit bison]

looks kind of random

[azure swift]

Wait so the perceived effect is vigorous shaking?

[fallow grotto]

Yes I can really make it accurate but yeah

[azure swift]

Is that what you’re seeking

[fallow grotto]

Vibration in X and Y sector so it vibrates around a point smoothly

[azure swift]

So in a sense you want them to behave independent of each other?

[fallow grotto]

Yes

[twilit bison]

kind of like brownian motion

[fallow grotto]

Similar to that yeah I guess so

but it would be more constructed rather than random

[azure swift]

You can add a bunch of sin waves together, with different phases, frequencies and amplitudes and you’d get that effect

Do this twice, but obviously differently to get the effect that they’re independent of each other

[twilit bison]

i guess it's different cause it stays in one spot

[fallow grotto]

yea

ok im going sleep this is too much for now. out of my knowldge tomorrow ill use all the info u guys gave me to get this made. thank you

this is deffo a step forward

appreciate it

[azure swift]

E.g. like this

[fallow grotto]

Yes, that looks similar. ill save this picture thanks

does this chat log save?

[azure swift]

Yes, as long as you can find it back by say searching your name

[fallow grotto]

ok perfect thanks @twilit bison and thanks @azure swift , have a good day/night

[cedar kiln]

@fallow grotto Has your question been resolved?

[cedar kiln]

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[fallen compass]

I’m confused should I get a negative number for x

[past wind]

no

x should be positive for sure

[fallen compass]

Are they vertical

[past wind]

supplementary

[past wind]

this is vertical

[fallen compass]

Oh thank u

.close

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[young anchor]

how do i solve for the particular solution here

[young anchor]

i set y~(t) = tAe^t

but im very clueless

[hollow trail]

and were you able to find an A for that to work?

[young anchor]

ive made it to

e^(t) (-9tA -2t) = 0

[golden owl]

Do you know how to solve linear differential equations?

[young anchor]

im learning them so idk

wouldnt really say i understand what im doing

just trying to follow along instructions

[hollow trail]

so then, what value of A makes that true?

[young anchor]

oh wait

if i write

te^t and then whats left in brackets is (-9A-2)

ah

and then 2/9 i get for A?

-2/9

[hollow trail]

yes

[young anchor]

hm

but what exactly did i find here?

[hollow trail]

if you go back to your original choice of particular solution, tAe^t, you found the value of A which makes that a particular solution to the ode

[young anchor]

ok then i guess i still dont fully understand, what is a particular solution?

when i was doing first orders it kinda made sense

[hollow trail]

a particular solution is one function which satisfies the ODE. so y = -2/9 te^t is one solution to the ode y'' - 2y' - 8y = 2te^t

[young anchor]

set the y' = 0 and to my understanding that particular solution is when y(t) converges to some value right?

[hollow trail]

that's an equilibrium solution

[young anchor]

but these ode's they have infinte solutions?

[hollow trail]

there are infinitely many solutions to the ODE, yes

[young anchor]

and ODEs themselves, "their goal" is to showcase how a function changes wrt some variable which is most commonly used for time?

[hollow trail]

the ODE relates the function and its derivatives (its rate of change), where the goal is to solve them (find a function which satisfies the equation)

usually either finding all possible solutions, or the solution satisfying some particular initial conditions

[young anchor]

with first orders, if y'(t) + a(t)y(t) = b

i can find the particular solution just by y(t) = b/a(t)

why cant i do the same thing here

with the problem

[hollow trail]

that strategy would only work if b is constant

[young anchor]

why

[hollow trail]

because what you are doing there is assuming that y = C where C is some constant, then solving for what C would be

which isn't really any different than what you did here, other than the fact that you assumed a different form for y to take

you can certainly try y = C for this equation, but you would end up with -8C = 2te^t and there is no constant C which would make that work

[young anchor]

okay i guess thats my problem, i just never fully comprehend what am i solving for. every time i see something like this -8C = 2te^t my first instncit is to just solve for C but what i have to solve these ODEs just feels like something completley new

like beging with assumptions that yh(t) will take the form e^rt

am i just plugging stuff into these equations untill they work?

[hollow trail]

you are assuming the solution is some particular form with some unknown constants, then solving for what constants would make that true. if there aren't constants which make it true then your guess was wrong to begin with

[young anchor]

am i correct in understanding that essentially im supposed to find some y(t) which just works for the field?

[hollow trail]

the field represents the differential equation, and the solutions are the curves traced out. so yes, finding the general solution would be finding the form of all of those curves, and finding a particular solution would be finding one such curve

[young anchor]

and theres, from what i see, various ways of finding such solutions?

