help-10

[royal basin]

have you actually worked out all the necessary details for my L and h? or did you blurt out "so both inclusions don't hold?" as a blind guess?

bc that is definitely not the conclusion that my example appears to suppose.

[static patio]

i wrote it down on paper

L = {ab} L^R = {ba}
h^-1 (L^R) = {ba}
(h^-1(L))^R = {c}

[royal basin]

h^-1(L) isn't {c} though.

h^-1(L) = {ab, c}

and thus [h^-1(L)]^R = {ba, c}

[static patio]

i thought that there might be problem with it

[static patio]

do i need to use available mappings when constructing this?

[royal basin]

what do you mean "available mappings"?

[static patio]

we have h(a) = a, h(b) = b, h(c) = ab

and when we want h^-1(L) we use just the last mapping

[royal basin]

are you trying to insinuate that h(a) = a, h(b) = b and h(c) = ab are meant to be taken as definitions of 3 separate functions???

because that is not the case at all

[static patio]

we want to have h^-1({ab})

we have word ab

how do we get {ab ,c} ?

[royal basin]

h^-1({ab}) consists of all words w such that h(w) = ab

do you understand that h(a) = a and h(b) = b, and the fact that h is a homomorphism, together imply h(ab) = ab and so ab ∈ h^-1(L)?

[static patio]

in the problem statements isn't homomorphism mentioned

just compare languages

[royal basin]

so it is not mentioned that h is intended to be a homomorphism?

are you sure?

most problems you've posted so far considered only those functions between string sets that are homomorphisms.

[static patio]

exciplitly not

probably it would be

[royal basin]

...can you post the full, un-cropped image just to be sure

because i am highly suspicious

[static patio]

it says, compare following languages

[royal basin]

of course if h is not known to be a homomorphism then you cannot say shit about h^-1(L^R) vs. [h^-1(L)]^R. they are two different sets and they have precious little to do with each other.

does it say elsewhere on the sheet anything about h

[static patio]

nowhere

[royal basin]

strange.

then perhaps you SHOULD say neither inclusion holds in general, by way of malicious compliance.

[static patio]

$let \ w \in h^{-1}(L^R) \implies h(w) \in L^R$ by the definition on preimage, $(h(w))^R \in L$ by reverse rule

[royal basin]

not inverse

reverse

still

[warm shale]

Michal

[static patio]

my bad

[royal basin]

why is the word "let" inside dollars

it doesnt belong there

anyway

[static patio]

im tex noob

[royal basin]

IF h is a homomorphism, then you can prove h^-1(L^R) subset [h^-1(L)]^R and the proof is fairly elementarty

elementary*

in fairness the fact h is a homomorphism comes into play at exactly one point

and it is when you say h(w)^R = h(w^R)

[static patio]

hmm

i have no idea

i dont get it how homomorphism can help here

[royal basin]

literally just definition-pushing

almost automatic

at every step there is really one choice to make

let w ∈ h^-1(L^R)
=> h(w) ∈ L^R
=> h(w)^R ∈ L
=> h(w^R) ∈ L (this is where h being a homomorphism matters!!!)
=> w^R ∈ h^-1(L)
=> w ∈ [h^-1(L)]^R

[static patio]

i got it, except this => h(w^R) ∈ L (this is where h being a homomorphism matters!!!)

[royal basin]

actually. hold on.

i might have fucked that up.

h(w^R) = h(w)^R actually does not always hold

my apologies

h(w^R) = h(w)^R only holds when the image of every single character under h has length 1.

[static patio]

hmm

then what to do with proof

[royal basin]

learn from my mistake, and attempt to conjure up a homomorphism for which h^-1(L^R) is not a subset of [h^-1(L)]^R

sometimes, though not always, the mistake in a failed proof can suggest a way to make a counterexample.

[static patio]

i will try

maybe

$L = {ab}$ and $h : {a,b,c}^* \to {a,b}^*$ defined by $h(a) = ab, h(b) = ba$

[warm shale]

Michal

[royal basin]

and h(c)?

or maybe next time you should not copy what i wrote without understanding it.

[static patio]

$L = {ab}$ and $h : {a,b}^* \to {a,b}^*$ defined by $h(a) = ab, h(b) = ba$

[warm shale]

Michal

[static patio]

removed c from set

[royal basin]

that's more like it...

ok let's take a look

h^-1(L) = {a}, L^R = {ba} and h^-1(L^R) = {b}

...yeah checks out

[static patio]

so neither of inclusions hold

[royal basin]

yes, neither inclusion holds in general

even if h is a homomorphism

[static patio]

interesting

but these problems with sets, functions and homomorphism are so difficult for me

thanks for help

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[silver plover]

could someone plz teach me a method on how to do these kinda questions, as ive never understood how to do them

[obtuse pebble]

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[little tinsel]

We constructed a circle around a regular triangle and a third circle that touches the other two circles. What is the ratio of the areas of the circles?

[untold flax]

hey bro

let the radius of the circle inside be 1

and see how far you can go

[little tinsel]

this one?

