help-13

[azure stirrup]

and that formula i assume is y-y1=m(x-x1) ?

[crimson sedge]

yea

[azure stirrup]

OHH I GET IT

then i’ll get the equations then graph it

[crimson sedge]

Yes

[azure stirrup]

ahhh icic i think i csn continue the rest of 7c myself

tyyy

how abt 11c?

[crimson sedge]

Lemme see

[azure stirrup]

i did those for 11a and b

sorry kinda messy i draw it with my finger

when asks to sketch the curve would it be f(x) or f’(x)

[crimson sedge]

f(x)

f'(x) is there to tell you about the slope at different points, f'(x) is a curve yes but right now we are only concerned with it to know the slope

[crimson sedge]

it should be 1/x not -1/x

[azure stirrup]

i cannot read my own writing ooops mb

[crimson sedge]

you don't have to sketch the derivative

[azure stirrup]

just gonna send it here again so no need to scroll up

[crimson sedge]

kk

[azure stirrup]

so we don’t need to sketch the tangents?

[crimson sedge]

the sketch of the derivative is not the sketch of the tangents

[azure stirrup]

oh-

[crimson sedge]

a tangent is a line and always a line, the curve of the derivative isn't necessarily a line

And

[azure stirrup]

icic

[crimson sedge]

The derivative at a point tells you the slope of the curve at that point, only the slope it doesn't give you a tangent and it can't give you a tangent it returns a number lol

[azure stirrup]

ahh okk

when it says all the tangents slopes downwards

like that?

is the slope of the tangents the derivative

[crimson sedge]

at a point yea

you can remove the -1/x² curve from the sketch

[azure stirrup]

what does it mean by x-> 0^+ ans 0^-1

[azure stirrup]

okk

what is 0^+ amd ^-1

[crimson sedge]

it means negative slope as it says in the qsn and negative slope means obtuse angle formed by tangent anti clockwise from the positive x-axis

[azure stirrup]

obtuse angle??

[crimson sedge]

it means as x approaches that value, so 0+ means as x approaches 0 from the right hand side meaning from the positive side and x -> 0- means as x approaches 0 from the LHS or negative side, so the question is what happens to the slope as x approaches those values

[crimson sedge]

greater than 90° angle

[azure stirrup]

ohhhhh

uhhh as they approach 0 uh it gets closer but then turns away from x??

that is worded very poorly

[crimson sedge]

away from x-axis yes but what does that mean for it's value

[azure stirrup]

it increases..?

sorta

[crimson sedge]

Yes but does it have a stopping point how much does it increases

?

[azure stirrup]

infinity(?)

[crimson sedge]

Yea

but but

you see the slope is already negative and it stays that way

And keeps increasing

So negative infinity

[azure stirrup]

ahhh right

thr solution paper says aa x ->0^+ and 0^-1 “the gradient decreases
without bound, so the tangents slope more and more steeply backwards.”

what does without bound mean

[crimson sedge]

without bound means without a stopping point it happens forever

[azure stirrup]

ohhh okk

[azure stirrup]

so as x approaches 0^+ and 0^-, it approaches away from the x axis as it increases to infinity and negative infinity?

[crimson sedge]

heyyyyy we are talking of the slope

Not the curve itself

The slope is always negative

in both cases visualise that using the graph

Imagine drawing a tangent and then moving the point on the curve closer to 0

And seeing how the slope changes

[azure stirrup]

o h whoops

[azure stirrup]

the tangents become sorta looking more vertically straight ish?

[crimson sedge]

Yes what's their slope then

[azure stirrup]

still negative…?

[crimson sedge]

that is right but what's the slope of a vertical line

[azure stirrup]

uhm

the gradient?

[crimson sedge]

slope and gradient are the same thing

Do you know slope of a line is tan(theta)

[azure stirrup]

yea

[crimson sedge]

where theta is the angle made by it from the positive x-axis in anti clockwise direction

what's the angle made by a vertical line with the x-axis then

[azure stirrup]

there’s no angle..?

[crimson sedge]

bruhhhh

vertical

VERTICAL

|

[azure stirrup]

oh 180

[crimson sedge]

no

[azure stirrup]

uhm

[crimson sedge]

The angle with x - axis

[azure stirrup]

90?

[crimson sedge]

• what's the angle in this

Are you guessing

[azure stirrup]

yes

[crimson sedge]

breh

angle ACB how much is it

[azure stirrup]

90

[crimson sedge]

yea so a vertical line with the x-axis, does it not form the same angle

[azure stirrup]

yes

[crimson sedge]

and what would be the slop then the tan of that angle

[azure stirrup]

tan(90)?

[crimson sedge]

Yea which is

[azure stirrup]

my calculator gave me math error

[crimson sedge]

Lol

[azure stirrup]

indefinite..?

[crimson sedge]

it's like as the line would become more and more vertical meaning as it goes from 89° to 89.5° and 89.9°, 89.99°

Put these values in calculator

[azure stirrup]

tan(89)=57.28996163

?

