help-49

[dreamy lichen]

the blue plane is plane x = 0

so anything with x coordinate 0 lies in it

its the green thing, i chose a bad color so its not very visible

you can see that the green vector is perpendicular tho

[molten bay]

Yes you are right

[molten bay]

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[twilit field]

So I'm trying to prove Monotone convergence theorem but I seem to have run into a problem.
\
Let M be the supremum of the sequence( which exists as it's bounded) and L the Infimum.
\
Let the sequence be increasing
\
Then $a_n≤a_{n+1}
\
$\forall n \in \N, L≤a_n≤M \implies L-M≤a_n-M≤0 \implies M-L≥a_n-M≥0$
\
So
\
$\abs{L-M}≥\abs{a_n-M} \implies 0≥ \abs{a_n-M}-\abs{L-M}$
\
Applying the reverse triangle inequality
\
$\abs{\abs{a_n-M}-\abs{L-M}}≤ \abs{a_n-L}≤0 \implies a_n \to L$

[twilit field]

So I'm trying to prove Monotone convergence theorem but I seem to have run into a problem.
\
Let M be the supremum of the sequence( which exists as it's bounded) and I the Infimum.
\
Then
\
$\forall n \in \N, I≤a_n≤M \implies L-M≤a_n-M≤0 \implies M-L≥a_n-M≥0$
\
So
\
$\abs{L-M}≥\abs{a_n-M} \implies 0≥ \abs{a_n-M}-\abs{L-M}$
\
Applying the reverse triangle inequality
\
$\abs{\abs{a_n-M}-\abs{L-M}}≤\abs{a_n-L}≤0 \implies a_n \to L$

[grand pond]

wai

[twilit field]

I should be L

[dawn dagger]

There is a typo in the 2. implication

it should be M-a_n

[twilit field]

ah yes

but abs sorts that out

[dreamy lichen]

|an - L| <= 0 is pretty sus

i dont get the chain of implications

are you just randomly throwing in absolute values?

[dawn dagger]

me too

[lyric charm]

wai is a known typoist 😔

[twilit field]

Yea, but that's what the reverse triangle inequality gives, right?

[dreamy lichen]

how did you even obtain it though?

It seems to me like you're just randomly throwing in absolute values wherever you want

which just doesnt work

[twilit field]

M-L≥a_n-M≥0

[dreamy lichen]

L is the lower bound on an btw?

[twilit field]

yea, that's my concern too

[twilit field]

then applying the mod to all sides

[dreamy lichen]

1 > -2

applying mod to all sides...

1 > 2

[twilit field]

it's all greater than 0

[dawn dagger]

yep

[dreamy lichen]

okay, then sure

[dawn dagger]

but it should be M-a_n

[twilit field]

|M-L|≥|M-a_n|

ao 0≥|M-a_n| - |M-L|

I think my mistake is using the reverse triangle inequality here

[dreamy lichen]

your mistake is applying modulus on both sides

if i understand it correctly

[twilit field]

yea

I get that now

[dreamy lichen]

0 >= -1
|0| >= |-1|

now it doesnt work anymore

[twilit field]

We could multiply across by -1 again to get something that makes sense, but I doubt that will gives us anything useful

[dreamy lichen]

what inequality are you even trying to achieve?

[twilit field]

|a_n-M|<0

[dreamy lichen]

or are you just pointlessly manipulating that inequaltiy hoping you will get something useful?

[dreamy lichen]

do you think this can be true?

[desert siren]

Bro trynna prove modulus can be less than 0

[dreamy lichen]

even if it was <=, do you think it could be true?

[twilit field]

Not really, that would mean it's a constant sequence

[dreamy lichen]

Yeah, if it was <=

so seems like you won't have much luck with that

try something else than just manipulating inequalities

[twilit field]

okay, thanks

.close

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[viral dagger]

let A and B be disjoint sets and that the union of A and B is the natural number set, prove there exists infinite distinct pairs (x,y) such that {x, y, x+y} are all in either A or B

[viral dagger]

well im guessing theres some sort of proof by contradiction

[oak nymph]

They do though , every pair in 2Nx2N does the job

[rare maple]

I will delete all my messages

Sorry for the trouble

[runic hamlet]

does the original wording say anything about x=y ?

otherwise thats where I would start probably

x in A, 2x in B, 4x in A, 5x and 8x in B and so on, maybe contradiction

[viral dagger]

oh my bad i missed that

[dreamy lichen]

so x must be distinct from y or not?

[viral dagger]

yeah

[dreamy lichen]

Suppose 1 is in A and take some n > 2 in A, then play with it for a while

derive as much as you can

and perhaps do some casework on other low naturals

[viral dagger]

ooh wait

if 1 is in A then if m is in A then m+1 is in B

so either A is odd and B is even, or A is even and icl 1, and B is odd exl 1, or viceversa

[dreamy lichen]

there is much more you can derive

[viral dagger]

the one with the evens can always have infinite of the x,y,x+y

[dreamy lichen]

not necessarily

[viral dagger]

oh why not?

