help-19

[silent zephyr]

Uhh a/h??

[honest turtle]

Sine looks more easier.

[hushed island]

Oops I thought you were solving for the distance

[silent zephyr]

Yea solving for B

[odd edge]

@silent zephyr Has your question been resolved?

[honest turtle]

Use law of sines.

[silent zephyr]

Ojay

so sin(220)/130 or something?

[silent zephyr]

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[sharp oak]

There's only like 6 cases. You could enumerate them all

[near zealot]

you do a=3k ;3k+1 ; 3k+2 same for b?

[sharp oak]

Yeah basically. Or you just sub in a = 1, noting this covers all cases where a reduces to 1 in (mod 3)

Also, ab(b² - a²) = -ab(a² - b²)

So if the left reduces to 0, so does the right

A quicker way to do this would use Fermat's little theorem. Is that okay for you to use?

[near zealot]

yes

I tried it actually but it didn't work

[sharp oak]

Basically a² = 1 (mod 3)

So ab(a² - b²)
= ab(1 - 1)
= ab(0)
= 0

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[merry finch]

if i have A = {1, 2, 3} and i want to identify {1} and {3}, how do i write that, A/{{1}, {3}}?

[ivory raven]

A/1~3 is sometimes used in topology

otherwise more formally define ~ to be the equivalence relation such that 1~3 and nothing else, then write A/~

[merry finch]

👍

[modern sundial]

I would never write A/1~3 personally, that looks horrible to me compared to A/~

[merry finch]

this is from the edge contraction deletion theorem

but this looks like it's gonna be real ass to write

[modern sundial]

I mean, as long as you define your notation and it's clear what you are doing, you can commit whatever warcrime you want

[merry finch]

tilde on a tilde???

this is an extreme warcrime

[ivory raven]

I basically only ever see this for I/0~1 for S^1

but yeah it's gross

[merry finch]

i need an equivalence relation for the 2 vertices being identified

[modern sundial]

The problem is when people don't define their notation and just assumes the reader knows what they are talking about

[merry finch]

then i need another equivalence relation to identify all the edges together

or maybe i dont need that?

maybe i just need to remove the edge that goes from a to b where [a]_~ = [b]_~

[modern sundial]

One way to define an equivalence relation and make people hate you is to use the symbol $\xi^{T}$

[ivory raven]

I don't really often see contractions in graph theory written as quotients

[clever fjord]

JessicaK

[modern sundial]

i.e. ~ transposed

[ivory raven]

just state that you're constructing a new graph via contraction

[merry finch]

no but i also need to replace all the (z, a) and (z, b) to be written instead as (z, [a]) and (z, [b]) which would be the same edge

maybe i just use some words

the notation seems like it'll do nothing more than confuse the marker

[ivory raven]

yeah

though if you wanted to explicitly say what you're doing with the edges, you could just say that any edges that were adjacent to either of the vertices adjacent to the contracted edges are now adjacent to the new vertex

the wording/notation is clumsy anyhow

graph theory has its own specialised vocabulary that avoids this for a reason

[merry finch]

[modern sundial]

Just write a paragraph explaining the procedure in plain words and give it a name.

[merry finch]

yeah that's what i did

thanks guys

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[dull fossil]

,rotate

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[strange musk]

i got 682957314

[static totem]

it's almost that

[strange musk]

but idk how to explain how i got it

[static totem]

try to get higher

[strange musk]

wait

[strange musk]

do i have to start with the corner pieces

i cant find a way to use the middle pieces

[static totem]

yes

you start at 6

[strange musk]

683147592

[static totem]

definitely that

i mean, i definitely got that, i don't know if that's what they expect

and yeah i would have trouble explaining it

[viscid flint]

starting at 9 puts you into a corner

you need to use all 9 numbers

[strange musk]

did i get it

or is there a way to go higher than that

@viscid flint

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@strange musk Has your question been resolved?

[odd edge]

@strange musk Has your question been resolved?

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[round radish]

I don't understand what the question. If someone could word this in a easier to understand way that would be great

[molten vector]

Looks like it wants you to simplify that expression down to some function of only tan theta

Using double angle formula

[round radish]

ok thanks

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[mortal mirage]

what have you tried

[sharp flume]

i think x = sqrt(2)^sqrt(2) y =sqr(2)

[sharp oak]

√3^√2 is either rational or irrational.

Assume it's rational. What can we say?

Assume it's irrational. What can we say?

[mortal mirage]

that's a good instinct, however, do you know if sqrt(2)^sqrt(2) is rational?

[sharp flume]

thats the thing idk and how do i find out if its rational or not

[mortal mirage]

well consider the case where it is

is sqrt(2) rational?

[sharp flume]

no

[mortal mirage]

then let x = sqrt(2), y = sqrt(2)

if sqrt(2)^sqrt(2) was rational, then you found an example

[sharp flume]

example of what

[mortal mirage]

of irrational numbers x and y so that x^y is rational

[sharp flume]

how does that logic work

[mortal mirage]

well we are just considering the case where sqrt(2)^sqrt(2) is rational

maybe it is, maybe it isn't

but we're just checking if it was

[sharp flume]

if it was what does it tell us

[mortal mirage]

that x = sqrt(2) and y = sqrt(2) are examples of irrational numbers so that
x^y
is rational

which is what we're looking for

so if sqrt(2)^sqrt(2) is rational, then there are two irational numbers that satisfy the required condition. what other case do we need to consider?

