help-23

[limber jewel]

To approximate the bigger whole area

[lean otter]

We are taking the rectangles to be points why can't we do the same for polar rectangles?

brain not working

[limber jewel]

@lean otter actually we don't make the rectangles into points. We make the rectangles smaller and smaller to reduce the error in the approximation

[lean otter]

which eventually become points ?

[limber jewel]

Hmm do you know what a * $limit$ * is?

[lean otter]

If this isn't true my whole foundation of double integrals will collapse

[lean otter]

Oh yeah

[flat frigate]

Kino

[lean otter]

Close put not that

Okay

[limber jewel]

Not precise enough but that will do

[lean otter]

Its not an arc

👍

Still

We will always be left with this

Wait

Wait

OH

I think I get it

[limber jewel]

@lean otter we use limits here, we dont get the exact area when we go in step size (by making the rectangle smaller in discrete units)

But we are approximating the area as we decrease the area of the smaller rectangles , hence we can get as close to the original area as desired

[lean otter]

There is absolutely no difference between rectangles and polar rectangles ?

Just that the dimensions for the first are x and y where we are measuring the length and breadth

While in this case it's theta and r

But there is no difference in the shape at all

Am I right?

[limber jewel]

@lean otter there is a difference. The polar rectangle's area is approximated by an trapezoid's area while a normal rectangle is approximated by the area of a rectangle itself. The difference is because of theta existing, which makes the length look curved, that is the area looks more close to a trapezoid

Do you get it?

[lean otter]

Kind of

[limber jewel]

Hmm just imagine a polar rectangle getting smaller and then look at this

[lean otter]

Now I see the slant heights

[limber jewel]

The curved arc would look straight enough if you make the polar rectangles small enough right

[lean otter]

I get it

Where I made a mistake was
I thought the region for small theta and r would look smth like and arc join theta = alpha and theta = beta

I understand it nkw

[limber jewel]

So you add all these smaller trapezoidal areas together. If you keep on making the area of the trapezoids smaller ( ∆A becomes smaller) then you get your original area

[lean otter]

Yeah

[limber jewel]

Yes

Remember limits are very important and defining integrals usually involve using the concept of limit as a base

[lean otter]

Thank you

[limber jewel]

Welcome

[safe radish]

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[summer vortex]

Suppose you are rolling a fair six-sided die, and you aim to roll a six. However, there's a catch: you can only consider the game complete if the six appears on an even-numbered roll. What is the expected number of die rolls required to successfully roll your first six on an even-numbered trial?

[pseudo scroll]

Maybe you should try drawing a table for a random variable

[summer vortex]

.close

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[jaunty root]

Wouldn't f just be y+C?

[safe radish]

@jaunty root Has your question been resolved?

[last wren]

yes

but you technically don't need the constant, just a function f whose gradient is F

so just y works fine

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[safe radish]

Please don't occupy multiple help channels.

[teal thicket]

Please take a look at this image for a clear solution… sorry for the handwriting-

A summation or product in which all the terms in between cancel out except the first and the last is generally known to be a telescopic summation

this is one of them…

if given an infinite sequence, check whether you can slipt them like this… that should be your first priority..

if any doubts, ping me

[clever pawn]

@teal thicket how did the 2n+1 become (n+1)^2-n^2

[teal thicket]

oh…

i split the 2n+1 like that, because doing that would make me get a chance to cencel them out

Check if that happens, like if the products in the denominator do happen to be the numerator’s difference

It comes by practice🙂

[teal thicket]

these kind of models, you will see in calculus, inverse trig, sequences and many more

ok…

what are you planning to do with that??

also, you got A,B,C,D values if n=-1

What if they depend on n

[clever pawn]

idk we learned it this way

but

with 4 parts is really desperate

[teal thicket]

Is the chapter “limits” over for you?

Atleast “sequence and series”

[clever pawn]

yeah

[teal thicket]

alright…

So umm… Idk if that problem is from your teacher’s to do list or some assignment… But the best way to approach infinite summations or products is Telescopic method…

I dont know if this sounds mean.. sorry-

[clever pawn]

no no its fine

the only thing

i dont get it

[teal thicket]

ohh haha

Just add an n^2 and subtract

[clever pawn]

holy

how could i think of that

i guess practice XD

am 1 hour here

[teal thicket]

well, as i told, look if the numerator is the difference of the products

[clever pawn]

wdym difference of products

sorry my mathematical english are bad

[teal thicket]

I get it i get it

So notice the n^2 and (n+1)^2 below?

[clever pawn]

ye

[teal thicket]

Look if the difference of them both is the numerator

like subtracting them

[clever pawn]

u mean subtracting the bot with the top i get the top

or between them n^2(n+1)^2

[teal thicket]

Yes

between them

[clever pawn]

so n+1^2 - n^2 = 2n-1?

[teal thicket]

yes

these models help out in lot of other chaps

[shadow prawn]

Hi can anyone help me with this

[clever pawn]

so in telescopic series if this case is true

[teal thicket]

what’s wrong with it??

[teal thicket]

Mhm?

