help-19

[acoustic grotto]

i don't understand this part

how is tan(90-a) equal to cot(a)?

[tepid pelican]

it's an identity

do you know identities for cos(90-a) and sin(90-a)?

we can work with that too

[acoustic grotto]

yes, i know those

[tepid pelican]

cool

[acoustic grotto]

haven't memorised for tan and cot yet

[tepid pelican]

so note that tan(90-a) = sin(90-a) / cos(90-a)

[tepid pelican]

its not so necessary, you can simply derive it using the above

[tepid pelican]

Now plug in the identities you have memorised

[acoustic grotto]

okay i see

[tepid pelican]

what will it become?

[gritty oar]

were you initially suggesting this approach

[acoustic grotto]

cos(a)/sin(a)

... which is cot(a)

got it

[tepid pelican]

perfect

you can actuallly leave it in the form cos(a) / sin(a)

[acoustic grotto]

makes sense

[tepid pelican]

so we initially had $\frac{\cos\left(a\right)}{\tan\left(90-a\right)}=\frac{\sqrt{3}}{3}$

[clever fjord]

MethIsAlwaysRight

[tepid pelican]

now it's $\frac{\cos\left(a\right)}{\frac{\cos\left(a\right)}{\sin\left(a\right)}}=\frac{\sqrt{3}}{3}$

[clever fjord]

MethIsAlwaysRight

[tepid pelican]

because of the newly discovered identiity

[acoustic grotto]

yes

[tepid pelican]

it can be simplified a bit more

[tepid pelican]

not really but its actually quite interesting

this is alternative approach by cheby btw

it uses triangle and definitions of trig functions inside triangle

[acoustic grotto]

into sin(a) = sqrt(3)/3?

[tepid pelican]

yes

and this is where the problem comes

there is no nice value of a that would make sin(a) = sqrt(3) / 3

[acoustic grotto]

yes

[tepid pelican]

so ig you will need to just leave it in form sin^-1(sqrt(3) / 3)

[acoustic grotto]

can i just give this value?

[gritty oar]

you can also

but arcsin(sqrt3/ 3) just nice cause it's closed form

[acoustic grotto]

alright

how do i find tan(b)?

[tepid pelican]

it's not necessary actually

note that the sum of all angles is 180 degrees

[acoustic grotto]

yes

[tepid pelican]

one angle is right, that is 90 deg

one angle is alpha, that is arcsin(sqrt3 / 3)

and the last angle is just 180 - 90 - arcsin(sqrt3 / 3)

thats 90 - arcsin(sqrt3 / 3)

[acoustic grotto]

woow

it works out

thank you both so much for the help

i hope your pillows are cold both sides lol

[tepid pelican]

yw lol

[acoustic grotto]

i close this right?

[tepid pelican]

yep, you can

or ill do it for u

.close

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[normal snow]

im trying to find the solutions for x in
3sinx - 2sinx^2 -1 = 0

[normal snow]

but im not sure how to simplify this further

so I can factorize

[tepid pelican]

sounds like a good idea

it's essentially a quadratic equation

[normal snow]

oh I still didn't take quadratics with trig functions, this is the original question

domain is 0..2pi

[tepid pelican]

dw its simple

-2sinx^2 + 3sinx - 1 = 0

now factorize it like if it was a completely normal quadratic equation

imagine that it's -2u^2 + 3u - 1 = 0

[normal snow]

ahh alright, give me few secs

ok im back

i apparently forgot how to factor

[normal snow]

(2sinx + 1)(-sinx - 1)

is that right

[tepid pelican]

almost

[open crown]

@tepid pelican Can u tutor me on DMs or call? I need help with my master's thesis defense (Satire)

[normal snow]

is it something with the signs

[tepid pelican]

indeed

in the first factor

oh

and in the second too

[normal snow]

hmm

[tepid pelican]

what method did you use?

[normal snow]

-sinx(2sinx+1) - 1(2sinx + 1)

[tepid pelican]

-2sin^2(x) + 3sin(x) -1

-2sin^2(x) + sin(x) + 2sin(x) -1

i suppose you had it like this, right?

[normal snow]

-2sin^2(x) + 3sin(x) -1

-2sin^2(x) - sin(x) - 2sin(x) -1

i had them oppposite signs

[tepid pelican]

oh, thats wrong then

3sin(x) = sin(x) + 2sin(x)

not -sin(x) - 2sin(x)

this method works by rewriting the middle term to something equivalent

so 3sin(x) can be rewritten as sin(x) + 2sin(x)

because sin(x) + 2sin(x) = 3sin(x)

that's like apple + 2 apples = 3 apples

[normal snow]

i just thought to myself what two numbers multiply to 2 and add up to -3

so I chose -2 and -1

[tepid pelican]

they actually have to add up to +3

and you'd therefore choose 2 and 1

instead of -2 and -1

[normal snow]

isn't -2 - 1 = -3

i probably did something stupid

[tepid pelican]

it is

but they have to add up to **+**3

not **-**3

so you're looking for numbers which add up to +3 and multiply to 2

[normal snow]

is that because I had to factor out the negative at the beginning

[tepid pelican]

no

it's standard

the equations were

-2sinx^2 + 3sinx - 1 = 0
or alternatively
-2u^2 + 3u - 1 = 0

now you need to find numbers that add up to the B term (+3) and multiply to a*c ((-2) * (-1))

that's how the method works

it's +b, not -b

[normal snow]

ohhh I forgot...

my bad

let me start again

[normal snow]

(2sinx + 1)(-sinx - 1)

[tepid pelican]

thats the same

isnt it?

alright so let's look at it together

-2sinx^2 + 3sinx - 1 = 0

we have this

we are therefore looking for numbers which add up to +3 and multiply to (-2)(-1), that is +2

what numbers are these?

