help-0

[gritty pine]

i thought that i could just sqrt(9*2 x^6 y^10)

[alpine sable]

no because 2 is not a perfect square ?

[gray isle]

the end goal is to simplify the radical

[gritty pine]

may im not understanding this because idk what is a perfect square

[alpine sable]

eventually you would write it as (3x^3y^5)²(2y)

[gray isle]

how would you simplify something like
$$\sqrt{8}$$

[ocean seal]

ℝamonov

[gritty pine]

8/2

2^2

[gray isle]

not quite sure what you mean by that

[gritty pine]

4*2

sry

[gray isle]

well 8 = 4*2

so what would sqrt(8) simplify to

[gritty pine]

sqrt(4*2)

then it should became 4sqrt(2)?

[alpine sable]

he wants more than that

[gray isle]

no

sqrt(4*2) isn't 4sqrt(2)

[gritty pine]

sqrt(4sqrt(2))

i just remembered now that

[gray isle]

$\sqrt{ab} = \sqrt{a}\sqrt{b}$
(for $a,b \geq 0$)

[ocean seal]

ℝamonov

[gritty pine]

idk what is a perfect square or perfect cube

[alpine sable]

perfect square is a number, when you take root of it you get an integer

[gritty pine]

2 in this case? as 9*2?

[alpine sable]

no 9

[gritty pine]

2

[alpine sable]

because sqrt(9) = 3

[gritty pine]

when u take the 2

[alpine sable]

sqrt(2) = 1,3.....

[gritty pine]

oh ur right

[alpine sable]

not an integer

[gritty pine]

lol

[alpine sable]

hahahahahahha

1.4

1,4.......

point is, you dont get an integer

[gritty pine]

i have Dyscalculia so i have a bad process when i have to make calculations like that

i don't figure it out easily division/subtraction

[alpine sable]

cool

but you get the answer ?

[gritty pine]

i just took that 2 as garbage and gave to 2y because 9 is a perfect square?

[alpine sable]

yes

[gritty pine]

then i understood that thanks.

[alpine sable]

np

[gritty pine]

so i have just to give to another variable when there is not a perfect square

[alpine sable]

end result would be , i'll try latex : $3x^3y^5sqrt{2y}$

[ocean seal]

Mona 2.0

[alpine sable]

$3x^3y^5\sqrt{2y}$

[ocean seal]

Mona 2.0

[gritty pine]

it is

[alpine sable]

yes

[gritty pine]

thanks

[alpine sable]

np

[sonic kelp]

Hi! If $\bar{X}=\frac{\sum_{i=1}^{n} X_{i}}{n}$ where $X_i$ are iid with $Poisson(\lambda)$ distribution, what distribution $\bar{X}$ has? im confused because the limit central theorem says that normal distribution, but i know that the sum of poisson is a poisson then i dont know

[ocean seal]

alef0

[alpine sable]

How can I solve this operation? | -3 × 8 |

[sonic kelp]

this channel is occupied D:

<@&286206848099549185>

[loud iris]

how did the log k arrive ????

when i have only 1/x

[vale wigeon]

it's a constant

[loud iris]

like the c ?

[vale wigeon]

yes

which they're writing in a somewhat weird way here because doing it this way will be somewhat handy later

[loud iris]

but how did they convert the c to log k?

[vale wigeon]

any number can be written as the logarithm of something if we so choose.

after all, $c = \log(e^c)$.

[ocean seal]

Ann

[loud iris]

so its just an assumption ?

[alpine sable]

no log should be ln

[vale wigeon]

if you're wondering why they went with log(k) instead of any other function whose range is the whole number line,

i believe that it'll be made clear in the next few lines

[loud iris]

no no im not wondering that

[loud iris]

is it ?

[vale wigeon]

you could say it is, if you really want it to be.

@alpine sable some people, myself included, use log without a base for the natural logarithm, and log_10 when decimal log is required.

[loud iris]

oh ok

[alpine sable]

Okay

[loud iris]

thanks

:)

[pseudo barn]

Hey I had a question regarding abstract alg

[arctic wren]

what i'm doing wron ?
$e^{e^{x}}$ the derivative of this function according to wolfram alpha is $e^{x+e^{x}}$
But i'm getting $e^{x²+e^{x}}$

[ocean seal]

Guilhotina

[arctic wren]

my steps is

[pseudo barn]

Say we have a linear vector space of square summable sequences which does not have an orthonormal basis set. If we do Gram Schmidt orthonormalisation does it still stay an l2

Please help

[vale wigeon]

ah the curse of #help-0...

don't make me tap the sign.

y'all should go move to other questions channels that aren't as busy as this one.

all of you, to make things fair.

[pseudo barn]

Fine

[arctic wren]

ok

[alpine sable]

lol hahahaha

All of you get out of here

[loud iris]

what happened to the 1/2

in this case

they have assumed y=vx

[vale wigeon]

they multiplied everything by 2 but i think they forgot to do that on the right hand side fsr

it should have been log(k/x^2) at least

or log(k^2/x^2) but k vs k^2 doesnt matter as much since k is an arbitrary constant and can absorb some algebraic things done to it

so it's a mistake on the part of whoever wrote this

[loud iris]

oo

[loud iris]

also can i simplify the lhs more ?

i need to match the ans w tan y/x

[flint python]

a cheetah can run up to 60 miles per hour, how many inches per second is this?

Thats the Q

[alpine sable]

Why does factoring solve quadratics?

[flint python]

<@&286206848099549185>

[alpine sable]

It solves every polynomial, not just quadratics, because it exposes the roots

[alpine sable]

Figure out how many inches per second 1 mph is and go from there

For that, figure out how many inches per hour 1 mph is

[ionic jewel]

"solves"

[alpine sable]

Yeah alright fair, that premise was introduced by the asker not me

[icy trail]

why did bunny say "solves"? is it because the factor is (x-a) but the solution is actually x=a ?

[alpine sable]

Yeah it’s more a method for finding the roots

[ionic jewel]

"solving" a polynomial makes no sense, although it would solve P(x) = 0, which is equivalent to finding the roots

like if an assignment said "Solve 4x^2" it would be nonsensical

[alpine sable]

why does finding the roots work?