[clever pawn]

i add and substract on top

[teal thicket]

yes

But what you need to add depends

[shadow prawn]

So it's Reasoning by recurrence I don't really understand the part when we have to prove p(n+1)

[clever pawn]

thank you

[teal thicket]

show me your work

[shadow prawn]

I'm stuck here

Wait

[teal thicket]

alright

[shadow prawn]

Idk which part I messed up on

[teal thicket]

you got the answer right..

[shadow prawn]

What !! Wait let me finish it rq

[teal thicket]

??

is it solved?

[shadow prawn]

Nope:(((

Well chat gpt said I have to add (n+1)³ not (n+1) I don't understand why?

[teal thicket]

Well, the function is k^3 right

[shadow prawn]

Yes

But aren't we supposed to do it like
K³+(n+1)
Then we replace K³ with the first expression?

[teal thicket]

here

Take a look at this

i have to go… sorry

dm me if you have any doubt

[shadow prawn]

It's fine!!

[teal thicket]

Ill explain it later

[shadow prawn]

I'll thank u

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[whole maple]

!help

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To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[whole maple]

mathematicians

whats your profession!

oh wait

its math

das is the question

Findd C and B

[safe radish]

@whole maple Has your question been resolved?

[whole maple]

HEELP

AAGH!

.close

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[hot portal]

For the first one this is what Im thinking:

for part b im not sure

for c) it has to be possibvle since the total sum of degrees is even right

acc idk if c is possible

2 vertex with degree 5 and one vertex with degree 1 🤔

that cant be possible

now the only part im totally clueless about is part b. How should I approach this one

no. of vertex with degree 3 has to be even right. Thats what I have so far

midterm in 20 mins 😭 🙏

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@hot portal Has your question been resolved?

[inner tapir]

hmm not sure but b might have an argument with K_4 and adding enough vertices of degree three will exhaust your options before reaching 13 edges

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[inner tapir]

oh wait i think i drew one lmao

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[eager elm]

If I have the function -x^4 + 8x^2 - 8, how do I determine the absolute minimum value over the entire domain? Specifically, is there any way yo algebraically determine if the function has no absolute minimum since it goes to -inf?

[eager elm]

Should I just take the limits of each end of the domain?

wtf?

[clever pawn]

adolfmatics\

no u cant

[eager elm]

<@&268886789983436800>

[clever pawn]

thats illigal

[kind tinsel]

gone

[clever pawn]

lol racist got racist on

[past orbit]

If the function has a limit to -inf, then it has no minumum, yes

However your function does have a maximum

[eager elm]

but my question is taking the limit of the function as it approaches the lower bound of its domain and the upper bound of its domain a good way to determine if there is no abs max or min?

[past orbit]

There is no one-size-fits-all solution

That's one thing you may want to try, yes

[eager elm]

okay thx

thats all i rlly needed

appreciate it

.close

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[safe radish]

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[next gate]

Any idea?

Im not sure of that answer btw

Ok thinking about it and assuming theta with different values all this transformation does is squish the vectors into the -ve y axis, now i only dont get why didnt we multiply thw transformation matrix by the constant k

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[sullen scroll]

I'm blocked in question ii)
I know i'm supposed to get a lambda variable times the transpose of the eigen vector but I feel like there'S a theorem or property I don't know.

[sullen scroll]

Clean page

Messy previous one

I’m just blocked

<@&286206848099549185>

thats the book in which you can also find the exercise:
https://ucsb.app.box.com/v/LecturesNetworkSystemsSildes

[safe radish]

@sullen scroll Has your question been resolved?

[sullen scroll]

i feel like no one is in masters 😦

(I,m not either idk why i have this course)

[sullen scroll]

@robust river can you help me ?

sorry to ping you

(i helped 3 people while i was waiting 🙂 )

i can pay you if you want

if you want i can legit help people for you, while u help me 🙂

[robust river]

Unfortunately it seems like this exercise is not from my field of study, and it's not a standard mathematical knowledge, so I'm likely not the right person to ask

[sullen scroll]

https://tenor.com/view/donald-duck-ducktales-ducktales2017-house-of-the-lucky-gander-gulp-gif-21467706

now i'm scared

[polar lynx]

Offering money for this sort of thing is against the rules.

[safe radish]

@sullen scroll Has your question been resolved?

[safe radish]

@sullen scroll Has your question been resolved?

[safe radish]

@sullen scroll Has your question been resolved?

[safe radish]

@sullen scroll Has your question been resolved?

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[rare ember]

I have a question related to disk and shell method

[rare ember]

y=0.27x^3 3<=x<=4

[teal thicket]

what do you need help with?

[rare ember]

Hi

This is for a calculus project

[rare ember]

I need to find the volume using disk and shell method

When revolved around y axis

[teal thicket]

Ohh..

[rare ember]

But the volumes of both disk and shell method are not matching

[safe radish]

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

[mortal sandal]

!show

[safe radish]

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

[rare ember]

I'm required to use an online calculator for this so I used that

and I double checked the equations as well

This is how I set up for the shell method

This is for the disk method

<@&286206848099549185>

<@&286206848099549185>

Yes

College calculus

I want to see if I set up the equation correctly

Because I'm not getting the same answer for disk and shell method

[rare ember]

These are the equations I set up

Original equation is y=0.27x^3 and x=3,4

Yeah sure

[safe radish]

@rare ember Has your question been resolved?