[hollow trail]

yes, they depend heavily on what type of ode you're working with

[young anchor]

okay and final quesiton, i saw an example of say the homogoenous part of my solution is e^2t and my particular solution is e^2t. This is not possible, why?

i understand its something to do with linear independence, so if both homogenous and particular parts are the same yh + yp = y doesnt hold?

[hollow trail]

well if you were to plug in that solution to the left side of the equation, then either you would get 0 or you would get the right side (something not 0), but not both

[young anchor]

i dont follow

[hollow trail]

it's not possible for a particular solution to an ode and a solution to the corresponding homogeneous ode to be the same

for example with the ode
y'' - 2y' + 4y = e^(2t)
then a solution to the homogenous part would be a solution to the equation:
y'' - 2y' + 4y = 0
and a particular solution would be a solution to
y'' - 2y' + 4y = e^(2t)

if you were to take a function and plug it into the left side (which is the same for both), then you would come out with some function, which is either 0 or it isn't

so it can't possibly satisfy both equations simultaneously

[young anchor]

aha

okey

i see

makes sense

thank you for the help

also final final

wolfram doesnt make mistakes when solving these right?

i can rely on its answers to check myself?

[hollow trail]

you can take wolfram alpha as a reliable ode solver

[young anchor]

great, thank you very much sir

.close

[cedar kiln]

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[sturdy ermine]

Why am I wrong

[sturdy ermine]

answer^

i used the partial fraction

which is defined as:

[coral jewel]

?

[sturdy ermine]

ive done this problem 3 times now

yea i meant to do u=x+1 du= dx

i ended up not changing the bounds and pluggin back x+1 for u

so.5ln(x+1) | 0 to 1

[coral jewel]

why are you not changing the bound

[sturdy ermine]

becuase u dont need to, u can plug U back in after u solve for the integral

and keep the original bounds

[coral jewel]

i mean sure

[sturdy ermine]

my prof taught both

[cerulean sail]

(As a comment though, if you do that, it’s worth labelling the bounds as being with respect to x and not u)

[sturdy ermine]

i usually "forget the bounds for now"

and when i have 1/2 * ln|u|

i plug it bavck in

lemme write it out

[cerulean sail]

My comment is more that for the bottom integral here you have, you could denote that as e.g. $\frac12 \int_{x = 0}^{x = 1} \frac1u \dd u$ (bearing in mind the integral is wrt $u$ rather than $x$)

[wraith dagger]

@cerulean sail

[sturdy ermine]

yes

but regardless its still gives u the same answer no?

it should still lead me 1/2*ln|2|

👀 bro writtig a whole essay

[cerulean sail]

It does get the same answer in the end if you do so correctly, though the comment is that it isn't really correct to write the bounds as per the picture I quoted, and that just labelling them as wrt x basically does enough to, at the very least, serve as a mental reminder that the bounds are for the original variable for both you and who's reading

[sturdy ermine]

Ill keep that in mind for the future

but again my question in coming here isnt the bounds its about

why my answer is so off

👀 radio silence

Ive on the question 3x and i get the same answer

each with a different approach to the problem

[cerulean sail]

I'm reading through it, for which, did you change the bounds for this one? (where does tan(1) come from?)

[sturdy ermine]

yes changing the bounds

i did x=tan(theta)

so x(0) = tan(0) = 0

and x(1) = tan(1) = ugly number

is that not how u change bound?

than?

[cerulean sail]

Not here it isn't you want the thetas that correspond to x = 0 and x = 1, so if x = tan(theta) then theta is arctan(x), from there you can find the corresponding thetas

[sturdy ermine]

wait its arctan(0) and arctan(1)

yea

not what x= we want theta is since its a d(theta) integral

i forgot

ok gimme a few minutes and let me recalulate these last steps

arctan1= ugly numbner so imma keep it arctan(1)

[cerulean sail]

pi/4 is ugly?

[sturdy ermine]

idk how to convert these things into terms of pi

and my prof said as long as i give like 5 decicmal places he doesnt care

so im not gonna bother learning the unit circle rn :D

i need to learn the rest of calc2 rn and

remember calc 1 which i took in 2021

and remember everything else that is needed

[cerulean sail]

Don't the calculators give you the exact value though, they really should if you're in radians

[sturdy ermine]

not in term so pi

it does in decimal place

[cerulean sail]

Sad

[sturdy ermine]

ig the ti-84 plus

isnt so plus afterall

[cerulean sail]

Agreed

[sturdy ermine]

asnwer still no match :(

[sturdy ermine]

it still doesnt match the prof answer

[cerulean sail]

Where's that sec^2(theta) coming from

[sturdy ermine]

1/2 * integral from 0 to 2 of -x/x^2 +1 dx

x=tan(theta)

to get -tan/sex^2 sex^2(theta) d(theta)

x= tan dx = sec^2 d(theta)

[sturdy ermine]

why no?

