help-26

[woeful drift]

Anyway, <@&268886789983436800>

[stable minnow]

mods?

[charred ivy]

ig ima wing it

[stable minnow]

fair fair

dont konw why this guy pinged mods but

[urban grove]

quiz as in, something you're supposed to be working on alone?

[charred ivy]

i dont need it, its for extra credit

so i can ask for help

[topaz sinew]

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[topaz sinew]

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[slender sonnet]

a boat is in a rive it takes 5 hours to go from point A to point B. And it takes 7 hours yo go from point B to A. How long would it take a log to go from point A to B. Keep in mind a wood log doesn’t have its own speed its dependent on the river

[topaz sinew]

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.
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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

[slender sonnet]

Well ik j need to make a equation

But not sure how to

I need to find the rivers speed somehow

[ocean fjord]

The distance is same in either case so one has to be upstream and one has to be downstream. try identifying which is which

[slender sonnet]

Well less 5 hour one is with the stream

Cuz its faster

[ocean fjord]

suppose the speed of the stream is v, and speed of the boat (assumed constant) is b

what would the speed upstream and downstreams be respectively?

[slender sonnet]

b+v

And b-v

[ocean fjord]

suppose the distance is d

will the log be going upstream or downstream

[slender sonnet]

Upstream

[ocean fjord]

A to B is the lesser time

[slender sonnet]

Ok?

I mean with the stream

[ocean fjord]

lesser time => higher speed => downstream

yes

[slender sonnet]

Mixed words mb

[ocean fjord]

so d = tv you want to find d/v

try setting up the equations with the boat now

[slender sonnet]

d=5(b+v)?

[ocean fjord]

yes

also, d = 7(b-v)

eliminate the b

[slender sonnet]

Its a system now right?

[ocean fjord]

yes

[slender sonnet]

d=5b+5v

d=7b-7v

5b+5v=7b-7v

d=2b-12v

[ocean fjord]

no you dont want to do that
you dont want to eliminate d since d/v is what you need to find

notice d/v is independent of b

try eliminating b instead

[slender sonnet]

7d=35b+35v

[ocean fjord]

+35v*

[slender sonnet]

5d=35b-35v

Yes

[ocean fjord]

yep

subtract the eqns you should be done

[slender sonnet]

2d=70v

d=35v

Now what tho?

@ocean fjord

[ocean fjord]

you need d/v

simply rearrange the eqn to d/v = 35

thats the answer

[slender sonnet]

So how long dose the log take from A to B?

35 hours?

[ocean fjord]

35

yes

[topaz sinew]

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[woeful drift]

!status

[topaz sinew]

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[hallow cape]

2

[woeful drift]

i'm not quite sure I follow your work

[hallow cape]

brb ill redo it

[woeful drift]

thank you

[hallow cape]

when I redid it clearly i solved it correctly thanks lol

[woeful drift]

i take all of the credit

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[lyric grove]

Can someone help me with this please

[terse abyss]

!status

[topaz sinew]

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7. None of the above

[lyric grove]

sorry

i am getting the answer as 10/43

there are three cases:

1. If the ball drawn from bucket 1 is red
2. If the ball drawn from bucket 1 is black
3. If the ball drawn from bucket 1 is white

Probability of case 1 happening is 2/10
Probability of case 2 happening is 3/10
Probability of case 3 happening is 5/10

i guess i solved it

wait

hmmmmmmmmmm

1 sec

lesssgoooo

is it

20/43

can someone confirm please

[balmy roost]

fr I don't know

I myself think the answer would be 5/18

but I'm not sure

[lyric grove]

thats not in the options

ig its not that most probably

[balmy pendant]

What are the options

[lyric grove]

I am getting A) 20/43

[balmy roost]

oh wait I thaught the ball taken from bucket 2 is white

look

I think it doesn't matter

because it says that the ball drawn from bucket 1 is random

so that means that it will be 5/10

always

[lyric grove]

ig not 😅

[balmy roost]

why

[lyric grove]

there is a condition here now

that the ball drawn from bucket 2 is black

[balmy roost]

yes but that condition applies after the ball is drawn right?

maybe I'm wrong

[balmy pendant]

I'd do it with bayes' theorem

[lyric grove]

did it with that only

i just want to confirm my answer

[lyric grove]

but see there could be a case that the ball drawn from first bucket is black or red

that would affect our probability of selecting a black ball from bucket 2

and here we are given that we slected a black ball

[balmy pendant]

Yes it is right

[lyric grove]

think it in this way:
if the ball drawn from bucket 2 had been white, then there would have been more chance that the ball drawn from bucket 1 is white

[balmy pendant]

Dit it mentally took me ages ...