[rare ember]

<@&286206848099549185>

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@rare ember Has your question been resolved?

[safe radish]

@rare ember Has your question been resolved?

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[south lynx]

how do we do this question?

[vagrant ice]

Double angle identity

[south lynx]

sin2(pi/4n)?

[vagrant ice]

Yes, times n/2

[south lynx]

okay, thank you

.close

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[warped venture]

Hi, I'm stuck with a geometry problem. I got 2 points of a circle and I need to find the radius

[livid igloo]

I have a question can someone help me out

[versed pendant]

Not here

[safe radish]

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).

[warped venture]

Can you take a other channel please

[tall bluff]

2 points won't be enough

I mean

If they something like they are "diametrically opposite"

Then yes

Hm? @warped venture

[warped venture]

No, it's in a electromagnetic problem, we put electron on a magnitic field and take 2 points of his trajectory

We know it's supposed to be a circle

[tall bluff]

But well

U cannot find

Equation of circle with 2 points

And radius also not possible because there is no information if they both are end points of diameter or something

[warped venture]

We know gor sure that one point is the 0, 0 and it's the lowest

[tall bluff]

Mhm

Did u manage to find the highest then?

[warped venture]

And the other is 9, 2 and it's somewhere between the lowest and the one at the right

[tall bluff]

Well then do we know the centre?

[warped venture]

No, it's to easy that way

[tall bluff]

Mhm then

We can't decide

[warped venture]

The teacher said to use the theorem of Pythagoras

[tall bluff]

Hmm

U need a triangle tho

Wait u can get locus of centre of circle tho

[warped venture]

Locus?

[tall bluff]

Locus is like

Set of possible values of centre

[warped venture]

It's at x=0

[tall bluff]

Centre at x=0?

[warped venture]

I tried the equation (y-r)^2 + x^2= r^2

[tall bluff]

Mhm

[warped venture]

Yep

[tall bluff]

Ok then

Use distance formula

Consider centre to be (0,x)

Distance from this point to (0,0) and (9,2) should be same

Ull get it

[warped venture]

Ok I see

[tall bluff]

Mhm

Should be

x² = 81 + (2-x)²

Solve and ull get

[warped venture]

Ok thanks you

.close

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[tall bluff]

Anytime!

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[lean otter]

Find the coefficient of x^99 in:

[lean otter]

Fancy notation.

[thin bridge]

what ahve you tried

[nocturne wraith]

that looks like a straight polynomial

alr its just a polynomial written in terms of its factors, but because there are 100 factors and it'd be too long, its just written as a fancy notation

i remember this uses nCr or something

[desert juniper]

this particular one is a lot simpler

(x-1)(x-2)(x-3)...(x-100). You're asked for the term on x^99

[tall bluff]

Yea

[noble kiln]

you can actually pair the things up

[thin bridge]

we should first wait for the op to respond to see what issues they're actually having with this

[desert juniper]

dont use walpha for this...

[nocturne wraith]

@desert juniper i want to confirm but this uses pascal's triangle right? i remember about it being used here

[tall bluff]

Wot

[radiant osprey]

I don't believe so

[tall bluff]

It's simpler than that

[noble kiln]

w a t

[nocturne wraith]

my bad

[noble kiln]

is not that complicated

[nocturne wraith]

i might've messed up in memory. but i remember pascal's triangle and nCr being used. it's just the first thing that popped up. that's why

my bad

oh my bad. it was in binomial expansion of a single term to nth power

i messed it up in my memory

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

I have found a pattern:
Let the product in the image denote p(n).
Then p(100) = x^100 + ...............
p(99) = x^99 + .............
Then p(99) * (x-100) = p(100)
(x^99 + .........) ( x-100) = p(100)
Cof. x^99 in p(100) = -100 + Cof. x^98 in p(99) = -100 - 99 + Cof. x^97 in p(98).
Continuing in this pattern, we get that Cof. x^99 in p(100) = sum(-100-99-.......-3-2) - 1 = the summation provided in the image

Is this correct?

[tall bluff]

Yes

Correct

[lean otter]

Ok

Thanks

.close

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[lean otter]

It's -5050

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[prisma swift]

i need help understanding whats going on here

i get that Rn(x) needs to approach 0 as n goes to inf

so that the taylor series converges to the function

but how did they go about proving that

[prisma swift]

specifically how we got this

[safe radish]

@prisma swift Has your question been resolved?

[plucky elk]

Which line in the screenshot do you get lost first

The purple?

[prisma swift]

yes

[plucky elk]

Do you see the definition of R_n

[prisma swift]

yes

its the first unused term in the taylor expansion

n+1th term

but for some c not a

c is between a and x

[plucky elk]

Do you see why e^c <= e^x

[prisma swift]

yea cause e^x is +ve

and c < x

if we assume that

[plucky elk]

Do you see why e^c <= 1 when x <0?

[prisma swift]

yea cause a is fixed at 0

and c is between x and a

e^0 = 1

x < c < a = 0
e^x < e^c < 1

right?

