help-13

[twilit jasper]

I don’t know how to set it up

[radiant fjord]

Start by drawing a diagram

Of the base and the projections

[cedar kiln]

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[agile turret]

can anyone help me with this i will pay you back in steam

[cedar kiln]

Please don't occupy multiple help channels.

[dire geode]

don't offer money for math help

stick to your #help-23 channel

.close

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[final glade]

ln(9,3x10^-6/k) = -100x10^3/8,31 x (1/5250)

How can I solve for k?

[tough aurora]

with a calculator for sure

exponentiate then isolate k

[cerulean sail]

• also can you write it on paper or something please? a bit ambiguous on what you want

$\ln \left( \frac{9.3 \times 10^{-6}}{k} \right) = -\frac{100 \times 10^{3} }{8.31} \times \frac{1}{5200}$

[wraith dagger]

chartbit

[cerulean sail]

Is that what you mean?

[cedar kiln]

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[dusk oyster]

aye if anyone can help me figure out why i got 3 questions wrong on my quiz please help me out

[cedar kiln]

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[dusk oyster]

oh

[vale willow]

@dusk oyster Not 100% sure what the first question is trying to ask you, but for the second photo, you did almost everything right, but you solved for SQ not SR, 4(12) + 2 is the lengh of SQ not SR. See if you can figure out what SR must be since you know SQ. Then in the third photo it seems you solved for x wrong. It is hard to read your hand writing, but it seems to me that x should be 7. Then you know that AM + MB = AB so use x to get the lengths of AM and MB and then add them to get the result.

[dusk oyster]

I figured out the first photo

Ill go onto the second one and ill tell you what happened @vale willow

[dusk oyster]

Wait

so for the second

I would do 2x+1

2(12)+1?

[cedar kiln]

@dusk oyster Has your question been resolved?

[dusk oyster]

@vale willow

<@&286206848099549185>

[dusk oyster]

<@&286206848099549185>

specifcally the 2nd and 3rd pic

alr got the first one

[nimble veldt]

second photo: yes, 50 is the full length SQ, but you need just SR, which is the half.
third: you started right with 2x-3 = x+4, but then you had a mistake.

Add +3 add both sides: 2x = x +7
then subtract x on both sides: x = 7

[dusk oyster]

so SR would be 25

[nimble veldt]

yes

[dusk oyster]

2(12)+1 = 25

thats right

then its 2x + x+7

x = 7 AB=22

@nimble veldt

[nimble veldt]

yes

[dusk oyster]

shit alr thanks bro

.close

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[thin hedge]

Can someone explain this quadratic

I'll take a pic of my work

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[cobalt leaf]

can someone breakdown how they simplified it from the red star

[crimson delta]

a fraction is 0 when the numerator is 0

then move the second part to the other side

square both sides

substract x^2(a-x)^2 from both sides

divide by b^2 on both sides

and the last one is a clever observation

[cobalt leaf]

i got it upto here

after that im confused

[crimson delta]

what are you left with now

[cobalt leaf]

b^2a^2=2ab^2x

[crimson delta]

how did you get that

you have $x^2(a-x)^2+x^2b^2 =(a-x^2)x^2+(a-x)^2b^2$

[wraith dagger]

Denascite

[crimson delta]

after the squaring

I multiplied out the {} bracket

[cobalt leaf]

got it

ty

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[carmine elk]

how do i integrate this?

[south tundra]

By parts

[cedar kiln]

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[crimson sedge]

can someone help me figure out where this one went to?

[dull oxide]

inequality

[crimson sedge]

can you elaborate?

[dull oxide]

The +1 becomes +0

And that satisfies the inequality between those two lines

[foggy merlin]

of course, n² + n + 1 ≥ n² + n

[dull oxide]

https://tenor.com/view/wtf-confused-look-gif-24835232

[foggy merlin]

im tired

[crimson sedge]

hm but how can you say that without any justification / proof

doesn't make any sense

[dull oxide]

0<=1

start there

[timber plover]

Puk is this boi 0<1 i get it why is there equal sign

[crimson sedge]

wym

[foggy merlin]

basically,
n² + n + 1 ≥ n² + n
-(n² + n +1) ≤ -(n² + n)
2-(n² + n +1)≤ 2-(n²+n)

[crimson sedge]

ah this helps

[surreal aurora]

@crimson sedge come to vc and Ill explain

nvm here's got u

[timber plover]

It should be n²+n+1>n²+n , there shouldn't be ≥

Equality won't be equal for any value of n

[foggy merlin]

yea i know, i was just in the context of the exercise

[surreal aurora]

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[jovial sandal]

There were 240 pears and apples at a stall. 8/15 of them were pears and the rest were apples. after some pears were sold, the number of pears left was 4/11 of the pears and apples left. How many pears were sold? Could anyone explain to me how to solve this, as my tutor didn't do a very clear explanation? Thanks.

[cedar kiln]

@jovial sandal Has your question been resolved?

[timber plover]

You have learned to assume variables yet?

[cedar kiln]

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[twilit nest]

Hey

can someone tell me what i've done wrong to get the red marked ^2 in last line ?

the first line is my goal to achieve but its exactly what i got now but with 2n instead of n

[crimson delta]

(a^b)^c = a^(b*c)

[twilit nest]

yes i know but this wont get rid of the 2n aint it ?