[keen venture]

P(B1 White | B2 Black)
= P(B1 White AND B2 Black) / P(B2 Black)

[lyric grove]

haha thankyouu

[keen venture]

Or no sorry, let me edit that

[lyric grove]

reverse

[lyric grove]

yup

[keen venture]

P(B2 Black) is the tougher case here, but law of total probability can get it.

In other words, use a tree diagram

[lyric grove]

yea i got it

[lyric grove]

made these 3 cases

and calculated P(black) for each case

2/10x4/10 + 3/10x5/10 + 5/10x4/10

[keen venture]

Yep, exactly. Sorry I must have misunderstood your earlier work. Didn't mean to retread.

[lyric grove]

np np

P(B1 White AND B2 Black) = P(B2 Black| B1 White) x P(B1 White) = 4/10 x 1/2

P(Black) = 2/10x4/10 + 3/10x5/10 + 5/10x4/10 = 43/100

P(B1 White | B2 Black) = P(B2 Black| B1 White) x P(B1 White) /P(Black) = (4/10 x 1/2)/(43/100) = 20/43

@keen venture right?

[topaz sinew]

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[lyric grove]

.close

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[unborn hornet]

(B)?

[lyric grove]

A

[topaz sinew]

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[timber kindle]

This is pretty straightforward, I already knew how to do the derivative. However, this kind of questions take such a long time to do and I won't have enough time to finish it in the test. Any tips to deal with it?

[timber kindle]

<@&286206848099549185>

[wooden osprey]

!15min

[topaz sinew]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[wooden osprey]

kết hợp quy tắc đạo hàm tích và chain rule

[timber kindle]

làm sao làm nhanh hơn

trong mấy cái đáp án nó còn ko rút gọn nữa

[wooden osprey]

có thể ko có cách nhanh hơn

[timber kindle]

;-;

[wooden osprey]

,tex .diff rules

[timber kindle]

Ok thanks

[timber kindle]

.close

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[sterile comet]

hey quick question, can someone explain to me how we get from (2^(n)-2)/(2^n) to 1 - 2/2^n and then eventually to 1 - 2^(1-n)

[oak fjord]

$2^n - 2/2^n = 2^n/2^n - 2/2^n simp 2^n/2^n =1 so 1- 2/2^n$

[thorny flame]

!♛ Илья

[sterile comet]

i dont get it

[oak fjord]

which part?

[sterile comet]

oh wait one sec i think i am getting there

the middle part

2^n/2^n - 2/2^n

we have 3 divisions?

[oak fjord]

just see 2^n/2^n now something/ something =1 so...

[sterile comet]

oh yeah no i understand that part. the problem for me is, i dont understand how we "get rid" of the -2 to do so.

[oak fjord]

you mean sec part of simplifying ?

the one 1-2^1-n?

[sterile comet]

yes

i mean the step to get to there

[oak fjord]

if yes you can move powers in this case (n) From the denominator to the numerator by time it to -

so if we say 2 = 2^1 in 2/2^n we can move n to numerator and it become -n so we would have 2^1-n

simple as that.

[sterile comet]

so 2/2^n becomes to 2^-n/2 and then 2^1-n?

ok np

[oak fjord]

ok that's true. look x/x^n = x.x^-n and we have 2/2^n = 2^1. 2^-n = 2^1-n

[sterile comet]

what does the . stand for?

multiply?

[oak fjord]

time

yes

[sterile comet]

ah aight ok

ok now i think i understand but my question is

now i understand what you did there with x/x^n being x * x^-n

is it the same if we have x/x^1-n

would that just make x * x^1-n?

[oak fjord]

no

[sterile comet]

🥹

[oak fjord]

x*x^-(1-n)

[sterile comet]

oh

SO x*x^-n right?

[oak fjord]

right

[sterile comet]

so would that be correct?

[oak fjord]

yes

[sterile comet]

ok just last check to see if i understood:

and then we can turn the 2/2^n to 2^1-n for some reason?

[oak fjord]

when you say for some reason, this is a bad sign

let's reccap, x/x^n= x*x^-n

2=2^1 so 2/2^n = 2^1*2^-n = 2^n-1

[sterile comet]

yea it just isnt or wasnt logical for me

the only thing i dont understand is the -1

i mean now i understand since 2^n is 2 * 2 * 2 * ... *n, and dividing would just subtract one from these so n -1 ?

[sterile comet]

but why is it 1-n then

[oak fjord]

What grade are you?

[sterile comet]

university

[oak fjord]

ok so mybe i dont get it explain more where is your problem

why 2^1*2^-n = 2^n-1 ?

if it is, when you multi two same numbers with different powers the answer will be the number to power of The sum of their two powers

[sterile comet]

i mean why it becomes 2^1-n and not 2^n-1

[oak fjord]

ok i get it

so

[sterile comet]

yes understood

[oak fjord]

very well

[oak fjord]

and about this it would be 2^1*2^-n = 2^-n+1 and it same as 2^n-1

[oak fjord]

not n-1 , -n+1

[sterile comet]

oooh

i think i got it

thank you!

wait

[oak fjord]

any time

[sterile comet]

i see the analogy yes -n+1 = n-1

but its not 1-n, thats something different right?