[plucky elk]

No

e^0 = 1 and e^c is positive for all c

c<=0 implies e^c <= 1

Also e^c is an increasing function

Once you substitute e^c <= 1 into the definition of R_n, and take absolute value signs, you get the purple circled equation

[prisma swift]

makes sense

oh wait

i see it now

[plucky elk]

Sorry this should be the reverse order. You take abs val first, then substitute e^c <= 1

[prisma swift]

e^c <= 1
|Rn(x)| = e^c(|x|^n+1)/(n+1)! <= 1(|x|^n+1)/(n+1)!

correct

?

god that looks hard to read

[safe radish]

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[safe radish]

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[brave pollen]

Everything i attempt this problem i get a nonsense answer that isn't allowed or possible

[full gust]

is it c?

where did u get this question from

ive solved it before

[brave pollen]

It's from a practice sheet for the acedemic decatholon

How did you solve it?

[full gust]

just input the values in the formula for standard deviation

and simplify

σ = √(1/n * Σ(x_i - x̄)^2)

[severe pond]

eh, use the shortcut formula

this doesn’t help

[full gust]

ye its a little long but it gets the job done

[severe pond]

not really, you’d have to derive the shortcut formula

[full gust]

after expanding, we get σ = √(1/n * (Σx_i^2 - 2x̄Σx_i + n*x̄^2))

[severe pond]

yea i know but you’re just deriving the shortcut formula

you should know it

by memory

[full gust]

ye, but i personally find it easier to use these general formulas, just less stuff to rmemeber

[severe pond]

it’s a good exercise to derive it though

[full gust]

also, i havent really studied it much, im in class 10,i just do a little higher maths because its fun

[brave pollen]

You mean the
(sum of the squares of the terms - (sum of the terms)^2/ # of terms)/ # of terms
Formula?

[full gust]

@brave pollen u should get σ = √1512.5 after simplification

[severe pond]

yea

should be n-1 though

[full gust]

isnt that variance? @severe pond

it should be the root of that i think

[severe pond]

$s = \sqrt{\frac{n\sum x_i^2 - (\sum x_i)^2}{n(n-1)}}$

[severe pond]

yea but they had the right idea

[flat frigate]

knief

[severe pond]

that

[brave pollen]

So this?

[severe pond]

yea

[brave pollen]

Wouldn't the inside become a negative though?

[severe pond]

oh wait

[full gust]

σ = √(1/8 * (450 - 2.12.96 + 8.12.2))
it should be this i think

[brave pollen]

The inside of the square root would still be negative???

[full gust]

why?

σ = √(1/8 * (450 - 2176 + 13824))

σ = √(1/8 * 12100)

σ = √1512.5

have i done it wrong?

wait

[brave pollen]

Oh. I didn't see your decimals

[full gust]

not decimals, theyre multiplying, but discord just bolds the numbers in between astrisks

so it wasnt clear

[brave pollen]

Allso that is not one of the awnsers???
It's supposed to be a multiple choice question and sqrt(1512.5) isn't one of the avaliable awnsers

[full gust]

σ = √1512.5 is approximately 3√6

should be c

[brave pollen]

Oh okay. Thanks

.close

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[grand shoal]

PLEASE CAN SOMEONE HELP TO PROVIDE A PROOF I GOT A TEST ON MONDAY AND I AM EXPECTED TO KNOW THIS PROOF AND BE ABLE TO RE WRITE IT

[brave pilot]

I don't get how to finish off the equation.

[lean otter]

is it ok if u prove the reciprocal ?

[grand shoal]

well the only other thing that was mentiones is it follows from the contrapositive of another statment

[lean otter]

i mean that u prove that if it converges then Un -> 0

[grand shoal]

but i think i need an actual proof simply saying its the contrapositive of another lemma wont suffice

[grand shoal]

i dont know ho w

[lean otter]

but is it enough if u do this ?

[grand shoal]

would that prove the statement?

[lean otter]

yes

[grand shoal]

then i imagine yes

[lean otter]

ok so u have ur sum from 0 to inf that u can split into 2 sum

from 0 to n and from n+1 to inf

2 sec

no i do shit

u have that Un = Sn - S(n-1)

with Sn = sum from 0 to n of Un

[lean otter]

and u do the limit with n tend to inf in this expression

[grand shoal]

got it

then what

[lean otter]

uh it is finished

[grand shoal]

what would the limit be

[lean otter]

0

since Sn converges

lim Sn = lim S(n-1)

[grand shoal]

ah right

thanks

[lean otter]

np

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[merry quest]

does anyone know how to find AD

[merry quest]

i been trying for a bit and i have gotten no where

[safe radish]

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

[merry quest]

ans is 3 thank you for helping

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[plain arrow]

ok so like i thought understood this but appernatly not cus im getting it wrong

[severe pond]

yea you don’t understand it

this is the graph of f’

not the graph of f

you’re supposed to describe what happens at f given what happens at f’ on the graph

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[digital swallow]

can someone please help with this one

i have no idea how it works

<@&286206848099549185>

[lucid stream]

You should have been given formulas that help you make the conversion to spherical coordinates

[plucky elk]

!15m

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[digital swallow]

ok sorry

[lucid stream]

Do you have in your notes the formulas?