[ionic jewel]

if a polynomial has a root c, at x = c, then P(c) = 0

basically it's a number you can plug into the polynomial to get 0

[alpine sable]

I'm just not registering this

is it possible to explain quadratics/factoring equations without prior knowledge of formulas/rules/procedures

like a pure intuition explanation

nothing I am finding online is making it click in my head

because I slacked off in math class all throughout school

[ionic jewel]

a root of a polynomial is where the polynomial touches or crosses the x axis

does that make sense?

[alpine sable]

like the center?

[ionic jewel]

presumably you know what a polynomial is and what a graph of one could look like?

[alpine sable]

yeah

[ionic jewel]

lets try x^2 + 3x + 2

,w plot x^2+3x+2

[ocean seal]

Results provided by WolframAlpha

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

Requested by 兔子#6562

[ionic jewel]

do you see that it crosses the x axis twice?

[alpine sable]

yeah

[ionic jewel]

where?

[alpine sable]

-2 and -1

[ionic jewel]

yes exactly

so -2 and -1 are the roots of x^2+3x+2

does that make sense?

[alpine sable]

yeah

[ionic jewel]

okay so now the problem is, what if you don't have the graph?

for example, you couldn't tell me the roots of y= x^2-3x, could you?

[alpine sable]

yeah no idea

[ionic jewel]

so that's where "factoring" comes in

theres a variety of techniques, but they always split the polynomial into something that looks like (x-a)(x-b)..., where a and b are numbers

for example, the one we did before

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

following?

[alpine sable]

yeah I know that's the form

how does that get the roots?

[ionic jewel]

so you see how it's (x-a) and (x-b)

well the roots are just a and b

[wary wolf]

I am not shure how to do this, someone please help

[ionic jewel]

so notice it's also (x-(-1))*(x-(-2))

so the roots are -1 and -2

more questions?

[slender forge]

Still need help?

If so think about the equation for finding the average of 5 solve for what the total is then think of the formula for average in terms of 6 games and solve

[mossy siren]

Determine the zeros of the function h(x) = x² - 8x + 12

can someone help me?

[slender forge]

Any ideas on how to proceed?

[mossy siren]

no, i did it wrong and then saw that i had to determine the zeros, i don't know how to do

[slender forge]

Do u know how to factor?

[mossy siren]

yes

[slender forge]

U can just factor it

[mossy siren]

ok wait a minute

i did it

(x - 6) (x - 2)

[slender forge]

Yep good job so what would our zeros be

[mossy siren]

what do you mean?

[slender forge]

Ok so u have (x-6)(x-2) right?

[mossy siren]

right

[slender forge]

Let (x-6)(x-2)=0

What x(s) would make this statement true

Hint: 0 multiplied by anything is 0

[mossy siren]

the first one?

[slender forge]

U want the value of x that could make that statement true

Let’s do it like this:

What value of x makes this statement true

x-6=0

And what value of x makes this one true

x-2=0

They are simple don’t overthink them

What minus 6 equals 0

[mossy siren]

x = 0

lol

[slender forge]

Nope

6-6=0

X=6

[mossy siren]

fr?

[slender forge]

What’s the other one

[mossy siren]

2-2 = 0?

x = 2

[slender forge]

Those are ur two zeroes

U can plug them into the original function and get 0

Do u see why this works?

It’s better to understand than get the answer

@mossy siren

[mossy siren]

yes

i thought it was hard

thank u so much

[slender forge]

U got the hard part on ur own

(Factoring)

Now if u can’t factor u can just use the quadratic formula

[wary wolf]

Thanks

[slender forge]

Did u get it?

[mossy siren]

determine the domain of the function:
f(x) = 1
-------
x^2 - 1

x < -1

[remote heron]

how did you arrive at the answer you did?

what condition did you check?

note just for clarity you have $f(x) = \frac{1}{(x+1)(x-1)}$

[ocean seal]

jan Niku

[remote heron]

Sd7 is right though 😄 you need to figure out what you're not allowed to plug in here

division specifically has a lot of problems we have to watch out for

[slender forge]

Those and square roots

[remote heron]

logs

[mossy siren]

-1 < x < 1

i guess

[remote heron]

if youre into those

so youre identifying -1 and 1 for what reason?

[mossy siren]

i thought that was the purpose of the exercise

[remote heron]

it is yea, your answer is still not correct but i think youre really close

are you able to identify the problem values for x?

what are they? can you list them all? why are they a problem?

is 1 a problem value? what happens if you plug it in for x

[mossy siren]

x ≠ -1

x ≠ 1?

[slender forge]

Yep

Do u see why?

[mossy siren]

yes

[slender forge]

Do u use interval notation in ur class? Or do u just write it like that?

[mossy siren]

i just write it like that

thanks a lot

[slender forge]

Np

[flint python]

2-6x+9=4x-1+2x solve for x

[kindred hull]

@flint python start by simplifying

Then isolate all the x terms to 1 side

[flint python]

11-6x=6x-1

[kindred hull]

Good

Now get all the terms in x to one side of the equals sign

[flint python]

ok

so

-6x+6x=11-1

[kindred hull]

Are u sure u wrote the question right

[flint python]

yeah

[kindred hull]

U can’t find x from that I’m pretty sure cos u end up with 0

[flint python]

yeah

i did the same steps

and go 0

i came here cause i thought i did something wrong

[glass field]

you typed 11-6x=6x-1 and that has a solution

looks like you made an error when putting the x's on the same side

[kindred hull]

Oh yeah

U didn’t change the sign

Ur subtracting 6x from both sides not just moving one term over

@flint python so rewrite it

Remember to change the sign on the 11

Too

[flint python]

wait where did we get the 6 from

[kindred hull]

From adding 4x and 2x

[flint python]

oh

so when we simplify

[kindred hull]

Ye

[flint python]

ok

[kindred hull]

11-6x=6x-1 is right so far

[flint python]

so do we do

[kindred hull]

U wanna get all the terms with x on one side

[flint python]

11-6x+6x=6x+6x-1

?