[sterile comet]

so mine is wrong?

[oak fjord]

no becuse 1 + (-n) = 1-n and if you say (-n)+1 it would be -n+1 = n-1

[sterile comet]

oh yeah right

my brain is so melted from this task i cant even do basic math

thank you man

[oak fjord]

no problem

[sterile comet]

this is what i have and understood

[oak fjord]

great

[sterile comet]

have a good day mate, thanks alot really appreciate your help

actual heros

[topaz sinew]

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[safe thorn]

Hello, I wanted to get help with my project

[safe thorn]

.close

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[tiny rover]

Need help checking if the answers with the blue underline are correct and need help solving problem e

[split cloak]

In e use quotient rule

[fervent crag]

I don't think c is correct

[split cloak]

For c use product rule

So ig only h is right

[tiny rover]

.close

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[tiny rover]

.reopen

[topaz sinew]

✅

[tiny rover]

thank you

.close

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[fair dagger]

I have these 2 circles
There is point M, and from it there are 2 tangents to both circles, and the distance between M and the tangent point on both circles is the same
I need to find the locus of M

[fair dagger]

I know the answer

this is the locus

I have no idea how to get there

I tried to find the equations of the tangents, but apparently I was wrong somewhere

[abstract wadi]

You don't want that.

You can assume the point M is (a,b) or something...

You can write out the length of tangent to a circle from a point, without writing down the tangent equation.

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[scarlet zenith]

pls help

A circuit in an undirected graph which uses every edge of the graph precisely
once is called an Euler circuit.
(a) (4 marks) Consider the city of Königsberg circa 1730, modelled by a graph whose
vertices are the landmasses and whose edges are the bridges. Find the minimum
number of bridges that you would need to build so that this graph contains an Euler
circuit. Draw this updated graph with the new bridges clearly identified.

[scarlet zenith]

is gpt correct>

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@scarlet zenith Has your question been resolved?

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[exotic arch]

could someone solve this without substituion please

[exotic arch]

this is for no homework

[opal vault]

why are you against substitution?

[exotic arch]

because in our course, we do not use substitution for this integral

however our lecturer did quite the poor job explaining how it's solved without it

and I cannot use substitution in exams

so I want to know how it's solved without it

[opal vault]

well uhm

[ionic oar]

[ \f{\dd{x}}{\sqrt{x}} = 2\dd(\sqrt{x}) ]

does this make sense

[thorny flame]

Sadie Carnot (η > 1)

[exotic arch]

yeah

[ionic oar]

alright

we can do the same in the integral

[ 2\int \f{d(\sqrt x)}{1 + (\sqrt x)^2} ]

[thorny flame]

Sadie Carnot (η > 1)

[ionic oar]

agreed?

[exotic arch]

oh

right

[abstract wadi]

Substitution without calling it substitution. 😂

[ionic oar]

and [ \f{\dd{x}}{1 + x^2} = d(\tan^{-1} (x) ) ]

[thorny flame]

Sadie Carnot (η > 1)

[ionic oar]

do you agree to this too?

[exotic arch]

yeah I get it

ty

.close

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[ionic oar]

alright

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[empty galleon]

Greetings. I wonder did i made mistake or..? I just need explanation how did they got that solution. Thank you

[empty galleon]

i see

wht bout ln2/ln3

[half edge]

,w (ln 2)/(ln 3)

[thorny flame]

Results provided by WolframAlpha

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Requested by northsteve

[half edge]

,w (ln(1.5))^(-1)

[thorny flame]

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[empty galleon]

okay?

[half edge]

Ur answer is incorrect

[empty galleon]

i see

okay

thank you

.close

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[obtuse garden]

can anyone help me out?

[clever wedge]

is that 0 to x/2

[obtuse garden]

pi/2

[clever wedge]

are you aware of f(x) = f(a+b-x) rule

in definite integrals

[obtuse garden]

tried

didnt work

[clever wedge]

oh umm wait

[obtuse garden]

I tried that once I realized there's no indefinite integral

[hasty smelt]

is it e^x(the whole thing)

[obtuse garden]

It's e^x multiplied by the thing inside

[hasty smelt]

there's a parenthesis missing

[obtuse garden]

Yea at the end

[hasty smelt]

okay

have you tried distributing maybe

[obtuse garden]

Yup

[clever wedge]

e^sin does cancel, but it leads nowhere

[obtuse garden]

Yup

It simplifies to e^x(1-e^-2cosx)

[clever wedge]

ye

[obtuse garden]

Is this impossible to solve?