[sharp flume]

OHHH

so two numbers of irational make up a rational number in this case

[mortal mirage]

two irrational numbers x and y form a rational number x^y (in this case), yes

but what is the other case

[sharp flume]

if its irrational

[mortal mirage]

right, if sqrt(2)^sqrt(2) is irrational

but what do you do then

[sharp flume]

that means both is irrational numbers make irrational

[mortal mirage]

which irrational numbers

[sharp flume]

sqrt(2)

[sharp flume]

so this consists of 2 irrational numbers would form a irrational number?

[mortal mirage]

im not sure i understand your question

[sharp flume]

im just clarfying

is that right

[mortal mirage]

well i don't understand what you're saying

can you rephrase it

[sharp flume]

sqrt(2)^sqrt(2) is rational, then there are two irational numbers that satisfy the required condition. if sqrt(2)^sqrt(2) is irrational then that number consists of two irrational numbers

[mortal mirage]

yes it does

[sharp flume]

so thats the answer to this

[mortal mirage]

well sqrt(2)^sqrt(2) is now assumed to be irrational

sqrt(2) is also irrational

refer to what you did at the beginning

[sharp flume]

wait doesnt that make x^y rational

[mortal mirage]

right

that's the idea

[sharp flume]

i dont get why we stated if sqrt(2)^sqrt(2) is rational

[mortal mirage]

because if it is, then you can't use
x = sqrt(2)^sqrt(2)
y = sqrt(2)

because we need x and y to be irrational

[sharp flume]

if we cant use it why say it is rational then?

[mortal mirage]

because we don't know if sqrt(2)^sqrt(2) is rational or not

so you have to consider the case where it is

[sharp flume]

what about if it is irrational

doesnt it contradict the statment

if sqrt(2)^sqrt(2) is irrational then that number consists of two irrational numbers

[mortal mirage]

well the point is that if it is rational, then
x = sqrt(2), y = sqrt(2)
give
x^y = sqrt(2)^sqrt(2) which is rational

so the statement works, given that sqrt(2)^sqrt(2) is rational

now you consider what happens when sqrt(2)^sqrt(2) is irrational

then you say
x = sqrt(2)^sqrt(2), y = sqrt(2)

and compute x^y, show it is rational

[sharp flume]

im losttt

[mortal mirage]

what is the question asking you to prove

[sharp flume]

two irrational numbers x, y, such that x^y results in rational number

[mortal mirage]

right

now we want to use the example

x = sqrt2^sqrt2

y = sqrt2

[sharp flume]

yedah

[mortal mirage]

but we don't know if we can, because we don't know if x is irrational

[sharp flume]

yeah

[mortal mirage]

so we consider the two cases

1. x is rational
2. x is irrational

[sharp flume]

yes

[mortal mirage]

if x is rational, then we have (sqrt2)^(sqrt2) is rational

note that sqrt2 is irrational

this proves the statement

[sharp flume]

hows it rational

[mortal mirage]

because we assumed it was

for this case

remember

we are doing 2 cases

[sharp flume]

so why u say if x is irrational

[mortal mirage]

1. x is rational
2. x is irrational

because we need to consider both cases

[sharp flume]

do u mean is irrational if x is irrational

[mortal mirage]

oh i see the confusion

that's my fault

the correct statement is

if x is rational, then we have (sqrt2)^(sqrt2) is rational

[sharp flume]

ok yeah and sqrt(2) by itself is irrational;

[mortal mirage]

so irrational^irrational = rational

[sharp flume]

yup

[mortal mirage]

that's if x is rational

[sharp flume]

by case

[mortal mirage]

now we consider the other case

[sharp flume]

if its irrational^irrational = irrational

[mortal mirage]

yes, the point is that now x = sqrt2^sqrt2 is irrational

so we go back to our original example

x = sqrt2^sqrt2, y = sqrt2

and x^y = (sqrt2 ^ sqrt2) ^ sqrt2 = sqrt2 ^ (sqrt2 * sqrt2) = sqrt2 ^ 2 = 2

[sharp flume]

so how is sqrt2^sqrt2 is irrational by this case

[mortal mirage]

we just assume it is

we considered what happens when it is rational

and showed we can make a rational x^y out of irrational x and y

now we consider what happens when it is irrational

[sharp flume]

i kinda get it now

[mortal mirage]

what we just did

consider the case where it is rational, and irrational

[sharp flume]

is it just this if sqrt(3)^sqrt(2) is rational, then there are two irational numbers that satisfy the condition. if sqrt(3)^sqrt(2) is irrational then that number consists of two irrational numbers

?

[mortal mirage]

yes

but then you consider x = sqrt(3)^sqrt(2), y = sqrt(2)

they are both irrational, but x^y is rational

[sharp flume]

do we just reject the case that two irrational x^y = irrational, because we proved that x^y is rational

[mortal mirage]

no

we dont care about irrational^irrational = irrational

we are trying to look for two irrational numbers so that x^y is rational

if sqrt(3)^sqrt(2) is irrational, then x = sqrt(3) and y = sqrt(2) are not an example of that

but now we can use x = sqrt(3)^sqrt(2), because we assumed it was irrational

[sharp flume]

oh and y in this case sqrt(2) which is irrational which proves irational^irational is rational

[mortal mirage]

yes

[sharp flume]

does this tell us if sqrt2^sqrt3 is actually irational?