[digital swallow]

yes i think so

but im lost i dont understand it

[lucid stream]

Can you show me what your notes say the formulas are?

[digital swallow]

yeah

might be a bit hard to read

[severe pond]

bro you literally have the solution

they held your hand the whole way

[digital swallow]

i tried doing it for this problem

@severe pond

using the same method but its wrong

this is a diff problem tho

[safe radish]

@digital swallow Has your question been resolved?

[digital swallow]

<@&286206848099549185>

[digital swallow]

pls help

[safe radish]

@digital swallow Has your question been resolved?

[hot rampart]

hi everyone

[safe radish]

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[rustic goblet]

really dumb question: is this a typo?

[rustic goblet]

surely it's $B \in \mathcal{B}$ and not $B \subseteq \mathcal{B}$, right?

[flat frigate]

higher!

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[ruby tulip]

writing...

[ruby tulip]

In this image:

I don't understand how the last matrix shown is equal to the the previous matrices. When I try to work it out myself, I always end up with a transposed matrix. So, if I have a matrix that looks like this:

and I first transpose the matrix:

and then I take the rows, I endup with transposed matrix

If I take the row of the non-transposed matrix, and then transpose that row into a column, I still end up with a transposed matrix.

Where am I going wrong?

[median vigil]

it says that $\vb m_{i,}$ represents the $i$th row vector in column vector form

[flat frigate]

cloud

[ruby tulip]

oh, yeah

I got it now. Thank you.

.close

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[kind beacon]

Guys How do I solve for t here?

e^6+2 = ln(t)+t^2-1

I asked my teacher for help and he said this

"There are two terms on the left, and two terms on the right. Maybe, start by studying one of the terms and see what it would need to be to work. Then, check if it satisfies the whole equation. I'm not trying to be too cryptic, but it's hard to answer without giving the answer away"

BUt I dont understand still

[kind beacon]

$e^6 + 2 = \ln(t) + t^2 - 1$

[flat frigate]

usernamephobic

[median vigil]

maybe you can rearrange it so that you have all of the constants on one side?

[kind beacon]

$e^6 +3=lnt + t^2$

[flat frigate]

usernamephobic

[median vigil]

yes

now you have two terms on the left and two on the right like your teacher said

[lean otter]

hint: e^c will turn into c when run through ln

[kind beacon]

omg

i am so stupid

this is ridclous

thanks guys

@lean otter

[lean otter]

you can prove that t is unique by noting that the right hand side of this is strictly increasing

[kind beacon]

wait

[lean otter]

if you haven’t learned that, don’t worry about it too much

[kind beacon]

nvm im still lost, this is the furthurst i got

$t = e^(9-t^2)$

[flat frigate]

usernamephobic

[austere cypress]

$t = e^{9-t^2}$

[kind beacon]

$t = exp(9-t^2)$

[median vigil]

you have two terms on each side

[flat frigate]

kaue

[median vigil]

if you are lucky then you can have a solution where the first and second terms are equal individually

(depending on how you arrange things)

[rotund cloud]

huh tf

ima ned help

[kind beacon]

what do you mean by first and second terms are equal indiviually?

like t^2 = e^6?

$t^2 = e^6$

[flat frigate]

usernamephobic

[kind beacon]

$t = e^3$

[flat frigate]

usernamephobic

[kind beacon]

ln(t) = 3, subbing in e^3

ln(e^3) = 3

so is it e^3?

i dont know if thats what you meant

[median vigil]

well if you substitute in t = e^3 to the entire equation, are both sides equal?

[kind beacon]

yes

but i dont think i learned this

what is the method for this called?

nvm

lol

thank you, ive been stuck for a long time

.close

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[viral oyster]

Im aware i can use the sum of first n integers ( n(n+1))/2 ) to solve the first one but im clueless for the second one

[safe radish]

@viral oyster Has your question been resolved?

[queen saddle]

you can find it here https://math.stackexchange.com/questions/836006/limit-of-lim-n-rightarrow-infty-sumn-i-1-frac1-sqrt-n2-i?noredirect=1&lq=1

[viral oyster]

thanks man

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[bright sphinx]

i need desparate help for this questionp lease

[bright sphinx]

there answer states that there are two sets of dimensions possible and i cannot get 1

i did earlier but now as i revisit the question i cannot do it

[quick crater]

could you draw it

[bright sphinx]

done that

but when i did cosine rule for some reason i got BC being 3.7

[eternal carbon]

vould you draw it

• a romanian, probably

[bright sphinx]

which is wrong

lemme get my phone so i can show my drawing rq

[quick crater]

ABC=35°

[bright sphinx]

oh fuck

makes more sense now lemme try it

i still don't understand how there are two possible dimensions though in the answer

[vagrant ice]

search up sine rule ambiguous case

[bright sphinx]

yeah i understand it

i just forgotten how to use it since i did i many many moons ago

right so since ACB is = 60

it could also possibly be 120 because it still allows enough space for 35 and whatever the other angle is

[vagrant ice]

it can also be 120 then, cause sin(60) = sin(180 - 60)

sin(x) = sin(180 - x) in general yes

[bright sphinx]

right because sine is also positive in the second quadrant

[quick crater]

i just chug and plug cos rule in desmos lmao

[bright sphinx]

right and then i find BAC

and then do cosine rule with both of the possible angles?

or is that wrong

wait since cos(120) doesn't exist does it

[vagrant ice]

it does

[bright sphinx]

oh wait nvm

[vagrant ice]

it's just negative

[bright sphinx]

yeah okay

right so it would be -1/2

okay thanks

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[vagrant ice]

no worries!