Or am I dumb

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@obtuse garden Has your question been resolved?

[obtuse garden]

No

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[normal pilot]

How to (any unit)^2 ?

[normal pilot]

How do i multiply the velocity of something by itself?

its somewhat a basic i forgot

https://prnt.sc/LNZJ0KwX59P8

is that correct?

[loud oasis]

that looks right

[normal pilot]

aight thanks, thats about it.

.close

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[native fern]

hi I need help with a math question

[glossy arrow]

Hi, what is your question?

[native fern]

The surface integral of the cylinder (1/2)x^2+y^2=1 with 0<=z<=1. and the vector field u=(x/(x^2+y^2),y/(x^2+y^2),z^2-z)

Using gauss theorem the triple integral becomes 0 when 0<=z<=1. I'm supposed to use this to conclude that every cylinder at 0<=z<=1 has the same flux going through it. But I don't understand.

If we have a field F that is not defined in the origin but otherwise is and divF=0 then hollow spheres that encloses origin will have the same surface integral. But I think with cylinders it is different because connecting two arbitrary cylinders (with 0<=z<=1) makes a top and bottom which I don't know the flux in or out.

[topaz sinew]

@native fern Has your question been resolved?

[native fern]

<@&286206848099549185>

i have posted the question here too https://math.stackexchange.com/questions/4913375/calculating-a-surface-integral-of-a-surface-by-another-surface?noredirect=1#comment10492626_4913375

[topaz sinew]

@native fern Has your question been resolved?

[radiant marlin]

the flux where the two cylinders meet cancel out because one's going in and the other's going out. it's like a more complicated version of this

[native fern]

what do u mean with where two cylinders meet?

[radiant marlin]

oh maybe I misinterpreted what the 2 arbitrary cylinders would be

not on top of each other but like different base shapes?

[native fern]

not on top no

they just need to satisfy 0<=z<=1

so yes they can have different shapes

any cylinder ax^2+by^2=R^2 with 0<=z<=1 will give the same answere

[radiant marlin]

the first two bits of u x/(x^2+y^2),y/(x^2+y^2) are conservative in 2d, so any path at a z value gives 0

[native fern]

what do u mean with any path at a z value?

[radiant marlin]

like a circle at height z=1/2

line integral will be 0

[native fern]

hmm

[radiant marlin]

I don't really see the z part yet but presumably it's because z^2-z is 0 at z=0 and z=1 for the bases

[native fern]

yes

that with line integral i think that is it

[radiant marlin]

oh I missed that the flux wasn't zero just some constant value

the main idea is that the spike at the middle line x=0,y=0 is what creates the flux and has a nonzero divergence

[native fern]

I think so

I need to think about this

[radiant marlin]

back to your actual first question lol the tops and bottoms have the z-component 0 so they don't have flux

[native fern]

so any circle that goes around the z-axis will have line integral that is 0?

[radiant marlin]

yea maybe scrap what I said about that, it's like a dimension under and about curl so it doesn't matter

[native fern]

if it is true then the top and bottom will have zero flux

i think

[radiant marlin]

I was thinking stokes theorem but that's 2d but this is 3d gauss

here u is 0 in the z-direction at the top and bottom because z^2-z

so the dot product with the flux is 0

it's pointing horizontal at the caps

[native fern]

stokes theorem can be used in 3d

im not sure

[radiant marlin]

it's like line to surface instead of surface to volume

[native fern]

but u know what im saying right

if u have two cylinders u need to connect them to use gauss theorem

since to use gauss theorem u need to have a 3d body

and the cylinders are just surfaces

[radiant marlin]

I don't understand no

like this blue region? between two cylinders

[native fern]

yes

[radiant marlin]

you can just do it for one cylinder instead and it's fine

but since that blue doesn't touch the middle, all divergence is 0 and there's no net flux

and the top and bottom has 0 flux by symmetry

[native fern]

can u explain that last part

ur saying something different every time lol

but it is good I will through this

but what do u mean with flux is 0 by symmetry?

the field is different on top than at the bottom

[radiant marlin]

like the field lines go flat against the top and bottom

but flux needs lines to go through the surface to be nonzero

[native fern]

no they go up

[radiant marlin]

with z^2-z?