[crimson sedge]

yea and put all those after that

And tell it here

[azure stirrup]

tan89.5 = 114.5886501

tan89.9= 572.9572134

tan89.99 = 5729.577893

[crimson sedge]

What about 89.9999999

[azure stirrup]

5.72953173 x 10^10

[crimson sedge]

So that's like 57 billion

what do you think tan 90 will be then

[azure stirrup]

a lot more than a billion

[crimson sedge]

Lel yea

Infinite

[azure stirrup]

oh-

[crimson sedge]

bruh see how fast it grows

Like at first it's so slow

tan(0) = 0, tan(45°) = 1

tan(89°) = 57

[azure stirrup]

yesyes

[crimson sedge]

so yea as the line becomes vertical it's slope in magnitude goes to infinity but it's negative so negative infinity

[azure stirrup]

ahhh okk

[crimson sedge]

do you see why it's negative though

[azure stirrup]

sorta

[crimson sedge]

if it makes an angle greater than 90° from the positive x-axis then the slope is negative

[azure stirrup]

so for as x approaches infinity the slope becomes more horizontal?

[crimson sedge]

Yes

[azure stirrup]

the line gets closer to 0 right

[crimson sedge]

the slope yes

[azure stirrup]

okkk

i think i got it

thank u for the help!!

[crimson sedge]

You're welcome

[azure stirrup]

.close

[cedar kiln]

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[small wave]

How would I simplify this?

[rotund jasper]

multiply and divide by the conjugate pair

[small wave]

.close

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[crimson sedge]

If ܽa ⊗ ܾb = ܽa− 2 × ܾb find the value of 7 ⊗ (5 ⊗ 2) anyone have any idea?

[cedar kiln]

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[cedar kiln]

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[mystic fiber]

Solve the equation $\frac{\sqrt[3]{7-x}-\sqrt[3]{x-5}}{\sqrt[3]{7-x}+\sqrt[3]{x-5}}=6-x$

[wraith dagger]

mendeleevpro

[vestal bear]

have you gotten anywhere on thisd

[cedar kiln]

@mystic fiber Has your question been resolved?

[mystic fiber]

No

[vestal bear]

[south tundra]

Try rationalizing the denominator

Might be messy at first but everything gets simplified

[vestal bear]

i was thinking that but im not sure how to rationalize cube roots

[south tundra]

To rationalize with square roots we multiply a - b by a + b (or vice versa) because (a - b)(a + b) = a^2 - b^2

And, similarly, (a + b)(a^2 - ab + b^2) = a^3 + b^3 can be used for cube roots

[vestal bear]

oh interesting

[cedar kiln]

@mystic fiber Has your question been resolved?

[cedar kiln]

@mystic fiber Has your question been resolved?

[mystic fiber]

<@&286206848099549185>

[lapis beacon]

@mystic fiber what is your question??

[mystic fiber]

Solve the equation $\frac{\sqrt[3]{7-x}-\sqrt[3]{x-5}}{\sqrt[3]{7-x}+\sqrt[3]{x-5}}=6-x$

[wraith dagger]

mendeleevpro

[lapis beacon]

is the answer 6

[mystic fiber]

???

[lapis beacon]

value of x=6

[gritty coral]

What?

[mystic fiber]

Explain why?

[lapis beacon]

solve th given equation you will get the answer..

[mystic fiber]

But how lol

[cunning forum]

It is obvious that x=5,6,7 work
Note that if you expand it and do all the messy algebra it will become a cubic(not necessary to actually do this)
Cubics have at most 3 roots, but you've already found 3
So 5,6,7 are the only roots

wait is it a cubic?

doesnt matter should still have no other roots

[mystic fiber]

only cube roots

[lapis beacon]

@mystic fiber ??

[cunning forum]

what do you mean

[mystic fiber]

nvm

[lapis beacon]

no the only answer is 6 which is in positive and all other rest are negative..

[cunning forum]

no, 5,6,7 all work

check it

[gritty coral]

5 is not a solution

[cunning forum]

no it is?

cbrt 2/cbrt 2=1=6-5

so is 7

[lapis beacon]

yes you are right all the 3 are solution.

[mystic fiber]

$\frac{\sqrt[3]{7-5}-\sqrt[3]{5-5}}{\sqrt[3]{7-5}+\sqrt[3]{5-5}}=\frac{\sqrt[3]{2}}{\sqrt[3]{2}}=1=6-5$ ...

[wraith dagger]

mendeleevpro

[gritty coral]

Sorry mb

[frozen path]

woah this place is active

[gritty coral]

They are solutions

[cunning forum]

not as active as MODS

yeah lol

[gritty coral]

Lmao

[mystic fiber]

bruh

[lapis beacon]

Next question please

[zealous quail]

We can use componendo dividendo

Here

It will make it much easier

[mystic fiber]

Can u pls solve it

[cunning forum]

I did, the answer is x=5,6,7...

[waxen plume]

Do you need help?

[mystic fiber]

Yes

[waxen plume]

Alright do you know any way to remove the cube roots?

[mystic fiber]

Show ur work

[cunning forum]

i did

[cunning forum]

here

[mystic fiber]

No (Idk where to start)

[waxen plume]

What's the given equation?

[cunning forum]

equations here

@waxen plume

[mystic fiber]

$\frac{\sqrt[3]{7-x}-\sqrt[3]{x-5}}{\sqrt[3]{7-x}+\sqrt[3]{x-5}}=6-x$

[wraith dagger]

mendeleevpro

[mystic fiber]

No one will understand it bro

[cunning forum]

no

its not that hard to understand

imagine expanding it it should become a cubic right

[waxen plume]

$(\frac{\sqrt[3]{7-x}-\sqrt[3]{x-5}}{\sqrt[3]{7-x}+\sqrt[3]{x-5}})^3=(6-x)^3$

[wraith dagger]

literatus

[waxen plume]

Do you get what I just did?