[bright sphinx]

i understand it now

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[lucid atlas]

im still struggling with this graph, im not sure why i cant quite understand how the x affects it

[noble kiln]

,rccw

[noble kiln]

Hmm

What exactly do you need help with

[lucid atlas]

as x tends towards infinity y tends towards 0 for y=1/x

[lucid atlas]

understanding what the final graph looks like ie how y=x affects y=1/x when u add it

[unkempt venture]

hus

[noble kiln]

Desmos time!

[lucid atlas]

haha

[thin bridge]

as you said,as x→inf, 1/x →0
so for large values of x, y = x + 1/x
behaves close to y = x

[lucid atlas]

yes

[thin bridge]

this would be descibed as an oblique/slant asymptote

if in doubt, make a table of values

[lucid atlas]

and for x becomes small?

y=1/x with a small x

so y becomes large

but asymptote at x=0

this doesnt change right

[thin bridge]

yeh

[lucid atlas]

okay ty and for the negatives, y=-x? for x tends towards negative inf

[thin bridge]

no

[lucid atlas]

for y= 1/x +x

[thin bridge]

how are you getting y=-x

[lucid atlas]

sorry i think i miswrote that haha

as in it looks like y=-x no?

[thin bridge]

no

[thin bridge]

how are you getting y=-x

draw y=-x on there and see what happens

[lucid atlas]

oops 😭 not quite what i meant

haha

so for x tends towards negative infinity y=1/x + x 1/x tends to 0 and x to negative infinity, y also to negative infinity so y=x

[thin bridge]

reasoning is a bit off

both approaching negative inf doesn't necessarily mean that they'd be equal

[lucid atlas]

oh okay, do you mind explaining your reasoning

[thin bridge]

for large x, 1/x approaches 0
you'd be left with y = x + (something very small)

[lucid atlas]

right okay

and for small x

x approaches 0

leaving 1/x

[thin bridge]

yeh

[lucid atlas]

do we need to consider turning point

[thin bridge]

depending on what course you may be expected to find them

[thin bridge]

they are considered meaningful components of the graph

[lucid atlas]

would we do this by just differentiating and equating to 0

[thin bridge]

yeh

for the x coords and determine the y-coords from there

[lucid atlas]

awesome thank you so much legend

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[wraith swift]

If two curves intersect at a point and then the tangent is possible from that point if both curves have same slope

[wraith swift]

so i can equate teh derivatives at that point ?

[pseudo scroll]

I feel like this is xy

[safe radish]

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.

[wraith swift]

there was no problem immediately with it

i was just thinking about it

[pseudo scroll]

alright

But I don't get what you mean

If two curves intersect their derivatives may or may not be equal at the point of intersection

y = sin x and y = cos x have them different, and y = x^3 and y = -x^3 have them the same

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[golden compass]

hi, im trying to make a 3d project with filled triangles, but i know how to calculate the incenter in 2d but i want to make a 3d position for to do the painter's algorithm, i just want to know the Z coordinate in the incenter (also, the image is not accurate to the positions)

[safe radish]

@golden compass Has your question been resolved?

[queen saddle]

You can do it in vectors. Make unit vectors b=AB/|AB| and c=AC/|AC|. Then a=b+c is a vector towards incenter O from A. You just need to find the length |AO|, which is plane geometry.

So, O=A+a*|AO|/|a|

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[kind basalt]

A 1.20 kg box is being pulled with a force of 3 N as shown in the figure.
What is the acceleration?
What is the normal force on the box?

[balmy goblet]

oh

[kind basalt]

can anyone help? i believe this is abt angle of repose

[balmy goblet]

skibidi

i like it

a sec

[balmy goblet]

do you know Newton's second law

[kind basalt]

mhm

[balmy goblet]

say it

[kind basalt]

A net force causes acceleration on an object andacceleration is directly proportional to the mass

[balmy goblet]

?

sigma ( vector(F) applied on sys) = m x vector(a)

so vector(F) + vector(P) = ma

do they say there is friction ?

[kind basalt]

theres no friction given

[balmy goblet]

ok

[balmy goblet]

do you know this formula

[digital sparrow]

Break F_A into two forces

[balmy goblet]

nah

[digital sparrow]

And draw FBD

[balmy goblet]

i was explaining

let him think a bit

[kind basalt]

hold on AHAHAHAAH

[digital sparrow]

What is P?

[balmy goblet]

poids

[digital sparrow]

Poids?

[balmy goblet]

force associated to his mass

[kind basalt]

whats poids?