[kindred hull]

Yes now simplify

That

[flint python]

11=12x-1

[kindred hull]

Ok now you can do something to both sides to get rid of the -1

[flint python]

right

12x-1+1=11+1

[kindred hull]

Yes

[flint python]

12x=12

x=1

[kindred hull]

Correct

[flint python]

ohhhhh

ok

ty

[kindred hull]

@flint python it’s good to check the answer by substituting the value of x into the original equation

And checking if the equation still holds true

[obsidian marsh]

what does rot mean?

i know S is a surface and F is a vector field

[sleek elbow]

https://letmegooglethat.com/?q=what+does+rot+mean+math+vector

this probably

@obsidian marsh

[obsidian marsh]

so it's the same as curl?

[sleek elbow]

seems to be

[obsidian marsh]

ookie

thanks

[sleek elbow]

np

[celest anvil]

using u substitution i got u=sin(x) and u^24-2u^12

du=cos(x)dx

[obsidian marsh]

have you tried doing sin(x)^2=1-cos(x)?

[celest anvil]

no

i'll try that then thx

[obsidian marsh]

i'd give it a go

[alpine sable]

cos(x) ^ 2

Right?

[obsidian marsh]

yep, sorry

$sin^2(x)=1-cos^2(x)$

[ocean seal]

bortoletto

[celest anvil]

ok

[obsidian marsh]

i'm trying to do a surface integral

and my surface is this

$S=x^2+y^2+z^2=4, \ z \leq 0$

[ocean seal]

bortoletto

[obsidian marsh]

i need to convert this to the parametric form in order to calculate the integral

i got that part

but how will i set the limits of integration?

my surface is obviously a sphere

and since Z \leq 0, i want it's bottom half

[obsidian marsh]

this is my question

the surface S, parametrized is:

$r(\theta, \phi)=\sqrt(4)sin(\phi)cos(\theta)i+\sqrt(4)sin(\phi)sin(\theta)j+\sqrt(4)cos(\phi)k$

[ocean seal]

bortoletto

[obsidian marsh]

what should i do now

does the k part must be between 0 and -2?

so i make $\frac{\pi}{2} < \phi < \pi$?

[ocean seal]

bortoletto

[obsidian marsh]

and $0<\theta<2\pi$???

[ocean seal]

bortoletto

[obsidian marsh]

i just want the bottom half of the sphere....

[slate cargo]

r = 2 and does not move
theta goes from 0 to 2pi ok

And phi from pi/2 to pi also agree with you

[obsidian marsh]

yep r=2

@slate cargo if i do phi from pi/2 to pi wont i only get the left half of the sphere?

dunno

i just thought about it

and thought the bottom half might be phi from pi to 2pi

[slate cargo]

No with your parametrization of the sphere you have the bottom

[obsidian marsh]

i was thinking of the unit circle

[obsidian marsh]

oh ok

but i have this constraint:

the surface is oriented downwards

how should i proceed?

[slate cargo]

I dont see what are the changes

[obsidian marsh]

uuh

what is the difference if i do this instead:

$\pi<\phi<\frac{3\pi}{2}$

[ocean seal]

bortoletto

[obsidian marsh]

i still have the bottom half, right?

or do i have something completely different?

[slate cargo]

In fact phi can only take values between 0 and pi

[obsidian marsh]

what?

you meant phi?

oooh

i get it

so pi/2 to pi it is

[ionic jewel]

yes

[obsidian marsh]

but the thing that bugs me is the downwards oriented part

whats the difference if the bottom half of the sphere is 'oriented upwards' or 'downwards'

i was actually think of a normal unit vector along the surface

[slate cargo]

Upwards would be" towards inside the sphere"

[obsidian marsh]

yea

but how do i include this is my integral?

[slate cargo]

It is not common to do this

[obsidian marsh]

multiply by -1?

in this case i want it towards the outside of the sphere

[slate cargo]

No I think you have to compute a flow

In fact the vector "er" is the good normal vector to the sphere

[obsidian marsh]

er?

[slate cargo]

This

When you write your integral you have something like

$\int_{\phi =\pi/2...\pi}\int_{\theta = 0...2\pi} f(\theta,\phi) * Rsin(\phi)d\theta d\phi$

[ocean seal]

Boomer

[obsidian marsh]

yeah

i got that part, i think

i just did the integral without accounting the orientation of the surface

and i got 24pi

actually

-24pi

[slate cargo]

It is included in Rsin(phi)

Otherwise it would be -Rsin(phi)

[obsidian marsh]

how come?

[obsidian marsh]

this means it's oriented upwards?

[slate cargo]

Rsin(phi) =>> oriented towards outside

For your half sphere : outside = backwards

The surface element is |dS|.n with n towards outside

Here dS = R.sin(phi)dphidtheta

[obsidian marsh]

ohhhhh

i get it

thank you!

[modest gulch]

here it mentions "equate the coefficients of x^2"
i dont understand how they 1 = A+C

its a partial fractions calculus question

😦

<@&286206848099549185> if anyone can help

[slate cargo]

In fact I would multiply both parts by x and make x tend to infinity

You would have at the limit 1 = A+C

[modest gulch]

uhh

im sorry but thats not much help to me xD im a bit confused

what do you mean by both parts

it mentions equating coefficients of x^2

only coefficient of x^2 i know is just 1

[slate cargo]

The original fraction

Multiply it by x

And look for the limit at +oo

What is the limit?

[modest gulch]

theres many fractions

can you be more specific

[slate cargo]

The original (x^2+x+1)/((x+1)^2(x+2))

[modest gulch]

ok yes

but it says the coefficients of x^2

so what is that supposed to be

because the coefficients of x^2 is just 1

i dont think limits are used at all in this equation

[slate cargo]

Its another method and it gives quickly the same equation

In x^2+x+1=A(x+1)(x+2)+B(x+1)+C(x+1)^2

If you take the coefficient of x^2

[modest gulch]

what is this coefficient of x^2 you keep talking about

its literally 1

what am i supposed to do with 1

[slate cargo]

In the left side : 1
In the right side A+C

[modest gulch]

how is it A+C?

what

[slate cargo]

Expand the expression

[modest gulch]

i did

its not A+C at all

[slate cargo]

Your result is wrong then

[modest gulch]

your telling me this equates to A+C?

makes no sense

both Ax and Cx exists

[slate cargo]

No I tell you to expand and identify the coefficient before x^2

[modest gulch]

i have no idea what you mean

[slate cargo]

Method by "identification"

[modest gulch]

you are saying ti expand 1?

how am i supposed to expand 1

[slate cargo]

If ax^2+bx+c = dx^2 + ex + f for any value x then a=d b=e and c=f

[modest gulch]

what

isnt the coefficient 1?

i didnt use e and f in here

or d

so im not sure what youre trying to say sorry

[slate cargo]

A(x+1)(x+2) + B(x+2) + C(x+1)^2 = ...

