help-39

[fair creek]

respectively

and resistors and capacitors do not consume any current

[midnight haven]

No, I can give you what I have

If you can correct me

I'd appreciate that

[fair creek]

okay

[midnight haven]

So, gotta consider this

And I found the number of currents to be 3

I'm just trying to understand which direction to label them in

And how they relate to each other (in the equations)

[fair creek]

can you show what currents you have labelled?

[midnight haven]

I have this so far

My friend told me to flip either I2 or I3

[fair creek]

yeah this is pretty much it

[midnight haven]

But I don't understand why

Or if this is fine

Tryna be careful LOL

[fair creek]

Kirchhoff's first law dictates that incoming current = outgoing current in a junction

and per standard convention

[midnight haven]

Yeah

[fair creek]

currents are produced by cells only

so use that :3

[midnight haven]

Whwat does this mean

I'm not sure

this is okay? the labeling and everything

[fair creek]

yes but

do note taht I3 = I1+I2 (it's better to write the latter)

[midnight haven]

sorry just tryna check, have to leave in 7 min

I have I1+I2=I3

[fair creek]

yes this is the correct way

[midnight haven]

That's the junction

Equation

And there's 2 loop equations, right?

For the smaller loop (V2 and R1), we have V2 - I3R1 = 0

For the "greater" loop (whole circuit), we have V1 - (Q/C) - I1R2 = 0?

Do we neglect the "smaller loop" for the second equation? And its resistor, i.e., R1?

I think the first one's ok

but not the second one

[fair creek]

yes, but they're unnecessary here if we're only concerned about the currents

[midnight haven]

It's asking me to apply kirchoff to write one junction eq and two loop equations for this circuit

without solving them

@fair creek

[fair creek]

yes, there's one junction (two but they're the same) and two loops so use that

this is pretty straight forward if you know it

[midnight haven]

yeah

so

V2 - I3R1 = 0

within the loop

that's fine right

and for the other one?

[fair creek]

yes

[midnight haven]

For the other one, would it be

V1 - (Q/C) - I1R2 - I3R1 = 0

or

V1 - (Q/C) - I1R2 = 0

[fair creek]

i think you ignore the smaller loop because it's equivalent resistance (via the cell) is zero

[midnight haven]

so this right

[fair creek]

i believe so

but i think you should

think of it this way

i believe i'll just show you what the loops should look like

i think the problem was intended to be approached this way

so for the larger loop, it goes from V1, C, V2 and R2

@midnight haven

[midnight haven]

yeah, stuck to I1+I2=I3, v2-I3R1=0 and V1- Q/c - I1R2 = 0

i think it makes sesne

though i "neglect" the smaller loop in my second equation

[fair creek]

i made a tiny error 😔 if you rotate the junctions, you can form a loop including R1

[midnight haven]

it's kinda like it doesn't exist

[midnight haven]

what does that mean

[fair creek]

your figure is equivalent to this

then apply the loop so you get R1 in it

[fair creek]

basically, this is the expression for the larger loop

pardon for my initial error

[midnight haven]

No v2 in it?

[fair creek]

from the smaller loop, V2 and V2=I3R1 so they mean the same thing :3

[midnight haven]

You sure?

man, i don't know

[fair creek]

yes

[fair creek]

did you forget about this?

[midnight haven]

no but like are you sure there is a v2 in the larger loop equation

@fair creek

[fair creek]

yes

like i just showed, you apply a loop and whatever is in the loop comes into the equation

hope that helped anyhow

[midnight haven]

so this right

[fair creek]

yes

[pearl pond]

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[pearl pond]

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[fast pilot]

for this qu, what would the projection lines for all the other vectors look like? I did one for -i + j but im not sure how the projection is meant to look for the others

[wild fable]

hsc nice

u do 4u as well?

what on earth are you talking about

[frank finch]

lolcow
you dont know about lolcows?

[wild fable]

no.

[frank finch]

it is a term for people online who make a fool of themselves

[sharp vigil]

please take conversations not about the question to #discussion

[frank finch]

basically the lowest of low

oh jesus i thought i was on discussy

[toxic lichen]

!redir

[pearl pond]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[toxic lichen]

yeah uh. you're not

[pearl pond]

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[pearl pond]

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[sharp smelt]

Part(b)

[sharp smelt]

part(b) is confusing the heck out of me

[rustic tendon]

you're gonna have to pull some trickery

i need paint

[pearl pond]

Show your work, and if possible, explain where you are stuck.

[sharp smelt]

part(a) is doable without green's

[sharp smelt]

no idea on part (b)

[rustic tendon]

The encircled bit is the origin

What I want you to notice is the decomposition of the path such that the line integral over C can be written as one over C1 and C2

C1 is the upper half, including the semicircle, and C2 is the lower half, also including the semicircle

[sharp smelt]

hmm

[rustic tendon]

Notice that their traversal ends up cancelling the straight lines in between

[sharp smelt]

hmm

okay

[rustic tendon]

I've added some more labelling so that we can get explicit

[sharp smelt]

hmm

okay

My concern is that the path is arbitarry

[rustic tendon]

C is still arbitrary

The cyclic integral on C1 can be written as the integrals of:

1. B -> A (over C)
2. A -> P (straight line)
3. P -> Q (upper half of the semicircle)
4. Q -> B (straight line)

I'm sorry I drew the arrows the wrong way on the inner circle

[rustic tendon]

Do you agree to this?