[normal snow]

sinx and 2sinx

1 and 2

[tepid pelican]

cool

-2sinx^2 + sinx + 2sinx - 1 = 0

so now we are here

[normal snow]

alr

we factor then

[tepid pelican]

yes

[normal snow]

first will be

-sinx(2sinx + 1)

[tepid pelican]

wrong

if you factor out -sinx from sinx

it will be negative 1

-1

[normal snow]

oops

yeah

[tepid pelican]

so -sinx(2sinx - 1)

[normal snow]

ok

and the next

[tepid pelican]

so now we have

-sinx(2sinx - 1) + 2sinx - 1

or -sinx(2sinx - 1) + (2sinx - 1)

or -sinx(2sinx - 1) + 1(2sinx - 1)

[normal snow]

yup

[tepid pelican]

can you factor that?

[normal snow]

(2sinx - 1)(-sinx+1)

+1

[tepid pelican]

+1, indeed

[tepid pelican]

as it is written here

okay so that's basically all

(2sinx - 1)(-sinx+1) = 0

this happens when either of the factors equals 0

so we have 2 equations, 2sinx - 1 = 0 and -sinx + 1 = 0

I'll give you some time to solve them, if you need help ping me

[normal snow]

thanks ❤️

I solved the first

sinx = 1/2

x = pi/6

[tepid pelican]

cool

[normal snow]

or 5pi/6

[tepid pelican]

yes

[tepid pelican]

indeed

that's first 2 solutions

[normal snow]

alright the next one is

-sinx+1 = 0

-2sinx = -1

[tepid pelican]

?

how did you get this?

-2sinx = -1

[tepid pelican]

hint: try adding sinx to both sides of this

[normal snow]

mixed both equations somehow

[tepid pelican]

I see

[normal snow]

sinx = 1

[tepid pelican]

yep

[normal snow]

so x= pi/2

that's the only one

[tepid pelican]

mhm

[normal snow]

i think

nice

[tepid pelican]

yes

only one in the interval 0, 2pi

so in that interval, there is total of 3 solutions

pi/6, pi/2, 5pi/6

[normal snow]

awesome, thank you so much for taking me through this step by step, it was very helpful

[tepid pelican]

you are welcome :]

[normal snow]

.close

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[zenith tartan]

is there a way to write P and C in desmos?

[zenith tartan]

i cant find anthing on google

[thin kelp]

?

what is P and C?

as in permutations and combinations?

[zenith tartan]

yea yea

[thin kelp]

probably just use the formulas

who decided a png would be good for this

what is the goal of using combinatorics in desmos?

[zenith tartan]

man idk im just trying to graph this function for fun

nvm

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[subtle perch]

I have to calculate the Area of this shape. i used pythagorean theorem to calculate EB and then used the Cosine theorem to calculate BC. Then i wanted to use the Sines Theorem to calculate the Angle ECB but when i do my calculations, the result is wrong. i will not attach my calculations as they are SUPER messy, but you can see my results on the Image.

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@subtle perch Has your question been resolved?

[subtle perch]

@dusk warren

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@subtle perch Has your question been resolved?

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[viscid meadow]

hey all! wondering how i can find convergence for this series:

[mystic saffron]

I'm not sure how this server works but my question isn't being solved.

[viscid meadow]

i want to use the divergence test but that ends up indeterminant

why is there

[viscid meadow]

ur question was asked 8 minutes ago

give it another 10

then ping helpers if no one has responded by then

[mystic saffron]

I had 2 ppl respond and they ghosted me, I even asked if they don't know its ok

[pastel steeple]

diverges by ratio test

[viscid meadow]

when we do ratio test, doesnt it end up at 2/3 ?

[viscid meadow]

patience is a virtue

[viscid meadow]

because it ends up as root n ((3-1)/3)

so that would state it converges. no?

[mystic saffron]

It's been almost 20 minutes since I asked

[pastel steeple]

doesnt nth term test work

[viscid meadow]

then @ someone

[viscid meadow]

whats the nth term test? sorry im not familiar with that one

oh divergence test

[pastel steeple]

i think you just called it divergence test

[viscid meadow]

yeah but when we take the limit of that we get indeterminate form no?