[untold flax]

that's the one

[little tinsel]

right

i did this in paint

measured by eye

thats the lenght of the radius

so if we say area is $\pi r^2$

[warm shale]

DanielCsocsik

[little tinsel]

as how you said

let radius be 1

so the are will be

$\pi$

$4\pi$

$16\pi$

right?

[untold flax]

uh what are you talking about

where'd you get 2 r and 4r from

[little tinsel]

i grabbed the radius from here

cuz i dont have a ruler cuz i made this in paint

and then i grabbed the biggest one

that one

and then the smallest one

i put these next to each other so they are parallel to each other

[untold flax]

bruh

use maths

[little tinsel]

what math

area is pir^2

[untold flax]

use maths to find the other radii of the circle

[little tinsel]

isnt the answer16:4:1?

[untold flax]

no

and even if it was your reasoning isn't valid

[little tinsel]

well that was epic then

radius of the circle is $a/sqrt3$, where a is the lenght of the triangle

[warm shale]

DanielCsocsik

[untold flax]

oh nvm this is correct but how you did it was not mathematically

[little tinsel]

logic

ahaha

[untold flax]

this is true

it's a bit unnecessary though

[little tinsel]

bro

eye measure is always handy

[untold flax]

lol

they wouldn't be to scale

in an actual exam

but I guess you could construct it yourself

[little tinsel]

if I can draw a perfect drawing on plaid paper

[untold flax]

yeah but still

you wouldn't have mathematical reasoning to back it up

[little tinsel]

thats why i am happy that they only ask for the answer

if they ask for the answer of 1+1

or answer and proof 1+1

[untold flax]

bruh this is painful for me as a mathematicain

please don't do that 😭

proof by eye measure lol

[little tinsel]

its not much

but its honest work

[untold flax]

if you wanna solve it properly use this construction

that's your hint, you should be able to get everything from here

[little tinsel]

is that kite

deltoid*

[untold flax]

maybe

just see all the angles you can find

[little tinsel]

eye meassured 120degree

[untold flax]

no not eye mesaure

360/3

[little tinsel]

ahahah

[untold flax]

😠

[little tinsel]

i'm dead bro

[untold flax]

bro eye measure

are you an engineer or smth

[little tinsel]

maybe

anyway

thanks for the help bro

i have my answer

thank you for that

[untold flax]

no problem

[little tinsel]

see ya later :))

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[little tinsel]

x= 1,5 - 1/y

[little tinsel]

.close

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[dark orchid]

How to solve number 1?

[final thunder]

Just take out the common factor

[dark orchid]

?

[final thunder]

Whats the GCF of 5x^2 and 50x

[dark orchid]

5x

[final thunder]

Yes

So factor it out

[dark orchid]

Uhh

[final thunder]

Like ab + ac = a(b+c)

[dark orchid]

5=a
50=b

Also, how would i solve number 2 with cmmon factor? 1?

[final thunder]

${\color{red} 5x} \cdot x -{\color{red} 5x} \cdot 10$

Er

[warm shale]

Pure

[final thunder]

Factor out the red bit

[dark orchid]

5x(2x*5)

Wait

F.

[dark orchid]

How?

Is there a way that woulds for all of them?

Works*

@final thunder

F.

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[boreal cipher]

!help

[boreal cipher]

Is this appropriate working?

1460cm -> 0.0146km
23m -> 0.023km
P = 0.0146 + 0.0146 + 0.023
P = 0.0522km

[fading quest]

yep thats right

[boreal cipher]

Thanks

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[tardy solar]

I know the answer

[boreal cipher]

What is it?

[tardy solar]

If 1460 = 14.60m Right?

[boreal cipher]

1460 cm = 14.60m

[tardy solar]

Ok

Let me continue

[boreal cipher]

Sure

[tardy solar]

Then do 14.60 Multiply by 23 You get 335.8 Meters If 1000 meters = 1km then the answer is 0.03358

Of course youve not got the answer yet,

They said the perimeter

[boreal cipher]

I'm sorry, to find the perimeter you add the sides.

[tardy solar]

Yep'

So whats the perimeter? Hands up

[boreal cipher]

Was my answer wrong

[tardy solar]

Just say the perimeter

[boreal cipher]

0.0522km

[tardy solar]

Correct ✅✅

[boreal cipher]

🥳

[tardy solar]

Then do

[obtuse pebble]

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[grave wyvern]

i really need help here i do not even understand where to start

[obtuse pebble]

@grave wyvern Has your question been resolved?

[grave wyvern]

<@&286206848099549185>

[obtuse pebble]

@grave wyvern Has your question been resolved?

[slim leaf]

@grave wyvern Can you translate?

[grave wyvern]

let's consider the set E=........

show that: N⊂E

@slim leaf

[slim leaf]

Is $0 \in \mbb{N}$ according to your book?

[warm shale]

Sup?

[grave wyvern]

yup 0 belongs to N

0,1,2,3,4

[slim leaf]

Well what happens if u fix n = 0

[grave wyvern]

E becomes m

cuz n equals 0 so it removes sqrt 3

[slim leaf]

So E contains N

it contains a lot more stuff

[grave wyvern]

but-

are we going to replace it for every number

[slim leaf]

Just fix n=0

and let m be any natural number

You will get m+0 = m, a natural number

which means, every natural number is in E

that means E contains N

[grave wyvern]

hmmmmm

[slim leaf]

For example, list some elements of E

you will see E contains N

[grave wyvern]

i see

so we have the power to make n whatever we want?

and we choose to make it =0 right?