[balmy goblet]

idk english word

like the weight ig

[digital sparrow]

Pseudo?

[balmy goblet]

weight in Newtons

[digital sparrow]

Oh

Then a should be vector too

[balmy goblet]

yep

oh yh youre right i forgot to put it

[digital sparrow]

And it is easier to think of this in X and Y dimensions rather than vector form

[digital sparrow]

^

[balmy goblet]

nah vectors are goated

[kind basalt]

whats an FBD 😭

[digital sparrow]

Free body diagram

!nosols

[safe radish]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[kind basalt]

oh i see

[balmy goblet]

how can you find the horizontal component of F ?

[digital sparrow]

I'll leave him to you then

I have to go

[balmy goblet]

alr

[kind basalt]

okay i think i kinda got it

[balmy goblet]

@kind basalt have you studied vectors

[kind basalt]

yeah im figuring this out rn

[balmy goblet]

ok

Fx is the vertical component of F

do you know what that means

[kind basalt]

isnt the Fx the horizontal compoentn

[balmy goblet]

oh yeah sorry 😅

[balmy goblet]

do you understand what i did here ?

ask your questions if there are

[kind basalt]

alrighty

[balmy goblet]

so have you understood what i did

[kind basalt]

trying to 😭

this is different to what is taught to us

[balmy goblet]

where are you from ?

[kind basalt]

Philippines

[balmy goblet]

what formulas have you learnt

[balmy goblet]

oh cool

i've been there once

[kind basalt]

oh? thats so cool:o

[balmy goblet]

damn

that is a strange way of teaching it

[kind basalt]

this is the angle of repose but the object is the one being pulled by two chords i guess

[balmy goblet]

basically what i use is this formula

oh but you havent learnt about vectors right ?

[kind basalt]

we did learn about vectors

yea im familiar with that formula

[balmy goblet]

ok great then

what is cool about it is that you can decompose it into two coordinates

x and y

[kind basalt]

ah yes

ill try to do that

[balmy goblet]

if you need help with that tell me

[kind basalt]

okayy

thank you!

[safe radish]

@kind basalt Has your question been resolved?

[safe radish]

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[craggy leaf]

anyone can help me with this?

The Lagrange polynomial that interpolates the Points (1,3),(3,5)and(4,2)

a.
P(x)=12(x−3)(x−4)−52(x−1)(x−4)+23(x−1)(x−3)

b.
P(x)=52(x−3)(x−4)−12(x−1)(x−4)+43(x−1)(x−3)

c.
P(x)=52(x−3)(x−4)−32(x−1)(x−4)+13(x−1)(x−3)

d.
P(x)=12(x−1)(x−4)−12(x−3)(x−4)+23(x−1)(x−3)

[safe radish]

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[rain dagger]

there are n students in class, and n is divisible by 3. how many different groups of 3 students can we arrange the class in?

idk how to this problem

[rain dagger]

I tried using the binom to do binom(n, 3), etc but it doesn't work

[knotty anchor]

Why doesn’t B(n,3) work?

Wait don’t you need a value for n to work this out?

[rain dagger]

nope
it's class arrangements
like abc and xyz are 2 different groups of 6 students

[peak estuary]

how many options are there to pick the first group

how many options are there to pick the second group

how many for the third

[rain dagger]

I cant do it without overcounting

whats the solution with the multiplication rule

[peak estuary]

well how much are you overcounting

overcounting doesnt mean that it wont work

it just means you need to account for it

[rain dagger]

binom(n, 3)binom(n-3,3)..*binom(3,3)

which is overcoming it

I cant fix it

[peak estuary]

how is it overcounting it

[rain dagger]

I'm very sure there is

[peak estuary]

if you write that with all the factorials then you can simplify it

[rain dagger]

I'll try an example

[peak estuary]

yes you are overcounting

I am asking you how much you are overcounting

[rain dagger]

I think I need to divide each one by 3! but idk

each binom I mean

or maybe something like n/3, n/9, n/27

but I just don't understand even what to do now

[peak estuary]

an example is a good idea

[rain dagger]

n=6
there's 6 arrangements of students in trios (using x1x2x3, y1y2y3)

my solution would be in total: 6!/(3!*3!)=5!/6

how would I reduce that? I don't see a pattern

[peak estuary]

I dont know what you mean by your first sentence

[rain dagger]

like the actual correct answer for n=6 is 6

[peak estuary]

if there is only 6 then you can list them, right?

[rain dagger]

yeah using x1 x2 x3 y1 y2 y3 as students

[peak estuary]

do it

[rain dagger]

waitf there's more now

3(3+3+3) 3*9 +1=55?

[peak estuary]

how

[rain dagger]

i just took out xi every time and switch yj and did it for 2 numbers too and the original is 1

can I just walk me through a solution ffs

[peak estuary]

I dont know how you get 55

how many ways are there to choose the first group

just the first, not caring about everything else

[rain dagger]

in the original questions? binom(n, 3)

[peak estuary]

and for n=6 ?

[rain dagger]

here I just switched the triplets of Xi and yi

[rain dagger]

binom(6, 3)

which is

[peak estuary]

which is?