[native fern]

0

wait what

[radiant marlin]

like u at the top is going to be like (3,5,0) or something, no z

[native fern]

oh u are right

now i get it

thanks

kind of anti climatic though

but thanks

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[neon iron]

I have trouble making conclusions, given AB = O, where A, B are square matrices, and O is a null matrix

[neon iron]

there’s 3 cases my teacher discussed, but I don’t follow them

1. When A is non zero, |B| = 0

2. When both A and B are non zero, then |A| , |B| = 0

3. When |A| is not zero, then B is zero matrix

Can someone please help me understand

[dense crescent]

we know that |AB| = |A||B|

as AB is null matrix |AB| = 0

so |B| = 0 if |A| not zero

if A and B are both not null

then |A| and |B| = 0 as it is the only possibility

[neon iron]

Why

Tag me

[dense crescent]

ok so assume A is singular and B is not

so B can have its inverse right

multiply AB = O with B^-1 right side

u will get ABB^-1 = OB^-1

that is A = O

but that is not possible as we cnsidered A and B are not null matrices

@neon iron

[neon iron]

Wait let me process all this

[dense crescent]

okay sure

just ask if u dont understand it

[neon iron]

Oh are these three cases just a variation of this

[dense crescent]

yep

[neon iron]

Omg

I understand

My teacher could never

🫡

Thank you so very much

I hope you have a nice day

[dense crescent]

No problem!

[dense crescent]

you too

[neon iron]

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[mystic dragon]

how come its 5?

{(a,a)} can also be such relation, no?

[dense crescent]

a function is reflexive if and only if every element in set is mapped to itself

that is (a,a), (b,b) and (c,c)

then only it is reflexive

[mystic dragon]

oh okay.

thanks

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[turbid linden]

How does this work? Why does x have 2 Y values

[turbid linden]

And does that mean that equation isn't a function

[grave gale]

Yes you are correct it is not a function in the traditional x and y sense

Because one x maps to multiple ys

One input to multiple outputs

and at first glance, it appears that x has two y values due to the property of the exponent

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@turbid linden Has your question been resolved?

[turbid linden]

How does it work

[turbid linden]

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[turbid linden]

Thanks anyways

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[keen cosmos]

hi! i've been trying to solve part (ii) of this questions for NEARLY 3 MONTHS now, and i was wondering if anyone could help or give a little insight?
i've tried many things, but if it helps, part (i) was solved using the substitution x = atanθ.
i have really tried everything i could, can someone pls try it? i would appreciate it so so much. thank you!

[dense crescent]

by parts?

[keen cosmos]

also, one method that ive noticed to solve this is to essentially show that this integral is true... but wolfram|alpha doesn't give me a step-by-step solution.

[keen cosmos]

hmm i've already tried using ibp... though my working may have been not the correct way to do it.... could you elaborate? thanks!

[dense crescent]

ill try to solve

[keen cosmos]

thank you so so much!!

[dense crescent]

yep got it

@keen cosmos

by parts is the way to go

[keen cosmos]

OMG REALLY??

send working please!!

[dense crescent]

yea lemme send

[keen cosmos]

okay thank you!!

[dense crescent]

here

@keen cosmos

[keen cosmos]

hii thank you so much!! let me take a look rqq

[dense crescent]

welcomee

[keen cosmos]

okay hi i think you're right!!! thank you so so muchhh

ur amazing <3

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[frigid skiff]

Help i need your help
Let $X_{1}, X_{2}, \ldots, X_{n}$ be a random sample from a Poisson distribution with mean
$\lambda .$ Thus, $Y=\sum_{i=1}^{n} X_{i}$ has a Poisson distribution with mean $n \lambda .$ Moreover, by the Central limit Theorem, $\bar{X}=Y / n$ has, approximately, a Normal $(\lambda, \lambda / n)$ distribution for large $n$. Show that for large values of $n,$ the distribution of
$$
2 \sqrt{n}\left(\sqrt{\frac{Y}{n}}-\sqrt{\lambda}\right)
$$
is independent of $\lambda$.

[thorny flame]

TomTheCat

[frigid skiff]

Anyone?

[topaz sinew]

@frigid skiff Has your question been resolved?

[reef fjord]

My guess is that you use the fact that barX=Y/n is approximately a normal distribution for large values of n, and write it out as a normal distribution formula. Then find limit as n->infinity...transforming it via the square root though...hmm...

@frigid skiff

[frigid skiff]

Thanks for the help
I'm still confused how to rewrite the equation to become a normal distribution as I still dont understand the question itself..
But yea i'll try more

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[dusty burrow]

AB=AC and AD=BC. Show that AD=4HD

[dusty burrow]

Anybody?

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[real dagger]

For cartiesian to spherical, why is phi undefined when x and y are 0 ??? what about (0,0,1) ??? isn't that obviously valid?

[regal charm]

arccos(x/0)?

in fact acos(0/0) since x = 0

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@real dagger Has your question been resolved?

[iron harness]

any is valid

jsut pick any phi you want

they say undefined because there is none that is better than any other

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[latent snow]

How can I solve this system for x,y,z in R*

[craggy haven]

there are infinitely many solutions

[mellow arrow]

two equations and three variables?