[grave wyvern]

but what will you respond if someone says what if you put n = 1 or any other number that belongs to N

[slim leaf]

You can do that

And you'll get irrationals

You have to put every natural number in place of m,n

so somewhere along the way, you have to plug n=0 and m=1,2,3,...

and this way you get all the naturals again

[grave wyvern]

hmmm

[grave wyvern]

so if its irrationals then N doesnt contain E

[slim leaf]

you mean E doesnt contain N?

[grave wyvern]

no its not E⊂N

[slim leaf]

N ⊂ E means E contains N

[grave wyvern]

oh yah

they got messed up in mi head

[slim leaf]

E contains irrationals, but it also contains the naturals

[grave wyvern]

hmmm

we have a rule in our book that confuses the living hell out of me

can i show you?

[slim leaf]

Lets say you fix m=0, you get 0 + 0sqrt(3), 0 + 1sqrt(3), 0 + 2sqrt(3), 0 + 3sqrt(3),....................

Then its time for m=1

[grave wyvern]

ah

[slim leaf]

1 + 0sqrt(3), 1 + 1sqrt(3), 1 + 2sqrt(3), 1 + 3sqrt(3),..................

Then it's time for m=2

[grave wyvern]

so if i were to answer the question

[slim leaf]

2 + 0sqrt(3), 2 + 1sqrt(3), 2 + 2sqrt(3), 2 + 3sqrt(3),....

[grave wyvern]

i would say fix n=0

[slim leaf]

You can see the first term in every sequence is in N

Yeah

[grave wyvern]

and every m belongs to N

so E contains N

[slim leaf]

Fix n=0, let m be any natural number. This way we get m + 0sqrt(3) = m.

That means m(any natural number) is in E.

[grave wyvern]

Here is what we have in our book

It confuses the hell outa me

[slim leaf]

YEah this is the same

[grave wyvern]

For all x belongs to E n belongs to A

[slim leaf]

To show N ⊂ E, you have to take any element in N, and show it's also in E

[grave wyvern]

So for all x belongs to N, x belongs to N?

[slim leaf]

that's what we've been doing tho

[grave wyvern]

Yeah it just confuses me

[slim leaf]

Take any natural number m from N.

[grave wyvern]

Lemme just write an answer and see if it's good

[slim leaf]

If we fix n=0 in E, then we get m + 0sqrt(3) = m in E.

Both ways are same actually

The justification I showed above by listing elements above is correct, dw

[grave wyvern]

I see

Something like this?

I wanna try to write it in the same manner as in the book

To guarantee that the teacher gives the point

[slim leaf]

Fix n=0, so we get m + 0sqrt(3) = m in E.

But since m is any natural number, that means N is in E.

[grave wyvern]

Okay

so A⊂B is A is part of E
so N⊂E means that N is part of E

it confuses me and i always think E⊂N is how its supposed to be lol

since is E is just a sub-set and N is the big guy

[slim leaf]

A ⊂ B means A is a part of B, yes

N ⊂ E means E is the big guy, N is a part of E

And you can see why E is the big guy

It contains all naturals + a lot of irrationals

[grave wyvern]

oh so E can have N+R right?

it can have all kinds of sets in it

[slim leaf]

Not really

Not every real number is in E

[grave wyvern]

oh wait irrationals

what set is that

i forgor 💀

[slim leaf]

pi, e, and so on

[grave wyvern]

Q?

[slim leaf]

Q is rationals

[grave wyvern]

Q' i guess

ok so i myself understand it but i want to write it in a way that's in the lesson

im trying to figure out how to turn it into the implication that i send in the picture of the lesson

[slim leaf]

Okay so first.

List some elements in E.

Can you do that?

[grave wyvern]

hmmm

[slim leaf]

Fix n=0, then put m=1,2,3,4..

[grave wyvern]

ah

[slim leaf]

Fix n=1, then put m=1,2,3,4

[grave wyvern]

1,2,3,4,5,6,7,8,9,10

[slim leaf]

Uh no

Look above

Like I did it

Do till n=4 atleast

[grave wyvern]

yes

then

[slim leaf]

Do it first

[grave wyvern]

okay

ok i did it

@slim leaf

[slim leaf]

What did u do

[grave wyvern]

i did it till n=4

[slim leaf]

can u show

[grave wyvern]

i might have screwed but one sec ill send a pic

That's what I understood

[slim leaf]

Yeah

so if you look at the top where its written 1,2,3,4

you can see that list will go on

Yeah?

[grave wyvern]

yessir

[slim leaf]

So now

To do it the "book way"

[grave wyvern]

yup

[slim leaf]

Choose any natural number.

Call it m.

[grave wyvern]

okay

[slim leaf]

Looking at our list, we can see that it will appear there somewhere

[grave wyvern]

indeed

[slim leaf]

Because the list 1,2,3,4,.. will go on.