[dreamy lichen]

if there is A, then it follows but B

but B doesnt have to follow by A

you can have ABBBBBBBBABBBBBABABBB

[dreamy lichen]

but you can take this further

there are more numbers for which you can know the group

[viral dagger]

oh

[dreamy lichen]

if you find out 5 more, you will have a contradiction

you are currently here

you can do more

[viral dagger]

if 2 is in A, then 3 is in B

uhh

[dreamy lichen]

thats true

but you can do this without casework

look at n-1

what can it be?

[viral dagger]

B?

[dreamy lichen]

Yeah, it can be B

can it be A?

[viral dagger]

oh so 2 mus tbe in A

[dreamy lichen]

Yep

[dreamy lichen]

and it cant be A, because we would have {1, n-1, n} all in A

[dreamy lichen]

oh we are currently proving kind of weaker version of the statement, we are only proving existence of any (not necessarily infinitely many) such pairs

but i think we should be able to fix that later

since n is arbitrary, we can make it arbitrarily large

[viral dagger]

hm

[dreamy lichen]

I'd say dont worry about it for now (it can be fixed later by making n sufficiently large, much larger than any of the finitely many pairs which dont have the property)

[dreamy lichen]

any other number whose group you can determine?

[subtle blaze]

Is this supposed to be (… are all in A) or (… are all in B)?

[dreamy lichen]

yes

[runic hamlet]

would be quite boring otherwise

[subtle blaze]

Yeah

[viral dagger]

3?

[dreamy lichen]

okay, what must 3 be?

[viral dagger]

in B

[dreamy lichen]

i cant phrase what i mean

i mean so that {1, 2, 3} arent all in A

[viral dagger]

n+2 and n-2 must be in B aswell, but 3,n-2,n+1 is in B

like this?

[dreamy lichen]

how did you get n-2 in B?

[viral dagger]

2,n-2,n

[dreamy lichen]

oh, interesting

right

so basically everything around n must be B

[dreamy lichen]

now it might be time to fix this

So the issue is we are assuming that there are no pairs without the property

while the negation of there being infinitely many such pairs is that there are at most finitely many pairs without the property

so we should actually be assuming that there are at most finitely many pairs (x, y) with the property that {x, y, x + y} is either fully in A or fully in B

is that clear?

we can kind of get around that by choosing our n to be significantly larger than anything in those finitely many pairs

most of what we derived will then stay true, except for 3 in B

because (1, 2, 3) might be one of those finitely many pairs with exception

idk if what i wrote makes any sense, if it doesnt make sense, please just tell me it doesnt make sense lol

[viral dagger]

so basically just casework the fact that weather 3 is in A or not?

[dreamy lichen]

you can find out what 3 must belong to

but use the things with n instead

not 1 and 2

use those numbers that are large enough to not have an exception

its similar to how we found out 2 is in A

idk if you noticed it already, but we are basically doing the same thing over and over again to find out which numbers belong to which group

once you spot the pattern, youll be able to generalize and finally conclude the proof

we found out that 2 is in A, because if it was in B, {2, n-1, n+1} would all be in B

how can we find out where does 3 belong

where does 4 belong

...

and where does n+4 belong, n+5, n+6...

what we have is enough to determine all naturals (even with some contradictions)

[viral dagger]

actually cant you use the fact that both A and B must be infinite to get that there exists infinite elements (which also means infinite n) and we determined that if there exists n then there exists the (x,y) pair

[subtle blaze]

They have to be infinite yes

Otherwise one of them has a max

[dreamy lichen]

t if there exists n then there exists the (x,y) pair

have we done that already?

[subtle blaze]

Then you start from the max for the other set and you’ll be fine for finding infinity many pairs

[viral dagger]

the (n-1,3,n+1) no?

[dreamy lichen]

3 doesnt necessarily have to be B though

3 can be A

[dreamy lichen]

and if it's A, there is no contradiction here

you can have {1, 2, 3} all in A

it can be one of those finitely many exceptions

[viral dagger]

oh damn right

[dreamy lichen]

3 actually must be A

because {n-1, 3, n+2}

and n-1, n+2 are in B

if we look at the current trend:

1, 2, 3,... all seem to be A

n is A

[viral dagger]

can we do the small natural number conclusion with 4 or is it the same problem with 3

[dreamy lichen]

and n+1, n+2, .... all seem to be B

[dreamy lichen]

same problem as with 3

[dreamy lichen]

you will need to prove these 2 trends

and then you will get a contradiction

[viral dagger]

err

like say every positive integer up to i-1 is in A, then if n+i-1 is in B, then n-1,i,n+i-1 is a pair, so we do casework and assume i is in A as if i is in B then we get that if there exists n then there is the x,y pair

[dreamy lichen]

That's the general idea which proves that i is in A, you will also need to prove that n+i-1 is in B

it should be kind of a zig-zag thing

prove that n+i-1 is in B, prove that i is in A, prove that n+i is in B....