[mystic fiber]

Yeah

[waxen plume]

Then what do you get

As the fraction is ^3 the same as the square root of 3, the square just dissapear

This is what you should get after

$\frac{{7-x}-({x-5})}{{7-x}+({x-5})}=(6-x)^3$

[wraith dagger]

literatus

[waxen plume]

Maybe this helps you out solving this in order for x

[mystic fiber]

The truth

[cunning forum]

No, this is wrong

There is indeed a common term but it is not the same

Even if it was it wouldn't cancle like that

[mystic fiber]

$(a+b)^3=a^3+3a^2b+3ab^2+b^3$ ...

[wraith dagger]

mendeleevpro

[cunning forum]

expanding SHOULD give you a cubic i think

all that matters is that its a cubic

then it has at most three roots

which must be 5,6,7 since they work

[mystic fiber]

Notes : continue to solve it from 16:00:00 (4 p.m) - 23:00:00 (11 p.m) UTC + 00:00 (23:00:00 (11 p.m) - 06:00:00 (6 a.m) UTC + 07:00) because I'll offline at that time (I'm from 🇻🇳 )

[cedar kiln]

@mystic fiber Has your question been resolved?

[cedar kiln]

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[tulip harness]

Let $A={1,2,\cdots,200}$ and $B\subseteq A$ such that $|B| = 101$. Show that there are two numbers in B such that one divides the other.

[wraith dagger]

casiel368

[tulip harness]

No idea on where to start here

[cedar kiln]

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[signal kiln]

Which one is a simple event?

[cedar kiln]

@signal kiln Has your question been resolved?

[signal kiln]

<@&286206848099549185>

[bold vine]

Which event looks the simplest?

[signal kiln]

tossing two coins

[bold vine]

Which has just one thing

That's 2 things

Idk the defn, but I have a guess

[signal kiln]

none of them have just one thing

[bold vine]

One does

[signal kiln]

drawing two hearts?

.close

[cedar kiln]

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[cedar kiln]

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[eager marsh]

Hello! Im trying to confirm that the math I have done is correct,, im slightly unsure about the compass heading but I am confident about the asnwer

[eager marsh]

this is kinda what im plotting

except the degrees on the second diagram are incorrect

[cedar kiln]

@eager marsh Has your question been resolved?

[cedar kiln]

@eager marsh Has your question been resolved?

[cedar kiln]

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[wild belfry]

https://i.imgur.com/1zvn11j.png

[wild belfry]

I can figure out if its increasing easily, but any time I got this question before I would just use my graphing calculator to see if it was concave up or down 😅

[cunning forum]

Try differentiating

[wild belfry]

Now I gotta actually learn how to do it haha

[wild belfry]

okok 1 moment I will do it

[cunning forum]

Then find dy/dx when x=-4

that’s your increasing decreasing

then differentiate again etc

[wild belfry]

I got a negative in the discriminant...

[cunning forum]

Why is the discriminant involved in differentiation

[wild belfry]

Oh

[cunning forum]

Discriminant is for solving the thing

[wild belfry]

Wait what is differentiation again

Im getting mixed up haha

[cunning forum]

the rate of change of a function

[wild belfry]

ah yeye sry

[cunning forum]

if you’ve been using your GC to do these questions I’d suggest you revise your differentiation

[wild belfry]

I got 3x^2 - 6x + 8

[wild belfry]

for this I got 80

[cunning forum]

So is it increasing or decreasing

[wild belfry]

increasing

[cunning forum]

Now do the same for concave

[wild belfry]

Thats what I was confused about

How do I figure out which it is, up or down

[cunning forum]

Oh sorry was busy

[wild belfry]

npnp

[cunning forum]

You take d^2y/dx^2

at x=-4 and you see whether it’s positive negative or zero

[wild belfry]

Oh ok

1 moment I will do that

[cunning forum]

If it’s positive the graph is concave upwards at that point

Otherwise concave downward

[wild belfry]

https://i.imgur.com/psE9jj1.png

Yep got it!

Thank you so much, you explained it really well

❤️

.close

[cedar kiln]

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[wild belfry]

does f(x)=1/2 have a horizontal asymptote

[vestal bear]

yeah

f(x)=1/2 is the same as y=1/2

sactually

[wild belfry]

Okok

[vestal bear]

no\

[wild belfry]

Oh?

[vestal bear]

thats not an asymptote

asymptote is when a function approaches a value but never reaches it

1/2 reaches it

,w graph y=1/2

[wraith dagger]

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[vestal bear]

,w graph 1/x +2

[wraith dagger]

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[wild belfry]

https://i.imgur.com/Ybl3TQm.png

[vestal bear]

horizontal asymptote at 2

[wild belfry]

This was why I was asking

It says a=0 has an asymptote

[vestal bear]

did you find the limit as x appraches infinity and -infinity

[wild belfry]

But I guess the website is wrong? Idk anymore hahaha

[wild belfry]

It will be 1/2 right?

[vestal bear]

theres not even an option for a=0

you can use desmos for this, make a a variable with a slider'

and see how it affects the equation

[wild belfry]

https://cdn.discordapp.com/attachments/903481102591750184/1128022844907999282/2023-07-10_19.00.00.00.png

[crystal raptor]

This just depends on how that website defined a horizontal asymptote

[vestal bear]

yeah thats not an asymptote

[wild belfry]

What do you think is a safe bet for me to put in my exam

Asymptote or no asymptote?