[iron harness]

How about x=y=z=1

[thin adder]

let z = t, then solve for x and y in terms of t

that will give you a general form for any solution (of which there are infinite)

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[vestal arch]

Gnggg

[vestal arch]

How tf does

This minus

I mean

This plus

Turn to minus

Itttt don't make sense

To me

Help please😁

[wheat grove]

You subtracted (AB)^2 on both sides

[vestal arch]

Holon

dumb it down

for me

[wheat grove]

Say x + y = z. What is y=? In terms of x and z

[vestal arch]

so ab is 9 sooo
9 - 9?
is that right

9^2
so
9 x 9
is uh
81
So its 81 - 81

alr

ALR

okay

nicceee

okayy

i get it now

wait

to get the answer

225 - 81 = 144

and how do i root it

to be 12

what is the best and easiest one

crazy how we in the same grade and i dont know anything

OHHH wait holon
alr
alr
i understand

okayy

thxx

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[upper zealot]

is this a tree?

[sleek gate]

those are numbers, next

[upper zealot]

what

[sleek gate]

thats a tree

[cursive patrol]

what

[sleek gate]

im trolling lol

[upper zealot]

that is, but is this a tree?

[sleek gate]

sorry

[upper zealot]

okay

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@upper zealot Has your question been resolved?

[normal tapir]

No

it has 4 edges and 4 vertices

[upper zealot]

but it doesn't have a cycle though, i still don't understand why

1 is root, 4 is a leaf

[normal tapir]

What is your definition of a tree

dags isn't enough

[upper zealot]

i am not too sure, i was given a task but it only had a definition for trees with edges that can be travelled both directions

but it never said that two edges can't lead to one vertice

[normal tapir]

Ok well in general, if you have a directed graph, turn every edge into an equivalent undirected edge

and if the result is acyclic connected then you have a directed tree

and that isn't the case here

Since you'll get a cycle

[upper zealot]

okay, thank you

[normal tapir]

Alternatively, there may only be one path between any two vertices

and 1,4 has two different paths

[upper zealot]

oh, i didn't know there can only be one path between two, thank you

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[tulip mountain]

Hey guys i need some help with this one:

[tulip mountain]

Let f(x) = 2cos(13π+x)-2sin(π/2 - x)

i) Prove that: cos(13π+x) = -cosx
ii) Prove that F(x)=-4sinx
iii) Solve the equation: f(x) = -2

[ornate magnet]

Simplify f(x)

[tulip mountain]

with what?

[ornate magnet]

Did you do no.1?

[tulip mountain]

no i have no idea how to do the whole thing

[ornate magnet]

use cos(a+b) formula to prove for 1

[quartz magnet]

cos(13pi + x) = cos(pi + x) which is in fact -cos(x)

if that's where you were stuck at

[tulip mountain]

okay thanks guys i think i got the idea

[thin adder]

you should definitely memorise these

[tulip mountain]

youre the man

[quartz magnet]

i still can't memorize these tbh lmao

[thin adder]

you can simplify them into two identities using $\pm$

[thorny flame]

Obotron

[thin adder]

like this

[tulip mountain]

great thanks

[quartz magnet]

that's incredibly helpful ngl thank you

[thin adder]

no worries guys

[quartz magnet]

yo did you write F instead of f intentionally?

[tulip mountain]

no just a mistake

[quartz magnet]

bcz i got f(x)= -4cos(x)

[tulip mountain]

you are prob right

i am still being bothered by the first

[quartz magnet]

the question says F(x)= -4sinx

uhhh?

[tulip mountain]

yeah my bad it was cos

its in greek and we say cos as συν

[quartz magnet]

makes sense

[tulip mountain]

yep

can you help me with the first

like write it for me

because i expanded the cos thing but still doesnt makes sense

[quartz magnet]

.

[tulip mountain]

how did the 13 got deleted?

[quartz magnet]

cos(13pi+x) = cos(12pi + pi + x) = cos( 6(2pi) + pi + x)

2pi is an entire turn on the circle so it's useless to count cos

[tulip mountain]

okk then thank you

[quartz magnet]

pi is half a turn on the circle

which means cos( pi + x) = -cos(x)

[tulip mountain]

great

thanks man

[quartz magnet]

np

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[fair dagger]

help, I have no idea where to start

[near crypt]

rationalize them

[fair dagger]

wdym

[near crypt]

for first expression, multiply and divide by ai+2b

do you know what is rationalization ?

[fair dagger]

english isn't my native language, so maybe I know but I do not recognize the word

[near crypt]

ohk

[fair dagger]

but I think I got it

[near crypt]

ok then

[fair dagger]

thanks

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[real dagger]

ahhh yeah that makes sense

[iron harness]

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[proud ridge]

is this correct?