[modest gulch]

where do you get that from

[slate cargo]

Sorry

[modest gulch]

oh okay

its all g

so this right here

[slate cargo]

Expand the expression!!!!!

[modest gulch]

i did do that

[slate cargo]

(x+1)^2=x^2+2x+1

[modest gulch]

i just told you

but i get 2A 2C Ax Cx

its why its making no sense

[slate cargo]

Put these parts apart you dont need them

You just need to compare x^2 monoms

[modest gulch]

but how do i know which one is the x^2 ones

[slate cargo]

Because you have an x^2 before 😆

[glass lichen]

they... have x^2 in them...

[modest gulch]

there is none with an x^2

[slate cargo]

Then your result is wrong

[modest gulch]

its why im asking for help

[glass lichen]

$x^2+x+1=A(x^2+3x+2)+B(x+2)+C(x^2+2x+1) \ x^2 = (A+C)x^2 \ 1=A+C$

[ocean seal]

Mosh

[modest gulch]

so in this example

i would use only the x^2?

[glass lichen]

no

[modest gulch]

nvm

its the only par tim stuck on

[glass lichen]

you'd use all the terms..

[modest gulch]

its not making any sense to me

[glass lichen]

and get your 3 eqn's in A, B, and C

[modest gulch]

yeah but when you expand them they cant just cancel eachother out

[glass lichen]

$1=A+C \ 1=3A+B+2C \ 1=2A+2B+C$

[slate cargo]

Its an important property of polynomials

[ocean seal]

Mosh

[glass lichen]

from x^2, x and constant respectively

[modest gulch]

where did you get that from

[glass lichen]

read.

[modest gulch]

im talking about the texit

[glass lichen]

yeah

it's the same answer...

[modest gulch]

but why is A and C in the x

they have x^2 not an x

[glass lichen]

.....

do you know how distribution works?

[modest gulch]

i mean obviously not if im struggling with it

[glass lichen]

distribute $3(x+2)$

[ocean seal]

Mosh

[modest gulch]

do your saying this term is used

for x^2

and x

[glass lichen]

No

that was just the x^2 terms

hence the middle line of that

[modest gulch]

i just dont understand how you just remove the insides of the equations

like where does the x^2+3x+2 go

is it some kind of rule?

[glass lichen]

$$x^2+x+1=A(x^2+3x+2)+B(x+2)+C(x^2+2x+1)$$ By comparing the $x^2$ terms, cause I need to say it explicitly... $$1x^2=(A+C)x^2$$

[ocean seal]

Mosh

[modest gulch]

then how am i to do it for the x terms

[glass lichen]

same way I did for x^2..........

just, with the linear terms

[modest gulch]

but they all have an x in it

[glass lichen]

so??

[modest gulch]

it would just be x = B then

how does that even tell me anything

i dont know what B is

[glass lichen]

$x^2+x+1=(A+C)x^2+(3A+B+2C)x+(2A+2B+C)$

[ocean seal]

Mosh

[glass lichen]

since you need the algebra done for you...

[modest gulch]

look if you don't want to help

thats fine

but you're overwhelming me with being passive agressive

im already struggling

i dont need someone putting me down for trying to get some help

it may be obvious to you, but its not for me

if someone wants to help

im trying to understand the process of equaling the coefficients together

[icy trail]

are you comfortable with how we got to this point?

[modest gulch]

im gonna try it again to see if i get it

but form my understanding you take the A+C as the "coefficient"

so like

A+C=1

but the form there is only gotten after expanding and organizing it like its a quadratic equatio right?

or i guess like the numerator

cause in other questions the numerator is a number and not this equation

[icy trail]

ohh i see

but then wouldn't it be
0x^2 + 0x + number

[modest gulch]

exactly what i was thinking

here let me grab the example i was just working on

this is the next questions solution

where the numerator is 26

omg!

wait

that explains why A+B = 0

[icy trail]

nice!

[modest gulch]

thank you

how do i give rep again

[icy trail]

ngl i can do partial fractions but i can't integrate them so I'm glad you got it before that

[modest gulch]

the integration part is only thing i can do

[stable leaf]

isn't the integration just a u sub?

[modest gulch]

yeah for some parts

[stable leaf]

and possibly factoring?

[modest gulch]

or you use the 1/atan-1(x/a)

like how it does here

[icy trail]

i do not know u sub yet

[modest gulch]

oo

Its very helpful

[stable leaf]

wait

[modest gulch]

will make a huge difference

[icy trail]

I've seen it in the wild

[stable leaf]

you learned partial fractions

[icy trail]

yes

[stable leaf]

unless that has another application

[modest gulch]

it does

[stable leaf]

outside of this

can't remember what though

i mean i see what it does

[icy trail]

uhh i think it's so we know how to do them when it comes to integrating

[modest gulch]

thats cool that they teach you early

the integration part is a piece of cake if you know the first

[stable leaf]

i don't really like trig subs though :/

espeically when you have to memorize all of them

[modest gulch]

im very hit and miss with them

[icy trail]

you can't work them out on the go?

[modest gulch]

its why im taking this class now because the lack of memorizing due to it being online lol

depends on what clases you take, some dont teach you the proofs

[icy trail]

this is what I'm relying on to remeber the like 5 trig identities i know atm

[stable leaf]

i mean i guess you kinda can with the unity circle if you know what they are in terms of trig like (sin, cos , tan....

[icy trail]

the unity circle is a cooler name

[stable leaf]

oops

[icy trail]

i think I'll start calling it that

[modest gulch]

i like that

thanks guys, ya'll cheerd me up

hope you have a great night

[stable leaf]

someone was way ahead of the game

[spiral kernel]

Quick question

How hard do you need to hit the ground to damage something high up in the air?

[remote heron]

damage?

what do you mean

by what mechanism

[spiral kernel]

Via vibrations i would assume

[ionic jewel]

very

matter of fact it might be more efficient to hit the other side of the earth to push it into your airborne object

[icy trail]

what like a shockwave from the ground back up to the air?