[slow wing]

asymptote means it approaches a value not that its actually at the value

[vestal bear]

,w define asymptote

[wraith dagger]

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[wild belfry]

Damn didnt know you could do that haha

,w define cringe

[wraith dagger]

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[wild belfry]

😄

[vestal bear]

im trying to use latex and the bot functions more

[crystal raptor]

You should ask your lecturers, there are valid arguments for allowing a straight line to be asymptotic to itself or not

[wild belfry]

Unfortunatly I dont have any haha I am just doing self study

[foggy estuary]

the asymptote exists when the limit is a real number but it doesn't mean the function can't be equal to this limit 🤷‍♂️

[wild belfry]

So what do you guys think is the safest bet to put in an exam?

It probably wont come up but just to be sure haha

[crystal raptor]

Find your courses definition of asymptote

[wild belfry]

Okok I will email the examiners

I will do it now haha

Thank you everyone

❤️

.close

[cedar kiln]

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[twin flax]

Find the coordinates of the absolute minimum for the function f(x) = sqrt(x-3)

I got (3,0)

Is this correct?

[cunning forum]

Yes

Sqrt always nonnegative output

So it can’t be any smaller

There’s also only one equality case

[twin flax]

= 0?

For the case

[cunning forum]

No =0

Equality case

[twin flax]

I have another question, would this conclusion be correct?

For this function, f'(x) > 0 on (-infinite, -1) U (0, infinite) right?

[cunning forum]

Don’t the two ranges you gave overlap

also no

Look at [-1,0]

[twin flax]

Oh, it's increasing, for the y-output

[slate lintel]

i think you have a typo or something, check your answer again

like I'm pretty sure you have the right idea but there's a typo

[twin flax]

Which typo?

[slate lintel]

you say f'(x) > 0 on (-infinity, 1) U (0, infinity)

[twin flax]

My other conclusion was this:

[twin flax]

This kinda made sense to me because it was concave down till -1

Then its concave down from 0 to infinite

[twin flax]

Yeah

[slate lintel]

the function isn't even visible at x=1 so i'm not sure where you're getting that

[twin flax]

It's going up exponentially then

It would mean that it's a polynomial function

Which would explain that massive jump

[slate lintel]

exponential functions are not polynomial functions but ok i think i see what you're saying

still though, why do you think it stops increasing at x=1?

or maybe -- could you highlight the regions where f(x) is increasing?

[upper abyss]

You mean -1, not 1.

[twin flax]

Yeah

Ohhhh

I see what u mean

I see the typo now

Mb

[twin flax]

@slate lintel would this be correct then?

I corrected that 1 to -1

Mb

[slate lintel]

yehh

[twin flax]

Alr, thx for the help

.close

[cedar kiln]

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[viscid pendant]

How do I compute the limit?

[long arrow]

divide both the top and bottom by n

[viscid pendant]

Do I divide everything by n^2

[viscid pendant]

isn't it highest power?

[long arrow]

sqrt(n^2) = n

for n >= 0 (here n tend to +inf)

[viscid pendant]

what do I do afterwards

[long arrow]

not like that

[viscid pendant]

divide everything by sqrt n?

I mean sqrt n^2

[long arrow]

$$\lim_{n \to \infty} \frac{\sqrt{n^2+4}}{n+8}=\lim_{n \to \infty}\frac{\sqrt{n^2\Big(1+\frac{4}{n^2}\Big)}}{n\Big(1+\frac{8}{n}\Big)}=...$$

[wraith dagger]

modus7591

[long arrow]

and now sqrt(n^2) = n

[viscid pendant]

the n cancel out?

and then the limit is 1?

[long arrow]

yes

[cedar kiln]

@viscid pendant Has your question been resolved?

[viscid pendant]

do you always do this method for these types of problems?

[long arrow]

yes, similar way to solve other problems like this

[cedar kiln]

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[granite shuttle]

not sure wat im supposed to do here

[long arrow]

start from something simpler

can you solve x^2 = 25?

(using this method)

[granite shuttle]

um not sure how i would write that

are you asking me to start from here

[long arrow]

noo, that was an other example

to ilustrate how it works

but

I mean

we take square root

but you have to remember to do +/-

like with x^2 = 25

we have

x = +/- 5

and same thing happens with yours

[granite shuttle]

oh

idk if i did this right

unless you meant just this

[cedar kiln]

@granite shuttle Has your question been resolved?

[granite shuttle]

figured it out

[spare ore]

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[cedar kiln]

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[lucid adder]

I want to find the minimal polynomial of sin(2 pi / 7) = cos(3 pi / 14) over Q[X], I think it should enter galois theory in somewhat moment but I already have finded the minimal polynomial of cos(2 pi / 7) and no uses nothing on galois theory.
Any Hints?

[cedar kiln]

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[hollow patrol]

how is the first one increasing?

[hollow patrol]

heres a better pic

since if you plug in any number lower than -3 in the s'(x) u still get a neg answer

so how is the behavior increasing?

is because its the derivative so its going the opposite?

[main sage]

show me your first derivative number line

first derivative test number line

[hollow patrol]

they didnt give me one, ill show you everything they showed me

[main sage]

you have to make one to solve the problem

[main sage]

"determine the behavior of s on each interval"

behavior of s and s' are different

when first derivative is positive, that means s is increasing

I think that is where you are getting confused

Read carefully

when s' is pos, s is increasing

s' is neg, s is decreasing

just a fundamental relationship between first derivative to original

[hollow patrol]

yes i understand that part, so are you saying whatever the answer is for the derivative you have to flip the sign?