[pastel steeple]

the limit is 1 ≠ 0

[viscid meadow]

and l-hopitals would go infinite

[viscid meadow]

this is what i though the limit was just looking at the series

im struggling just putting it on paper

[random yew]

hmm

[pastel steeple]

wait

💀 oh shit

[viscid meadow]

yup

wonky question

it should diverge anyways so im just putting that down and moving on with my life

now how do we go about solving this one:

i dont think we can use limit comparison here right

unless we can? can we just use e^n for limit comparison?

because it cance;s to 1/ (n^2+1)^3 which is a p-series

[pastel steeple]

i just realized this is a sequence not a series

the sequence converges to 1

[viscid meadow]

yeah

[pastel steeple]

iirc you can only use p-series if it is in the form 1/n^p otherwise you need DCT or LCT

[viscid meadow]

welli kind of goofed in my thinking

bc it would have required LCT anyways

but you can (i think) use DCT here because this series is less than just e^n definitively

.close

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[kind basalt]

how do I evaluate this? i know it has to do with using log, and i know the square root of 81 is 9, but i don’t really know how to even start

[cold sage]

try express everything as an exponent of 3

[kind basalt]

like this?

[cold sage]

uh ish

why do your exponents have exponents

(a^b)^c is not a^(b^c)

[kind basalt]

ohh ok sorry I just don’t know what to do with the variables

[blissful musk]

(a^m)^n = a^(mn)

[kind basalt]

so like this?

[blissful musk]

Yea

Then you can simplify it

[kind basalt]

sorry but do i simplify it using log? sorry I just don’t really know what to do with the exponents

[blissful musk]

No you don’t need log

[kind basalt]

ohhh

[blissful musk]

Do u agree 3^(12n+8) is the same as 3^(12n) x 3^8

[kind basalt]

yeah

[blissful musk]

Then cancel out that -5n wit the 12n

[kind basalt]

by subtracting 12n by -5n?

[blissful musk]

double negative right

Or ykw

Let’s not complicate it

[kind basalt]

oh so 18n?

[blissful musk]

17 yea

Basically get rid of the negative exponent

[kind basalt]

oh yeah

oh ok

[blissful musk]

By bringing to the numerator

Now you can equate and ta da

Done

[kind basalt]

oh ok thank you sm

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[mystic saffron]

Pls help I have a unit test on Friday

[mystic saffron]

And I’m so confused on this

[kind pendant]

What’s confusing you?

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[mystic saffron]

The whole topic

[mystic saffron]

.reopen

@kind pendant

[kind pendant]

Are you familiar with grouping like terms?

[lost hull]

i think thats were you transpose

the same terms

[odd edge]

@mystic saffron Has your question been resolved?

[mystic saffron]

Yes

[kind pendant]

Group the ones with variables at the LHS and the ones without at the RHS and make the variable the subject

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[mystic saffron]

I can do these

[mystic saffron]

For all liner equations right

?

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[stiff kettle]

im a little confused on the final answer

It asks to find the exact value

Thats what I did

[cold swift]

so let’s rewrite the expression in terms of sine and cosine so that we can evaluate them easily

[stiff kettle]

The answer key says its square root 3 over square root 2

so idk what i did wrong

[cold swift]

yeah because $\frac{cot(30)}{sec(45)}=\frac{cos(45)}{tan(30)}$

[clever fjord]

y0shi

[cold swift]

not what you have up there

[stiff kettle]

o

[clever fjord]

y0shi

y0shi

[stiff kettle]

ok but either way

i flipped it and now i have

3 (√2) over 2 √3

so does the three and two just cancel out

yep ok i got it ty

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[real nexus]

How do I do seperation of variables on this

[narrow cove]

A little factorising ought to do the trick

And recalling that $(xy)^a = x^ay^a$

[clever fjord]

TayBee

[real nexus]

Omg I was factoring it wrong this whole time

I was factoring it as if there was addition between the x and the y

so i was getting like x^2 +2xy +y^2 or sum

[narrow cove]

oops lmao

Well yes, you might find it becomes a little more simple without the addition shenanigans XD

[real nexus]

ya ty

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[lament cargo]

yo

[lament cargo]

9/12 multiplied by 100/1

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@lament cargo Has your question been resolved?

[lucid shadow]

do u know how to multiply fractions?

with lesser numbers

[lament cargo]

yes

but i aint getting the answer bruh

i reduce them

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[civic hound]

how can i find inverse function of this? the domain is R^2 \ L, where L = {(x,y) | x=0 and y <= 0}

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[tender oak]

https://math.stackexchange.com/questions/1711535/finding-an-inverse-of-fx-y-ex-cos-y-ex-sin-y-on-a-neighborhood-of-a-gi

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[distant flame]

find all primes $p$ such that there exists integers $a,b \geq 2$ satisfying $p^a-1=5*2^b$.

[clever fjord]

chloë

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@distant flame Has your question been resolved?

[distant flame]

<@&286206848099549185>?

[kind pendant]

!original

[odd edge]

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

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[open shadow]

Can I use power rule to find the derivitive of $y=x^2-3x$

[clever fjord]

guy

[open shadow]

we haven't learned chain rule yet, so we cant use that, but we have learned power rule, so can that be used to find the deriviative of this?

or do i need to use limit deffinition

[static mulch]

you can use power rule

[open shadow]

but what about the 3?