[mortal mirage]

no

[sharp flume]

ok ty for da help

[odd edge]

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[formal cave]

what happened in the 2nd line that it turned into the 3rd line?

[forest sky]

power reduction identities

[formal cave]

oh its juts that

thanks

.close

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[formal cave]

.reopen

[odd edge]

✅

[formal cave]

actually

you bring out 1/8 because u squared the other 1/2 right?

[amber schooner]

(1/2)^2 (1/2)=1/4 * 1/2=1/8

[formal cave]

thanks for confirming

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[last vector]

idk how to explain it in english

but

basically

[next mist]

?

[last vector]

the angle that the function crosses the x axis with is tan(a)

[cursive field]

find a?

[next mist]

see.

[cursive field]

wdym? didnt exactly get ur def of a

[next mist]

are u to find slope?

[last vector]

so I thought that the Steigung (idk the English word for it)

[last vector]

Yeah

[cursive field]

??

[last vector]

Slope

[cursive field]

of the line?

[last vector]

yea

[cursive field]

pretty easy

[next mist]

slope in tan ( angle)

[cursive field]

you don't need a

[last vector]

wouldn’t it be tan(80)?

[cursive field]

slope = change in y/change in x

[next mist]

nope bro

[last vector]

the answer says tan(-80)

I don’t get why

[next mist]

yes cause

angle is between line and positive direction of x axis

[last vector]

what do you mean by positive direction of x axis?

[next mist]

um east

ig

[last vector]

you mean this?

[next mist]

no

[last vector]

I don’t get it ☹️

[last vector]

ohhh

why is it negative tho?

[next mist]

bcoz angle is more than 90

see

[next mist]

can u now tell whats this angle equal to?

[last vector]

wouldn’t that angle be 110

[next mist]

bingo

and

answer is tan (100)

[last vector]

nah 100 sorry

[next mist]

yea

100

[last vector]

how is tan(100) equal tan(-80) tho?

[next mist]

do u know whats tan(180-x)

[last vector]

nope

[next mist]

well tan(180-x) = -tan(x)

[last vector]

ohh

[next mist]

now x is what here kid?

[last vector]

x is 80

[next mist]

well 100

[last vector]

wiat

so

[next mist]

angle u got is 100

[last vector]

tan(180-100)= -tan(80) ?

[next mist]

yes

[last vector]

and -tan(80) = tan(-80) ?

[next mist]

yes

[last vector]

ohhh alright

I get it now

thanks a lot

🙏🙏

[next mist]

also sin(-x) = sin(x)

[next mist]

howd u figure out im indian

[last vector]

I didn’t

:p

[next mist]

wow

nvm

now type

. close

[last vector]

oke

.close

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[lost iris]

In each of 3 bags there are 4 white and 6 black chips. We randomly select a chip from the first bag and transfer it to the second, then randomly select a chip from the second bag and transfer it to the third. Finally, we select a chip from the third bag.

a. What is the probability that the selected chip is white?
b. We see that the selected chip is white. What is the probability that we only transferred white chips?

[lost iris]

anyone knows how to solve these types of questions ?

I tried my ways, but the results don't seem right

[odd edge]

@lost iris Has your question been resolved?

[tiny hazel]

!show

[odd edge]

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

[tiny hazel]

what you tried

[lost iris]

ok, but its in my language so I have to explain anyway

so the probability of taking white and black chip from one bag to another always seems to be the same

because I just multiply the probabilities of events if one is dependent on another as they taught us

the probability of taking white from BAG1 to BAG2 is 0,4 and black 0,6 obviously

so then if we took white there is 5/11 white chips in BAG2 and 6/11 black ones

[slender rampart]

Try to find out all the events that need to happen for the selected chip to be white

[lost iris]

i think I did

didnt I ?

I calculated probabilities of all the events but I just didi it with wrong numbers I think the mistake that I multiplied something that I shouldn't or something

[odd edge]

@lost iris Has your question been resolved?

[lost iris]

<@&286206848099549185> please

[marble lintel]

sorry no

[odd edge]

@lost iris Has your question been resolved?

[lost iris]

Should I use Bayes theorem on this one ??

anyone willing to provide guidance ?

[lost iris]

nevermind then

.close

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[static mulch]

when doing line integrals such as this one how would i do the 2nd curve that goes back towards the origin?

[static mulch]

this is the integral i currently have $ \int_{0}^{1}\left(6t+3\sqrt{t^{2}}\right)\sqrt{1+4t^{2}}dt+\int_{1}^{0}\left(6t+3\sqrt{t}\right)\sqrt{2}dt $

$\int_{0}^{1}\left(6t+3\sqrt{t^{2}}\right)\sqrt{1+4t^{2}}dt+\int_{1}^{0}\left(6t+3\sqrt{t}\right)\sqrt{2}dt$

[clever fjord]

Jaxx

[static mulch]

for 2nd curve my parametrization was r(t) = <t,t>

[sly badger]

is this the curve y=x?

and u want it to go backwards?