[crimson delta]

I mean the whole exponent is wrong

not sure why they only marked the 2 in red

[twilit nest]

can u tell me in what line the exponent started to go wrong ?

i mean isnt it exactly what ive done? (x^2)^(n+1) = x^(2[n+1]) ?

[crimson delta]

https://i.imgur.com/ohP3JCc.png

[twilit nest]

isnt this exactly what the rule says?

[crimson delta]

x^a^b means x^(a^b)

not (x^a)^b

[twilit nest]

Oh

So this would be the correct next step? @crimson delta

[crimson delta]

yes

[twilit nest]

okay but how would i get rid of the ^2 and instead have n+2 ?

wouldn't it still give me ^(2n+2) at the end ?

[crimson delta]

well now you can apply the law I wrote down earlier

$\left(x^{2^{n+1}}\right)^2 = x^{2\cdot 2^{n+1}}$

[wraith dagger]

Denascite

[crimson delta]

and what is 2*2^(n+1) ?

[twilit nest]

Ohhh and then shorten it to n+2

Yesss i get it

Thats correct aint it ?

[twilit nest]

Question about that one, why does the 2 multiply with the 2 and not with the (n+1) ?

because the brackets are around the x and not around the 2?

[crimson delta]

the exponent is 2^(n+1)

that's one thing

set a=2^(n+1)

then (x^a)^2 = x^(2a)

[cedar kiln]

@twilit nest Has your question been resolved?

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[olive heart]

How do u do number 38

I just need a refresher

[livid hound]

38. is pretty much already in vertex form

[olive heart]

really?

[livid hound]

yes, don't get intimidated by the lack of expressions

[olive heart]

ok how about 39

[livid hound]

complete the square to get vertex form

you should already have an idea of how to complete the square if you've done Q17-37

[olive heart]

ye

after u complete the square?

@livid hound

[livid hound]

after you complete the square, you'll have the vertex form as desired

from which you can identify the properties being sought

[olive heart]

can you show me wih 39

so its $g(x)=-x²+2+1$

[livid hound]

no

[wraith dagger]

Hehehehaw

[olive heart]

how

i completed the square

[livid hound]

-x^2 + 2 is not the same as 1 being added to itself

[olive heart]

ohhhhhh

mb i thought it was 2x

[livid hound]

oh and technically 39 was also actually already in vertex form

[olive heart]

ye

number 40 is not

$x²-5x+25/4$

[wraith dagger]

Hehehehaw

[olive heart]

right?

[livid hound]

you can't just add 25/4 like that

[olive heart]

wdym

[livid hound]

doing just that, you'll have something different to what you're starting with

[olive heart]

i completed the square

[livid hound]

which isn't quite what you want

[olive heart]

hmm

[livid hound]

adding 25/4 completes the square,
but you'd also need to subtract 25/4
effectively adding 0,
to ensure you have something equivalent to what you started with

[olive heart]

so y-25/4=x²-5x?

[livid hound]

no

[olive heart]

can you do 40

i wana see how u do it

havent done this in a long time

[livid hound]

didn't you do 17-37 recently?

[olive heart]

ye but winter break screwed me up

[livid hound]

so you did 17-37 several weeks ago?

[olive heart]

so for 40 whats the first step

[olive heart]

ye

[livid hound]

read this

$\cts$

[wraith dagger]

ℝamonov

[olive heart]

ohhhhh

so u basically add or subtact it to both sides?

[livid hound]

do something that keeps the original equation balanced

[olive heart]

so it would be y-25/4=x²-5x+25/4

right?

[livid hound]

no

[olive heart]

whatt???

[livid hound]

you subtracted one side by 25/4 while adding 25/4 to the other

[olive heart]

ye

[livid hound]

it's like going from
1 = 1
to
1 - 1 = 1 + 1

[olive heart]

hmm

[livid hound]

which gets 0 = 2 (something clearly false)

there's no issue if you apply the same operation to both sides

or doing something that effectively adds 0 to one side

[olive heart]

yea

[livid hound]

which isn't what you've been doing

[olive heart]

g(x)=x²⁻5x

[livid hound]

since ideally you want to keep
g(x) on the left side by itself,
you'd want to go with this option

doing something that effectively adds 0 to one side
adding 25/4 completes the square,
but you'd also need to subtract 25/4
effectively adding 0,

[olive heart]

y+25/4=x²-5x+25/4

right?

[livid hound]

is now valid

[olive heart]

ohhh

so the sign never changes

essentially

[livid hound]

but there's no real point of changing g(x) to y

and doing stuff to the left side

[olive heart]

my teacher said its easier

[livid hound]

not really

[olive heart]

cuz your gonna find the axis of symmetry

[olive heart]

how?

[livid hound]

if the end goal is to convert to vertex form,
you don't want to do anything to the left side, on focus on manipulating only the right side

note that this is not the same as solving an equation using completing the square

[olive heart]

yea

[livid hound]

if you were solving equations like in 17-37,
then adding stuff to both sides may be more efficient

but this is different

[olive heart]

yea i see

[livid hound]

y+25/4=x²-5x+25/4
in the end, you're gonna want y or g(x) by itself,
so you're gonna need to subtract 25/4 from both sides
so why not just start with
g(x) = x^2 - 5x + 25/4 - 25/4
immediately

[olive heart]

hmmm

good point

but

were trying to find the coordinates of the vertex

[livid hound]

yeh,

[olive heart]

so u wouldnt find g(x) by itself

[livid hound]

wdym

[olive heart]

because after y-25/4=x²-5x+25/4

y-25/4=(x-5/2)²

[livid hound]

missing - sign

my point still stands

[olive heart]

the coordinates are (5/2, 25/4)