3x

[static mulch]

first try $x^2$ what do you think that'll be

[clever fjord]

Jaxx

[open shadow]

that would be $2x$

[clever fjord]

guy

[static mulch]

yes now explain to me what you did by doing the power rule

what happened to the power of x

[steady tide]

,tex .diff rules

[clever fjord]

فطر

[open shadow]

i subtracted it by 1

and moved the power over to the side

[static mulch]

so with 3x

what is the power of x

[open shadow]

1

[static mulch]

so do the same thing now

[open shadow]

but my main worry was isnt there a rule for like power rule that you cant do it with something with multiple terms

or something like that?

and thats where chain rule comes in?

not sure

[static mulch]

reference the rules above

specifically sum rule

[open shadow]

(1)(3)x^0=3

[static mulch]

$f(x) + g(x)$ = $f'(x) + g'(x)$

[clever fjord]

Jaxx

[static mulch]

yes

[open shadow]

ok i see, thank you

.close

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[steady tide]

this is not true

[static mulch]

$\frac{d}{dx}(f(x) + g(x))$ = $f'(x) + g'(x)$

[clever fjord]

Jaxx

[steady tide]

there we go

[static mulch]

[odd edge]

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[fair cloak]

How do I take the limit

[fresh ruin]

you'd put the whole thing in the 1/2 exponent to see it's 1

[mystic saffron]

intuitively speaking, n + 1 shouldnt be all that different from n when you are approaching infinity

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[gaunt frigate]

how do i find the minimum and maximum values that each variable can take in a augmented matrix?

[gaunt frigate]

for example a matrix of
1 0 1 0 1 | 1000
0 1 1 0 1 | 1000
0 0 0 1 1 | 800
0 0 0 0 0| 0

i want to find the minimum and maximum values that each of x1, x2, x3, x4, and x5 can take

<@&286206848099549185>

[crisp sierra]

Are there any other constraints on x1,...,x5? like they are all non-negative or something?

[gaunt frigate]

no constraints as far as im aware

i just gave a random example matrix, i just want to know in general how i would find it

[crisp sierra]

i suppose, if you have additional constraints you can just set each variable to the maximum allowed by the constrtaint, like in row 1 you set x1=1000 and x3=0 x5=0, then continue like that, but if there are no non-negativity constraints alongside the given matrix, you could set x1 to be as large as possible and x5 is a free variable so it can be as low as possible so every variable has an unbounded min and max at the moment.

[gaunt frigate]

hm ok

ok wait i think the constraints are that the solutions to the equations must not change

[crisp sierra]

yes, the 1000, 1000, 800 are fixed

but for example, i could make x1=1000000, x3=1000, and x5=-1000000, then do some algebra and find what x2 and x4 would be but you can do this for each variable

[gaunt frigate]

actually yeah i believe the constraints are that they cannot be negative

[crisp sierra]

that would make a lot more sense, then you look at each variable individually and choose the highest value it can be while still satisfying every equality it is part of

so x1 is only in r1 so you can set x1=1000 which would be its maximum value

same thing for x2

but x3 is in r1 and r2, but both are equal to 1000 so the maximum value for x3 is also 1000

x4 would be 800 for similar reasons, but x5 is present in r1, r2 and r3, where the largest it can be is 800 without violating the constraint from r3

and the minimum for each variable would be 0

[gaunt frigate]

alright i think i get what you mean

thanks

@crisp sierra sorry just 1 more thing

for example what if the 1st equation in the matrix was something like 1 0 -1 0 1 | 1000

because im pretty sure the only constraint is that the variable itself cannot be a negative number

[crisp sierra]

correct, so x3 would still be bounded by 1000 from r2, but if i took x3 to be 1000 i could take x1 to be 2000 and x5 to be 0 which would make x2 also 2000 and x4 800

but if you would make the first row -1 0 1 0 1|1000, then you are unbounded since x1 does not appear in any other row

[gaunt frigate]

yea because x1, x2 and x4 are pivots and x3 and x5 are free

[crisp sierra]

correct

ehh actually that example might be a bad one

[gaunt frigate]

ill just show u what my matrix is

maybe that would be easier to explain

[crisp sierra]

fdair

fair

[gaunt frigate]

1 0 -1 0 1 | 1000
0 1 1 0 -1 | 1000
0 0 0 1 1 | 800
0 0 0 0 0 | 0

[crisp sierra]

so the maximum value for x1 is 2000, since i could make x3=1000 and x5=0

the maximum value for x2 is 1800 since i could make x3=0 and x5=800

x3 has a max of 1000

x4 is 800

and x5 is 800

[gaunt frigate]

ok but why i cant i just say x1 is 10000, since you could make x3 = 9000 and x5 = 0. now it seems to me that there is no upper limit

[crisp sierra]

since that would break the other equations in the other rows

r3 would not be satisfied in that case

r2 sorry

[gaunt frigate]

r2 as in row 2 or equation 2?

oh wait same thing

[crisp sierra]

same thing yea

[gaunt frigate]

so we would be saying x2 + 9000 - 0 = 1000, which doesnt make sense

[crisp sierra]

correct since x2 is positive

[gaunt frigate]

because x2 must be positive

oh ok that makes so much more sense

thank you

[crisp sierra]

np

[gaunt frigate]

is the minimums all still 0?