And so thats it.

[grave wyvern]

OH YEAH

[slim leaf]

We took an arbitrary element m from N and showed that its in E.

Hence N ⊂ E.

[grave wyvern]

so N ⊂ E only if n=0

[slim leaf]

You dont need to say only if n=0

n=0 will happen

[grave wyvern]

ah

[slim leaf]

But yes, you get the naturals once you fix n=0

But its not a restriction, because we know n=0 will happen

[grave wyvern]

so can i write the book way and then show you?

[slim leaf]

Sure

[grave wyvern]

i couldnt do it lmfao

yeah i definetly understood it but couldnt get it to work with the book way

[slim leaf]

What I wrote basically

just that

cause u already did the working

of the list

[grave wyvern]

N⊂ E <=> (for all m belongs to N)

going on the right track?

[slim leaf]

Whats this?

[grave wyvern]

the book way-

[grave wyvern]

.

[slim leaf]

Yeah dont worry about that

All it tells us is that

If you take any element in N

It must also be in E.

And we did that.

[grave wyvern]

okay

[slim leaf]

We took an arbitrary element from N( call it m )

We listed the elements of E.

We saw that the list of elements in E (for n=0) contains every natural number(1,2,3,4,....) and irrationals(for n other than 0)as well

And so m must also be in that list.

That means m is in E.

[grave wyvern]

i see

[obtuse pebble]

@grave wyvern Has your question been resolved?

[obtuse pebble]

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[willow mantle]

How do I approach this question?

[willow mantle]

this is what i have done so far

A = (2, -4, 5)

Then, I found the direction vector AP = (0, 2, -4)

Then, I am stuck

I think i need one more point, B

so i can then find the normal

[tardy epoch]

https://math.stackexchange.com/questions/2606873/find-the-equation-of-a-plane-containing-a-point-and-passing-through-a-line

[willow mantle]

Thank you : )

.close

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[minor belfry]

why does this turn into (-x)

[knotty crow]

maybe you have restriction x < 0?

[minor belfry]

oop

ur right

fuck

ur a genius

ok ty ahahhahhaa

<3

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[minor belfry]

.reopen

[obtuse pebble]

✅

[minor belfry]

actually

@knotty crow also, when u have 2ln|x|

is that

ln|x|^2

or

ln|x^2|

[knotty crow]

|x^2| = |x|^2

[minor belfry]

oh i see

but it doesnt matter with ln?

[knotty crow]

it's same thing, ln|x^2| = ln|x|^2

[minor belfry]

ok easy

ty

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[silver plover]

how do i find max min sup inf of this

[solar trellis]

let f(x) = x - sinx

differentiate

[fossil crag]

Yes

[silver plover]

to find max?

f'(x) = 1 + cosx

[fossil crag]

It's not necessarily going to be a max. It's going to be an extremum (local)

I recommend that you draw the table of signs of f'

Then use this to draw the table of variations of f

[silver plover]

wdym table of signs

[novel knoll]

Huh? Not needed here

Notice anything about the derivative?

[silver plover]

has a range of 0 and 2

[novel knoll]

So what does that say about our function?

[silver plover]

has a max of 2?

[novel knoll]

Thats the derivative having max 2

[novel knoll]

What if I wrote this as f’(x)>=0

[novel knoll]

Can you say a property about f now?

[silver plover]

it also has a max of 2?

[novel knoll]

What?

[silver plover]

idk tbh

[novel knoll]

If I plug in x=4 is 4-sin(x) not greater than 2?

[silver plover]

yh idk

[silver plover]

it is

[novel knoll]

What does derivative tell us about a function?

[silver plover]

tells u the gradient of a function

[novel knoll]

Okay and we have that the gradiant is >=0 always

[silver plover]

ye

[novel knoll]

So our function is increasing?

[silver plover]

yes?

[novel knoll]

Can’t we use that

[silver plover]

so

does that mean that

max and sup dont exist?

[novel knoll]

If you don’t allow inf as a value

Then they don’t exist, yes

[silver plover]

so what would inf and min be then?

[novel knoll]

if its increasing

there is only gonna be one point such that its equal to 0

and that is gonna be min

[silver plover]

so inf and min = 0?

[novel knoll]

yes since 0-sin(0)=0

[silver plover]

so max = sup = dont exist

inf = min = 0

?

[novel knoll]

max and sup don't exist yes

and yes inf = min = 0

[silver plover]

hmm ok

gonna be honest

i found that quite hard

im a very visual student, so this logical stuff takes time to understand

[novel knoll]

(btw just because a function is increasing always doesnt mean limit is inf)

if thats why u said it

[silver plover]

yh that is why i said it

rip

[novel knoll]

consider say

,w plot arctan(x)

[warm shale]

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[novel knoll]

,w plot 1-1/x from 0 to 100

[warm shale]

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[novel knoll]

both are increasing but they approach a finite value

[silver plover]

ye

i seee

so how u know

for our question

that sup max dont exist

then

[novel knoll]

well easiest is just note that x-sin(x)>=x-1

and x-1 doesn't have sup

[silver plover]

where has x -1 come from

[novel knoll]

-1<=sin(x)<=1

[silver plover]

oh right

ok i think i sort of it get it now

[silver plover]

ive got another example question, would it be ok if we could go through that as well

got an exam monday on this so gotta make sure i get it

[novel knoll]

sure

[silver plover]

this scares me

lol

ive go a theory

[novel knoll]

so know how arctan(y) looks?