@sharp smelt

[sharp smelt]

yea

I do

so the only bit of the integral that survives is the integral from P to Q to P?

[rustic tendon]

A similar decomposition may be performed on C2:

1. A -> B (over C)
2. B -> Q (straight line)
3. Q -> P (lower semicircle)
4. P -> A (straight line)

[rustic tendon]

in essence yeah, over the circle

Add the integrals over C1, and C2, and what you have is:

1. (B -> A) & (A -> B) (complete integral over C)
2. (A -> P) & (P -> A) (cancels out to 0)
3. (P -> Q) & (Q -> P) (complete circle enclosing the origin)
4. (Q -> B) & (B -> Q) (cancels out to 0)

You get a very interesting result here now

[jolly parrot]

dyxn

[rustic tendon]

@sharp smelt are you with me so far?

[sharp smelt]

uh, one min

tea

okay

[rustic tendon]

now C1 and C2 DO NOT enclose the origin

Which means that the integrals are 0

[sharp smelt]

that's for part(b), right

wait, what

but the answer says 2π

[rustic tendon]

I'm using the result from (a)

does not enclose ==> integral is 0

[sharp smelt]

yea

[rustic tendon]

I fixed the arrows to make it easy to follow

to summarize:
C1 is the contour that starts at B, moves to A along C, then to P, along the upper semicircle and back to B.
C2 is the contour that starts at A, moves to B along C, then to Q, along the lower semcircle and back to A.

is that clear now?

C1 and C2 both DO NOT enclose the origin

[sharp smelt]

yeah

okay

[rustic tendon]

now we have this result here

and since C1 and C2 do not enclose the origin, the right side is 0

agreed?

this follows from what you have proven in part (a)

for an arbitrary contour that does not encircle the origin, the integral is always 0

[sharp smelt]

yea

!occupied

[rustic tendon]

delete your message and post it in #help-18

[pearl pond]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[heady bane]

ok, but it was school maths

[rustic tendon]

so now we have $\oint_{C} = -\oint_{\text{circle}}$

[jolly parrot]

dyxn

[rustic tendon]

a note on orientation, if C runs clockwise, the circle runs counterclockwise and viceversa

so now can you compute the integral over the circle with a simple parametrization

we never specified a radius, so you can take it to be 1

[sharp smelt]

but now this circle contains the origin, right

[rustic tendon]

yes it does

you can just compute the line integral

[sharp smelt]

How can I assume radius 1 though

[rustic tendon]

what would you like the radius to be

[sharp smelt]

the radius coule be $e^π-1$ for all we know

[jolly parrot]

What a wonderful world !
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[rustic tendon]

sure do that then

[sharp smelt]

just saying

[rustic tendon]

it doesn't matter

1 is easy for computation

while constructing C1 and C2 we didnt care for the radius

[sharp smelt]

the line integral still won't be 2π

wold it

[rustic tendon]

why not

isn't that the perfect differential for arctan(y/x) or something

[sharp smelt]

Oh right

my bad

yea

[rustic tendon]

$- \int_{2\pi}^0 \arctan (\frac{r\sin \theta}{r\cos \theta}) \dd{\theta}$

[jolly parrot]

dyxn

[sharp smelt]

Got it

[rustic tendon]

eh or actually just parametrize y = sin t and x = cos t

dont do this perfect differential bullshit

you might have to cry about domains and stuff

[sharp smelt]

fair

yea

okie

[rustic tendon]

To run it by you one more time, what we did was select 2 contours (both NOT enclosing the origin) such that they sum up to C (arbitrary) and another contour on which parametrization is easy (the unit circle, traversed in the opposite direction).

Since the contours also sum up to 0 we get a handy relation between the arbitrary contour C and our cherrypicked contour

and the integral over the cherrypicked contour is 2pi

which is equal to any arbitrary contour C as long as it encloses the origin

I hope all that made sense

[sharp smelt]

got iot

• got it

[rustic tendon]

and I know Im not supposed to say contour, its just easy

[sharp smelt]

Thanks!

Yeah, it's fine!

Can't wait to do contour integrals for real :D

Will probably do them in my third year

.close

[pearl pond]

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[rustic tendon]

is this a pure math thing

i'm in my first year, and i study them 😂

(engineering though)

[pearl pond]

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[sharp smelt]

yea

[sharp smelt]

oops

I'll just finish what I mean

I thought that was integration over $\C$

[jolly parrot]

What a wonderful world !

[sharp smelt]

Contour integrals

[rustic tendon]

yeah, they are

a closed simply connected curve on C

[sharp smelt]

Yeah, that will probably be covered in a CA course for me

If I take it

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[earnest flint]

guyss sorry, ai isnt working rn so i cant ask it. this isnt homework im js studyinggg. isnt this wrong bcs the one in red should be positive when the lcd is taken or is the guy right?

[earnest flint]

lke how did it become negative?

[brittle tinsel]

they distributed the negative to the x and the Δx.

4 - (x + Δx) = (4 - x - Δx)

[cobalt hinge]

Depends on where you’re asking.

[cobalt hinge]

^

[earnest flint]

that happens if i put 4 inside the parenthesis? does it apply to all signs too, if its originally negative it will become positivee?

[cobalt hinge]

Yes.

It is the distributive law, $a(b+c)=ab+ac$.

[jolly parrot]

;(

[cobalt hinge]

In this case, a=-1.

[earnest flint]

Okayokay, thank you!