[fiery badge]

how did you get that?

[proud ridge]

i meant to type b=6

but i still got b wrong

[fiery badge]

when you replaced tan as sin/cos

why is cos to the third power

and sin not?

[proud ridge]

but i wrote the ^3 in the last line

sin^3 (x)/cos^5(x)

[fiery badge]

oh

notice how cos^6 is in the bottom of the fraction

while in the answer you are given it's in the top

so what happens to the exponent?

sin^5(x)/cos^6(x) is not the same as sin^5(x)*cos^6(x)

[proud ridge]

becomes negative?

[fiery badge]

ye

[proud ridge]

ohhh so a = 5 and b = -6?

[fiery badge]

ye

[proud ridge]

thank you

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[proud ridge]

do i need to multiple 7 x 6? so there's 42 solutions?

because 6pi is 6 times larger than pi?

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[cyan spoke]

I need help

[fiery badge]

Should I call 911?

[cyan spoke]

haha if i cant figure this out you might

[fiery badge]

Please state the nature of your mathematical emergency

[cyan spoke]

I need help finding the average rate of change between two points on a function

f(x)=2^x+1

x=2 and x-6

this is my final and im loosing it

ive actually spent 40 minutes on this question

[abstract wadi]

do you know how you'd typically find the average rate of change for a function f?

[ivory sorrel]

do you mean this is an exam question

[cyan spoke]

no

final question

lmaoo

[ivory sorrel]

oh

ok

[ivory sorrel]

yeah, do this

[cyan spoke]

not off top of my head but i am able to do it within 1-2 google searches

[abstract wadi]

Well, I'm not trying to belittle your efforts here but

If that was an option, I think 40 minutes is about when you should have done that already.

Nevertheless, if you get it, do mention it here.

[cyan spoke]

i just dont know how to do it between 2 points on a function

the rate of change for the funcition i belive is 2

[ivory sorrel]

$\frac {f(b)-f(a)}{b-a}$

[thorny flame]

ƒ(Why am. I here)=I don't Know

[cyan spoke]

so Plug in the function where the f's are and plug in x=6 for b and x=2 for a correct?

[ivory sorrel]

I believe so yes

[cyan spoke]

god bless

[ivory sorrel]

unless I'm forgetting teh defn of average slope

[cyan spoke]

lets hope youre not

I got 2

[ivory sorrel]

,w (2^6-2^2)/4

[thorny flame]

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[ivory sorrel]

should be 15 I think

[cyan spoke]

where did the 1/4 come from

[ivory sorrel]

b-a

6-2

[cyan spoke]

i got 2 again

i see that 15 is likely the right awnser obviously but i just dont know how i keep getting 2

[abstract wadi]

Do you have any work to share on how you got 2?

[cyan spoke]

I can kind of type it out

2^x+1(6)-2^x+1(2)

[abstract wadi]

but you replace 6 or 2 with x

Not the constants.

f(2) would be 2^2 + 1

[cyan spoke]

oh i see and the first f(1) would be 2^6

ok yeah let me do that rq

i got 17

that should be easier to read

[abstract wadi]

that's still wrong

[cyan spoke]

Yeah ik

[abstract wadi]

((2^6 + 1) - (2^5+1))/4

Is not the same as what you have.

Also

Why are you multiplying 6 or 2 with that 1

That 1 is a constant.

When you have f(x)

And someone asks you to evaluate f(a)

You replace all the x with a.

But ONLY the x.

Not constant numbers such as 1.

[cyan spoke]

Ok i think i see

Thank you let me plug it in one more time because i want to be able to atleast figure out how to do the equation right

[abstract wadi]

Take your time.

[cyan spoke]

Got it and understand now

thank you

love learning math just hate doing it sometimes lol

have good day appriciate you

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[silk gate]

.reopen

[hazy valley]

would the answer for this be E

[glossy arrow]

Yes

[hazy valley]

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[prime mulch]

would i start with multiplying both sides by 17?

[noble laurel]

no, I'd start by finding alpha

[prime mulch]

how would i go about finding that

[neon pulsar]

arcsin

[prime mulch]

okok

[vernal matrix]

Question seems a bit sus tbh, unless I'm fried: alpha being between pi/2 and pi implies that 2alpha is between pi and 2pi, yet 8/17 is positive...

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[frank token]

for 13a how do i know which angle to take

[frank token]

ik that 11pi/6 is right but why isnt 5pi/6 right

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<@&286206848099549185>

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[gilded veldt]

can someone help me with (b)? I am getting zero for some reason

[gilded veldt]

this is my working

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[tardy stone]

I know this is a really basic question but why is this C and not E?