[cerulean sail]

Because e.g. choosing u = x^2 + 1 would be a better choice

[sturdy ermine]

but isnt x=tan(theta) also mathematical valid>

if not

why

[cerulean sail]

I mean, "valid" sure, but whether it actually gets you anything useful that you can work with...

[sturdy ermine]

it does tho doesnt?

the integral of tan(theta) = sec^2

[cerulean sail]

That's supposed to, spoiler, get you ||a -ln(2)/4|| which is where the solution gets theirs from...

[sturdy ermine]

but im tryna think why my method, even tho its accurate math

it's "wrong"

simply cuz

no reason ¯_(ツ)_/¯

simply cuz "haha u sub better" .. and no im not mockin u , im mocking math cuz where does "oh math is logical go"

[cerulean sail]

Also you differentiated rather than integrated here, integral of tan(theta) is actually -ln|cos(theta)| (or equivalently, ln|sec(theta)|, if that's easier to work with)

[sturdy ermine]

thats stupd

but thanks tho

i need to get 80 on this finak otherwise i fail this class

[cerulean sail]

Awwww, best of luck hopefully you do manage it!

[sturdy ermine]

then i cant take 1/2 my classes winter semester

im already at the point where i changed majors as a junior, this wouldve been my senior yr

half my friends already graduated and have been married

other half graudate this fall

im still 4 yrs away from

graduation

IM STIL NOT GETTING AN ANSWER

[cerulean sail]

If you choose this one, you hopefully should get it "easily"

[cerulean sail]

If you e.g. notice that ln|sec(theta)| is ln|sec^2(theta)|/2 = ln|tan^2(theta) + 1|/2, then that should hopefully get you there too

[sturdy ermine]

🙂 🔫 i was in degree mode

Thanks

im sure ill be back

my exam was monday but i got it extended till thurs day

i think my prof is pittyig me since i got below 40s on all 3 exams

23/100 for one of them..22 of those points from a question the professor accidentally made an impossible question

otherwise id get 1/100 cuz i left that question blanks anyways

anyways imma close this for now 😭

.close

[cedar kiln]

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[cedar kiln]

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[dreamy thicket]

I do not understand where the x+20 came from.

“Than” literally indicates that we have to subtract not add

[vestal bluff]

well

the shorter side is defined as x

so the longer side will be x+20

it depends on what side youre defining as x

[dreamy thicket]

Alright

,close

.close

[cedar kiln]

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[cedar kiln]

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

Where do i start?

[dusk goblet]

what’s this for

[sturdy ermine]

Practice exam

for final on thurs

[dusk goblet]

prove

[sturdy ermine]

and i have answers

but i wanna know how to do

[dusk goblet]

you compute the integral

[void sand]

,, \int_{0}^{\infty} (t + 1)e^{-wt}dt

[wraith dagger]

higher!

[void sand]

just gotta compute this c:

[dusk goblet]

you can distribute then do IBP

[gritty viper]

Why distribute?

[honest rune]

No need to distribute, u can use IBP directly

[dusk goblet]

you don’t have to

but it’s just what i’d do

[honest rune]

yea, it's oka, it's just one more step, it's not a very complicated step either

[cedar kiln]

@sturdy ermine Has your question been resolved?

[cedar kiln]

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

does anyone know the formular for Riemann Hypothesis

[silver ibex]

The Riemann zeta function?

Wdym

[surreal frost]

idk how to do it its hard..

[silver ibex]

What are you trying to do

Prove the Riemann hypothesis 💀💀💀

[mighty shuttle]

it's still an open problem

[surreal frost]

𝜁
(
𝑠
)

∑
𝑛

1
∞
1
𝑛
𝑠

[silver ibex]

Good luck bro🙏

[surreal frost]

Bruh

..

I CAN'T COPY THE

WORKIN OUT

so does anyone know?

[honest rune]

are u going to prove the Riemann hypothesis?

[silver ibex]

Brother the Riemann hypothesis is still unsolved

[surreal frost]

Ik

so solve it

[silver ibex]

…

[honest rune]

he is cooking

[silver ibex]

Wait

HOLD UP

[surreal frost]

if yall don't know, ask people for help

[silver ibex]

I’ve discovered a truly wonderful proof of the hypothesis

I don’t have enough space to write it down tho

Sorry

[surreal frost]

soo

[honest rune]

Fermat syndrome

[surreal frost]

..

so yall don't know?

[silver ibex]

😱🙊

[honest rune]

out of joke, what do you need?

[surreal frost]

Can't yall just edit math?