[earnest flint]

wait wait sorryyy, to follow up, in this law it means that it should all be positive/negative if i take/put inside a parenthesisss?

[cobalt hinge]

Yes, according to the distribution of signs, hence the name.

[earnest flint]

OHHHHH

OKIII

THNKIUU

[pearl pond]

@earnest flint Has your question been resolved?

[pearl pond]

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[mossy remnant]

hey rizzlers

[light helm]

Do you intend to seek help here?

[mossy remnant]

no nerd

[brisk scarab]

.close

[pearl pond]

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[toxic lichen]

theta just runs from 0 to 2pi

r = 0 happens never

precisely because cos(θ) = 7 has no solutions

...

[rough stream]

Do you have any formulas for the area of a polar graph?

[toxic lichen]

i have no idea how to answer that, i'm so sorry!

[frozen lantern]

!nogpt

[pearl pond]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[frozen lantern]

oh

thats old

nvm

[rough stream]

Haha I do that a lot

Still, we must laugh at you. I hope you understand.

[pearl pond]

@spring crystal Has your question been resolved?

[pearl pond]

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[frigid eagle]

The midpoints of the sides of a parallelogram are connected in order. Find the area of ​​the resulting quadrilateral, if the area of ​​the parallelogram is 24 cm^2.

[pearl pond]

@frigid eagle Has your question been resolved?

[nimble socket]

Just $\int_0^{2\pi}\frac12(7-\cos(\theta))^2d\theta$...?

[jolly parrot]

ooaa

[north fractal]

u can think of it by dividing the figure in 4 parts

[frigid eagle]

?

[north fractal]

im drawing it wait

[frigid eagle]

Okay

Wouldn’t all sides of the quadrilateral inside be equal?

[north fractal]

i dont think they would

[frigid eagle]

What about opposite sides

[north fractal]

if u observe, upon cutting through the dotted line, u get 4 parallelogram and the half of each parallelogram is what is required

[north fractal]

they would be, but that information is not essential

sry i hv to go sleep

good luck

[frigid eagle]

Yea thanks

<@&286206848099549185>

[pearl pond]

@frigid eagle Has your question been resolved?

[frigid eagle]

.close

[pearl pond]

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[tender flax]

so uhh i cannot find the constant for thi s cause there arent enough values

[merry carbon]

There are

[tender flax]

okayy leme try it once more

failed thrice already

[merry carbon]

Actually wait a moment-

Am I tripping, or is the question strangely worded

[chrome plank]

it's very stragely worded

I'm having a difficult time parsing it

I think it's a first-order differential equation?

[merry carbon]

I'm thinking it's meant to mean/say "the rate of change of the population of..."

[tender flax]

i think the first line is t ryi ng to say rate of change

[tender flax]

yeahh

so is it stil possible ?

Lsob

😭

wha t

what does this ha ve to do with math s

[pulsar lark]

<@&268886789983436800>

[merry carbon]

Also the questions are worded-

[chrome plank]

Translation.

The population of a town grows at a rate that is directly proportional to the square root of its current population. If the initial population is 20,000:
(i) What will be the population after 10 years?
(ii) How long will it take for the population to double?

[merry carbon]

Was this translated from another language?

[spare lark]

From my english kekw

"How much the population"

10pounds

[tender flax]

are u high dude

[tender flax]

is C constant for the entire equation?

or its different at different time intervals

[merry carbon]

The reason I'm pointing that out is that I'm wondering if it's meant to say "how much does the population change" or something

[tender flax]

yeah it is

how do i find the pop after 10 years

i dont have k

[chrome plank]

maybe they want it as a function of the proportionality constant?

[tender flax]

nopee

the asnweris 66860

after 10 years

and the time is 14 years for double the pop

bro wtf

bro what

how does it take 10 years to reach 66860

ye t it takes 14 years to reach 40000

huh

[chrome plank]

this problem is terrible

[spare lark]

Its probably the city where theres that one guy buying 2456 oranges for reasons

[chrome plank]

thus causing a food shortage, starvation, and therefore it does indeed take 14 years to actually double

[spare lark]

Problem solved

It was a third hidden variables

[pearl pond]

@tender flax Has your question been resolved?

[pearl pond]

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[dire swift]

not exactly math ig but coudl someone explain the KCL at node A for me?

[pearl pond]

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[alpine vessel]

How can i REALLY find pi
Ex:finding all the numbers without memorizing them

[compact ridge]

there's a spigot algorithm for pi that gives you the nth digit of pi

[alpine vessel]

I meant like manually

[compact ridge]

https://www.gavalas.dev/blog/spigot-algorithms-for-pi-in-python/

[compact ridge]

there's infinitely many digits though with no pattern

[alpine vessel]

True, like find an aproximation JUST with the circunference nothing else

No algorithm just my mind

[compact ridge]

oh mathematicians have done this

[compact ridge]

the solution to do complex stuff in maths is to learn more maths

[alpine vessel]

Ty

.close

[pearl pond]

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[compact ridge]

search up 'Archimedes method pi'

nw!