[main sage]

no

no sign flipping

You found out that the first derivative at any number less than -3 is positive right

you plugged say -4 into derivative, got a positive result

[hollow patrol]

i did this and got confused

[main sage]

that means that the original s will be increasing at that interval

[hollow patrol]

let me do it again

[main sage]

the question is asking you to determine the sign of s', so you have to plug in

just plug in -4 and tell me what u get

[hollow patrol]

i got a neg tho

[main sage]

the sign of the number is whats important

[hollow patrol]

since u have to plug it into the denom too right

[main sage]

did u use parenthesis?

[hollow patrol]

for the denom?

[main sage]

it should be positive mate

plug it into a calculator carefully

in this case (-4)^2 = 16

top is positive

[hollow patrol]

ok ill do them separately: for the numerator if i plug in -4 i get 819

[main sage]

and denom

[hollow patrol]

yes

[main sage]

no

you have to plug into the whole thing

not one by one

[hollow patrol]

i was going to show u how i got the demon neg tho

if i do the whole hting i get a domain error on my calc

[main sage]

no

bottom is positive

i promise you it's positive

u are not inputting into the calculator correctly

first derivative at -4 is positive

do you know what fractional exponents mean

[hollow patrol]

but if i plug in 5(-4)^2/5 in my calc i get domain error

[main sage]

basically, if ur pluggin negative 4, the denominator will be 5 times 5radical (16)

convert it to a radical

5 times root 5 radical 16

5th root 16

[hollow patrol]

oh so its 1.7?

[main sage]

also, never assume domain error means negative

[main sage]

the value doesn't matter in this case, the sign does

it's positive

i didn't calcualte it, but i'll tell u denom is positive

[hollow patrol]

yeah ik the value but im jst sayin that its pos

[main sage]

pos / pos = pos

[hollow patrol]

wow thanks for this, i never did this for these questions

[main sage]

u have to

don't plug in exponents like that calcualtor can get confused if u don't use parenthesis correctly

[hollow patrol]

so how would you use parenthesis correctly?

[main sage]

5 * (4^(2/5))

or whatever it was

[hollow patrol]

it was neg 4 tho is that fine?

and if i plug the neg 4 in my calc i still get a neg answer even with the parenthesis

[hollow patrol]

ill practice doing questions like these

instead of using the calc

thx again

[main sage]

i know it's positive because the 5th root of 16 is

1.7

positive

ur doing something wrong with ur calcualor

i guarentee it

[hollow patrol]

yeah i got that too

[hollow patrol]

but when i inpunt something like this but with -4 i get a neg idk what im doin wrong

but doesnt matter anyways ill do the radical thing, thx

[main sage]

idk either

hard to tell unelss i see ur calc

use a ti - 89

best calc

[hollow patrol]

gotcha will check that one out

thx

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[wicked brook]

Is there a nice intuition for the definition of $D_4$ compared to the other $D_{2n}$?

[wraith dagger]

zander_a

[main sage]

what do you mean intuition

Oh sorrry, i never toook logic classes yet

cannot help

[wicked brook]

Like $D_{2n} \cong \Sigma(\pi_n)$, where $\Sigma$ is the symmetry group of an n-gon

[wraith dagger]

zander_a

[wicked brook]

Of course, there is no 2-gon and using a line with two points fails to be group

So why do we bother with $D_4?$

[wraith dagger]

zander_a

[upper abyss]

Why not?

I don't see how there's less value in recognizing this is a group

Even though it's just Z2

[wicked brook]

I guess my question is why is it called like this

[upper abyss]

Or wait, it's 4 elements isn't it

[wicked brook]

Yes (1) and thing of the form (1 2)(3 4)

In S_4

[upper abyss]

Okay yeah, so it's just V4

[wicked brook]

yea, but why's it called D_4?

Is there some what to view it nicely so it fits?

[upper abyss]

The presentation is probably the best way to see why it works

{r, s | r² = e, s² = e, rs = sr^-1}

[wicked brook]

Ah ok

[upper abyss]

That's clearly a dihedral group, the way I've presented it, but it's obviously just V4 lol

[wicked brook]

Yes yes this makes sense

tyty

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[abstract relic]

so i basically have no idea how to begin, the only thing i noticed is that the 1/5 corresponds to the 5th root that the question wants. other than that im pretty lost. also probably know it has something to do with the (1+x)^n formula but no idea how its being applied

[austere hull]

,tex .exp rules

[wraith dagger]

Akira

[austere hull]

@abstract relic

[abstract relic]

oh i know that 5th root equals ^1/5

but i dont really know how youd work it out

i did this question on an exam and sorta winged it so wanna knwo for sure if i got it right

i set X to 30 and then substituted it in every term on the series

[runic garnet]

That was wrong?

That’s the approach I would’ve taken

[abstract relic]

oh i dont know yet

i just did the exam i havent gotten the test back yet

[runic garnet]

Oh ok

Well yea that should be right

[abstract relic]

ahh its that simple?

damn

thanks

is this highschool level math?

[austere hull]

Yes

[abstract relic]

ahh, im learning it in college

aight thanks guys, ill close the channel

.close

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[deep rivet]

I was looking through my syllabus, and came across this:
Describe the radian measure of an angle as the ratio of the length of an arc intercepted by a central angle and the radius of the circle
Could someone explain what this means? My interpretation is; Describe a degree measure as a radian (I don't understand the rest)

[deep rivet]

Could Someone give a visual exapmple (if possible)?

ping @deep rivet when answering

[flat oracle]

@deep rivet it s just θ=L/r

[deep rivet]

oh alright! thanks

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[spiral star]

How do I do this? Thanks

[vestal bear]

find BC

then you can find measurew D

[dim mist]

yes

find BC using SOH CAH TOA

THEN BC + 22 units = BD

then use pythagorean theorem to find AD

u don't need to find AD

then find Angle BDA

using soh cah toa

[spiral star]

Tq

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[true grove]

Did I do this correctly?