[silver plover]

for this

[novel knoll]

sure go ahead

[silver plover]

ye but only cuz u showed me above but otherwise i had forgotten

[silver plover]

whatever the max is for 2x - x^2 - 1

will it be the same for this whole thing

ie arctan (2x - x^2 - 1)

[novel knoll]

indeed

[silver plover]

nice

[silver plover]

max at (1,0)

@novel knoll

so max and sup of arctan (2x - x^2 - 1) is 0?

??

[obtuse pebble]

@silver plover Has your question been resolved?

[novel knoll]

yes

[silver plover]

Ok cool

Now what about inf and min

Now 2x - x ^2 - 1 doesn’t have a minimum

Does that mean that arctan (2x - x^2 - 1) doesn’t have a minimum as well?

Ie min and inf do not exist?

[obtuse pebble]

@silver plover Has your question been resolved?

[solar trellis]

limit of arctan x as x --> -infty is -pi/2

[obtuse pebble]

@silver plover Has your question been resolved?

[obtuse pebble]

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[coral marsh]

is this reasoning right?

[coral marsh]

(ab) | c

wait

/close

.close

[obtuse pebble]

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[opaque thistle]

could anyone explain number 18

[timber fox]

I would draw both the f(x) and f''(x) then go from there

[obtuse pebble]

@opaque thistle Has your question been resolved?

[opaque thistle]

how would you draw it

[obtuse pebble]

@opaque thistle Has your question been resolved?

[timber fox]

hmm

main rule is negative gradient = under the graph

that is going from f^(n)(x) -> f^(n+1)(x)

stationary points become roots

uhhh

yeah

[obtuse pebble]

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[unique forum]

Need help with Basic Geometry

[unique forum]

This is the problem

[timid silo]

use vertical angles

[unique forum]

Is it like this "5x-3=180"

.close

[obtuse pebble]

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[mental plaza]

vertically opposite angles brah

[unique forum]

.reopen

[obtuse pebble]

✅

[unique forum]

I have no clue how to do it 😭

[mental plaza]

u see the angles

in ur pic

see the angles opposite each other

[unique forum]

yea they are 180

[mental plaza]

opposite

what angle is opposite 5x-3

vertically opposite

which one?

[unique forum]

4x+9?

[mental plaza]

yes

nice

they’re equal

so 5x-3=4x+9

and u solve for x

same thing for y

[unique forum]

12y-3=6y+63

[mental plaza]

yes

[unique forum]

Is that the final answer or is there more steps

[mental plaza]

?

they want u to find the value

of x and y

so u have to solve

to find what x=

and what y=

[unique forum]

x is 11

i think

and y is 12

[mental plaza]

are u sure x is 11?

[unique forum]

yea

[mental plaza]

are u sure y is 12

show me ur work

how’d u get x as 11 and y as 12

[unique forum]

alr

x*

this is for y

[mental plaza]

yep

so y is 11 not 12

now do the same for x :)

[unique forum]

This is for x

[mental plaza]

yep!

nice

[unique forum]

is that the answer?

plz don't tell there more

[mental plaza]

yep that’s it

HAHAHAH

[unique forum]

Bytw I have a D in my geometry class and my all the grades are due toady : (

💀

[mental plaza]

atb !

[obtuse pebble]

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[indigo oriole]

does anyone know any resources to help with this question?

[indigo oriole]

or show me the working out, i don't know if I should add the two different youngs modulus of the two materials for finding the stress

[obtuse pebble]

@indigo oriole Has your question been resolved?

[obtuse pebble]

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[drifting badger]

Let Yasmin's age be x and Zayd's age be y. How can you describe Zayd's current age from the first sentence, in terms of Yasmin's age?

This but x instead of z

Yes! That's your first equation

Now, how can you describe the relation between Yasmin's age 10 years ago and Zayd's age 10 yers ago?

The first is Yasmin's age 10 years ago, yes, and the second is Zayd's 10 years ago

If you know Zayd was twice as old as Yasmin at this age, how can you describe this relation?

Okay, we'll do that

Ahh, yes, you're right

My bad. z = y + 4 is correct to describe Zayd's current age in terms of Yasmin's

We still have for the second equation that Yasmin's age 10 years ago is y - 10 and Zayd's is z - 10

You know Yasmin's age at this point in time is y - 10

And Zayd's is twice as big

Yes, that's what I meant by this point in time

Other way around

Si

Yes

[regal kindle]

take zaid age as x , then yasmin age x - 4 , 10 years ago , that is then zaid age was x - 10 and yasmin's age was x - 14 , 2(x - 10) = x - 14

[obtuse pebble]

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[strange whale]

Does anyone know when do you to ether common factor a negative number or positive number? when theres a negative number in the polynomial??