[livid hound]

one sec

yuo signs are wrong

[olive heart]

how would you get the coordinates doing it your way

[olive heart]

which ones

[livid hound]

should be
y + 25/4
if you insist on having stuff on the left side

y+25/4=x²-5x+25/4
in the end, you're gonna want y or g(x) by itself,
so you're gonna need to subtract 25/4 from both sides

[olive heart]

i meant +

mb i was typing fast

[livid hound]

note that on the right side

[olive heart]

but it would result in the same result

[livid hound]

x²-5x+25/4 something you know is a perfect square is still present

but like i said, you're just making more work for yourself

[olive heart]

true

but thats how we learned it

whats your shorter way of doing it

im curious

[livid hound]

$g(x) = \underbrace{x^2 - 5x + \frac{25}{4}}_{\br{x-\frac52}^2} - \frac{25}{4}$

[wraith dagger]

ℝamonov

[olive heart]

hmm

im a bit confused

[livid hound]

which part

[olive heart]

fraction

the bottom part

[livid hound]

that's just explicitly indicating something you already claimed to know

[olive heart]

whats the point of putting 25/4 if they cancel out

[livid hound]

+25/4 - 25/4 = 0
adding 0 doesn't change the value of the expression and you have something equivalent to what you're starting with

[olive heart]

ohh

[livid hound]

similar idea of adding 25/4 to both sides of the equation

[olive heart]

yeaa

alr i see what u did

thanks for the help

i have to go now

[livid hound]

its all outlined in that summary i made

[olive heart]

ye

thanks

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[upper thicket]

i understood the solution , but how do i get started with answering these kind of questions ?

[cedar kiln]

@upper thicket Has your question been resolved?

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[ember plaza]

I have to disprove this statement. How to do it? I’m a bit lost.

$$\text{Disprove } f\in \mathcal{O}(g) \text{ for}\ f : \mathbb{N}\rightarrow \mathbb{R}{\geq 0} \text{ with} \ f(n) = {2n^3+4} \text{ and} \ f : \mathbb{N}\rightarrow \mathbb{R}{\geq 0} \text{ with} \ g(n)=n^2 + n$$

[wraith dagger]

Summe

[ember plaza]

Egg

[sinful current]

wuh

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.close

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[lone shadow]

How do I prove the lower bound?

[lone shadow]

I can get a meh upper bound by saying let |S| = n, then |SxS| = n^2 and |P(SxS)|=2^(n^2) which is an upper bound

I imagine the 2^(n(n-1)/2) comes from some sort of pairing and doing n choose 2 but also unsure about this

[cedar kiln]

@lone shadow Has your question been resolved?

[cedar kiln]

@lone shadow Has your question been resolved?

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@lone shadow Has your question been resolved?

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[cedar kiln]

Please don't occupy multiple help channels.

[dense tangle]

.close

[cedar kiln]

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[dense tangle]

.reopen

[cedar kiln]

✅

[dense tangle]

!help

[cedar kiln]

Please read #❓how-to-get-help

[cosmic steppe]

Holy shit

Calm the fuck down

[dense tangle]

im sry

.close

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[dense tangle]

.reopen

[cedar kiln]

✅

[dense tangle]

.close

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[floral pagoda]

How do I find the derivative of f(x)= x*ln(6x)?

[floral pagoda]

Using the uv formula I got ln (6x) + 1/6

this is wrong

[south tundra]

Where's + 1/6 coming from?

Derivative of ln(6x) is 1/x

You seem to have forgotten about chain rule

(ln(6x))' = 1/6x * 6 = 1/x
or
(ln(6x))' = (lnx + ln6)' = 1/x

[dire geode]

another way to see it is log(6x) = log(6) + log(x)

[south tundra]

Yes

[floral pagoda]

I used this formula

[south tundra]

Again, derivative of ln(6x) is not 1/(6x)

[dire geode]

@floral pagoda

[floral pagoda]

yeah mb

Im still getting 1* ln 6x = ln 6x tho

[dire geode]

1 * log(6x) = log(6x) yes

[floral pagoda]

I rewrote it to be more clear

[floral pagoda]

the correct answer should be ln (6x) + ln (x) + 1

[dire geode]

show the "correct answer"

[floral pagoda]

.close

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[dire geode]

that's not what you typed

[cedar kiln]

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[floral pagoda]

yes my message was deleted

[floral pagoda]

I said I was re*ard

.close

[cedar kiln]

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[dull blaze]

What do i do next

[dire geode]

,rotate

[modern compass]

you shouldn't have (5x+2)^3

[dull blaze]

Why not

How do i change it

[modern compass]

the denominator on the left has (5x+2)^2

that's what you should multiply through by.

[dull blaze]

Don't you need to multiply C, by whats under A and B

[modern compass]

no. you multiply the RHS by the denominator on the LHS and cancel appropriately

[dull blaze]

Is that better

[cedar kiln]

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[chilly warren]

Still wrapping my head around continuous functor. Is there continuous functor other than hom functor?

[cedar kiln]

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[cedar kiln]

@chilly warren Has your question been resolved?

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[cedar kiln]

Please don't occupy multiple help channels.

[glacial owl]

I= 3872812500mm

So I'm getting confused with unit conversion and it's giving me really strange numbers

#help-15

[radiant fjord]

go type .close there

[dire geode]

done

[upper garnet]

How much 300 kN is mN?