[true grove]

A parabolic bridge over a river has a span of 100 meters and a height of 30 meters. What is the maximum height of a boat that can go under the bridge at a point 20 meters from the shoreline?

That’s was the question

[dreamy sleet]

Notice that in the question it says the height of the bridge is 30m

930m is way over that, so you can tell you didn’t do it right

The part you got wrong is the equation of the bridge

[cedar kiln]

@true grove Has your question been resolved?

[true grove]

How do I fix it?

How would I solve this question?

[cedar kiln]

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[true grove]

.reopen

[cedar kiln]

✅

[true grove]

<@&286206848099549185>

[crimson sedge]

😡

[dawn junco]

<@&268886789983436800> ban this guy already

[crimson sedge]

What he sent

[true grove]

Adult film

[crimson sedge]

In study group

[severe estuary]

Corn

[true grove]

.close

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[woven crater]

given a 3x3x3 rubik's cube with 1 edge piece (a piece with two colours) missing, what algorithm can you use to decide how to put that edge piece back in so that the cube is solvable?

[woven crater]

so for example how do you make an algorithm that decided wether the green and white piece should be oriented with the green or white sticker up in this cube?

[slate lintel]

i think this is equivalent to solving the edges

because you can ignore the corners

[woven crater]

yes. Is there a more efficient way though? for the corner pieces there is as you can give each piece a 'rotation value' of 0, 1 or 2, based on wether the sticker that should be horizontal is horizontal, 1 clockwise rotation away from being horizontal, or 2 clockwise rotations being away from being horizontal, and then if the sum of rotation values of all corner pieces is divisible by 3, the cube is solvable @slate lintel

@dreamy sleet is this maybe a question you can solve? Because this is also a parity issue

[dull oxide]

You can add the orientation number of each piece

You don't have to solve it

Total orientation must be even

[woven crater]

how do you find the orientation number of pieces if they're not on their original spot?

[dull oxide]

It's a bit tricky, but I recently devised a system for this when I was trying to prove the number of positions in a rubiks cube

This will be a longish explanation

[woven crater]

if you have time for it then yes please :)

[dull oxide]

I'm assuming you have a standard color rubiks cube. White opposite yellow, green opposite blue, red opposite orange, and blue orange yellow faces go clockwise

[woven crater]

yes

[dull oxide]

We will let green be the front face (F) and yellow will be the up face (U)

I'm assuming you know the faces U F B L R D

[woven crater]

yes

[dull oxide]

Okay

The orientation number of every edge piece is zero when it's in its solved state. If you flip an edge piece, you add one to it's orientation number. You only care if this number is even or odd.

The total 3x3x3 edge orientation number is the sum all of orientation numbers of the 12 edges

It must always be even

Now for how we change orientation number

Twisting the U F D or B faces does not change the orientation number of any edge

A 90 degrees L move adds 1 to the four edges on the L face. A 90 degrees R move adds 1 to the four edges on the R face

That's it

You can get the orientation number of any edge at any position by mentally twisting it to it's solved state and counting it's orientation number along the way

You have to do it 11 times, which is rough

With practice, you can identify orientation quickly

Usually 1 second per piece

[woven crater]

cool algorithm! I'll play around with it a bit to see if I can figure out why that's true, and then ill get back to you. Many thanks for telling me this. Also, was your attempt to prove the amount of combinations succesful? If not I may have a video you'd be interested in

[dull oxide]

It was

I validated every NxNxN scramble number

[woven crater]

cool, how about N^4? ;)

[dull oxide]

Yes actually

I'm #56 in 3^4 solve

I'm old school

[woven crater]

oh? theres a leaderboard for that?

I did try to find that once but couldn't

[dull oxide]

https://superliminal.com/cube/halloffame.htm

[woven crater]

thanks!

[dull oxide]

Np

There's also a hypercubers discord

https://discord.gg/6MNbu7YK

[woven crater]

thanks :)

[cedar kiln]

@woven crater Has your question been resolved?

[woven crater]

I think I proved that its true :)

[cedar kiln]

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[wanton summit]

5 random points are distrubuted on the area of the sphere.

[wanton summit]

im not sure where to start

[turbid mesa]

Okay, let's try to convert the problem first.

[wanton summit]

how so

[turbid mesa]

The maximum distance between two points in the same hemisphere is equal to the diameter. Can we use this?

[wanton summit]

Yes

two pidgeons go in diameter pidgenholes

?

[turbid mesa]

Hmm, idts. That's also the maximum distance between any two points on the sphere, so this does not provide us with any additional info

The only idea I can think of rn trying to prove that when the 5 points are written in Cartesian coordinates, we have to prove that there exists an isometry (only rotational ig), which when applied to all the 5 points give us atleast one coordinate of atleast 4 points to be the same (either + or -)

[wanton summit]

this is a pidgenhole principle question though

its part of thta excersise

[turbid mesa]

Should be ig

[wanton summit]

wait

i think i got it

1s

@turbid mesa see any issues ?

[turbid mesa]

Can you explicitly state out what you did?