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[compact ridge]

,calc 1/380902.777778

[jolly parrot]

Result:

2.6253418413841e-6

[compact ridge]

yes that is indeed correct then they rounded

,calc 1/(1.097 * 10^7 * (1/16 - 1/36))

[jolly parrot]

Result:

2.6253418413856e-6

[compact ridge]

they rounded wrong? lmao

ah, because you're doing $\frac{1}{3.80902 \cdot 10^5} = \frac{1}{0.380902 \cdot 10^6}$

,calc 1/1.097

[jolly parrot]

Result:

0.91157702825889

south

[compact ridge]

so convert 389020 to standard form first

then you can divide 3.80902 by 10, and then multiply 10^5 by 10 (=10^6)

dividing by 10 and multiplying by 10 doesn't change the number

it's the same as 2.625 * 10^(-6): they've made a rounding error

yes then 1 over that

yeah so you probably don't know what standard form is then

the 10^(-6) means that the decimal point is 6 places left of 2.626....

so move the decimal point 6 to the right

try it for yourself

you should get 2.626

you need to do 1/380902.777778

[pearl pond]

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[sharp smelt]

I'm wondering if there's any way other than brute force

the normal way would be to minimize $d=x^2+y^4 + \frac{4}{x^2y^4}$

[jolly parrot]

What a wonderful world !

[sharp smelt]

I suppose this can be done with AM-GM

using AM-GM,
\
\frac{x^2+y^4+ \frac{4}{x^2y^4}}{3} \geq \sqrt[3]{4}$
we find the shortest distance would be $3\sqrt[3]{4}$

[jolly parrot]

What a wonderful world !
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[sharp smelt]

Using AM-GM,
[
\frac{x^2+y^4+ \frac{4}{x^2y^4}}{3} \geq \sqrt[3]{4}
]
we find the shortest distance would be ( 3\sqrt[3]{4} ).

[jolly parrot]

What a wonderful world !

[sharp smelt]

.close

[pearl pond]

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[midnight haven]

whats intermediate value theorem

[midnight haven]

2nd point

[placid geyser]

Can you elaborate your doubt?

[midnight haven]

i didnt understand what the theorem means

and how it could be useful

@placid geyser the second statement

[placid geyser]

If that condition is satisfied, then there is a root of the polynomial in between alpha and beta. Sometimes, they ask "between this and this, how many roots does it have" or "between what and what does it have a root"

Which chapter does this belong to?

I remember having some example questions with it

[midnight haven]

quad equations

[midnight haven]

ah alright. after this theorem; the questions given didnt seem to apply this but ig that makes sense yes

but it could have more than 1 roots, right. how does this help?

[placid geyser]

I'm not sure. I can't find the book rn

[midnight haven]

alr. lemme know if u find it. ill do some hunting as well

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[midnight haven]

factorising does not help at all

[sharp smelt]

,w factorise x^2+6x+8

[jolly parrot]

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[rustic tendon]

all you need to do is extract the constant term when this is rewritten as a polynomial

do you agree to that?

[midnight haven]

i dont know what it means or how to do that

[sharp smelt]

,w factorise x^2+14x+48

[jolly parrot]

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[rustic tendon]

okay

look they're asking for the product of all x values

[toxic lichen]

would this not be a matter of hammering the equation until it looks like a polynomial

[rustic tendon]

which is always the constant term in a polynomial

[sharp smelt]

I suspect you made a mistake while factoring

[sharp smelt]

yea, It should be

[rustic tendon]

I mean we don't need to do much work at all

[sharp smelt]

,w factorise x^2+10x+24

[jolly parrot]

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[rustic tendon]

if we only need the constant

[sharp smelt]

a lot of factors do cancel out

[rustic tendon]

since the only way to get a constant is to only multiply the constant

[rustic tendon]

@midnight haven do you want to continue with this?

[midnight haven]

yes please

[rustic tendon]

okay

for starters just take all the terms to one side

there's a much faster way to do this in your head, for the jee, but I'm going to drag it out a bit so that you understand

$\frac 1{x^2 + 2x} + \frac 1{x^2 + 6x + 8} + \frac 1{x^2 + 10x + 24} - \frac 15 + \frac 1{x^2 + 14x + 48} = 0$

[jolly parrot]

dyxn

[midnight haven]

they're different terms yk; you'd have to make the demoninators common and stuff

[sharp smelt]

I know

That's the brute force way to do it

[rustic tendon]

so now you take the lcm or whatever

but you don't have to actually do any hardcore math

Because you know that the numerator is gonna be comprised of all terms BUT the one in the denominator

@midnight haven does that make sense?

[midnight haven]

yep

[rustic tendon]

okay cool

[midnight haven]

alrigt ill solve it further

[toxic lichen]

you have exceeded the 15 second time limit.

[rustic tendon]

right so just get the constant term from each of the terms and add them up

[rustic tendon]

easily made up for the 1 second spent on each chemistry problem

[midnight haven]

thanks

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[sharp smelt]

Le random physicla chemistry problem, with the most insane numbers

[rustic tendon]

skip

[pearl pond]

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[hallow zinc]

the above is equal to area of triangle

where a1 a2 a3 are lengths of sides of the three sides of triangle respectively

and R is radius of circumcentre

and p is radius of inscribed circle

i dont know where the 8 next to the p comes from

https://babel.hathitrust.org/cgi/pt?id=wu.89043163211&seq=26

above link is the notations

this is the full question

[frozen lantern]

is that an eight

[hallow zinc]

idk

[sharp vigil]

that is an s (the semiperimeter)

[frozen lantern]

ill go out of my way to say that might be the semi perimeter

[hallow zinc]

bruh

ok

[sharp vigil]

the link does establish this notation

[hallow zinc]

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[sharp smelt]

Trying to solve this using green's theorm

[sharp smelt]

I can convert this line integral into a double integral to make calculuations easier

so $\int_{0}^{2 \pi} \int_{0}^{2} -3 r^2\cos^2(t) - 3r^2\sin^2(t) dt dt$

[jolly parrot]

What a wonderful world !