[spiral kernel]

Yes

[icy trail]

damn ok o.O

[remote heron]

very hard

[ionic jewel]

that's what I said

[remote heron]

i think the answer would depend on what the things are and how high it is and etc

heres a dumb question

well, nvm

i should study

[icy trail]

the only dumb question is the one you don't ask

[remote heron]

what causes the shockwave

is it a normal force

i mean if you were to calculate what is the principal source

[spiral kernel]

A normal force

[icy trail]

when i think of shockwave i think of some plane breaking the sound barrier

and theres a shockwave

[spiral kernel]

Like a large hammer hitting the ground

[icy trail]

like a sonic boom is that what they're called

[remote heron]

so if you calculated it you'd figure like

you hit a piece of metal with a hammer, some or all of that force gets redirected back into the air

since the metal doesnt move or maybe you figure it doesnt even heat up

[surreal meadow]

damage something high up in the air?
how high is it? how much damage are you referring to? what are we damaging?

all important specifications to answer the question

[ionic jewel]

oh no have we summoned the theoretical physicists

[icy trail]

what are we damaging >:)

[remote heron]

why are we damaging

i like how they prefaced it with "quick question"

[spiral kernel]

Around 100 feet up

And a creature about the size of a fighter jet

[remote heron]

a creature

[surreal meadow]

sounds like a fluid dynamics problem

[alpine sable]

Ask Randall Munroe and he might make an xkcd about it

[surreal meadow]

i'm assuming it'd be more efficient (and safer) to mess with the air the fighter jett will fly around than to try and create a shock wave that will reach the fighter jet and affect it

although its a fighter jett so any perturbation in the air probably won't do anything 🤷🏻‍♂️

[alpine sable]

If you're gonna hit the ground with some object just lob the object at the jet

[obsidian marsh]

if i have a vector field F = Pi + Qj + Rk

and the partial derivatives Px, Qy and Rz are continuous in R3

does that mean that curl(F)=0?

oh that doesnt mean anything

that's only true for the second partial derivatives

Let C be a smooth closed curve which lies in the plane $2x-3y+5z=17$, with positive orientation, and A is the area of the region bounded by C. What's the value of the integral $\int_C(3cos(x)+z)dx+(3x-e^y)dy-3ydz$?

[ocean seal]

bortoletto

[obsidian marsh]

how would i go about solving this?

i don't know how to begin

can i get some directions pls

[ionic jewel]

@icy trail where did you get that reaction lol

[icy trail]

does it not say?

[ionic jewel]

private server

[icy trail]

discord.py

[ionic jewel]

ah i see

[icy trail]

🏃 before i get banned for advertising

[elfin snow]

uhh

wtf

lol

[inner sequoia]

why is this true

[ionic jewel]

flipped, vertex at (-2,-2)

[inner sequoia]

whoops didnt notice the +2 lmao

[obsidian marsh]

pls help

[remote heron]

might have better luck in #multivariable-calculus once its free 👀

[astral dagger]

after calculating the curl, maybe you can find its norm

then the integral evaluates to a double integral by Green's theorem

[ocean seal]

SubGui

[mild berry]

Hello! I'm doing a project and would love to get some help. Currently, I'm trying to find the area of a rounded-edge square but I want it as a function so I could integrate and find the area from there. So far, the closest function I found is x^4+y^4=1 but there isn't a integral for it; any ideas?

[icy trail]

woah i was just thinking about this function! #help-2 message

[gaunt magnet]

i got 3

check if im correct

[undone dock]

yes

[tame finch]

Hi I'm currently a rising senior in hs right now and I have forgotten a decent chunk of the topics I have learned in calc 2 are there any resources online i can use to remember the content i was taught?

[carmine lion]

khan academy?

[slender forge]

There’s not a lot that transfers from calc 2 to calc 3 if that’s what worries u

[violet sphinx]

Can anyone tell me what's the name of this topic?

Like the chapter name or terms for it

[glass lichen]

Linearization I suppose, however that would most likely fall under curve fitting

it doesnt have a specific name that im aware of

[violet sphinx]

Because I can't search topics related to this in the Internet

So I thought maybe there's a name or something

[glass lichen]

I mean, power functions linearize via logs for example

$y=ax^b \ \ln(y)=\ln(a)+b\ln(x)$

[ocean seal]

Mosh

[violet sphinx]

There was a question with the same equation you wrote🤣

[glass lichen]

yeah

[violet sphinx]

I'll search it then

Thanks@glass lichen

[bitter sage]

can anyone help with this

[viscid topaz]

What issue are you having with it?

Can you actually calculate the tangent line with that description?

I know the question says words, but like, are you able to tell me the slope of that tangent line? I'm just trying to establish a baseline here

[loud iris]

someone ??

[loud iris]

..?

[bitter sage]

ya i calculated the slope of the tangent

then solve for x = 2

but they want 5 steps and thats only like 3

[gray isle]

show your work

[bitter sage]

u want me to send a photo?

[gray isle]

whatever way is most convenient to you

[bitter sage]

this is what i wrote

[gray isle]

as in you wrote the exact same output from symbolab?

[bitter sage]

it is

im saying i have this written on my laptop

i feel like theres not 5 steps when explaining this

[gray isle]

perhaps they intended for you to differentiate 5^x without using a shortcut identity

[bitter sage]

we have learned about this rule alr so idk

[gray isle]

i mean if its just a description, you can do it in two steps

[bitter sage]

ik

but my prof wants 5

so idk maybe ill just try to extend it as much as possible

into 5 sdteps

[gray isle]

maybe split certain things into multiple steps

e.g.
substitute in x=2

simplify

(as two steps)

[bitter sage]

ya i will

thanks for help!

[cedar hearth]

Hello guys, can I ask what is a XOR-sum? I have this question: XOR-sum of $13579BDF_{16}$ and $FEDCBA98_{16}$. But I don't even know what a XOR-sum is. Google and stackexchange were not understandable at all. I know that XOR is an operator, and the binary modulus of 0 and 1 is also the XOR of {0, 1}, but that's it.