[rain dagger]

6!/36

5!/6

but that I already said

actually no

no actually yes

[rain dagger]

here

[peak estuary]

yeah there I didnt understand how you got 6

anyway, 5!/6 is 20

can you list them

[rain dagger]

it was my mitstake, didn't account for a lot of cases in my example

[rain dagger]

that's way too long, do you have a solution that I can understand how to solve the question with?

[peak estuary]

20 is not much

[rain dagger]

like I can't even list those cases at this point

[peak estuary]

it really isnt

[rain dagger]

I'm on phone bro

💀

[peak estuary]

having an actual list in front of you is hopefully much more likely to get you to understand stuff

so you are on phone and dont have paper or anything?

[rain dagger]

well maybe no

idk

[rain dagger]

yeah I'm out of the house

well it's alright I'll try later on my own again

[peak estuary]

123
124
125
126
134
135
136
145
146
156
234
235
236
245
246
256
345
346
356
456

those are the options

for the first group

in each of those, what are the options for the second group

@rain dagger

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@rain dagger Has your question been resolved?

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[vivid harbor]

if the numerator and denomiator of a faction are borth decreased by one the fractopm becomes 2/3 if thenumerator and deonmitaotr are both increasd by 1 it becomes 3 /4 what is the oringal frcation

[vivid harbor]

this is my working out and I seem to be wrong but why?

16 is my attemt

[upbeat swan]

what did you do at your first step

(x+1)(x-1)=6

how did you get that

[vivid harbor]

because x is the numerator so it -1 over the denomaitor (y) - 1 = 2/3

[upbeat swan]

ok and

?

[vivid harbor]

wdym I am wrong apparantly so I was wondering where my logic is flwaed

[upbeat swan]

(x+1)(x-1)=6
(y+1)(y-1)=12

this is wrong

[vivid harbor]

but why both denominators times each other is = 6 and both numerators timeseac other is 12

[upbeat swan]

the 2 fractions are independent

they are completely separate from eachother

you can't mix them

[vivid harbor]

but why

[mortal sandal]

just because $\f ab=\f 34$ doesn't mean $a=3$ and $b=4$

[flat frigate]

Dreyuk

[vivid harbor]

oh

how would u form an equation orinally then

[upbeat swan]

yeah and if $\frac{a}{b} \cdot \frac {c}{d} = \frac{3}{4}$ doesnt mean that $a \cdot c = 3$ and $b \cdot d = 4$

[flat frigate]

Oogy Boogy

[upbeat swan]

multiply the diagonals together in each equality

you'll get 2 equations with 2 unknowns

if $\frac{a}{b} = \frac{c}{d}$ it means that $a \cdot d = b \cdot c$

[flat frigate]

Oogy Boogy

[vivid harbor]

thanks

[upbeat swan]

np

[vivid harbor]

how do I close this

[upbeat swan]

.close

[vivid harbor]

thanks

.close

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[upbeat swan]

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[wooden umbra]

"Which of the following functions represents the correct linear approximation to f(x) at x = 0?"

A. L(x) = tan(0)
B. L(x) = x
C. L(x) = 2x
D. None of the above.

I am unsure if the correct answer is B. or D.

I am thinking it is D. because B. isn't at x = 0

[upbeat swan]

well

you got L(x)=x

which is B

[wooden umbra]

Yeah

[upbeat swan]

so that's the answer

[wooden umbra]

But it says at x = 0

[upbeat swan]

yeah

at x=0 you get L(x)=x

if it were at another point you'd get another approximation function

[wooden umbra]

But wouldn't it be L(0) = 0 at x = 0?

[wooden umbra]

Yeah

[upbeat swan]

yes

[wooden umbra]

I haven't answered a linear approx question before

So Idk how they want it to be phrased

[upbeat swan]

they want you to find a function

which means to plug in x=0 into the formula

that's it

[wooden umbra]

Okay

Thank you

[upbeat swan]

sry

L(x) = f'(a) (x-a) + f(a)

you plug in a =0 not x=0

mb

[wooden umbra]

Oh true

I knew what you meant so don't worry ^^

[upbeat swan]

[upbeat swan]

you're welcome

[wooden umbra]

.close

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[viscid crane]

Hi, I just wanted to ask in which situations I should use the chain rule (derivatives), like for example one of the questions was (5x+3)^4 and what I did was multiply the left side with 4 and so it became 4(5x+3)^3
So I guess my problem is I don’t know when to use the chain rule vs the normal rule, a comparison would also be great, thank you : )

[torpid citrus]

yo

[viscid crane]

Yoooooooo

[delicate sphinx]

Do you know what a composite function is?

[viscid crane]

I think yea

[delicate sphinx]

Basically a function in a function

[viscid crane]

Yea I know what that is now

I remember

The thing is (5x+3) is not inside a function

It’s just cubed

😭

[delicate sphinx]

So if you notice in your problem, (5x+3)^4, you can see that it's f(5x + 3) = x^4

[delicate sphinx]

But it is in a function

The outer function is x^4

[viscid crane]

Oh I didn’t know that

Is there like a list of stuff like this

[delicate sphinx]

Just practice honestly

[viscid crane]

Oh ok, well now I know thank you 🙂