[pulsar warren]

inflection is for the second derivative

d^2y/dx^2

[tardy stone]

oh wait

im stupid

but isnt the inflection point where dy/dx = 0?

[pulsar warren]

nope

its the local extremas

if it changes signs

inflection is defined second derivative = 0

and changes signs

[tardy stone]

so the second derivitaive would be dy^2/dx^2 = 2x

and therefore not inflection point?

[gilded edge]

yeeeeee

[tardy stone]

and since dy/dx =0 at theat point, we just check what the second derivitive does around the point?

to see if its max or min?

[craggy haven]

hmmm second derivative would be someting like

[tardy stone]

and since the second dervitive is negative at that point its a max?

[craggy haven]

d^2y / dx^2 = 2x + dy/dx

= 2x + x^2 + y

[tardy stone]

ye i was scared of this xd

[craggy haven]

still though, (-1, -1) is not a zero of that ^

[tardy stone]

but its still negative, is this a proper way of proving it?

[craggy haven]

ye

[tardy stone]

👍

thx

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[tardy stone]

.reopen

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✅

[tardy stone]

why is this not A?

1 - pi(2)^2/2 + 5

[pulsar warren]

because you are going backwards

in integral when you go backwards everything you should add becomes negative

[tardy stone]

omfg

i have bc exam on monday

i dont think im getting a 5 xd

[pulsar warren]

lool m2

u got this

[tardy stone]

🙏

thanks

ill be back soon ig

lmao

im stoobid

[pulsar warren]

[tardy stone]

do you just do x = rcos(theta), y = r*sina(theta), and then do dy/dx?

nvm i got it

[pulsar warren]

okay good

[tardy stone]

i feel that i know the answers, i just make random mistakes and mess up, or just misread a question

idk

i will go do more practice and be back soon

nice meeting you

thx for help

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[pulsar warren]

you too

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[wide bane]

My formula for the area of the surface doesn't include the jacobian determinant of the parametrisation, wtf?

[wide bane]

why do I need to use a cross product when I can just take the determinant of the jacobian ??

its all so confusing

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[tawdry storm]

how is f(2) 1

[tawdry storm]

isn't it like not differentiable?

[tardy stone]

its given basecase

@tawdry storm

[tawdry storm]

oh ok

[tardy stone]

ye

[tawdry storm]

oh yeah its f(2) not f'(2)

so maybe know what f'(x) is and then integrate it?

[tardy stone]

f'(x) is the graph

so you just have to calculate the area under the graph

[tardy stone]

i forgor this

[tawdry storm]

oh yeah true

[tardy stone]

thats where i messed up

ye

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[tawdry storm]

so find the integral from -5 to 0 of f'(x)?

[tardy stone]

from 1 to -5

[tawdry storm]

k alr

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[tawdry storm]

1 to -5 right?

[tawdry storm]

i think i got D

but i might just be stupid

[tardy stone]

thats what i got b4 to xd

you have to do 1 - (area under graph)

bcuz ur going backwards

[tawdry storm]

i got 3-pi-(pi-3)

wait it might be E

[tardy stone]

2pi + 5

bringing this down so i dont have ot scroll

[tawdry storm]

k

[tardy stone]

so first find area under graph

[tawdry storm]

yeah

total area=6-2pi

[tardy stone]

but dont you have to

like subtract?

bcuz one is negative?

[tawdry storm]

true

[tardy stone]

so its 1 - (2pi + 5)

[tawdry storm]

so 6-2pi if you simplify

[tardy stone]

but its backwards

[tawdry storm]

oh nvm

[tardy stone]

so its 1 + 2pi-5

[tawdry storm]

yeah -2pi-4

[tardy stone]

so 2pi -4

-2pi?

[tawdry storm]

i think

because the semicircle is beneath the x-axis

[tardy stone]

but theres the thing

one sec

lemme find imaeg

[tawdry storm]

alr

[tardy stone]

this thing

we are going backwards

so wtv area is

we just subtract that

[tawdry storm]

oh yeah from 1 to -5 so its negative from -5 to 1

[tardy stone]

yes

if we go from left to right

we get

5 - 2pi

so we make negative

2pi - 5

base case f(2)=1

so 2pi -4

[tawdry storm]

oh yeah

[tardy stone]

but thats not an opiton

lMFOAOO

or

2pi -6

[tawdry storm]

lmao

[tardy stone]

or

6 - 2pi

idk

im high

fr

[tawdry storm]

i think its 6-2pi

idk

[tardy stone]

sure

yay

u taking bc exam to?

[tawdry storm]

nahhhh

im only 13 lmao

[tardy stone]

ab?

lmoa

have fun

[tawdry storm]

ty

[tardy stone]

im 15

old af

[tawdry storm]

nah

[tardy stone]

16 in like

4 days

[tawdry storm]

nice