[glacial owl]

Beat me to it

0.3

Just getting onto structural design so not 100% familiar with units conversion

Am I right?

[cedar kiln]

@glacial owl Has your question been resolved?

[cedar kiln]

@glacial owl Has your question been resolved?

[native valve]

yuh

[cedar kiln]

@glacial owl Has your question been resolved?

[wicked orchid]

can someone help me with this plz

[cedar kiln]

Please read #❓how-to-get-help

[wicked orchid]

thx

[cedar kiln]

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[tidal cargo]

Trying to solve a difficult problem at a challenge room:

There are 4 buttons, 1, 2, 3, and 4.

There's some 4-number combination to win. We know for sure that number 3 goes last and there is no repetition.

How would I find a list of all possible answers to brute force it?

[lusty grotto]

u want to see in how many ways u can arrange 3 different numbers in 3 spaces

right

[tidal cargo]

Yep!

With no repetition

[crystal raptor]

There are only 6 possibilities, just try and list them

[tidal cargo]

124, 142, 412, 214, 421, 241?

[crystal raptor]

Yes

[tidal cargo]

Awesome thanks

[cedar kiln]

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[elfin marsh]

If an angle equals to its arc in a circle does it always have to be in center

[gaunt hamlet]

The arc of a circle is defined as the angle from the center

[elfin marsh]

So there is no other point in a circle from where the angle would be equal to the arc

?

. Close

. close

.close

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[dusty hazel]

Those coefficients seem like they came out of nowhere, I got nowhere.

[cedar kiln]

@dusty hazel Has your question been resolved?

[sacred grail]

looks like roots of unity filter

[crimson sedge]

FOURIER?!

Oh wait

[sacred grail]

yeah

discrete fourier kinda

[crimson sedge]

Oh that shit

Bleh

[sacred grail]

because of periodicity reasons

[dusty hazel]

I got it.

Thanks.

.close

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[grand canopy]

Just being curious since this is outside the scope of my course but, given a matrix let's say A. Is the diagonalisation of A unique or non-unique?

[cerulean sail]

when you say "diagonalisation", what do you mean? Are you asking whether the diagonal matrix you get is unique, or the transition matrix is unique?

[grand canopy]

Well, this sort of arises from a question I was doing

The question states " Find a matrix P that diagonalizes A "

I'm sort of comparing answers and used some online calculators and they are giving me different values

[tropic oxide]

how different?

the P in A = PDP^-1 should be unique up to permutations of columns (and corresponding permutations of diagonal entries in D)

[grand canopy]

The entries are correct, but the positions are different

Here is the Matrix A I am working with

[tropic oxide]

assuming there are no repeated eigenvalues -- if there are, then the same eigenspace can have different bases

[grand canopy]

Calculator 1:

Calculator 2:

In my case, there are no repeated eigen values

[tropic oxide]

right ok hold on

i think your image of A didnt go through

but we can recover it

[grand canopy]

Sorry

[tropic oxide]

okay

right

yeah this is easily explained

[grand canopy]

Please do

[tropic oxide]

the cols of P are eigenvectors of A, arranged in the same order as the eigenvalues appear in D

[grand canopy]

I concur

[tropic oxide]

the eigenvalues in D can be arranged any way you like

so long as they are the right eigenvalues

[grand canopy]

Yes I was thinking of the same thing but this is not explained in my course

[tropic oxide]

what does your course say? does your course prescribe a particular ordering of eigenvalues?

[grand canopy]

No necessarily, but I am currently doing an acceleration course and a lot of content is mushed together so they do not have time to explain it thoroughly

[cerulean sail]

Ah yeah, I remember your course, it was that crazy one

[grand canopy]

So for this example, the eigenvalues I got were 2, -1 and 1 and the corresponding eigenvectors are (0, 1, 0), (-1, -1, 1) and (-2, -1, 1)

[tropic oxide]

right

[grand canopy]

I have checked, SINCE the these three vectors are linearly independent

[tropic oxide]

i mean like, if you want to be formal about it re: uniqueness

you could scale any one of these eigenvectors by a nonzero amount and it'd still be an eigenvector

leading to even less uniqueness

[grand canopy]

What you are saying is that, I can arrange them in n type of ways? I'm assuming n = 6 ways?

[cerulean sail]

As long as the order of eigenvectors corresponds to the order of the eigenvalues

[tropic oxide]

n! ways.

but i didn't even say that.

[grand canopy]

Yeah I kind of just guessed since n is small in this case

Interesting stuff

[tropic oxide]

no, i just said that eigenvectors are (in your case) only determined up to a scaling factor.

[grand canopy]

btw everyone dropped after the census date, there was only like 3-12 kids across the two tutorials I went to today, there was like 350-400 at the start

This might be a dumb question but what does the "diagnolisation matrix" do?

[tropic oxide]

are you talking about P?

[grand canopy]

Indeed

I'm assuming P is the P in the A = PDP^-1

[tropic oxide]

yes, im sticking to the notation thats been used thus far

[grand canopy]

Also I guess the name has something to do with the function of the matrix

[tropic oxide]

'the function of the matrix'...?

what

[grand canopy]

I mean like

What it does

[tropic oxide]

oh you mean purpose

anyway, as said before, P holds the eigenvectors of A.