[ocean seal]

Memiya

[surreal meadow]

i believe you're meant to put both of those in binary, and do the xor for each corresponding bit in the numbers

so for 6^3 (that is the xor sum of 6 and 3) you'd have 110^011
1^0 = 1
1^1 = 0
0^1 = 1
therefore 110^011 = 101

[cedar hearth]

How do I do that as well? I only understand 0 XOR 1 is (0+1) mod 2. For some reason, I can't seem to get XOR at all.

[surreal meadow]

oh wait i flipped the xor sums

[cedar hearth]

Wait but then how do you do, say, 011 and 1011?

[surreal meadow]

0011 and 1011

[cedar hearth]

But binaries don't need to have same number of bits right?

[surreal meadow]

no but its analogous to saying 01 = 1

or 02351 = 2351

[cedar hearth]

So if I have 011 and 1011, I add a 0 bit to the 4th bit position in 011?

[surreal meadow]

i'd assume so

i can't give a certain answer, sorry about that, but its the logical next-step

[cedar hearth]

Then I XOR them by matching position indices?

[surreal meadow]

yes, that seems to be the idea

[cedar hearth]

So XOR sum of a and b is (a+b) mod 2?

Isn't that just XOR?

[surreal meadow]

i don't think thats the case necessarily

[cedar hearth]

What's the difference between XOR and XOR-sum then?

[surreal meadow]

take the case of 10^01

10+01 = 11

11mod2 = 01

but 10^01
1^0 = 1
0^1 = 1

10^01 = 11

so xor sum is not necessarily (a+b) mod 2

[cedar hearth]

So the first is XOR and the second is XOR-sum?

[surreal meadow]

i'm not entirely sure where you got (a+b) mod 2

that's just what i did for the first part to show its not equal to the xor sum

[cedar hearth]

It's a really confusing bit in my module slides.

[surreal meadow]

(a+b) mod 2 is the xor operation for single bit binary numbers

[cedar hearth]

Wait so is XOR the same thing as XOR-sum?

Sorry, it's very confusing to me. I have no background in modular arithmetic but got thrown this without explanation.

[surreal meadow]

so to my understanding the XOR sum is the application of the XOR operation to every corresponding bit between 2 equally sized numbers

[cedar hearth]

Is there an example where XOR is not the same as XOR-sum? I can't visualise it in my head at all.

[surreal meadow]

do you have the slide/doc for the xor sum?

[cedar hearth]

There isn't any.

[surreal meadow]

then looking at that last line, it looks like they are the "same" thing

[cedar hearth]

It goes from: mod, mod congruent then immediately jumps to this.

Which is why it's incredibly confusing.

Or just plain old opaque as heck.

So XOR-sum is bitwise XOR for binary. So there is no step-by-step by-hand method for XORing hexadecimal?

I have to convert, XOR by bit, then convert back to hexa?

[surreal meadow]

there's surely a formula for hexadecimals

i just don't know what it is 😆

[cedar hearth]

I did find it, but it required a massive, complicated table that's memorised.

Was completely incomprehensible to me.

[stable leaf]

formula for hexadecimals?

[cedar hearth]

Or method.

[stable leaf]

like to convert from something to hex?

[cedar hearth]

No uh

XOR-sum of $13579BDF_{16}$ and $FEDCBA98_{16}$

This.

[ocean seal]

Memiya

[stable leaf]

not up to the task but what is all this adding bits or whatever for?

i have heard of things like bit manipulation and how you might shift bits but never heard of this

[cedar hearth]

It's for my tutorial. But I have no background in modular arithmetic. So I'm trying to figure out how to calc XOR or XOR-sum by hand.

It's one of those annoying as heck 'hidden' prerequisites in courses.

They don't tell you you need it, and they insist that it is taught in the course itself, but it's practically assumed you've already mastered it.

[stable leaf]

what aspect of modular arithmetic? isnt modular arithmetic about remainders ?

[cedar hearth]

Yeah, but

this slide came immediately after modulus.

Using modulus to do XOR for single bit binary.

[wary stream]

I'm pretty sure bitwise operations work best with binary numbers so you can convert these to binary then do XOR

[stable leaf]

but if that is in base 16

what is that in binary

[cedar hearth]

XOR each bit with corresponding bit in the same bit position index?

[stable leaf]

wouldn't that be pretty long

[wary stream]

32 bit number

[cedar hearth]

I found a method online, but completely alien to me. Incredibly complicated massive table required.

Stackexchange is no help at all in this case.

[wary stream]

As I said, XOR is best with binary numbers, as shown in that chart kn the screenshot you just sent, so convert the hexadecimal to binary

[cedar hearth]

Then XOR by position index?

Then what's the difference between XOR and XOR-sum?

[wary stream]

Honestly it wouldn't take long at all, take it one hexadecimal bit at a time, so from right to left you have $F_{16} = 1111_{2}$ and $8_{16} = 1000_{2}$, XOR the two and you get $0111_{2} = 7_{16}$

[ocean seal]

dldh06

[cedar hearth]

Yeah but what's the difference between XOR-sum and XOR?

[wary stream]

Don't recall

I just remember XOR because I'm CS

[cedar hearth]

This is doing XOR between bits position wise, so is that then XOR-sum? Doing XOR for each bit?

So for a decimal number, I should also XOR it by converting to binary then XOR each bit then back to decimal?

[wary stream]

I don't think you have to do XOR sum for this because XOR sum looks like the XOR a summation

https://stackoverflow.com/questions/17284337/what-is-an-xor-sum

That's what I found about XOR sum, and I think your goal is to do XOR operation on the two values

[wary stream]

Yes

[cedar hearth]

Ah yeah I saw this post as well, but I could not really understand the difference. I think with you guys' help I can understand that post better now.

[wary stream]

From that link, say you want the XOR sum from 1 to 4, you do 1^2, take that result, XOR it with 3, etc

That's how I understood it

[wary stream]

There is a huge difference

So another example is XOR sum from 4 to 8, you do 4^5, take that result XOR it with 6, then so on

[cedar hearth]

No doubt, just that I have no background so I didn't even know where to begin. It's hard to ask if you don't even have the basics, which I didn't even know I needed.

[wary stream]

Just use XOR with those two values

[wary stream]

These two

[cedar hearth]

I have calculus, linear alg and now doing probability and statistics, but didn't do modular arithmetic or rings and groups.