[static mulch]

yeah on y=x going from 1,1 to the origin

let me isolate the integral for that actually

$\int_{1}^{0}\left(6t+3\sqrt{t}\right)\sqrt{2}dt$

[clever fjord]

Jaxx

[static mulch]

im not getting the correct answer im wondering if its because my bounds from 1 to 0 arent correct?

[sly badger]

well

if your paramtrisation was r(t) (t, t)

what are you bounds with that

1 and 0 right?

[static mulch]

yeah

t=1 to t=0

[sly badger]

yea

if t = 0, we start and 0 and go up to 1

which is not what we want

[static mulch]

yea so i start at 1 which gives 1,1 and end at 0 which gives 0,0 but it's incorrect

[sly badger]

so try r(t) = (-t, -t) from -1 < t < 0

[static mulch]

so is my issue that t should start from the lowest number?

[sly badger]

Yes

I dont do maths, but from all the examplses ive done

t should always start lower and go bigger

[static mulch]

i recall this being done with a different parametrization so that t went from the lowest to highest number but what is wrong with this one?

it starts and ends in the right places i cant find whats wrong with it

[static mulch]

also this wouldnt work because the 2nd integral would put -t inside the sqrt

i got it with 1-t parametrization, also what i got from my initial integral was $\frac{15\sqrt{5}-3}{4}-5\sqrt{2}$

[clever fjord]

Jaxx

[static mulch]

the correct parametrization got $\frac{15\sqrt{5}-3}{4}+5\sqrt{2}$

[clever fjord]

Jaxx

[static mulch]

i assume the reversal of the integral bounds that i did initially ended up subtracting instead of adding so perhaps just arithmetic issue

[sly badger]

ahh

so we're in a pickle

[static mulch]

in the future ill just end up adding/subntracting in reverse to account for this i suppose

¯_(ツ)_/¯

[sly badger]

you could yea

what is the actual question tho

[static mulch]

just to integrate it over those two paths

[sly badger]

Ahhh

ok

Yea no clue mate

[static mulch]

no worries thanks for the help

.close

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[mystic saffron]

Do rational functions have maxima and minima?

[mystic saffron]

Like for example, if f’(x) is positive from -3 < x < -1

and f’(x) is negative when x > -1

then is -1 a max?

for a rational function

so -1 is like an asymptote

[alpine pewter]

what does it look like on a graph

[odd edge]

@mystic saffron Has your question been resolved?

[forest sky]

do you have a specific context? if all you know is that f' is positive when x<-1 and negative when x>-1 it could be an asymptote or a local maximum

[mystic saffron]

It is 100% an asymptote

from the question

but is the asymptote also a max?

[forest sky]

an asymptote is an asymptote, the sign of the derivative can change at either a maximum/minimum, or a discontinuity like an asymptote

[mystic saffron]

ohhhh so its not an absolute extrma

ok

[forest sky]

if the function goes to +infinity anywhere then it doesn't have an absolute maximum, and similarly for an absolute minimum

[mystic saffron]

i found the photo

[mystic saffron]

ohhhh. Thanks

[odd edge]

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[mystic saffron]

Can someone draw for me the representation of this?
$arccos(z / hyp) * (180 / pi)$?

[clever fjord]

Cut

[fallen solar]

like this? $\arccos \left( \dfrac{z}{hy\rho }\right) \cdot \left( \dfrac{180}{\pi }\right)$

[clever fjord]

SK

[mystic saffron]

No, like... Drawing it in a 2D space

I'm trying to

Like

The math i want to understand is

The Z divided by hyp

Is it giving to me an angle?

Or what

.close

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[stoic girder]

i need help with question b

[mental ridge]

w..why do they tell you to do a by substituing in numbers

[stoic girder]

i do not know😭

[mental ridge]

anyways

expand out (a+b)(a+b)

[stoic girder]

what does that mean...

[mental ridge]

expand it out

like

multiply it out

[stoic girder]

like ab squared?

im sorry if im too dumb😭

[mental ridge]

wait

b or c

[hushed island]

What is the product of (a+b)^2

[stoic girder]

question b

[mental ridge]

ah okay

so do you know how to expand a multiplication

what is (1+2)(3+4)

[hushed island]

Do you agree that this is multiplication

[stoic girder]

OH WAIT

I MEANT C

IM SORRY

[mental ridge]

oh alright

well what is the definition of sqrt(ab)

that sqrt(ab)^2=ab

now square sqrt(a)sqrt(b)

what do you get

[stoic girder]

um pls give me 5 minutes

[mental ridge]

what is (ab)(ab)

you did this in question 1

[stoic girder]

um is it ab squared

[mental ridge]

Yes

But like

Expanded out

What is it

[stoic girder]

um (ab)(ab)?

[mental ridge]

Mate

You did this in question a

[stoic girder]

um i actually...

didnt do question a

i have only done b..

should i do a now......

[mental ridge]

Yes

[stoic girder]

ok ok

should i close this

[mental ridge]

No

Keep it open

[stoic girder]

OH AB

is it right

[mental ridge]

yes

[stoic girder]

YES

[mental ridge]

So sqrt(ab)^2=ab by definition

And sqrt(a)sqrt(b)^2 also is ab

So they are equal

[stoic girder]

wait so they are

oh yea that makes sense

so they are equal

OHHHHH

I GET IT NOW

TYSM

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[hallow current]

so i have question when i plug the integral of sqrt(x^2-x^4) i get a diffrent result of x*sqrt(1-x^2) and idk where is the diffrence (from -1 to 1 in the integral)

[plucky owl]

Could you link your work? What answer did you get for each?