[grand canopy]

Yes, sorry for the confusion

[tropic oxide]

you can view P as the change-of-basis matrix from the standard basis to an eigenbasis for A.

[grand canopy]

Hahah sorry my LA is weak so I don't even know what an eigenvectors/eigenvalues actually "do" I just know how to compute them etc

But sure, I will take what you have said

I am only asking this because for the second part of the question we are given three simulataneous equations, with each coefficient corresponding to the entries of A, the one I have provided

And we have been told to solve the "ODE"

So I am assuming there is some trick for this

[tropic oxide]

nyeh...

have you seen 3b1b's essence of linear algebra?

[grand canopy]

Haha it has been strongly suggested but I have not yet found the time to watch the videos

[tropic oxide]

the eigenvectors/values chapter in that video series does a p good job of introducing eigenthings from a conceptual pov

[grand canopy]

I see

[tropic oxide]

in brief, an eigenvector of A is a vector v ≠ 0 such that Av = λv for some scalar λ

in other words, sometimes the effect of a linear map on a particular vector is to just scale it but not change its direction. and those vectors are called eigenvectors.

[grand canopy]

Are the videos "heavy" as in, would I need to think a lot while watching them, I've just been introduced to LA although we are covering a whole lot of it in 3 weeks

If it's not really heavy, I guess I could watch it more for fun than treating it as a lecture

[tropic oxide]

3b1b himself these videos should be used as a supplement to an actual linalg course.

not sure about heaviness.

[grand canopy]

Perfect! Thank you so much

And thanks for the help once again @cerulean sail

have a nice day guys

.close

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[prime karma]

hi can someone help me go through this?

[south tundra]

How'd you get 1 in iv?

[prime karma]

i made a piecewise function with sinx>1 = 1, sinx=0 f(x)=0, and sinx<1=-1

and then i just tried to match that with this but

[crimson sedge]

what does sgn() mean again

[prime karma]

i dont think i did it right

[tropic oxide]

yes you in fact didn't do it right

[south tundra]

Sign

[manic acorn]

sgn(f)= 1 f>0, 0 f=0, -1 f<0

[tropic oxide]

as x approaches pi from the right, sin(x) is positive.

[crimson sedge]

oh okay gotcha

[prime karma]

how do i find that out

do i just plug in points

pi/2, pi/4

[tropic oxide]

i'd just sketch the graph (or make a mental sketch of the graph)

[prime karma]

.close

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[stuck summit]

BE ⊥ AC AB = AC SABC = a^2

i need to find AB AC and BC using a

[stuck summit]

AD + BC + AC + AB

[crimson sedge]

SABC ?

[stuck summit]

the surfuce of ABC

[crimson sedge]

Oh

Do u know the formula for area of a triangle ?

[stuck summit]

yes

[crimson sedge]

And they have already given u area right ?

Assuming that's what u meant by surface

[stuck summit]

yes

[crimson sedge]

$A = \frac{1}{2} {BC×AD}$

[stuck summit]

but they dont say that <ADC = 90

[crimson sedge]

U already have BC in it

[stuck summit]

yes but BC isnt parrallar to AD

at least they dont give you tha

[crimson sedge]

Hmm, interesting

Then why not calculate the area still then if not AD

U have BE

Which is perpendicular to AD and is given

$A = a² = \frac{1}{2} {BE×AC}$

[stuck summit]

yes after that you get
AC = 2a^2/BE
AB = 2a^2/BE
BE = 2a^2/AC

[crimson sedge]

So u have that then ? Do u want like simplify it even more ? Or find BC ?

[stuck summit]

how do i find BC?

[cedar kiln]

@stuck summit Has your question been resolved?

[cedar kiln]

@stuck summit Has your question been resolved?

[cedar kiln]

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[rose rose]

Are these solutions sane? I don’t have an answer sheet so I’m unsure

[cedar kiln]

@rose rose Has your question been resolved?

[rose rose]

<@&286206848099549185> hate multivar, but I'll still try a ping 🙃

[cedar kiln]

@rose rose Has your question been resolved?

[cedar kiln]

@rose rose Has your question been resolved?

[rose rose]

[steep meadow]

I would have helped if I didn't delete flux stuff from my memory 😂

[soft owl]

Ex 8 is correct :
flux = 2πrh = 20 π

[rose rose]

Found the answer sheet. Ex8 is wrong supposed to be 10pi. Ex9 is right 😭

[soft owl]

i'm sure it's 20 π

[rose rose]

How sure? I'm pretty sure too, but the answer sheet is fooling me

[soft owl]

because the Ex 8 produce a cylinder so :
the flux of a cylinder (if i still remember) is : 2πrh
which is 2 π * 1 * 10 = 20π

[cedar kiln]

@rose rose Has your question been resolved?

[rose rose]

It has to depend on the vector field, no?

[soft owl]

You could say it's a special case :
total flux of a cylinder is 2πrh
which is the flux of the curved surface area of the cylinder
the flux of the bases canceled each other

[rose rose]

Huh? How could it possibly not depend on the vector field? That's bizarre

[cedar kiln]

@rose rose Has your question been resolved?

[sacred grail]

integral of r is r^2/2

wait no i'm blind

your calcs look right

[cedar kiln]

@rose rose Has your question been resolved?