[crisp sierra]

sould be if you talk about each value individually

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[glacial timber]

hi chat

pls help

i tried engineer's induction like i take the sum of the first term, then the sum of the first and second term, the sum of the first second third term, and so on till i find a pattern, and i did not find a pattern

[true nest]

chat here

Try rationalizing the denominator

[glacial timber]

okay chat

[true nest]

then apply difference is squares

[glacial timber]

i forgot what it's called

okay

[true nest]

show me what you get!! 👁️

[glacial timber]

like squareroot(a)+squareroot(b) so i multiply by squareroot(a)-squareroot(b)

is that what u meant

i forgot what it's calleddd

conjugate

yes

[true nest]

yes

[glacial timber]

i remembernow

okay

did not help

i think

@true nest

pls help again

🥺

[true nest]

damn

uh

whatd u get?

[glacial timber]

nothing

[true nest]

uh

like

(2sqrt(1)-1sqrt(2))/2 is the first term?

[glacial timber]

yes

2nd term is (3sqrt(2)+2sqrt(3)/30

[true nest]

yeah

wait

[glacial timber]

can't see shit ngl

there's probably some magical pattern i can't see

[true nest]

/3 not /30 but yes

also - not +

[glacial timber]

oh yeah

wait

i'll try agian

i forgor

tnx

[true nest]

😔

[glacial timber]

bruh i still can't see shit

😭

[true nest]

bro what

[glacial timber]

😦

[true nest]

seperate the terms in each feminization

err denominator

[glacial timber]

wdym

[true nest]

like

2sqrt(1)/2 - 1sqrt(2)/2

and then factor ig?

and then u see amazing patterns like as noted in color out of space by h. p. lovecraft

[odd edge]

@glacial timber Has your question been resolved?

[true nest]

where are u getting stuck?

[glacial timber]

nah i went out to do smth

i'll try it agian

[true nest]

that’s wild

[glacial timber]

fr

wat am i doing wrong

?

the answer is this btw

but no solution to learn from

[true nest]

ya

nonoooo wait

[glacial timber]

okay

[true nest]

do not merge the denominators

[glacial timber]

oki

[true nest]

seperate all terms and factor

[glacial timber]

sry idk what u mean

[true nest]

(2sqrt(1)-1sqrt(2))/2 becomes 2sqrt(1)/2 - 1sqrt(2)/2

which becomes 1 - sqrt(2)/2

this is also known as telescoping I think

[glacial timber]

okay

i'll figure shit outg

[true nest]

so real

[glacial timber]

i'm uneducated

[glacial timber]

WOAH

papa

ur smart

ye i found it

tnx

i'll kiss u goodnight

[true nest]

please

never

[glacial timber]

okay

thanks still

i appreciate u

[true nest]

I am fine with this but don’t call me papa pls

[glacial timber]

okay

[glacial timber]

mb

[true nest]

are you ok

[glacial timber]

yes

thanks

.close

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[hardy verge]

Don't mind the spss. hehe

[hardy verge]

Guys can you please help me?

[honest quest]

This seems like a standard question. Was there anything in particular you didn't understand.

Probability and Statistics for Engineering Sciences by Jay L Devore is good reference.

[hardy verge]

All of them 😭 I didn't understand a thing in online classes

[honest quest]

The first thing is know how things are defined

Take your time to understand things

Like conditional probability, what is mean/standard deviation

You can find fully worked solutions on Quizlet

I will leave the rest to you

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[forest thicket]

Is the bottom question 314

[forest thicket]

I have a test first period and I just want to double check if I’m doing these right

[nimble blaze]

no, show your work

[odd edge]

@forest thicket Has your question been resolved?

[forest thicket]

It’s wrong? Bruh

[mystic saffron]

show your attempt to answer lol

[forest thicket]

Yesterday I showed the formula I used to get the final surface area but nobody answered

[nimble blaze]

probably did it ata bad time
explain your reasoning/ show wrok again

[forest thicket]

I can’t

I’m in the car right now on the way to school

And the test is there

Wait what, I did it in my head and it’s right to me still

[nimble blaze]

type your thoughts

and name the propeties you're applying

and not a good idea to do stuff in your head for stuff you're having issues with

[forest thicket]

Okay so what I think I need to do is add the area of both

Correct me if I’m wrong

[nimble blaze]

of both what

[forest thicket]

So the area of the cone is 226.67

[forest thicket]

Both the cone and the cylinder

[nimble blaze]

how are you getting 226.67

[odd edge]

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[shy bison]

Hello, can someone help me with no. 4?

[shy bison]

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[tiny portal]

I need help

[odd edge]

@tiny portal Has your question been resolved?