[sharp smelt]

so $\int_{0}^{2 \pi} \int_{0}^{2} -3r^2$?

[jolly parrot]

What a wonderful world !

[inland ivy]

no

there's a lot wrong here

firstly, what is dt dt?

and also this is a line integral, not a surface integral

[sharp smelt]

Yea, I'm using green's theorm

$\int_{0}^{2 \pi} \int_{0}^{2} -3r^2 dr dt$

[jolly parrot]

What a wonderful world !

[rustic tendon]

how did you get this exactly

[inland ivy]

you also missed the jacobian term

[rustic tendon]

what

[sharp smelt]

yea

oops

[sharp smelt]

Tried a change of variables

but messed up

[sharp smelt]

$\int_{0}^{2 \pi} \int_{0}^{2} -3r^3 dr dt$

[jolly parrot]

What a wonderful world !

[sharp smelt]

This is probably it

which is $-12π$

[jolly parrot]

What a wonderful world !

[rustic tendon]

,calc (2^5)/4

[jolly parrot]

Result:

8

[rustic tendon]

24

[sharp smelt]

oosp

right

[rustic tendon]

-24π

yeah other than that it's right

i was blanking mb

[sharp smelt]

I've reallly got to work on calculations

Thanks

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[sharp smelt]

Got to code for a bit now

[pearl pond]

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[quasi turtle]

this is stating that the identity for multipication is 1, while the idenity in R_2pi is 0 right?

[quasi turtle]

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[sharp smelt]

Trying to solve this using green's theorm

[sharp smelt]

so I have $\int_{C} \mathbf{F} \cdot dr =\iint_{S} -2xy+2xy dA =0$

[jolly parrot]

What a wonderful world !

[sharp smelt]

is that it?

[rustic tendon]

how did you get the RHS

[sharp smelt]

Green's theorm

[rustic tendon]

yeah but like

shouldn't it be -(x^2 + y^2)

It's Q_x - P_y right?

[sharp smelt]

oops

yea

[pseudo oxide]

...

[sharp smelt]

got it

so $\int_{0}^{2 \pi} \int_{0}^{5} -r^2dr dt$

[jolly parrot]

What a wonderful world !

[rustic tendon]

rdrdθ

[sharp smelt]

r^2 because of the jacobian

[rustic tendon]

x^2 + y^2 = r^2 and dA is rdrdt

so -r^3

[sharp smelt]

yea

oops

so so $\int_{0}^{2 \pi} \int_{0}^{5} -r^3dr dt$

[jolly parrot]

What a wonderful world !

[rustic tendon]

yeah

doable from here

[sharp smelt]

very doable , yea

thanks

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[white sparrow]

hey can someone check? im not done yet but i feel like i did it wrong, was i supposed to add the prices without sales tax first?

[umbral star]

wouldn't taking tax before or after adding give the same result?

[light helm]

t(a + b + c) = ta + tb + tc
doesn't really matter,
you do seem to have some rounding issues though

[white sparrow]

yea im bad at maths......

[white sparrow]

i dont know either

but i figured now thank you tho!!

[umbral star]

trial and error is your best friend, gl!

[white sparrow]

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[fading raft]

Since the result of (a+2)!-(a+1)!-a! is divisible by 128 and the result of (b+2)!+(b+1)!+b! is divisible by 196 without remainder.
What is the smallest value that the sum a+b can take?

[fading raft]

İ did some things but they seemed wrong

Tbh dont know what to do

[west sapphire]

is there a constraint on a? clearly it's not true if a=1, for example

[fading raft]

Hmm true but i didnt see any constration

[brazen narwhal]

i think the divisibility part is the constraint, it is required to find smallest a,b

[west sapphire]

oh, i see, we're supposing the divisbility part

[brazen narwhal]

could u show what you have tried

[fading raft]

👍 1 min

[brazen narwhal]

yeah this is correct so far

[fading raft]

But what to do now

[brazen narwhal]

i think from here its just a bit of educated guessing

now we need to think about what values of a we can put so we can factor out 7 2's from that

[fading raft]

Hmmmmm🤔🤔🤔

[brazen narwhal]

||if we put a=2 we can factor out 4 2's, if we put a = 4 we get 6 2's, so a=6 will work||

im not sure if theres a better method but we can do this and its small numbers so it shouldnt take much time

[fading raft]

Yes you right a should be 6

[fading raft]

İ think there is no short method we have to write a bit

Lemme look at it bit more

[brazen narwhal]

👍

[fading raft]

Ye you right a is definietly 6

İdk i solved like that ik this is a mesy and long way but this is the only im able to do

[brazen narwhal]

yes i think this is what was expected

i cannot see any better way to do it

[fading raft]

Thx❤️

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[wild mantle]

Anyone help me please

I know the basics of parabola

[lethal mango]

wwhat grade are u in

[placid geyser]

If the angle of the line is theta and the line passes through (0, 0), what should be the line equation?