[rose rose]

Eeeeeh, maybe the answer sheet is wrong? It's always a scary thought 😄

[sacred grail]

could be

[soft owl]

,w flux of a cylinder

[wraith dagger]

Results provided by WolframAlpha

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

Requested by Mehdi_Moulati#1210

[soft owl]

the flux of a cylinder is the area of the curved surface

which is 2πrh

so your answer is correct 99.99%

[rose rose]

But even the phrase “flux of a cylinder” sounds wrong to me. Flux through a cylinder of all vector fields is the same?

[sacred grail]

like thats just straight up not right

the flux depends on the vector field

if a certain vector field produces flux k then twice that vector field will produce flux 2k just by linearity

[rose rose]

You took a look at the steps right? If nothing stands out to you I guess I’ll just close it. Unlikely I’ll get anyone else to take a look at it

[sacred grail]

i checked it

seems right

[rose rose]

Tysm

.close

[cedar kiln]

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[sacred grail]

wait

no actually dont wait

[cedar kiln]

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[rose rose]

.reopen

[rose rose]

Decide 😄

[sacred grail]

it says my name

no its fine

we good

[rose rose]

Now you close it 😛

[sacred grail]

.close

[cedar kiln]

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Channel closed

Closed due to the original message being deleted

[pine idol]

Hello!

I need help memorizing multipulcation facts.

[cedar kiln]

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[solar tendon]

How do I prove this?

[solar tendon]

This is where i got stuck

idk what i did wronf

proving fundamental identities btw

[cedar kiln]

@solar tendon Has your question been resolved?

[solar tendon]

<@&286206848099549185>

[scenic palm]

you messed up

cosa * cosa is not 2cosa

check your second step

@solar tendon

[solar tendon]

Oof

its supposed to be cos^2

right

[scenic palm]

yes

[cedar kiln]

@solar tendon Has your question been resolved?

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[cedar kiln]

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[drowsy glen]

sorry for the msg helpers

[drowsy glen]

.close

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[drowsy glen]

.close

[cedar kiln]

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[rose rose]

This is teeechnically physics/circuit theory but I’m having math trouble

[rose rose]

Can I somehow do a algebramagic to get i_1 + i_2 so I can get v in terms of i?

[foggy merlin]

$$V_1 = L_1 \frac{di}{dt}$$
$$V_2 = L_2 \frac{di}{dt}$$

[wraith dagger]

Ｈｅｒｅｌｓ

[foggy merlin]

you are doing a) or b) ?

[rose rose]

B

[short magnet]

v is already in terms of i_1 and i_2 in your last bottom right equation no?

[rose rose]

A was simple enough

[rose rose]

Now I need to get rid of i_1 and i_2 using i_1 + i_2 = i

[foggy merlin]

you cant, unless L_1 = L_2

[rose rose]

Mh, but since I know two inductors in series is equivalent to one with L = (L_1 * L_2)/(L_1 + L_2) it must be possible

Do I maybe need some other property of the inductor other than it’s characteristic v = L di/dt?

[foggy merlin]

wat is this

[floral forge]

use $\pdv{t}$ instead, it will save you a lot of time

[wraith dagger]

Duh Hello

SkyTwX

[lyric jungle]

(Use \pdv{i_i}{t} for partial derivatives)

[short magnet]

What package even is this XD

I lost 200% of my life time because of this

[rose rose]

Oh that actually works. Can I ask how you spotted that?

[cosmic steppe]

Does that actually work or is that a package

[floral forge]

pretty sure its amsmath?

[cosmic steppe]

$\pdv{a}{t}$

Ain't no way

[lyric jungle]

It's from the physics package

[floral forge]

oh yeah

its physics package

[short magnet]

I tried to make i_1+i_2=i appear from the two above right equations, and this came naturally by isolating the respective partial derivatives and expressing them in terms of v

[floral forge]

why doesnt the math package have that lol

[short magnet]

actually a good question, also why are there no proper dx is a mystery for me
Thank you physicists hahaha

[lyric jungle]

I mean there's $\dd x$

[wraith dagger]

Castroploiin¹

[floral forge]

ngl i prefer $dx$ over $\dd x$

[short magnet]

$\int_0^1x^2\dd x$

[wraith dagger]

Duh Hello

SkyTwX

[lyric jungle]

$\dv{y}{x}$

[short magnet]

Spacing is kinda odd :p

[wraith dagger]

Castroploiin¹

[lyric jungle]

Yeah so you use something like \; before it

[rose rose]

I do $, dx$

[wraith dagger]

Learath2

[cosmic steppe]

I like the slanted dx

[short magnet]

yeah, that's what I usually do but also using \mathrm{d}x instead of \dd x... Is \dd also physics package?

[cosmic steppe]

$\dv{yeiejdis}{hshshdhs}$

[wraith dagger]

Umbraleviathan

[lyric jungle]

\dd is from the preamble

[cosmic steppe]

Ehhhhh

I don't like the dd

[lyric jungle]

It's defined originally in the physics package I think

[floral forge]

can easily just redefine it yourself tho

[short magnet]

I actually don't because of differential geometry

[cosmic steppe]

Too lazy

I have too many preamvles

[rose rose]

I guess I’ll close this

[lyric jungle]

It's from the physics package

[rose rose]

Tysm @short magnet

.close

[cedar kiln]

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[drifting crypt]

Alrighty I need some helllllppp

[drifting crypt]

I suck a math and I'm trying to learn..

I got this question and Idk how to begin answering it

[cosmic steppe]

Set up your average value equation

[drifting crypt]

Is there a yt video or something I could use?