And make it so 0 can be divided

because technically

if you times a number by 0 it will become 0

so if you do that

then you could maybe increase in division

[silver ibex]

Brother

[versed jasper]

💀

[surreal frost]

but then u can

increase

[silver ibex]

That’s not possible

[surreal frost]

cuz of negative numbers

[silver ibex]

If 0/0 = x

That implies 0 = x multiplied by 0

[surreal frost]

yeah

[silver ibex]

But x can be any value because any number times zero is zero

So x has no definite value therefore 0/0 is undefined

[surreal frost]

I am a primary school kid..

[versed jasper]

._.

[silver ibex]

So why you trying to prove the Riemann hypothesis 😭

[surreal frost]

uh cuz I like physics?

[versed jasper]

he's every asian parent's ideal kid

[surreal frost]

Oh I have asian

parents

[versed jasper]

makes so much sense

[surreal frost]

so

Uh

Why don't yall just define something dividing by 0

[versed jasper]

i mean, there exists a branch of math that does allow it

,w Wheel Algebra

[wraith dagger]

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

Requested by shriihan

[versed jasper]

bru

[surreal frost]

what

Still

why can't you define

it

in base division

like 2/0

[versed jasper]

what do you want to achieve defining division by zero

[honest rune]

what happened

[surreal frost]

uh

Idk I just want to define something for fun

[honest rune]

so what did you want to force?

[surreal frost]

ah then I need help for quadratic formular then

So uh how do I work out

uh

[versed jasper]

this is what i meant

[surreal frost]

bro

Ok what about physics

physics is technically math

[versed jasper]

this is a math srever but ok

[surreal frost]

Theory of Everything.

give me the answer

[versed jasper]

of what

[surreal frost]

This is the quest to find a single, all-encompassing framework that unifies all fundamental forces of nature: gravity, electromagnetism, the strong nuclear force, and the weak nuclear force. Currently, general relativity describes gravity, while quantum mechanics describes the other three forces.

[versed jasper]

ok

but what's the question

[surreal frost]

so

something that has everything of forces of nature

that

Is compressed ALL

in one thing

[versed jasper]

it is supposed to

[surreal frost]

But its unsolved..

[silver ibex]

Kind of

The standard model lagrangian in nuclear physics is pretty good tho

[surreal frost]

actually

[silver ibex]

Here it is

[surreal frost]

I already know that

[silver ibex]

It has a few issues tho

[honest rune]

If you want to divide by 0 you must say goodbye to the field axioms

[silver ibex]

This one is outdated so it assumes neutrinos have no mass

[surreal frost]

[ \mathcal{L} = -\frac{1}{4} F_{\mu\nu} F^{\mu\nu} + \bar{\psi} (i \gamma^\mu D_\mu - m) \psi + \frac{1}{2} (D_\mu \phi)^\dagger (D^\mu \phi) - \frac{\lambda}{4} (\phi^\dagger \phi - v2)2 ]

[wraith dagger]

Force

[silver ibex]

And also it doesn’t incorporate gravity so 😛

[versed jasper]

^

[surreal frost]

I kinda simplified it

[versed jasper]

it doesnt have gravity, it has weird parameters that aren't real

[surreal frost]

so

uh

[versed jasper]

some of the coupling constants seem to come out of nowhere

[silver ibex]

Brother that’s not even the same thing 😭😭😭

[surreal frost]

Oh

I thought..

[versed jasper]

and you have "Ghost particles"

[honest rune]

derive it and equal to 0

[surreal frost]

So uh

[honest rune]

😢

[surreal frost]

GUYS

[versed jasper]

force

[surreal frost]

If u Drop a apple, what forces does it contain

[versed jasper]

depends

is it a gravity-free space, can we neglect air drag

[surreal frost]

no like when u have apple in ur hand and open ur hand

[versed jasper]

you have normal reaction along with the other two things

[surreal frost]

.

wrong.

the answer is air resistance, gravity, normal force, rotational force and orbital force, rotational force is the earth and apple rotating. and Earth orbiting

[versed jasper]

that is why i said "depends"

[surreal frost]

Dawg..

[versed jasper]

and you also said you have it on your hand and you open it

having contact with your hand implies there is a normal reaction

[surreal frost]

Am I the only one in the class maybe THE WHOLE WORLD Who understands the consept of gravity

bro

First

when u drop

[versed jasper]

why do i assume gravity exists

[surreal frost]

The apple does sping kinda

[versed jasper]

i could do the experiment in deep space

[surreal frost]

spin*

oh

in space there is still gravity

Fun fact

[versed jasper]

and there is no force named "orbital force"

[surreal frost]

Anything that has mass has gravity

[versed jasper]

it doesn't have gravity

it attracts other masses