[hallow current]

the first one i got 2/3 and the second i got 0

[plucky owl]

Your second one is correct - can you link the work you did for the first one?

[plucky owl]

This should be the indefinite integral you get

[hallow current]

yes i got that

[plucky owl]

Without seeing how you solved it, I can't say what went wrong

[hallow current]

but when i use calculater i get 2/3

[hallow current]

this is the question

[hot relic]

these are identical lol with the one that @plucky owl posted

[hallow current]

i m getting hella confused

[plucky owl]

Calculators do not "know how to do integration". They are given heuristics. The solution the calculator gave is identical when x > 0.

But your integration domains have negative values

This seems to be another piece of evidence why calculators aren't the 100% guarenteed solution

[hallow current]

so u dont need to change the sign or anything

[plucky owl]

Honestly, my guess is that this has more to do with bad programming of the calculator than anything else. It's definitely confusing though so I get the confusion

[hallow current]

tnx for the help

[plucky owl]

ofc

[hallow current]

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[nocturne glacier]

am i doing this correctly?

[nocturne glacier]

i’m having trouble with rational equations

[fallow tapir]

it was (2x + 6) - (5x + 12)

now expand the parantheses

[nocturne glacier]

so -3x-6

it would be extraneous right

since x would turn out to be-3

[fallow tapir]

yup

[nocturne glacier]

alr alr

imma keep the channel open just in case

[fallow tapir]

sure

[nocturne glacier]

what about this one

[mystic saffron]

this work

[nocturne glacier]

wym

?

[fallow tapir]

that's correct

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[quaint pewter]

can someone tell me why, for odd number of data, they find the position then whatever position that is, that would be the value for Q1/Q2... etc

but for even number of data, they do whatever this is, i don't even know what they did im ngl

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[sharp hill]

hello, can someone help me with this series problem?

[sharp hill]

when i solve, the final form i end up with is x-5

but it says radius = 1

oh wait

could it be like [6,4]

i may be stupid

.close

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[winter breach]

Hello, I'm grappling with a problem where a particle moves along an ellipse, given by $$\mathbf{r}(t)=(3 \cos 2 \pi t, 2 \sin 2 \pi t).$$ After finding the velocity, the speed equation is $$\text{Speed} = \sqrt{(-6\pi \sin(2\pi t))^2 + (4\pi \cos(2\pi t))^2}.$$ My challenge lies in maximizing this speed, especially because the (\sin^2) and (\cos^2) terms are weighted differently due to the coefficients. This mix makes finding the maximum speed analytically tricky for me. I'm seeking insights on how to approach the maximization given these conditions, particularly any mathematical techniques or properties that might simplify finding the speed's maximum value.

[clever fjord]

dghf
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

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@winter breach Has your question been resolved?

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@winter breach Has your question been resolved?

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@winter breach Has your question been resolved?

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[autumn ice]

can someone help explain the process behind this? my professors work from our notes. im just having a hard time understanding the whole process

[autumn ice]

acc i have a more in depth question on the implicit differentiation

on this part here

the 2(-2.5(dH/dt)/6.25)

first part i know is power rule, bring the power to the front

but the inside

before the derivative its: (10-H)/2.5

wait

is it quotient rule

[shy sundial]

ok from here $V(t) = \frac{1}{3} \pi r^2(t) h(t)$. Maybe the first step as finding dV/dt before solving the whole problem is better

[clever fjord]

sssssssssssvvvvvvvvvvccccccccccc

[autumn ice]

ok

thats what i was thinking cuz my professor seemed to make that process a little more confusing

😔

[shy sundial]

Ok what would be V' from there?

[autumn ice]

oh right, dV/dt is 6

[shy sundial]

i mean, no

i mean yes

but

[autumn ice]

not the question

?

[shy sundial]

I meant to find an explicit formula for dV/dt from here

[autumn ice]

ah

[shy sundial]

personally I would 1) state the formula 2) differentiate the formula 3) replace the known values

[autumn ice]

yeah

i tried doing it on my own but kinda

doing implicit differentiation with multiple variables is sometimes confusing for me

i will try

[native tinsel]

Is this even implicit diff

[autumn ice]

there is some of it in it

i cant say my professor is the best

[shy sundial]

honestly, more than imp diff is chain rule

[autumn ice]

so it comes across confusing for me at least

[shy sundial]

look at it this way

[autumn ice]

yeah

constant rule on 1/3pi

product rule on the inside

r^2 as f, h as g

[shy sundial]

after that is just plug the given values

[autumn ice]

.close

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[fiery wagon]

idk how to start this

[topaz lantern]

You want to find where the gradient is the highest and lowest. The gradient is the derivative of f(x). Start by differentiating f

Then you want to find the minima and maxima of f'(x)

Over that domain restriction

[fiery wagon]

those are stationary points right

[topaz lantern]

I'm not sure what you mean by this

they are the points in which f'(x) is highest and lowest

You can see them by looking at the graph of f'(x) = -3x^2 + 6x -2 (if you want to)

[fiery wagon]

im just confused on how to find maximas and minimas

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@fiery wagon Has your question been resolved?