Idk what that means

Cuz I pretty much only know the very basics of math

like 1 + 1 and 2 + 2 lmao

[cosmic steppe]

You've done algebra right

[drifting crypt]

Idk exactly what algebra consists of..I know basic multiplication, addition, subtraction, how to convert fractions to mixed numbers/division

Idrk much about percentages

[cosmic steppe]

Algebra consists of solving for variables

[drifting crypt]

But I do know a little about decimals

[cosmic steppe]

You don't need to convert the percentages since the answers are percentages

The average is defined as the sum of all the elements divided by the number of elements

Which might sound familiar

If we follow that statement, then:

$$80 = \frac{55 + 78 + 84 + 93 + x}{5}$$

[wraith dagger]

Umbraleviathan

[cosmic steppe]

And you're just gonna solve for x

[drifting crypt]

Oh okay

That looks manageable

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[magic steppe]

I would like help on how do we get from the first one to the second one, the last step is clear but Idk how to do the transformation

[sullen spire]

have you done

U sub

or

integration by substitution

[rose rose]

The first step is just splitting the fraction and rewriting it and an extra expansion by 2

$\frac{x}{(x^2+1)\ln{(x^2+1)}} = \frac{x}{x^2+1} \frac{1}{\ln{(x^2 + 1)}}$

[wraith dagger]

Learath2

[rose rose]

Now just take the entire fraction on the left and move it into the numerator of the one on the right

Honestly, this feels very very unnatural. Did some online tool generate this?

You can directly u sub u = x^2 + 1

[magic steppe]

no, this is from my teacher in a practice test

[rose rose]

She has a bizarre way of doing integrals than 😄 Anyway, do you get it?

[magic steppe]

Yes, thanks

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[scenic garden]

Can anyone pls help with c?

[scenic garden]

I’ve attempted it numerous times with no luck so working outs would be appreciated 😁

[violet night]

What have you tried?

[scenic garden]

I tried integrating by substitution originally but couldnt get to the overall solution because when i derive u i get a remaining x I then tried by substitution and by parts afterwards but wasnt able to work with that because a u value remained

Im able to get to the main chunk of the working out but i cant seem to get the final integral

Im talking about the top question btw! Didnt realise both were labelled C

[drifting crypt]

Okay I still need help with this (I was taking a break)
I took 310 (The total of the numbers given) and then I divided it by 5 which was 62

Now Idk what to do to get the answer

😅

[cedar kiln]

@scenic garden Has your question been resolved?

[drifting crypt]

hellpplplplplplpl

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[hexed shoal]

lets say a and b are playing drums person a taps the drum 10times per second , person b taps the drum 5 times per second . after 2 hours of training you get +1 tap per second. Person b trains 4hrs per week and person a trains 2 hrs per week? in how much time will a and b tap the drum same amount of times per second? Basically I just wanna know if any1 can make an equation for this I just wrote the weeks down but Im sure there must me an equation

[patent cape]

if b trains 2 hours per week that means the speed increases by 1 weekly, which can be written as w+5, where w is number of weeks
if person a trains for 4 hours a week that means the speed increases by 2 per week, 1 for each 2 hours
so you would obtain the equation 2w+10

if you want a generalised equation for any number of trainign hours and any beginning speed, we can do that too

let s1 be the beginning speed of person a. Since s1 increases by 1 every 2 hours of training, taking the training hours h1 and dividing it by the number of hours necessary to increase the speed (2) you would obtain w(h1/2)+s1

and for person b just change it to s2 and h2

you can also think of this as a liniar equation in terms of weeks, as the weeks increase so does the speed in a liniar fashion

[cedar kiln]

@hexed shoal Has your question been resolved?

[hexed shoal]

and what would be the equation for this ?

[hexed shoal]

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[nocturne jetty]

whats the answer?

[vivid furnace]

So

Since the dice is six sided

The probability of getting 3 on the first roll is 1/6

This is because your removing all possible outcomes except 3

When you remove just one possible out come like removing 4

You’ll have 5/6 as the probability

So 1/6 and 5/6

you can then multiply them to get the overall probability of this happening then

= 5/36

[nocturne jetty]

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[shut hound]

hello does anyone here familiar in boolean algebra? is (¬A ¬ B) and (AB)' are just the same? what i mean is are they just differ in symbols but they are still the same?

[earnest socket]

don't think they are the same

for example, A = true and B = false

[shut hound]

i'm really confused when the apostrophe is outside of the two variables i thought that im going to distribute it to them

so it means that (AB)' is A. ¬ B?

[dire geode]

(AB)' means the complement or negation of (AB)? If so, I guess you evaluate AB first

[shut hound]

oh got it thank you so much

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[orchid raven]

How do I use "e" on a ti-30xa calculator, all my other calc classes have been ti-84's, there is only e^x on this one and I am confused

[sage forge]

Can you sent a picture? Maybe we will find it then 🙂

[orchid raven]

Yes one moment

i might just be dumb lol

[sage forge]

Can you try the 2nd button and then LN?

[orchid raven]

Yeah I did

See I get thats what im supposed to use

[sage forge]

So does it work?

[orchid raven]

I just dont get how to get the solution from it

the problem is (3e^x-3)/x

and im solving for different values

[cedar kiln]

@orchid raven Has your question been resolved?

[cedar kiln]

@orchid raven Has your question been resolved?