[wanton summit]

Put two points anywhere on the sphere

and draw an equator

between them

it makes 2 hemisphere

s

remaining 3 points can go in those places

[turbid mesa]

Hmm, I see it working, but need to write it properly

[wanton summit]

in an exam im pretty sure thats fine

considering its nor really a formal proof

[wanton summit]

know how to solve for n ??

[turbid mesa]

Do you know the formula for sum of consecutive naturals?

[wanton summit]

no

[turbid mesa]

It's n(n+1)/2

Plug it in and you can solve

[wanton summit]

isnt that

nCn-2

[dire geode]

yes but it's not as useful as their formula

[wanton summit]

interesting

how math lines up like thta

[wanton summit]

no just an observation

cant do algebra with c and p

thank you

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[crimson sedge]

Show that for any natural number 𝑛 there exist two integers 𝑎𝑛 and 𝑏𝑛 such that (3+2sqrt(2))^n = an + bn sqrt(2). How can I solve this one ?

[inland fossil]

you can use induction

[crimson sedge]

ok thanks

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[hot sand]

Please help

[cedar kiln]

Please don't occupy multiple help channels.

[hot sand]

.close

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[dire geode]

you're getting help. don't open new channels

[hot sand]

Ok

Who are you?

[cedar kiln]

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[quaint spear]

How did i get this wrong? The answer is soppused to be 63/40

[long arrow]

,rccw

[long arrow]

2/3 - 1/9 isn't 7/9

let's check this out

[quaint spear]

How come

[long arrow]

$$\frac23 - \frac19 = \frac69 - \frac19 = \frac{6-1}{9}=\frac59\neq\frac79$$

[wraith dagger]

modus7591

[quaint spear]

Ooo i added im so stupid🥲thank u

[cedar kiln]

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[atomic vapor]

a coin is flipped 20 times, in how much combinations you can get side A 14 times and side B 6 times while you never get B side twice in a row

[atomic vapor]

I wanted to see how much posibble arengemnts are there while ignoring the second rule and then suntract the illegal combinations

so there are 20!/(6!x14!) different ways to arrange the 20 flips

[fair geyser]

it's actually not easier that way

[atomic vapor]

so how should I approch it?

[fair geyser]

nvm, maybe that's the right one

i don't know

[atomic vapor]

<@&286206848099549185> plz help

[fair geyser]

it's stars and bars basically

[atomic vapor]

?

can you elaborate?

[fair geyser]

you get 7 buckets between 6 Bs
but the outermost buckets can be empty

so you use 5 As to fill the insides

and then solve for 7 buckets that can all be empty

there's also a very specific trick, where instead you have 15 spaces between all the As

so that's 2 different ways

[atomic vapor]

I didnt really get the first one

[atomic vapor]

how can all of the seven be empty if we need the spaces?

[fair geyser]

we have 14 As, we can say we actually have 9, 5 are already placed

[atomic vapor]

oh I get it now

so we have to place 9 A's

[fair geyser]

somewhere among 7 buckets

[atomic vapor]

D(7,9) right?

[fair geyser]

like suppose you have 2 bars and 3 stars ***||
there's 10 permutations of this, and each one corresponds to placing 3 stars into 3 buckets
1+0+2: *||**

d for distribute?

never saw that but probably

[fair geyser]

yes

[atomic vapor]

this is very nice
thank you very much

Im going to close the room now

[fair geyser]

sure

[atomic vapor]

.close

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[atomic vapor]

A is a set with 2n+1 members, how many subsets of A have n or less members?

[atomic vapor]

I dont know how to even begin

[crimson delta]

if S is such a subset. what can you tell me about its complement

[atomic vapor]

If the universe is A than it should have more than n members right?

wait I got confused

A has 2^(2n+1) subsets

[crimson delta]

you might try an example

eg A={1,2,3} or A={1,2,3,4,5}

what are the relevant sets? why did I ask about complements?

[atomic vapor]

lets take A={1,2,3}

n=1

so the subsets are:
empty group,{1},{2},{3}

their complement are
{1,2,3}, {2,3}, {1,3}, {1,2}

[south tundra]

Hm, do you have the answer by any chance?

[atomic vapor]

yes

its 2^2n

[south tundra]

Oh

[atomic vapor]

the complements has n+1 or more members

they have between n+1 to 2n+1 members

[south tundra]

The problem becomes easy if you recall combinatorics

[crimson delta]

so, we have pairs

S and S^C

one has <=n elements, the other >=n+1

what does that mean for the problem

[atomic vapor]

n pairs?

[crimson delta]

how many pairs did we have in the example

where n=1

[atomic vapor]

4

[crimson delta]

so not n pairs

[atomic vapor]

yeah

[crimson delta]

we have 2^(2n+1) subsets. we split them into pairs

how many pairs

[atomic vapor]

2^2n

[south tundra]

Oh, I see now, pretty smart

I guess I went the harder way then

[crimson delta]

please dont tell me you added binomial coefficients

so, 2^(2n) pairs and in every pair there is one subset we want

so how many subsets

[atomic vapor]

how do we know for sure that no number was left without a piar?