[obtuse pebble]

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[timid silo]

I don't get it

Is the question asking u to find the inputs and outputs using the formula?

[timid silo]

ye

Oh okay that's cool

Just solve for whatever you want basically

Do u have any other questions tho?

im confused

so for the first boxes it would be like

40 = (20) + 20?

im confused

Yeah that's fine to be confused

Basically what u said is correct

You can also take the 20 to the other side by subtracting both sides so u have w = 40 - 20

Just think of that question as

"If you add a number with 20 and it gives you 40, what is that number? "

oh

thank you

Ofc are u confused about anything else?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[rotund rune]

how would i approach this question?

[alpine raven]

huh what is the condition for the numbers in A... ? what is between 2022 and 2021 ?

[rotund rune]

oh wait sorry

it should be 2011, 2012, ..., 2021

hold on let me fix it

[alpine raven]

Well, A_2 has numbers that A_1 and A_3 dont have.
A_1 and A_3 have only one number in common.
Every subset must have a prime number..

I want to brute force this lol

[rotund rune]

how would you brute force it?

[alpine raven]

well there isnt a lot of numbers in A

[rotund rune]

oh btw, the options are:
A. 6561
B. 13122
C. 19683
D. 26244
E. 39366

[alpine raven]

Jesus Christ...

[obtuse pebble]

@rotund rune Has your question been resolved?

[spring seal]

Ohh, this doesn't sound too bad. You'll want to use the N choose K function

And think about giving some number of the years to A_2, and then of how many years are left, how can you give those to A_1 and A_3 such that it satisfies those conditions

The only tricky part are the last two conditions, however I think this shouldn't be too hard to account for

Essentially you can think of the 2nd to last condition as having an extra year to give out

And the final one should just require you to first consider giving out the prime numbers in A before all of the other numbers (don't forget that A_1 and A_2 could share their prime)

[rotund rune]

don't forget that A_1 and A_2 could share their prime
i thought that A_2 must not have intersections with other subsets?

[spring seal]

Whoops, A_1 and A_3

MB

[rotund rune]

Essentially you can think of the 2nd to last condition as having an extra year to give out
i dont quite understand this part

can you please elaborate?

[obtuse pebble]

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[rotund rune]

.reopen

[obtuse pebble]

✅

[spring seal]

Sure. So say we gave 3 of the years to A_2, then from condition 1 and condition 2, there would be 7 years left to spread between A_1 and A_3

Right

[rotund rune]

wouldnt it be 8 years left for A_1 and A_3, since |A| = 11?

[spring seal]

Whoops, yes, fuck counting 😆

And we can just about say that |A_1|+|A_3|=8

[rotund rune]

yes

[spring seal]

But because |A_1| and |A_3| has to share an element, it is really that |A_1|+|A_3|=9

[rotund rune]

yes

[spring seal]

And so rather than thinking of dividing the remaining 8 years between A_1 and A_3 with one overlapping year, we can just think about it as diving 9 years between A_1 and A_2 with no overlap

[rotund rune]

wait let me process that

okay that makes sense

[spring seal]

Also, if you count the number of primes in A, you will see that the common element between A_1 and A_3 is a prime

Which simplifies the problem a little

[rotund rune]

yea i've noticed that earlier

let me try hold on

[spring seal]

👍

[spring seal]

I think this fact can mostly be ignored since we've realised that the shared year must be one of two primes

[obtuse pebble]

@rotund rune Has your question been resolved?

[spring seal]

How is it going?

[rotund rune]

i still dont have an idea...
the subsets
A_1 = {2014, 2015, 2016, 2017}
A_2 = {2011, 2012, 2013]
A_3 = {2017, 2018, 2019, 2020, 2021}
will work but where do i go from this point

[spring seal]

Ok, I think you're going about it from the wrong way. For this sort of problem, I find it easiest to think about the sizes of these sets

So, we know that A has 2 primes

[rotund rune]

yes

[spring seal]

As so because A_2 must have one of these primes, A_1 and A_3 must share the other one

And so after this, we have 9 years to distribute between the 3 sets

Right?

[rotund rune]

yes

[spring seal]

Furthermore, we can say that because A_1 and A_3 already share one number (the prime), these 9 other numbers must be distributed among these 3 sets with no overlap, right?

[rotund rune]

yes

[spring seal]

Ok, cool

[rotund rune]

i'm following to this point

[spring seal]

So now we're going to work sequentially distributing these 9 numbers between these sets. So first, (simple question) what are the bounds on the number of these 9 years we can give to A_1?

As in, if we just ignore that A_1 has a prime in it, what are the bounds on |A_1|?

[rotund rune]

im sorry but i dont quite understand what you mean by 'bounds'

[spring seal]

Like, ... <= |A_1| <= ...?

[rotund rune]

oh

[spring seal]

Let's suppose that A_2 and A_3 only have the prime in there sets so far

[rotund rune]

1 <= |A_1| <= 8 i think?