[wild mantle]

no idea

Like I do not have slope

[placid geyser]

Yk what the slop will be if the angle is theta

m = tan theta dude

[wild mantle]

tan(theta) = slope

ya

[placid geyser]

Yes, substitute it in for slope and get the equation of the line

[wild mantle]

so y - y(1) = m [ x - x(1) ]

[placid geyser]

Yes

[wild mantle]

right

[placid geyser]

x1 and y1 are both 0

[wild mantle]

y = tan(theta) * x

[placid geyser]

Yep

Then, as they r asking the length, it would be nice to find the other point

So, solve the parabola with this line

And get the other point of the chord and find the length of the chord

[wild mantle]

12th

can you explain a bit more

other points of chore in other words point of latus rectum right

if yes they already give end point of latus rectum right?

[placid geyser]

No

[placid geyser]

The chord they mention is that line

[wild mantle]

the equation of line which we find

[placid geyser]

So, the points of intersection of the line with the parabola would be the ends

[placid geyser]

Yes

[wild mantle]

so you are saying I have to find point on that line right

and the point is that where it intersect with parabola

[placid geyser]

Yes

[wild mantle]

now where does it meet with parabola by the way at origin?

[pearl pond]

@wild mantle Has your question been resolved?

[pearl pond]

@wild mantle Has your question been resolved?

[placid geyser]

Yes, and one other point

Just solve the equations

$y = x\tan{\theta}$

$y^2=4ax$

[jolly parrot]

Suika

[placid geyser]

We get $x=4a\cot^2{\theta}, y=4a\cot{\theta}$

[jolly parrot]

Suika

[wild mantle]

sorry for late reply

let me process it

[wild mantle]

Thanks for you help man

really appriciate it

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[safe prairie]

does int f(0) dx not always =0, but just a common lucky occurence?

[warm current]

It would be a linear function

[safe prairie]

since it at least multiplies every polynomial term by y=0

[warm current]

!xy

[pearl pond]

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.

[vapid cave]

if we set C=0 it is

[safe prairie]

then theres int sinx dx

[verbal whale]

$$\int\ f(0)dx = f(0)x + C$$

[jolly parrot]

Alberto Z.

[safe prairie]

no problem im thought experimenting

[verbal whale]

You meant this?

[safe prairie]

what does f(0)x mean

thats not what i mean when i say at least multiplied by x

[verbal whale]

f(0) multiplied by x

[safe prairie]

x becomes x^2/2, x^2 becomes x^3/3 and so on

[verbal whale]

Huh?

[verbal whale]

And what does "int f(0)" have to do with this??

[safe prairie]

so k becomes kx and so everything at least times x

[safe prairie]

plug in 0 and everything =0

[vapid cave]

bc we’re multiplying by x, when x=0 it seems like int f(x) dx should =0

[verbal whale]

I don't get you 🤷‍♂️

[vapid cave]

youre doing it without the +C btw

since you’re setting C=0 you’re getting 0

[safe prairie]

yes cuz im thinking about definite integration and areas

mainly im struggling with visualising int sinx dx

[warm current]

!xy

[pearl pond]

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.

[warm current]

You really should share the original, because everyone is just getting confused right now

[vapid cave]

there’s no original @warm current

since C=0 means “accumulating from x=0” taking int f(x) dx at x=0 is like taking an jntegral from 0 to 0

[warm current]

How do you know?

[vapid cave]

bc of this

[spare lark]

What are the bounds of the integral

[vapid cave]

hmm I see

[safe prairie]

pi and 0

[vapid cave]

since our accumulation is starting at 0 we’re implicitly adding a C so that f(0)=0

[safe prairie]

my headcannon is that int graphs are area distribution graphs of f(x)

this messes it up

[vapid cave]

but we just don’t account for subtracting since we’re subtracting that constant anyway

the area distribution graph you’re thinking of is -cosx+1

I think

[safe prairie]

if we know its 1 then that means we can find C from indefinite int?

[safe prairie]

just by using this?

[vapid cave]

well no thats only bc our bounds are from 0

[safe prairie]

so are int graphs not exactly these but are related to these

[vapid cave]

yes

[safe prairie]

alr thx

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[pallid fossil]

lowkey might be stupid but i keep getting the wrong answer for this

any help would be appreciated thank you

none of the numbers i find work and ive only got one attempt left on this question

[pearl pond]

@pallid fossil Has your question been resolved?

[pallid fossil]

nah

[pallid fossil]

!close

!end

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[grim glade]

can someone help me with my milestone, im having a hard time

[grim glade]

https://docs.google.com/spreadsheets/d/16aJJAoGcJN7LI1wiuRmNuLXxaJJI0FvN5ZdV1o2zrUI/edit?usp=sharing

heres a copy, u can check it

[hushed oracle]

Hmmmmmmmmmmm

[pearl pond]

@grim glade Has your question been resolved?

[grim glade]

<@&286206848099549185>

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@grim glade Has your question been resolved?

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[stiff wraith]

Honestly I just don’t understand arccos stuff

[unkempt yacht]

well by definition, arcsin(sin(x)) = arccos(cos(x)) = arctan(tan(x)) = x

well not really, but we will get to that later

[unkempt yacht]

what do you not understand?

[stiff wraith]

Well idk what to do

[unkempt yacht]

try evaluating them one by one?

like for 37, what is tan(11pi/6)?

[stiff wraith]

-1/2 idk

[unkempt yacht]

refer to the unit circle

,tex .unit circle

[unkempt yacht]

which angle does 11pi/6 correspond to?