[fiery wagon]

.close

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[eager schooner]

help

[odd edge]

Please don't occupy multiple help channels.

[eager schooner]

Here is my thoughts

[mental ridge]

You can try applying cramers rule if u wanr

[eager schooner]

I don't know what this is

[mental ridge]

Alr then we can do substitution

[eager schooner]

So

What to do

@mental ridge

[mental ridge]

Wait do you know gaussian elimination?

[eager schooner]

Kinda

[mental ridge]

Is it possible to apply that here

[eager schooner]

I don't think so

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[tranquil terrace]

how do i solve this

[odd edge]

Please don't occupy multiple help channels.

[leaden widget]

your goal is to get a normal vector to the plane then you can write an equation using a point and the normal vector (similar to how in 2d you can use point and slope to write an equation for a line), can you think of how you can find 2 vectors that are parallel to the plane so that you can get their cross product? Then you would have a vector normal to the plane

[odd edge]

@tranquil terrace Has your question been resolved?

[tranquil terrace]

ill try this out

[tranquil terrace]

it has to be parallel to the line though?

[leaden widget]

yeah, so a vector that is just using that line's slopes or whatever theyre called is parallel to the line and so also to the plane

so you can get 1 vector that way and then another vector between the 2 points

and presumably they won't be scalar multiples of each other

[tranquil terrace]

I’m not really sure what to do from here

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[spare hamlet]

can anyone help me with this ques "Find the area bounded by the curve x^2= 4y and the line x = 5y – 2"

[dark kraken]

Find the intersection points then compute the integral between them

[spare hamlet]

but i get complex roots is that prob or can i continue ??

[dark kraken]

What do you mean? There are two real intersections between the line and the parabola

[spare hamlet]

i hav equated y between the two equations and i get a quadratic equation wrt to x when i solve it i get the roots as (2+2√11)/5 and (2-2√11)/5

[spare hamlet]

sry i meant the real roots are complex i didnt mean the imaginary ones

[dark kraken]

Ok so those are the bounds on your integral, then to find the area between the curves, integrate f(x)-g(x)

[spare hamlet]

hmm thank you

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[primal wolf]

Hey, need help to see if I’ve done this right or if I’ve coped myself into it

[primal wolf]

My text book said to prove the top expression and I need to know if what I’ve done is the correct way of doing it or have I pulled some bullshit

I'm a high school drop out studying electrical engineering so the math is the part I'm struggling on the most

[sly juniper]

dont write ()^2 and the proof is correct

[primal wolf]

it's like, the square of that gets me the rt(a)-rt(b) * rt(a)+ rt(b) though right

or do i just claim it as ummm multiplying the conjugate??

these terms are all new to me

[sly juniper]

you multiply sqrt(a)-sqrt(b) /h by sqrt(a)+sqrt(b)/sqrt(a)+sqrt(b)

and thats how you get (sqrt(a)-sqrt(b))(sqrt(a)+sqrt(b))/h(sqrt(a)+sqrt(b))

and the rest of your proof is correct

[primal wolf]

dope

So I don't have to do the sqaure bs, I can just straight up multiply both the top and bottom by the conjugate of the top expression and it's all correct

ty

alright i'm gonna hit the close unless you have anything else to add I should know

[sly juniper]

yeah thats right

i have nothing else to say

[primal wolf]

thank you for that, I tried plugging in the formula to symbolab and it just wasn't helping

.close

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[analog swift]

need help w Q3

[odd edge]

@analog swift Has your question been resolved?

[sharp oak]

Any SUVAT question will involve 4 of the 5 variables.

When I start a SUVAT question, I think about which variable is NOT involved. Which variable won't be used in our problem?

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[manic bloom]

I got confused again :v

[manic bloom]

I don't understand that this is periodic and blabla

I understand this

But with this definition of the ArcCos

This shouldn't happen, even if the function is periodic

Do I explain myself?

[timid cairn]

Hmm?

[manic bloom]

Nvm

[timid cairn]

Does it make sense

[manic bloom]

I can say cos(2pi + x) = cos(x)

So I can just split and work with that

I was kinda confused with the variables

[timid cairn]

Yeah though it’s also worth noting cos(x) = cos(-x)

Kind of hard to understand the up and down graph without knowing that

[manic bloom]

My problem came here

For example, this wouldn't be true if x = 3pi

[timid cairn]

Yeah exactly

[manic bloom]

But cos(3pi) = cos(pi)

[timid cairn]

Kind of like how sqrt(x^2) isn’t always x

[manic bloom]

So doing that, now the inverse is defined (pi is in the domain taken for the definition of the inverse of Cos) and I can get x

Also, this seems to happen with all the periodic functions

This is the first example of the composition between of an inverse and normal function where the normal is periodic

The first example after 300 pages so... you can imagine how I felt seeing this :'v

I'll close, thx

.close

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[smoky pagoda]

i need help with part ii. I have done part i and got k = 6 but for part ii, i just put the values x=0 and y=-2 into the equation and got m=1. This question requires two values of m and is worth 6 marks. what am i missing here?

[leaden karma]

How did you get m=1?