[south tundra]

Yes, thanks to the symmetry of binomial coefficients you just half the 2^(2n+1) and get 2^2n

[atomic vapor]

2^(2n)

[atomic vapor]

but I still dont 100% get this part

[crimson delta]

the map S->S^C is a bijection

because it has an inverse

(itself)

and because we have 2n+1 elements, we cant have |S|=|S^C|

[atomic vapor]

so we made a func that gives us the complement of a subset, each subset is paired (not to itself)

it is inverseable therefor its bijection and surjective which means the number of the range and domain is equal and addition of them gives us 2^2n+1

so there are 2^2n pairs and 2^2n subsets with less than n members

took me pretty long but I think I got it

If this explanation is correct Im gonna close the room

thanks alot for the help 🙂

[crimson delta]

well range and codomain are both all subsets

so 2^(2n+1) elements each

the point is more that you can actually find pairs S, S^C and it doenst end up weird that you have a single set S without a partner and some triple maybe

because for each set you find one partner (S^C) which also has the original set S as its partner

[atomic vapor]

yeah thats why its invertable

[cedar kiln]

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[amber parrot]

I have zero clue how to even start these...

[slate lintel]

well the bottom one involves reading a graph and the top one involves using trig angles (sin/cos/tan) kind of like that problem you did yesterday I think

[amber parrot]

Oh okay so this switches into relations after the top question

[south tundra]

What's the name of the point at which the right angle is located? Can't see

[slate lintel]

R i think. PQ is the line going down the middle along the bottom to the right

[south tundra]

So, here RP + PQ = RQ, right?

[amber parrot]

right

[south tundra]

Can you calculate RP?

[frigid tundra]

guys, any one can explain what's the diffrene between the courbe and the graph of a function, or lets say an application ?

[south tundra]

Ask your question in an available help channel

[frigid tundra]

ohh right, thanks

[amber parrot]

Ill try give me one sec

[south tundra]

Sure, take your time

[amber parrot]

My calculator can be weird sometimes so I got 0.857...., and 7.349....,

[slate lintel]

how did you calculate those?

[amber parrot]

I used cos over 1 = 6.3 over h. I did cross multiply divide.

[south tundra]

Make sure your calculator is in angles mode

Oh

Why would you use cosine though? You don't know the hypotenuse

Perhaps using tan(31) could be helpful

[slate lintel]

(you don't know the hypotenuse and Bean didn't ask you to calculate the hypotenuse)

[amber parrot]

Using the same formula but using Tan(31) I got 3.78

[south tundra]

Yup, that's the length of RP

Now, can you calculate the length of RQ?

[amber parrot]

Calculating RQ is where I got confused.

[south tundra]

We know the length of RS and the angle RSQ, don't we?

[amber parrot]

RS is 6.3, angle I'm not sure would it be 43 degrees?

[south tundra]

Yeah

So we can just say tan(43) = RQ/6.3

[amber parrot]

so 5.87..?

[south tundra]

Yup

Let us recall the very first fact that I mention

RP + PQ = RQ

We now know the values of RP and RQ

[south tundra]

So we can calculate PQ using this equation

As it's the only unknown variable there

[amber parrot]

I'm a tad confused would you use Pythagorean theorem?

[south tundra]

No, you just imply that PQ = RQ - RP and plug in the values of RQ and RP that we calculated previously

And you will get the answer for PQ

[amber parrot]

Oh I see. So PQ is = to 2.09

[south tundra]

Yup

[amber parrot]

That helped a lot thank you!

Am I able to get come clearance on the second question?

[south tundra]

Generally to calculate a value of the function at a point like x = -4 given the graph, you could image (or draw) a vertical line crossing the x-axis at the point where the -4 is marked at and checking the y coordinate of the intersection point with the graph of the function

But here the graph just crosses the x-axis at x = -4, directly hinting at the fact that g(-4) = 0

And to solve g(x) = 4 you do something similar;
Image (or draw) the horizontal line crossing the y-axis at the point where the 4 is marked at and check the x coordinate of the intersection with the graph of the function

[amber parrot]

4?

[south tundra]

Yup

[amber parrot]

Hm okay, would this be same throughout the functions and relation questions? dealing with stating the domain and range, and the if the relation is a function or not..?

[south tundra]

Yup

[amber parrot]

.close

[cedar kiln]

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[silver mason]

No clue how to do this lol

[worthy leaf]

Have u done any working yet?

[silver mason]

Yeah I graphed both of the functions

But I'm stuck after that

[worthy leaf]

Okay so the two lines intersect right?

[silver mason]

Yup

[worthy leaf]

Okay so we want to find the area of the bounded reigon

[silver mason]

Gotcha

[worthy leaf]

Do u know how u would do this?

[silver mason]

Nope

[worthy leaf]

Okay have u done integration?

[silver mason]

Yeah

[worthy leaf]

Okay

So using integration u can find the bounded area

[silver mason]

How so?

[worthy leaf]

Okay so find the point where the two lines intersect

[silver mason]

(0.5,0.25)

[worthy leaf]

And use this as the top bound in the intergration for y=x^2

The x value

[silver mason]

Got it

[worthy leaf]

So this finds u the area under the quadratic

But we want the area of the bounded reigon

So what do u think we should do

[silver mason]

Find the antiderivative

[worthy leaf]

So find the area of the triangle (using the point found earlier) and take away the area under the quadratic

This will give the area of the bounded area

[silver mason]

I got 0.01308

[worthy leaf]

What is that as a fraction

[silver mason]

pi/240

[worthy leaf]

Im guessing u have rotated it by pi

[silver mason]

Whoops

Sorry I'm really bad at Calculus

[worthy leaf]

Its ok dw

[silver mason]

Wait so would it be -pi/240?

But that can't happen because area is never negative

Pretty sure it is pi/240

I did it again

[austere hull]

it's pre-calc not calc

[silver mason]

Well the course is Calc 1

[cedar kiln]

@silver mason Has your question been resolved?

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[cedar kiln]

Channel closed

Closed due to the original message being deleted

[cedar kiln]

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