[wary stream]

Never had to use XOR sum ever, this was the first time I've heard of it

[cedar hearth]

Alright, lemme try that. I got $FD8B2147_{16}$.

[ocean seal]

Memiya

[wary stream]

Doesn't sound right

Left most bit was 1 and F, so respectively 0001 and 1111, so XOR of that is 1110 or E

[wary stream]

You have F

As the left most bit

[cedar hearth]

Oh it should be E.

Checked with an online calculator, is same except for the first F, which should be E. Ah I made a mistake in converting 1 to binary.

Should be 0001, not 0000

[wary stream]

Yeah, 0000 is 0

[cedar hearth]

Thanks for your help. I'll try to see if I can;t do more practice with this.

@surreal meadow @stable leaf And thanks for your help as well.

[bitter sage]

can anyone check if ive done this right?

[astral dagger]

P(t) = 3t exp(-t/3)?

[bitter sage]

because it leaves me with a straight line

is that right?

[astral dagger]

the problem is

why did you differentiated wrt to x?

[bitter sage]

idk?

[astral dagger]

you should differentiate wrt to t

[bitter sage]

whats wrt mean

[astral dagger]

with respect to

[bitter sage]

what does the exp mean?

[astral dagger]

exponential

another form of writing powers of e

instead of using expoents

[bitter sage]

so 3te^(-t/3)?

[astral dagger]

no, that's the function

I asked you if it was correct because it's not too easy to see

[bitter sage]

ik its hard to see

i thought it was e^-2/3

u think it would be t?

[astral dagger]

yes, I think

[bitter sage]

makes more sense

mb

[astral dagger]

try differentiating with d/dt this time

[bitter sage]

i got it now

ya i got 3e^-t/3 - e^-t/3t

[astral dagger]

good

[bitter sage]

ok thanks

[astral dagger]

[pseudo summit]

is 0.9999... EXACTLY equal to 1?

or an approximation?

[ionic jewel]

if it has infinite 9's its equal

,w 0.9999... = 1

[ocean seal]

Results provided by WolframAlpha

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

Requested by 兔子#6562

[ionic jewel]

@pseudo summit

[pseudo summit]

oh

so I comprehended it right

thanks

[gaunt magnet]

do u get 3 for this

[alpine sable]

what are the odds of 315 to 50

so 50/315 ig?

like what %

[gaunt magnet]

go to another channel and read #❓how-to-get-help

[ionic jewel]

yes 3 is right

[ionic jewel]

theres a 315/(315+50)% chance of the 315 event happening

,calc 315/(315+50)

[ocean seal]

Result:

0.86301369863014

[ionic jewel]

86%

[alpine sable]

ty

[gaunt magnet]

Ok good when am i getting active role bcuz im active

[alpine sable]

,calc x3+y3+z3=k

[ocean seal]

The following error occured while calculating:
Error: Invalid left hand side of assignment operator = (char 9)

[ionic jewel]

It's moderated by a bot so im sure theres some sort of analysis, itll happen eventually

[ionic jewel]

okay but what are you even trying to do here

[strong furnace]

there's weights for channels and question channels sadly have low weights

[stable leaf]

1/3=0.333333 , 1/3 + 1/3 + 1/3 =3 (0.333333) = 1 does this show 0.99999 = 1?

[gaunt magnet]

Ok

this still my channel

so i ask 1 more question

[stable leaf]

like 0.999999..... equals 1 or 0.999..... approaches 1?

[ionic jewel]

1/3 + 1/3 + 1/3 is not equal to 3

[strong furnace]

equals 1 , its just another way of writing it as a limit or cauchy sequence

[gaunt magnet]

multiply and divide by (2+root(1-x)
so in numerator you will get (4-(1-x) = 3+x
and in denominator cancel the x+3 from the numerator
so u will be left with 2+root(1-x) in the denominator
so 1/(2+root(1-x)
now substituting you will get
1/(2+root4)
=1/4

[stable leaf]

i didn't say it was

[ionic jewel]

the proof ive seen is
let k = 0.99...
so 10k = 9.999...
now 10k-k = 9k = 9.999... - 0.999 = 9
since 9k = 9, k = 1

[stable leaf]

but they say 0.9999.... equals 1 or approaches 1?

[ionic jewel]

its equal

[strong furnace]

@stable leaf do you believe 1/3 = 0.333... ?

[ionic jewel]

you could write it as a limit and it would also be true i guess

[stable leaf]

i can't trust anything

can i

sure

[strong furnace]

the thing is when you take 1/3=0.333... to be true you are doing the same thing as taking 0.999...=1 to be true

so the argument that 3*(1/3)=1 does not really work

[gaunt magnet]

how do u write (2+root(1-x) on texit

$(2+root(1-x)$

[ocean seal]

Usman

[gaunt magnet]

how?

[strong furnace]

what would be better is if you work with the definition of 0.999... as the decimal representation of a real number and work out what 0.999... could possibly represent and this would lead to 0.999... = 1 (pedagogically speaking this would generally be covered in the first real analysis unit though)
the usual proof you will see in calculus would be using geometric progression and limits which is the same thing just not as rigorous if that makes sense

[stable leaf]

well saying the limit is true isnt exactly the same is it?

[gaunt magnet]

guys how

do u write (2+root(1-x) on texit

[strong furnace]

elaborate?

[strong furnace]

#latex-help

[stable leaf]

well maybe not the same example but like idk (x-1)^2/(x+1)

[strong furnace]

still don't understand what you're trying to get at

[stable leaf]

like a hole or whatever you calll it

about the idea of using a limit

[strong furnace]

Limit and value of function are not related in that sense

the limit will have some value yes ?

[stable leaf]

well that value there vs the limit there is what i am getting at

[gaunt magnet]

🙈

[stable leaf]

but yeah i agree 0.9999..... = 1 it just seems weird starting with it vs the idea of like saying

1/3 = 0.33333....

i was more curious

if like maybe there is a meaning with the ..... obviously it means the digits repeat forever

[strong furnace]

that is not obvious , it has to be defined in a particular way for it to be that

[stable leaf]

true i guess you might need to define what the .... means

[strong furnace]

yeah one of the popular initial exercises in analysis course would be to tweak the definition to get different decimal representation

[stable leaf]

so umm

how do you prove something like

1/3 = 0.333....

i mean sure when i keep trying to do long division i keep getting another 3

but

.-.