That’s not possible just from the point

[smoky pagoda]

the line passes through that point so when x=0 y=-2 ?

then solve for m

[leaden karma]

But you can’t

Try it

[smoky pagoda]

oh

its multiplied by 0

so it can be anything

i see

[leaden karma]

Indeed

[smoky pagoda]

what am i supposed to do then

[leaden karma]

Do you know what T=0 means?

[smoky pagoda]

nah

[leaden karma]

Okay

Never mind that

There’s an easier way

Do you know the formula for the distance of a point from a line?

[smoky pagoda]

no

[leaden karma]

Okay

Let’s say you have a line $ax+by+c=0$ and a point $(\alpha,\beta)$

[clever fjord]

kheerii

[leaden karma]

Then the distance d is given by $$d=\frac{|a\alpha+b\beta+c|}{\sqrt{a^2+b^2}}$$

[clever fjord]

kheerii

[leaden karma]

Basically place the point in the equation of the line, take the absolute value and divide by that square root

[smoky pagoda]

uhm

im not sure this is what im supposed to do

i havent been taught this

[leaden karma]

Well i mean

What else are you taught about tangents?

To a circle

[smoky pagoda]

uhh

discriminant = 0

i guess

[leaden karma]

Okay

So you can solve the line with the circle

And set the discriminant equal to 0

[smoky pagoda]

90 degrees from the centre aswell

[leaden karma]

So that there’s only one point

[smoky pagoda]

can you explain this

wdym

[leaden karma]

Basically, find the intersection points of the line and the circle

[smoky pagoda]

oh ok

ill do that

[leaden karma]

And find the value(s) of m such that only one such point exists

[smoky pagoda]

im a bit lost

i subbed y=mx - 2 into the circle equation

and when expanded i got x^2 + (mx)^2 -10(mx) +16 = 0

im not sure if i made a mistake or im missing something

[leaden karma]

No you’re doing it correctly

So you only want one value of x

In this equation

Because you only want one intersection point

[smoky pagoda]

yeah

[leaden karma]

That’s what a tangent means

[smoky pagoda]

how do i do that

[leaden karma]

So just set discriminant = 0

[smoky pagoda]

so do i need to get it into the form ax^2 + bx + c = 0

[leaden karma]

Yes

[smoky pagoda]

ok i think i got

it

m = -4/3 or m = 4/3

is this correct

@leaden karma

[leaden karma]

Idk

[smoky pagoda]

ah dw

[leaden karma]

Check your answer key

I don’t feel like doing the calculation part

[smoky pagoda]

the method makes sense

[smoky pagoda]

no worries understandable

thank you

.close

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[modern gate]

equation: e^1/2x= 2e -e^1/2x, why can't you take ln on both sides so you get: 1/2x = ln(2) + 1 -1/2x --> x= ln(2) +1

[zinc glacier]

because $log(a+b) \neq log(a) + log(b)$

[clever fjord]

GarlicBredFries

[modern gate]

could you say where I am using that? I don't get it

[royal herald]

itd be:

ln(e^0.5x) = ln(2e -e^1/2x)

which becomes

0.5x = ln(2e -e^1/2x)

cant really make the right side any simpler

@modern gate

[modern gate]

okay, I think I get it. Thanks for the help Stephen and garlic!

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[unkempt plaza]

Part ii

Dont even know where to start

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[orchid torrent]

Seems like it’s just using De Moivre’s followed by ||equating the real and imaginary parts||

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[modest dragon]

Gotta leave this in surd form

nvm

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[mystic saffron]

hello I need help with these really hard combinatorics proof

[mystic saffron]

look here is how i thought of it

but i am completely lost

because i cant move further in the proof

so like for example take a set of all natural numbers with n=5 and r =3. We have: A= {1,2,3,4,5,6}. So let the fixed subset be {1,2,4}. If the combination yields this fixed subset then it can be permutated in r! = 3! ways so there are 3! ways.

but now the second case i would assume is if it is not the fixed subset

but clearly the second part of the proof does not point to that

so maybe it is saying how many point of differences are there between the fixed subset and the subset we want to permutate

but that logic does not seem to hold tbh it seems complicated

i think im on the right track tho, at least for the r! part. Then there needs to be the 2nd part of the addition principle.

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[uncut plaza]

Can someone help me with this question? I've been trying to do this question for 3 hours and I still don't get it..

[uncut plaza]

I can't figure out the frequency density

i'm kind of desperate cuz I have a math exam tmrw and I wasted a lot of time already ;-;

[mental storm]

I'm not sure if there's a smarter way to do this unfortunately other than taking one "cell" as equalling 5 kids

terrible mouse drawing but you get the idea

[uncut plaza]

so one cell is one of those boxes u outlined

[mental storm]

well it said there are 10 kids for whom $m \leq 20$ yes?

[clever fjord]

jack

[uncut plaza]

yes

[mental storm]

it seemed reasonable to divide that into 2 squares

each square being 5 kids

since they didnt provide any other form of measurement I presume you need to find kids from $50 < m \leq 80$ by counting the areas...

[clever fjord]

jack

[uncut plaza]

yea but that's the last step

I have to find out the scale of the frequency density first

which i'm not so sure on how to do

I tried using the formula but my answer didn't match the correct answer in the markscheme

[mental storm]

isn't the goal to find the no. of children from the interval [50, 80]

how can you compute the frequency density if you don't know the frequency

[uncut plaza]

well, the frequency for the first bar was given as 10

[mental storm]

what about the other frequencies

[uncut plaza]

idk they didn't give them

[mental storm]

I'm confused as to why you think you need frequency density

it seems to me that it'd be far faster to simply determine the area

[uncut plaza]

ok then let's try that

um

hello?