[spring seal]

Almost, upper limit of 9 because we've already given out 2 of the 11 numbers (the primes)

[rotund rune]

but won't that causes intersection with A_2?

oh wait

nvm

[spring seal]

Ok, hang on

Let's just ignore the primes for now

They're easy to distribute

So if we ignore the primes in A, we have 0 <= |A_1| <= 9

right

[rotund rune]

yes

[spring seal]

Cool, so now suppose that |A_1| = n

Then how many non-prime numbers are left in A to distribute?

[rotund rune]

9-n

[spring seal]

Cool

Now we're distributing 9-n numbers between A_2 and A_3

So again, ignoring the primes, what are the bounds on |A_2| assuming |A_3| has nothing except the prime in it?

[rotund rune]

0 <= |A_2| <= 9-n

[spring seal]

Perfect

So now suppose that |A_2| = m

The since we have to distribute all of the non-primes between A_1, A_2, and A_3, how many non-primes must be in A_3

[rotund rune]

9 - (n + m)

[spring seal]

Eyy, awesome

Yep

So now, if we span across the bounds on n and m, we have essentially spanned all possible combinations of |A_1|, |A_2| and |A_3|, right

[rotund rune]

yes

[spring seal]

Cool, so we now just need to determine how many ways there are to pick n years from the non-primes in A and m years from the remaining non-primes (there's no choice as to what goes into A_3 as this is just what's left)

But we know how to do that, right?

[rotund rune]

so that would be $\binom{9}{n}+\binom{9-n}{m}$

[warm shale]

nichoals

[spring seal]

Umm, just about. Notice that we'll want to take their product and not sum

[rotund rune]

oh right
i forgot about multiplication rule

[spring seal]

Yep

[rotund rune]

$\binom{9}{n}\times\binom{9-n}{m}$

[warm shale]

nichoals

[spring seal]

And so we just want to do that for all n and m within our bounds

So a nested double sum

[rotund rune]

aahh i see

[spring seal]

And don't forget about about the 2 ways to distribute to two primes

Either 2011 to A_2 and 2017 to A_1 and A_3 or the other way around

[rotund rune]

that's essentially 2C1 right?

[rotund rune]

and that will be multiplied to this result?

[spring seal]

Yep

[spring seal]

Yeah, think about it and it should be pretty clear as to why this is the case

[rotund rune]

let me find it

[spring seal]

^ you should be able to do the double sum with a graphics calculator or desmos

[rotund rune]

wait i havent learnt double sum yet

how would i set it up?

[spring seal]

You know a regular summation, right?

[rotund rune]

yes

very familiar with it

[spring seal]

Cool, it's just a sum in a sum

so like $$\sum(\sum ...)$$

[warm shale]

Lachlan

[spring seal]

With bounds ofc

Note, the inner sum will have bounds dependent on the outer sum

[rotund rune]

ah i see

is this correct?

[spring seal]

Mmm, close

rn you're saying that 1<=n<=9 and 0<=m<=n

Which isn't correct

refer back to earlier

[rotund rune]

ah

[spring seal]

Also, use nCr() in desmos

nCr(n, k)

[rotund rune]

is this correct?

oh, and just one curious question, does the order of the arguments in a double sum matter?

[spring seal]

Why does n start at 1?

[spring seal]

WDYM?

[rotund rune]

oh im terribly sorry

[spring seal]

Haha, nope, it's all good

[rotund rune]

like, if i put nCr(9-n,m) before nCr(9,n) here, will it change the value?

[rotund rune]

here* i mean

[spring seal]

Nope, it's all inside the inner sum

[rotund rune]

ah okay

i got it now
it's 19683 x 2 = 39366 (E)

[spring seal]

Multiply by 2!

[rotund rune]

thanks a bunch

[spring seal]

Yep

Awesome

[rotund rune]

thank youu

.close

[obtuse pebble]

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[spring seal]

No problem!

GL

[obtuse pebble]

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[olive oracle]

How do you evaluate the definite integral given the integral of a function f(x)?

I have attempted 2(7)+1 which would give 15x and then I did (15(9))-(15(3))= 90, and this was incorrect.

I've thought of working backwards to find f(x), but I don't know how to do that. I've also thought it might be a transformation, but it didn't seem to meet the definition of an integral transformation.

I'd like some tips on how to evaluate such an integral please.

[royal basin]

@olive oracle it's $2\int_3^9 f(x)\dd{x} + \int_3^9 1 \dd{x}$

[warm shale]

Ann

[royal basin]

and the integral of 1 dx from 3 to 9 is 6, not 1

2*7+6 is what you would be looking for

[olive oracle]

ohh

so it was sum rule okay

[royal basin]

also it is not possible to find f(x) anyway - there are many many functions whose integral from 3 to 9 equals 7

[olive oracle]

that is what i thought too

okay thanks

.close

[obtuse pebble]

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[worldly parrot]

hi. can someone help me with this question?

it's about quadratic equation

[obtuse pebble]

@worldly parrot Has your question been resolved?

[royal basin]

@worldly parrot alpha*beta = p/2 and alpha+beta = -3/2

square both sides in the latter

[obtuse pebble]

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[dreamy canyon]

How do I write 83521 in a base 17

[dreamy canyon]

I know it’s 17^4 but that’s from searching it up

How do I get there?