[stiff wraith]

Square 3 Over 2

[unkempt yacht]

more particularly, what are the sin and cos value of the angle 11pi/6?

[stiff wraith]

Well idk

[unkempt yacht]

are you paying attention to the unit circle?

surely you can do more than that

[stiff wraith]

Well I tried -1/2 and square 3 over 2

[unkempt yacht]

okay, -1/2 is the sin value and sqrt(3)/2 is the cos value

good

is there a way you can express tan as some combination of sin and cos?

[stiff wraith]

Honestly idk

[clear mirage]

||tanx=sinx/cosx||

[unkempt yacht]

er, you might need to revise the trigonometry chapter

[stiff wraith]

Sin / cos

Square root 6 over 3 right

Neg

?

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[safe prairie]

is it true that pmf and cmf both find discrete probabilities, but for continuous, pdf cant, only cdf can?

[pearl pond]

@safe prairie Has your question been resolved?

[iron arch]

if cdf can be used to find discrete probability you can use this to find them using pdf too probably ?

[sharp vigil]

i think what you might be getting after is that out of those four functions, the pdf is the only one whose output can't be directly interpreted as a probability

[safe prairie]

and f(range) = F(top)-F(bottom) so f is never directly tied to the actual probability, F does

[safe prairie]

yeah

[pearl pond]

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[inland laurel]

hello

[inland laurel]

i didn't understand the highlighted part

[tardy reef]

Each of the terms corresponds to the area of the individual rectangles

the width is delta x for each rectangle

and the height is the value of the function at the bottom left vertex (coz the top left corner is a point on th function (t, f(t)))

[pearl pond]

@inland laurel Has your question been resolved?

[inland laurel]

but delta x is the distance from a to b?

it shows in figure

[safe prairie]

not A to B

[placid geyser]

No, $\Delta x$ is the width of each rectangle

[jolly parrot]

Suika

[inland laurel]

oh

[safe prairie]

it just so happens that every rectangle is within A and B

[inland laurel]

then what about lenght

[placid geyser]

f(a) for 1st, f(a + delta x) for 2nd, f(a + 2 delta x) for 3rd, and so on...

[inland laurel]

why height includes delta x in f(a + delta x)

delta x is the width

[safe prairie]

hwight doesnt include delta x

[inland laurel]

it is in the function

[safe prairie]

the stuff within f() includes delta x

but the stuff within () isnt height

[inland laurel]

and that function denotes its lenght

its not clear

[safe prairie]

f that () returns height

but the stuff in it isnt height

[safe prairie]

its clear to anyone who knows how functions work

[inland laurel]

i studied functions from scratch

[safe prairie]

it should be clear since like the first 2 lessons of functions

[inland laurel]

from set theory and relations and functions

[safe prairie]

set theory before functions???

[placid geyser]

The x coordinate of the thing would be a + delta x, so y = f(x) = f(a + delta x)

[inland laurel]

in sequnce it is set theory then relations and functions

[safe prairie]

my highschool doesnt teach set theory

but functions is highschool freshman level

[inland laurel]

first rect area = f(a)(delta x)?

[placid geyser]

No, 2nd

The height is the same as the value of the function at the left side of the rectangle, right?

[inland laurel]

yes

but for first it is f(a) and last it is f(b)

[tardy bay]

The area of the first rectangle is f(a) * Δx.

The area of the last rectangle is not quite f(b) * Δx because the rectangles are drawn from left to right. If you were to draw another rectangle it would have area f(b) * Δx.

The last rectangle instead has area f(b-Δx) * Δx.

[inland laurel]

i don't understand delta f = f(x + deltax) - f(x)

similarly in 1st picture i send

[plush moss]

The change in f from x for a given change in x is equal to the difference between the value of f at x+change in x and the value of f at x

[placid geyser]

@inland laurel is this understandable?

[placid geyser]

Last is not b

[placid geyser]

If u see here, it's the x coordinate less than b

[inland laurel]

why f(a + delta x)

since the height is increasing in y-axis it should be delta y

but y = f(x)

maybe thats why

[placid geyser]

Cause the x is a + delta x

[pearl pond]

@inland laurel Has your question been resolved?

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[inland laurel]

.close

[pearl pond]

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[hushed leaf]

hey guys for b) do i add up the binomial of 0,1,2,3,4,5?

or i can use 1 - (6,7,8) right?

[cosmic charm]

yes

[hushed leaf]

okay tq

wait how do i asking the first quesstion?

[cosmic charm]

what

[cosmic charm]

i dont understand lol

[hushed leaf]

sorry i mean how do i answer the first qn

[cosmic charm]

its asking what sort of distribution should you use

[hushed leaf]

binomial

[cosmic charm]

then the parameters of it

the parameters of a binomial distribution is n and p

which is?

[hushed leaf]

8 is n

p = 0.8

[cosmic charm]

then thats your answer

[hushed leaf]

why does their answer look like this form?

[cosmic charm]

because thats the standard form of writing a binomial distribution and its parameters

$X \sim B(n, p)$

brother pls

[jolly parrot]

Shioshi

[hushed leaf]

ohh icic

[hushed leaf]

what does the in stand for>?

[cosmic charm]

B or bin means the same

binomial

write whichever way that you are taught

[hushed leaf]

okie tq

for c) i got 0.009 for p(x=3)

should i believe the claim of United Medicine or no?

the probability looks kinda low

[cosmic charm]

yeah

[hushed leaf]

ok tq

.close

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[pearl pond]

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[analog girder]

can someone help me with this question, shouldnt the gradient of the normal be -√1-x^2 ? because isnt -1/√1-x^2 the gradient of the tangent?