[zenith sail]

Unfortunately with a calculator like that, you have to input one step at a time

You have to enter the number you want for x first, then press the e^ button and it will raise e to whatever power you entered

@orchid raven

So like if you're gonna do x = 2

Enter 2, then press the e^ button

Then multiply by 3, then subtract 3, then divide by 2

[orchid raven]

I assume there are no single line calculators that have a regular e

But i appreciate the help

[lyric jungle]

Casio has some.

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[karmic stirrup]

Hello, im currently doing some review.
I wanted to double check that a line with a slope of 3 and (3,1) as its point is y=3x - 8
(this is for y=mx+b) so i thought i would make sure i am correct? if not please let me know i should redo

[runic garnet]

yea thats right

y-1 = 3(x-3)

[karmic stirrup]

alright tysm!

[runic garnet]

y = 3x - 9 +1 = 3x - 8

np

[karmic stirrup]

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[dense fable]

does anyone know why P cant include the number itself or include 1?
ex:

I = [12, 15] # result = [[2, 12], [3, 27], [5, 15]] rather than [[1,27],[2, 12], [3, 27], [5, 15], [12,12], [15,15]]

[dense fable]

<@&286206848099549185> could I get some help please?

[dire geode]

1 is not a prime

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[orchid turtle]

FC and EB are bisectors

FA is 4 and AD is 6

I need to find area of BKC triangle

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

[vagrant pewter]

Bro the spam isn’t gonna get your question answered

Also geometry questions don’t get help here often idk why

[dire geode]

#geometry-and-trigonometry is partly why

[vagrant pewter]

Also what exactly do FC and EB bisect?

Do you mean that the angles are 45 degrees

[orchid turtle]

no i dont know the angles

[vagrant pewter]

Then what do they bisect?

[orchid turtle]

they are just halved basically

[vagrant pewter]

What does that mean?

[orchid turtle]

FEK=KED basically but idk what it is

[vagrant pewter]

Do you mean FAC=EDB?

Because what you wrote doesn’t make much sense

When you label three points it usually means a triangle or an angle

[orchid turtle]

im talking about the angles

[vagrant pewter]

So the angle above the line EK equals the angle below the line EK?

Well then the angles are 45

[orchid turtle]

ngl im tired af

i dont know how i missed that

[vagrant pewter]

Lol no worries

I hope that helps it should be sorta straightforward

[orchid turtle]

i found BC

[vagrant pewter]

Now you can use those angles to fill out all the values

[orchid turtle]

but how to i get the area tho

wait

i got it

yeah thanks for helping i appreciate it

[vagrant pewter]

Nw

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[dire tundra]

are these all the solutions to intervals of 0,2pi?

[dire geode]

you're missing $-\sqrt{3}$

[wraith dagger]

riemann

[dire tundra]

Yeah, besides that would all those angles make up the positive and negative?

I look at an answer sheet it only included the 1st and 2nd quadrant

[dire geode]

,w tan^2(4 * pi / 3)

[wraith dagger]

Results provided by WolframAlpha

Click here to refine your query online
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Requested by riemann#2529

[dire geode]

4pi/3 is not in the 1st or 2nd quadrant

[dire tundra]

OHHH

I see

Those other ones

[dire tundra]

are for Co Tangent

alr thank you

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[sharp adder]

I’m not sure if I’m supposed to use the answer from (i) or how. I’ve tried doing it with integration of rational functions but I don’t think it works with it

[dull oxide]

Use the substitution

You get a much easier integral

[sharp adder]

So just replace the root x with u and then integrate?

[steep meadow]

Not quite

You have to keep in mind the bounds too

[wraith dagger]

Yeetus

[sharp adder]

Okay thanks

[steep meadow]

No worries

[sharp adder]

But the root x stays the same outside the bracket? Or do I put it as u-1 ?

[steep meadow]

The substitution should get rid of all the roots

[sharp adder]

In my head that would just make 1/ u √ x or 1/u(u-1)

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[sturdy gale]

O am trying to find when particle crosses x or y axis

but I am really confused about how

thanks

[grand forge]

You know

Acceleration

And initial velocity

And distance

[sturdy gale]

so I tried integrating acceleration

[grand forge]

Distance is 6j

WOAH

i DONT think it’s necessary, since we’re only dealing with one dimension in this part a case

Coz we don’t care about the y position

[sturdy gale]

oh okay

[grand forge]

We just want x to be zero

[sturdy gale]

ahh alright 🙂

[grand forge]

Or Vice Versa

[sturdy gale]

yep

[grand forge]

So for part A we know it has move 6j

[sturdy gale]

yeah to get to x axis

[grand forge]

Yea

And u know acceleratatiob in that direction is 2j

And we know current velocity is j

So we can use suvat

[sturdy gale]

ooh cool

[grand forge]

s=ut + 1/2 a t^2

[sturdy gale]

this is for maths specialist so we are not allowed to use suvat

[grand forge]

And re-arrange for t = sqrt(2s/a)

[sturdy gale]

we have to use integrating

i know the suvat method

[grand forge]

Wtf

Why can’t we use suvat

[sturdy gale]

ik school is fed

[grand forge]

So we need to use calculus?

[sturdy gale]

mmhmm

thats what im learning now

Im in australia so it might be different for other countries

[mild arrow]

Logic dictates v=u+at , integrate that for distance

[grand forge]

^^

[sturdy gale]

ok ill try that

[grand forge]

With respect to t

[sturdy gale]

okay

I might get it now

will reopen if I need further help

thanks a lot

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[stark shell]

I kno this a math server but anyone know physics by any chance