[vale wigeon]

@gaunt magnet $2 + \sqrt{1 - x}$

[ocean seal]

Ann

[vale wigeon]

also #latex-help exists, for future latex questions

[stable leaf]

is it maybe one of this proofs of n+1?

type things

or whatever

[ionic jewel]

induction is the name

[stable leaf]

can that work for something to show 1/3 = 0.333 repeating?

[ionic jewel]

maybe? but definitely not the way you want to do it

[strong furnace]

here's something you can try to think of how the decimal representation is defined so that 0.999.... describes 1 using long division
when you do 1 divided by 1 write it as 0 instead of 1(how you usually would write it) and then just do it like you usually would do with decimal division

so you will have something like for the divisor, dividend , remainder pair (1,1,0) , (1,10,0._) , (1,10,0.9_)....

[candid thistle]

How can I find the side length of the square given the side lengths of the triangle (for example a 3, 4, 5 triangle)

Is it just half of the hypotenuse. (Hints appreciated but please don't give the solution)

[gaunt magnet]

yes

how do i become good in latex

[stable leaf]

maybe square inscribed in triangle also the square seems to make 3 triangees

[candid thistle]

Yes, all of the three are similar alongside the large triangle because of AAA

The square is inscribed I think. The side touches the hypotenuse perfectly (sorry for the bad drawing)

[ionic jewel]

yes, the trick (maybe multiple ways) is to find the area of the other 3 right triangles and subtract them from the full triangle area

[candid thistle]

Ahh, how can I do that though

[ionic jewel]

i thought you just wanted a hint

[candid thistle]

I don';t know the lengths of those

[ionic jewel]

now i actually have to solve it

[candid thistle]

:P i didnt want the entire answer

only part of it

[ionic jewel]

for starters, the right triangle with the original right angle is similar to the original triangle

wait found a link

[candid thistle]

wait whaat right triangle? all three of tthe smaller triangles are similar to the largest triangle, I know that

[ionic jewel]

https://math.stackexchange.com/questions/320367/square-inscribed-in-a-right-triangle

even has a drawing :D

[candid thistle]

thank you :D i need this

How did they get as/b and bs/a

in this

[ionic jewel]

right triangle

i mena

similar triangle

[candid thistle]

I really don't understand. I am having a massive brainfart :(

[ionic jewel]

so you see how the S side of the blue triangle is similar to BC on the big triangle

[candid thistle]

Yes

[ionic jewel]

and the side labeled as/b is using the ratio generated from that

[candid thistle]

ok

[ionic jewel]

like

[candid thistle]

I think I get it now a little bit

[ionic jewel]

maybe try drawing the triangles independently from each other

and calculate it yourself

[candid thistle]

Wait but BF corresponded with BC right?

[ionic jewel]

no

wait

[candid thistle]

Yeah it's cuz BF is shorter??

[ionic jewel]

appearently i cant read letters

yeah but draw the triangles yourself and rederive 👍

[candid thistle]

ok

[ionic jewel]

im just confusing myself and dont want to say anything wrong here

[gaunt magnet]

guys ive mastered latex and texit

[candid thistle]

I think I got something goign, If I need help I will ask here with my progress

[gaunt magnet]

wanna see

[stable leaf]

yes

[candid thistle]

Oh I got somthing! Note how BF (in a 3,4,5 triangle) is 3s/4 and AE is 4s/3

[gaunt magnet]

$\pm\sqrt{\left(\frac{2\left(\sum^\infty{k=0}\frac{x^k}{k!}\right)^{i}-2\frac1{\left(\sum^\infty{k=0}\frac{x^k}{k!}\right)^i}}{\left(\sum^\infty{m=0}\sum^m{j=0}\frac{x^m}{j!(m-j)!}\right)^i-\frac1{\left(\sum^\infty{m=0}\sum^m{j=0}\frac{x^m}{j!(m-j)!}\right)^i}}-\frac{\left(\sum^\infty{m=0}\sum^m{j=0}\frac{x^m}{j!(m-j)!}\right)^i-2+\frac1{\left(\sum^\infty{m=0}\sum^m{j=0}\frac{x^m}{j!(m-j)!}\right)^i}}{i\left(\sum^\infty{m=0}\sum^m{j=0}\frac{x^m}{j!(m-j)!}\right)^i-i\frac1{\left(\sum^\infty{m=0}\sum^m{j=0}\frac{x^m}{j!(m-j)!}\right)^i}}\right)\left(\frac{2\left(\sum^\infty{k=0}\frac{x^k}{k!}\right)^{i}-2\frac1{\left(\sum^\infty{k=0}\frac{x^k}{k!}\right)^i}}{\left(\sum^\infty{m=0}\sum^m{j=0}\frac{x^m}{j!(m-j)!}\right)^i-\frac1{\left(\sum^\infty{m=0}\sum^m{j=0}\frac{x^m}{j!(m-j)!}\right)^i}} + \frac{\left(\sum^\infty{m=0}\sum^m{j=0}\frac{x^m}{j!(m-j)!}\right)^i-2+\frac1{\left(\sum^\infty{m=0}\sum^m{j=0}\frac{x^m}{j!(m-j)!}\right)^i}}{i\left(\sum^\infty{m=0}\sum^m{j=0}\frac{x^m}{j!(m-j)!}\right)^i-i\frac1{\left(\sum^\infty{m=0}\sum^m{j=0}\frac{x^m}{j!(m-j)!}\right)^i}}\right)-\sum^\infty{m=1}\sum^m{i=0}(-1)^{m-1}\frac{x^{2m}}{(2i+1)!(2m-2i-1)!}}$

[ocean seal]

Usman

[candid thistle]

So the hypotenuse = (3/2 + 1 + 4/3)s

[candid thistle]

holy moly that must have been a pain to make

[stable leaf]

unless it was copypasta

[candid thistle]

yea ig

[gaunt magnet]

,tex
\newcommand{\pplot}[1]{
\begin{tikzpicture}\begin{axis}[hide axis,samples=50]
\addplot3[mesh,domain=-8:8]{#1};
\end{axis}\end{tikzpicture}
}
\pplot{x^2+y^2}

[ocean seal]

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

[gaunt magnet]

wtf