[safe prairie]

havent learnt normals in calc but dont perpendicular lines have this relation called m1 = -1/m2

how can both ur gradients be -

[frozen lantern]

I guess they're talking about how you can't just cancel out the x's because they be 0

[pearl pond]

@analog girder Has your question been resolved?

[analog girder]

wdym? isnt that the gradient for tangent though?

[analog girder]

oh i might have done a part wrong but eitherway isnt this the gradient for a tangent?

[frozen lantern]

you cant divide by 0

[analog girder]

but they cancelled out the x

[frozen lantern]

is that not your work

[analog girder]

i know thats not my work

but im just wondering isnt that the gradient for the tangent

and not the gradient of the normal?

because that's the given solution

[frozen lantern]

it is the tangent

[analog girder]

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[next ore]

$\binom{30}{0}\binom{20}{10}+\binom{31}{1}\binom{19}{9}+\binom{32}{2}\binom{18}{8}+...\binom{40}{10}\binom{10}{0}$

[jolly parrot]

∮Ē.dĀ = Qₑₙ꜀/ε₀

[next ore]

i need to find the sum

looking at just the first terms the index increases by 1 and the lower value increases by 1 too

my approach was with binomial expansion with -ve index

[placid geyser]

The first term is:
Select 0 from 30
Select 10 from 20
Right?

[next ore]

yes

[placid geyser]

2nd is:
1 from 30
9 from 20

[next ore]

no

1 from 31 and 9 from 19

[placid geyser]

Oh, sorry

Mb

[next ore]

its alright

[placid geyser]

1 from 31
9 from 19

Oh, nice.

[next ore]

9 from 19

[placid geyser]

I gotta the active tag just now

[placid geyser]

Yeah mb

[placid geyser]

And so on...

So I think it's the same as selecting 10 from 50?

The total cases

So 50c10?

[next ore]

no

its 51C10

[placid geyser]

The answer?

[next ore]

yes

$(1-x)^{-n}=\sum_{k=0}^{\infty}\binom{n+k-1}{k}x^{k}$

[jolly parrot]

∮Ē.dĀ = Qₑₙ꜀/ε₀

[next ore]

i wrote expansions like this for $(1-x)^{-31}$ and $(1-x)^{-11}$

[jolly parrot]

∮Ē.dĀ = Qₑₙ꜀/ε₀

[next ore]

when i multipled terms corresponding to the binomial coefficients in the question to one another in these 2 expansions i got that they are the coefficient of $x^{10}$

[jolly parrot]

∮Ē.dĀ = Qₑₙ꜀/ε₀

[next ore]

which will be the same as coefficient of $x^{10}$ in expansion of $(1-x)^{-31}(1-x)^{-11}$ which is the same as $(1-x)^{-42}$

[jolly parrot]

∮Ē.dĀ = Qₑₙ꜀/ε₀

[next ore]

plugging the values in i get 51C10

is my method correct?

[placid geyser]

Can u send the solution?

[next ore]

i typed pretty much everything from this

[placid geyser]

Everything is going over my head

So if I could see the steps and all

It would be helpful

[next ore]

i dont have a phone to take a photo rn

[next ore]

i took each binomial coefficient in this sum as the coefficient of the binomial expansion of a number with -ve index

i have written the formula for the expansion after that

i wrote two expansions which would correspond to those coefficients

multiplying those 2 i find what im trying to find is the coefficient of x^10 in (1-x)^-42

the (1-x)^-42 is the product of the two expansions i wrote earlier

[placid geyser]

I see

[placid geyser]

Yes, it seems to be correct

I did something like this a couple months ago so I don't quite remember it, but I think it is right

[next ore]

okaiii

thank youu have a nice day

[placid geyser]

Same to u

[next ore]

.close

[pearl pond]

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[pearl pond]

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[lucid moth]

This is the steps to a solution

I dont understand how they got positive 6 for 6cos(x)??

Isnt it supposed to be -6?

Cause cos(pi) = -1, no?

Wait

I read it wrong

.close

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[cosmic charm]

minus minus plus?

[pearl pond]

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[modest chasm]

Hi, I’ve done a lot of trial and error but I can’t seem to find the right variables that satisfy the graphs and it’s really hard for me to graph it on desmos or geogebra, if someone could provide me with variables that work and explain how it works I would really appreciate it!

[fair creek]

the question is kinda weird because you can't start at the original with a non-zero y-intercept 😔

unless it meant the y-axis

[modest chasm]

ohh

does this apply do all the horses?

to

**

[fair creek]

horses A,C, and D have constants so they have non-zero intercepts if the constant is non-zero

but question for horse B is a circle so it also has a non-zero intercept

[modest chasm]

ohh what if the question were to mean y-axis ?

[fair creek]

appears so

otherwise it wouldn't make sense

[modest chasm]

i’m sorry i’m really not good at maths but what difference does it make?

[fair creek]

starting with explicitly (0,0) vs (0,t)

t could be anything but for origin t is always 0

[modest chasm]

ohhh

so if t is 0 then the graph doesn’t work starting at the

origin?

[fair creek]

it's not that graph that doesn't work

it's that we're given that all the variables are non-zero

but if they're non-zero, they don't start at the origin

ugh this question is weird honestly 😭