[tiny portal]

<@&286206848099549185>

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[gentle timber]

This is a physics question

[gentle timber]

When is work done by a spring positive

I know that work is = the change in kinetic energy

[pulsar current]

Good, using that logic if it is increasing the kinetic energy of what's attached to it, it's doing positive work

Other way to think about it is dW=F.ds

Dot product ^^

[gentle timber]

If K finial is > K initial then Work is positive

But I guess what I’m asking for is for what position of the spring is work positive

Like if the spring is compressed or stretched

[pulsar current]

That you can figure out

If it's increasing the kinetic energy

It means it's increasing its speed

Which means its 'supporting' it's motion

[pulsar current]

This is definitely a better option in this case

Because you can easily relate the direction of force with the position of the spring

[pulsar current]

Anyway, this logic also works since we know it will apply a pulling force when it's elongated, so when it's elongated AND the block is moving TOWARDS the mean, it will be doing positive work

Or, when it's compressed AND and the block is moving away from the mean, the spring will be doing positive work inthis vase as well

[gentle timber]

It seems like the displacement and force both point in the same direction

[pulsar current]

To put it simply, when force and displacement are opposite work is negative

[gentle timber]

And in what scenario will that happen?

[pulsar current]

When the particle is going towards the extreme, and the mean has been crossed

[gentle timber]

What is mean

[pulsar current]

The mean position

Where x=0

Natural length

Lemme illustrate with a diagram wait

[gentle timber]

Oh

Ok

[pulsar current]

In these two scenarios

We have negative work done

Since velocity is opposite to the dirn of force

[gentle timber]

Oh i think i understand it now

After being released at the endpoints, extreme, Fs points towards the mean while displacement is positive

Before I was only considering the spring before it was released

Thanks

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[limber quarry]

what do you do here if i is zero?

[limber quarry]

ie the top part has a negative number?

also i forgot what the symbol was called

[gritty oar]

can you show original problem

[limber quarry]

its from the formula for newtons polynomial interpolation

[gritty oar]

ah

[limber quarry]

problem is i starts at 0 which means that there is a -1 on top of the multiplication symbol

[gritty oar]

that thing is defined for i >= 1

[limber quarry]

so ... what does that mean if the top has a negative?

[gritty oar]

well i >= 1 means that i - 1 >= 0, so the top is nonnegative

so the $x-x_j$ are indexing through the right points

[clever fjord]

chebyshev's infinite pee norm

[limber quarry]

but in the summation, i starts from 0

[gritty oar]

there's a base case for the basis polynomial for i = 0, n_0 = 1

their notation is misleading though yeah

[limber quarry]

ahh ok yeah thats what i meant

[limber quarry]

ill check out other formulas too then

thanks

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[latent topaz]

My math curriculum sucks, and is horrible at explaining how to graph hyperbolas, if i provide some problems can someone walk me through the steps on how to solve?

[split plover]

rectangular or diagonal

i mean either way the steps are the same

find the intercepts, find the asymptotes, trace through

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[forest kayak]

Hii

[forest kayak]

Does anyone know how to do this question

Its from m3

[fresh ruin]

you'd do conservation of energy, since it starts at the top with 0 kinetic, some gravity potential, and some string potential, and at B has speed v, less gravity potential, and less string potential

[odd edge]

@forest kayak Has your question been resolved?

[forest kayak]

@odd edge

No

[forest kayak]

@fresh ruin

Please see the marking scheme

They used EPE

AND L ROOT 2

Where did that come from

And in the b part the angle

They used 45

Can you please if possible do that question and send if you know

Thank you

[fresh ruin]

sqrt(2) is at B

that image shows the full solution so just let me know which part looks weird

[forest kayak]

They used pythagoras

Oh okay

I got the a part

What about the b part

That looks weird

@fresh ruin

They used the angle 45

Idk how did that come

[fresh ruin]

for b you know the horizontal total force must be mv^2/r to go in a circular path, so they write all the forces in the horizontal direction

which is the horizontal component of the diagonal string tension, and the ring normal pushing in some way

[forest kayak]

Yes

Oh they found the angle from that

Oh okay

@fresh ruin can i add you like in future to ask any questions if i have😅

M3 is like my biggest enemy so🥲

Thanks alot

I got this question now

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[inner socket]

I have the following inverse CDF function: N(m, sigma).inversecdf(tau) + tau = v. This function takes tau and returns v, how can I find the inverse of this function, aka the CDF?

[inner socket]

In other words, is there a way to solve for $\tau$ here?

$$
N(m, \sigma).inverseCDF(\tau) + \tau = v
$$

[clever fjord]

LucasYerz

[inner socket]

Still trying to solve for $\rho$:

$$
\Phi\left(x\right)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\frac{x-m}{\sigma}}e^{-\frac{t^{2}}{2}}dt
$$

$$
\Phi^{-1}\left(\rho\right)+\rho = v
$$

[clever fjord]

LucasYerz

[inner socket]

is there a shortcut here for trying to take an inverse of another inverse function?

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@inner socket Has your question been resolved?