[kind hawk]

well first step for writing anything in base 17 would be to compute 17^2, 17^3, 17^4, ... until you are bigger than your number

alternatively you can divide by 17 over and over again and the remainder at each step is the next digit, starting at the right

[stable rain]

check for division of all primes from the smallest?

lol

ull get 17 is the first n lalala

[drifting wraith]

you divide by 17

i don't get the first or the third idea

[drifting wraith]

it's just this

[obtuse pebble]

@dreamy canyon Has your question been resolved?

[obtuse pebble]

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[dreamy canyon]

I don’t get this one

.reopen

[obtuse pebble]

✅

[dreamy canyon]

Wdym next digit starting at the right

[kind hawk]

lets do a smaller example without so many 0 remainders.

lets write 431 in base 17

we divide 431 by 17 and get 431=25*17+6. so remainder 6

next we divide 25 by 17 and get 25=1*17+8, so remainder 8

and finally we divide 1 by 17 and get 1=0*17+1, so remainder 1

and now we put all these remainders together and get 186

and indeed 431 in base 17 is 186

[obtuse pebble]

@dreamy canyon Has your question been resolved?

[obtuse pebble]

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[quasi current]

How can I prove this statement? Currently got this done

[sage geode]

I'd expand the right hand side

[quasi current]

ahhhhhh

that would help

[sage geode]

How'd you get cos^2(theta * phi) though?

[quasi current]

product of two cos is cos^2 no?

[sage geode]

If they have the same argument

cosx * cosx = cos^2x
but cosx * cosy is just cosx * cosy

[quasi current]

ah K

ah is it just this loool

thank you thank you ❤️

.close

[obtuse pebble]

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[timid silo]

I have this problem:
$$\sin x = \frac{4}{5}, what's \tan x?$$

[warm shale]

smoobles

[timid silo]

how do I solve?

[tawny fog]

Do you know 3,4,5, triangle

[timid silo]

I was thinking that $\sin 2/\pi = 1$ so it has to be $2/\pi - y$ or something

[warm shale]

smoobles

[timid silo]

nope

[tawny fog]

Ok

So you have to visualise it in a triangle

Sin 53° = 4/5

And 5^2=4^2 + B^2

B=3

Tan x = 4/3

By Pythagoras

It is lot easier than you think

It will help you to visualise

In physics we use this ∆ in lot of questions so I know everything of it

[timid silo]

ah I see

thank you

.close

[obtuse pebble]

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[tawny fog]

,w plot x^2 + {y -3/4(x^2)^(1/3)}^2 = 1

[warm shale]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by 𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛#6985

[obtuse pebble]

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[timid silo]

$\hat{i}$ would just be the left column, and $\hat{j}$ the right column.

[warm shale]

3317

[timid silo]

You do know what a linear transformation is, right?

You get the two regular bases and transform them, transforming all coordinates.

We call them $\hat{i}$ for the $x$ basis and $\hat{j}$ for the $y$ basis.

[warm shale]

3317

[timid silo]

A base is a set of vectors that span all of said vector space.

Say you've got the regular 2d space, $\mathbb{R}^2$

[warm shale]

3317

[timid silo]

The bases are $\hat{i}=[1,0]$ and $\hat{j}=[0,1]$.

[warm shale]

3317

[timid silo]

You can write any point in the format $a\hat{i}+b\hat{j}$.

[warm shale]

3317

[timid silo]

Where $a$ and $b$ are real numbers.

[warm shale]

3317

[timid silo]

A transformation is something that transforms those bases.

Such that there's a mapping in between the regular points and the points after the transformation.

For problem b, for instance, it asks you to draw the image of the triangle $ABC$ after the transformation. This means you should take all the coordinates of $A$, $B$ and $C$ and transform their $x$ and $y$ values according to the transformation matrix passed.

[warm shale]

3317

[obtuse pebble]

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[toxic hollow]

If a polynomial with fractional exponent has solutions x_1,x_2….

[toxic hollow]

Does there exist a polynomial with integer exponents that has the same solutions?

[spice citrus]

yes, in general: given a finite number of solutions x_1, x_2 etc.. there will exist a polynomial with all those solutions

y = (x - x_1) * (x - x_2) * ....

[high lily]

polynomial with fractional exponent
uh...

[toxic hollow]

Yes I know polynomials can’t have fractional exponents but I couldn’t think of a way to word it better haha

[toxic hollow]

Thanks

[high lily]

function?

[limber elk]

whats 23 x 3?

i cant understand

[kind hawk]

haha 69 funny. don't troll in help channels. especially if its not even your own channel

[limber elk]

ok

[obtuse pebble]

@toxic hollow Has your question been resolved?

[obtuse pebble]

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[real acorn]

<@&286206848099549185> I sent this yesterday but didn't get a solution

(that why i tag)

[warm relic]

The answer is written right there

[warm shale]

nichoals

[obtuse pebble]

@real acorn Has your question been resolved?

[obtuse pebble]

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[glad stag]

Hello, is there any way to convert by algebra manipulation x^0.25 to something to the form x divided by something? Like x/a

[kind hawk]

no

$x^{0.25}=\sqrt[4]{x}$