[hearty yarrow]

It's the frequency divided by the class width.

The important point here is determining the scale of the y-axis so that you can find the other frequencies.

[uncut plaza]

wow the other person said the scale is unnecessary..

[hearty yarrow]

By "scale" I mean how much each "tick" in the graph paper represents.

[uncut plaza]

so frequency density?

[hearty yarrow]

If we focus on the first class, its width is 20 and its frequency is 10, so the frequency density is 10/20 = 0.5.
So the height of the rectangle representing this class will be equal to 0.5.
On the graph paper, this height spans 5 minor ticks, so each minor tick is equal to 0.1 in terms of frequency density.
Consequently, each major tick is equal to 1.
By "minor tick" I mean the thin horizontal lines in the graph paper and "major tick" is the thick horizontal lines.

[uncut plaza]

that first point goes against the formula tho, cuz frequency density should be frequency * width

[hearty yarrow]

From where did you bring that formula?

[uncut plaza]

ohh

this might have been why I was stuck for so long

but the scale doesn't work for all bars do they

so I count 25 small boxes each time to go up by 0.5

[hearty yarrow]

The scale is always constant in this case.

So if each large box is 1 unit, then every large box is also 1 unit.

[uncut plaza]

but the bars are all different

like one is wider than the other

[hearty yarrow]

I'm talking about the vertical gridlines.

We already know the scale of the horizontal axis.

[uncut plaza]

oh

[uncut plaza]

I got 2.9, 3.2 and 2 for those 3 bars

[hearty yarrow]

Which bars?

[uncut plaza]

the ones from 50 to 80

[hearty yarrow]

Yes that's correct

We're not going to use the third bar though.

[uncut plaza]

we're not?

[hearty yarrow]

Oh no you're right, I though the 80 stops at the third bar.

Now you can just find the total area, which gives the frequency.

[uncut plaza]

so I add the numbers I mentioned?

[hearty yarrow]

These numbers are the heights of the rectangles.
You need to multiply them by the correct widths to find the total area.

Some rectangles are cut off though, so their width is not going to be the full class width.

[uncut plaza]

519?

wait no

87

[hearty yarrow]

Yup 87

[uncut plaza]

but why area

shouldn't it be frequency

[hearty yarrow]

In a frequency density histogram, the area is literally the frequency.

[hearty yarrow]

If you look at this, you can see that frequency = density * class width.
But density is the height and class width is the width, and height * width = area.

[uncut plaza]

ohh

so it's the same thing

anyways I appreciate ur help Meadow

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[jade crypt]

Suppose that $\frac{1}{n} \sum_{k=1}^{n} \frac{x_k}{k} \rightarrow 0$, where $x_k \in \mathbb{R}$. Is it true that $\frac{1}{n^2} \sum_{k=1}^{n} x_k \rightarrow 0$? How do I prove/disprove this?
The part that's tripping me up is the fact that $x_k$ can be negative, so the absolute value of the second sum isn't automatically bounded by the absolute value of the first sum

[clever fjord]

boolean_coercion

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[jade crypt]

<@&286206848099549185> anyone have any idea? 😓

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[viscid flint]

uhhhhh what if ${x_k} = {1, -2, 3, -4, 5, -6, \ldots}$

[clever fjord]

new_nick: haprzTypo

[viscid flint]

no it still goes to 0 nvm

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[runic moon]

Please help

[runic moon]

I don't understand how I would go about answering this question.

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[harsh agate]

Thats an odd way to write them, you can plot the Phase Portrait or Vector Field to get an idea of how they flow qualitatively.

To get a quantitative answer you need to solve for eigen vectors and then get a solution

Unless I am mistaking what is being asked, that should allow you to plot the exact solution for initial conditions.

mistaken*

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[gleaming dew]

[
\sqrt{x^3} \cdot (f'(x))^2 - f(x) = 9\sqrt{x^3} - \sqrt{x} - 3x
]

[clever fjord]

Danango

[gleaming dew]

calculate integrate f(x) dx from 1 to 4

<@&286206848099549185>

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[proud spoke]

??

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[slate heath]

i dont know how to start this

[forest sky]

it's asking what size vectors are in the null space and column space of A

[slate heath]

is the first one six?

[forest sky]

yes

[slate heath]

second one is 5?

[forest sky]

yes

[slate heath]

thanks alot

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[granite lagoon]

hey anyone here?

[nimble blaze]

i don't know people with that name
but if you ask a math question, they might show up.

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[tulip thicket]

hi

[tulip thicket]

@mystic saffron

[mystic saffron]

yo

[tulip thicket]

how do i do this

i’m cooked

[mystic saffron]

i dont think i can help

[tulip thicket]

rip

[worldly rune]

gg wp

[tulip thicket]

i’m cooked

[worldly rune]

• in circle some xor operator?

[tulip thicket]

probably

[mystic saffron]

what is the question?

[tulip thicket]

this

[tulip thicket]

how do i do dis

@mystic saffron

[mystic saffron]

hmm it appears like you have no rules to do anything about xor