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[hollow abyss]

Where did it go awry? The answer is supposed to be x<0

[hollow abyss]

I lost the 3

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[hollow abyss]

.reopen

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✅

[hollow abyss]

Is there a way to solve in a way similar to what I did?

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[steady tide]

$\left(\frac{\dd y}{\dd x}\right)^2=(x+1)y$

[clever fjord]

فطر

[steady tide]

actually im xying

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[nimble lotus]

How would you do this? I tried to find the hypotenuse for the second triangle by using pythagorean. After that I found the scale factor but I keep getting 20 for CB but the answer key says it's 15. idk what I'm doing wrong

[nimble lotus]

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[desert marlin]

I assume there is an easy combinatorial way to count the amount of paths (possibly backtracking) of length 4 from v12 to v14. Could anyone enlighten me?

[bronze fox]

Paths of any length?

[desert marlin]

Of length 4

[bronze fox]

My bad

[green elm]

is v12 that black dot that is not connected to anything?

labeling is unclear

[desert marlin]

it is meant to be the dot next to the black dot

sorry

and same with v14

it's the nearest real vertex (all the real vertices are purple)

[summer cradle]

for any one of the the top or right "diamond" of vertices, there is exactly one path through them

so you can ignore those and simplify the picture a bit

[desert marlin]

this is not true though, you can go into the center, and then into any vertex of the top or right (or left) diamonds and then into the center and then into the graph we wanted

the paths can be backtracking

[summer cradle]

how does this disagree with what i said

you just described the unique path

[desert marlin]

There are multiople though depending on what vertices you pick

[summer cradle]

ok give me one second

can you tell me how many length 4 paths from v_12 to v_14 there are that go through the red vertex?

[desert marlin]

Just 1 ma'am

[summer cradle]

yea exactly

same with 7 of the other vertices

[desert marlin]

so we get 16 paths that hit the center twice

and now need to consider paths that only hit the center once

[summer cradle]

eh no

8 paths that use the top and right diamonds

[desert marlin]

16 total

[summer cradle]

huh?

[desert marlin]

We start at v12

we go to the center

we can go to any other vertex on the graph now

(besides the center)

and then back to the center

and then to v14

there were 16 options to choose from

when we were on the center

and the rest of our choices were determined

so 16 paths that hit the center twice?

[summer cradle]

yes ok but that wasn't my point

[desert marlin]

Sorry, what was your point?

[summer cradle]

it was that we can just count the paths on this graph now that we have the paths that use the ones i blanked out

[desert marlin]

Yes

Agreed ma'am

[bronze fox]

I think you should now count the number of paths that use one of the V12 K4 edges. Then count the number of paths that use 2 of the V12 K4 edges. Including the ones you just counted (center in/out). I think that’s all your paths

[summer cradle]

it's not hard to count cases from here

[desert marlin]

Agreed, but I thought there might be an easier way than counting cases

[summer cradle]

[desert marlin]

Is there not

[summer cradle]

maybe some kind of search but that's not going to be any easier to do by hand me thinks

[desert marlin]

Well

I'm going to next find paths of length 5

and then 6

etc..

[summer cradle]

oh

[desert marlin]

So I was hoping something would generalize for those

[summer cradle]

yea there are better ways then

[desert marlin]

I know that we can take the adjacency matrix

and raise it to a power k to get the amount of paths of length k

in the entries

But

I want to also know the vertices which are in those paths

so I thought that would not be helpful

I can show you what I am trying to compute

[summer cradle]

yea i know what you're trying to do

[desert marlin]

You do?

[summer cradle]

wait you need this?

[desert marlin]

Yes, see the formula

Summing over gamma paths of length k but then considering a product where vp are the vertices in those such paths

[summer cradle]

eh idk what that is honestly but if you just need the number of k length paths from v_12 to v_14 for each k it doesn't require anything too fancy

computational nightmare tho

[desert marlin]

I need the number of paths, but for each individual path I need to know the vertices that are in it

so I can compute that product, since if it hits the center the valence will be different than if it did not

[summer cradle]

what is the original question

[desert marlin]

Well, there is no original question, it's for my sir and I

we are researching

[summer cradle]

[desert marlin]

I can explain to you the gist

This formula gives you the resolvant matrix as a function of lambda for some graph

which is equivalent to

(L_G - lambda I )^(-1)

where L_G is the graph laplacian

And, my idea for my sir was that if we have a graph like the one in the image

where the center vertex is weighted very highly compared to the other vertices with uniform weight

that we should be able to approximate the entries in the resolvant matrix, by considering only paths that do not go through the center

as these will get killed off because of their weight being so low

So I'm just trying to work through an example and compare what happens with this

[desert marlin]

and as such I wanted to approximate the G_(v12, v14)(lambda) for this graph

using this formula

But then I'm getting stuck on computing it because the path shennanigans get complicated

That's my doubt

[summer cradle]

[desert marlin]

Do you have any ideas?

Maybe my sir @green elm has ideas

[bronze fox]

Is P_k the number of paths of length k?

[desert marlin]

yes

[green elm]

this looks disturbingly like graph theory

[desert marlin]

it is really combinatorics sir

I think

[bronze fox]

Have you done this replacing the K4s with individual vertices? That seems relatively simple since you can ignore the odd paths and might still be a good example of approximating the resolvent matrix.

[desert marlin]

yes

[bronze fox]

Also, it seems like you can simplify the counting significantly by counting the number of paths of length n that start and end at the center

[desert marlin]

That would only be one element of the resolvant matrix though

and we'd still need to compute the others

[bronze fox]

then you just count how many ways you can get from and to the center

[desert marlin]

I’m not sure how that would work but I’ll have to look into it

[bronze fox]

Then the sum splits into the three components for paths of length n=x+y+z. How many paths of length x stay in the first component before going to the center. How many of length y stay and return to the center. How many of length z leave and stay in the second component. This seems to get quite challenging to compute very quickly though, although there might be nice simplifications

For paths starting and ending in the center I think the sum is the same as the number of ways we can sum to y using integers >= 2 (going to any component requires 2 steps) times 4 (for the 4 components)

I’m pretty sure the paths staying in components are 3^(x-1) and 3*(z-1) respectively

[desert marlin]

x, z being what?

[bronze fox]

n=x+y+z for paths of length n (for the three components of the path)

[bronze fox]

From this message

[desert marlin]

I see

I’ll have to play around with this stuff

Thanks for the advice

[bronze fox]

Another potential idea. Call the number of paths of length $n$ that start and end at the center $\gamma$. Then, number of paths of length n from any two vertices $v,w$ (other than the center vertex) is $\frac{\gamma}{16^2}$. This is because we can imagine deleting the first and last edge from an initial path $p$, giving a new path $p’$. Then, $p’$ is a path between $v$ and $w$ if the fist and last edge of $p$ connect to $v$ and $w$. Thus, we divide by $16^2$.

[clever fjord]

Owen

[odd edge]

@desert marlin Has your question been resolved?

[odd edge]

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[odd edge]

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[acoustic grotto]

hii i need help with question 3 please, the translated text is: Determine the equation of the line inclined to the axis OX at an angle of 135°, which intersects the axis in point (-3.0).

[acoustic grotto]

,rotate

[acoustic grotto]

<@&286206848099549185>

[modern gyro]

Hi bro

[odd edge]

@acoustic grotto Has your question been resolved?

[odd edge]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[acoustic grotto]

I don't know where to begin

[acoustic grotto]

Hi again

[modern gyro]

Wooow

[acoustic grotto]

Sorry i left you on that last question i asked, i fell asleep 😭

[modern gyro]

Ok.np

But i cant understand that language

[acoustic grotto]

Translation here

[modern gyro]

Hmm

[acoustic grotto]

I have to eat rn, ill see ur msg once i finish, hope that's ok

[modern gyro]

Okk

tan 90 ° is undefined so you get (2/0) form

[acoustic grotto]

wait why do you say that x + y = -3

[modern gyro]

Thats the above question from that sheet i thought that was your problem

Ignore it

Got it ??

[acoustic grotto]

I was asking for question 3 tho

I think you solved for 5

[modern gyro]

Yes

You didnt mention

[acoustic grotto]

Bro i said it here

[modern gyro]

3 there 🙂

[acoustic grotto]

Thank uu

Aha i got it

Thanks for the help!!

[modern gyro]

🙂

Cool bro

[acoustic grotto]

@modern gyro i have one more question

[modern gyro]

Yeah

[acoustic grotto]

Yk these reduction formulas, for example sin (90+x) = cos x

[modern gyro]

Yes

[acoustic grotto]

How can i memorise them?

There's so many and idk how to memorise them quickly

[modern gyro]

See its simple ypu know which function +ve which quadrant

?

[acoustic grotto]

Yeah

1st - all +ve
2nd - sin +ve
3rd - cos +ve
4th - tan +ve
Right?

[modern gyro]

so see whenever you see a multiple of pi/2 change sin to cos , or tan to cot or cosec to sec

And keep in mind the sign

And accordingly put it

[acoustic grotto]

Wdym multiple of pi/2? Could you give an example?

[modern gyro]

Whenever you see multiple of pi just think abt the sign

[acoustic grotto]

Also I haven't done sec or cosec in school yet

[modern gyro]

I mean odd multiplw of pi/2

Lik 90

270

450 °

So on

So sin( 90 + x) gives you cos x

[acoustic grotto]

pi like 3.14?

[modern gyro]

Why because sin is positive in 2nd quadrant

[acoustic grotto]

I don't understand how 90 is a multiple

[modern gyro]

No i meant pi radian

180 ° = pi radian

You need to learn all this bro

[acoustic grotto]

90 is a multiple of 180 so I have to change sin to cos?

[modern gyro]

No odd multiples of 90

90×1 90×3 like that

[acoustic grotto]

Ohh so it changes only for 90, 270, 450, etc?

[modern gyro]

Yeahh

Not for even multiples of 90

[acoustic grotto]

And the rest like 180 I keep the same but I just change the signs according to the quadrant?

[modern gyro]

Yeah

Just change sign according to quadrant

Like sin 180 + x

Gives - sin x