[viral dagger]

we assumed that i-1 is in A, so if n+i-1 is in A then i-1,n,n+i-1 is a pair

[dreamy lichen]

and the idea is that eventually, i will become n-1

[dreamy lichen]

oh wait, n+i-1 isnt in A tho

[viral dagger]

huh

[dreamy lichen]

oh your pair is wrong

oh no its right

[dreamy lichen]

the numbers above n are in B

so is n-1

and so if i was in B, it would be a pair

which it cant, because n was chosen to be much larger than everything in the finitely many pair-exceptions

if n+i-1 is in B, then i must be in A (lest {n-1, i, n+i-1} would be a pair)

similarly, if i is in A, n+i must be in B, (lest {n, i, n+i} would be a pair)

we start with: 1 ∈ A, n ∈ A

n-1 ∈ B, lest {1, n-1, n} would be a pair

n+1 ∈ B, lest {1, n, n+1} would be a pair

2 ∈ A, lest {2, n-1, n+1} would be a pair

n+2 ∈ B, lest {2, n, n+2} would be a pair

3 ∈ A, lest {3, n-1, n+2} would be a pair

n+3 ∈ B, lest {3, n, n+3} would be a pair

4 ∈ A, lest {4, n-1, n+3} would be a pair

...

n+i ∈ B, lest {i, n, n+i} would be a pair

i+1 ∈ A, lest {i+1, n-1, n+i} would be a pair

...

continue until i+1 = n-1

then you have i+1 ∈ A, but n-1 ∈ B

contradiction

this could be formalized by induction

ill leave that as an exercise

namely, you should be able to prove that for all m > n, m ∈ B

and for all m >= 1, m ∈ A

which is a pretty clear contradiction

[viral dagger]

oo ok

.solved alr ill continue it later, i have to go now sorry! thanks btw

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[subtle blaze]

I gave up too

[subtle blaze]

.close

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[worldly pine]

by putting x and y in options i got option c which is right...
but i wanna know the basic

[dreamy lichen]

what's the point P you got??

[dreamy lichen]

,w 7x+11y=3, 8x+y=15

[grand pond]

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[dreamy lichen]

c seems correct to me

what are you asking about?

[worldly pine]

yes it is correct..but i wanna know the basic method

without using options

[desert siren]

He guessed, he's asking how to do it properly

[dreamy lichen]

You have to use options here

there are many different lines which pass through point P

you have to select one from the options

[worldly pine]

oh

[dreamy lichen]

the way you did it is perfectly fine

[desert siren]

That's usually the way you have to do it if you have a point and random equations

No other way to determine

[worldly pine]

ok

thanks

.close

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[narrow edge]

why are there 7 choices

[narrow edge]

shouldnt it be 5 choices?

because as choice 3,2,4,6,5 are eliminated donly 5 remain

[dreamy lichen]

why 2, 4, 6?

You have only used 3 digits:

3, 5 and one of 2, 4, 6

so there are still 7 digits remaining

[narrow edge]

oh nvm

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[molten bay]

I didn't understand this line how they got 2π?

[lyric charm]

radius is y so circumference is 2πy

[midnight plank]

@molten bay Has your question been resolved?

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@molten bay Has your question been resolved?

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[woeful turret]

when deriving the relation for
tan inverse x + tan inverse y
A=tan inverse x and B = tan inverse y
now one of the cases is
-pi/2 < A+B< pi/2
but here we are not considering the lower limit..why is that?

[fallen sparrow]

First off

You can only write cot inverse y = tan inverse (1/y) when y is positive real

[woeful turret]

yeah

i got that

i didnt understand why he didnt consider A+B>-pi/2

or why its "guarenteed"

[fallen sparrow]

It's not guaranteed

[woeful turret]

oh then why dont we consider it

[fallen sparrow]

Hold up howd you even get A + B less than pi/2

It should've been pi no?

[woeful turret]

no this is just one of the cases

from tan inverse tan x graph

[fallen sparrow]

Hold up imma solve this a bit later I'm not at a comfortable place to be writing rn

[fallen sparrow]

Yeah I know that

[woeful turret]

oh u dont need to solve anything

i just wanted an explanation

but okay if ur busy then 👍

<@&286206848099549185>

[fallen sparrow]

I think sign of x and y would also matter here

Just as a side note

[woeful turret]

yeah

we divide it into sub cases...this one is for x>0 y>0

[fallen sparrow]

Ohh that makes sense

Bruh isn't it fairly obvious why A + B will always be > -pi/2 then

@woeful turret

[woeful turret]

uhh 😔 not to me

[fallen sparrow]

Aree

X > 0 implies tan inverse x is also > 0

[woeful turret]

OH LMAO

ok now i get it

ty

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[tidal turret]

i have some ideas for 1

,w {{0,2,3},{-1,1,2},{-1,0,0}} * {{2,1,0},{1,1,0},{0,1,1}}^(-1)

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[tidal turret]

,w {{0,2,3},{-1,1,2},{-1,0,0}} * {{2,1,0},{1,1,0},{0,1,1}}^(-1) * {{a},{0},{-1}}

[grand pond]

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[tidal turret]

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[vague marsh]

Why in this example we can accept the limit as 3? Doesn’t the graph have to both meet at 3 in order for it to be considered the limit, as it approaches -2 from both sides?

Or, wouldn’t we have to specify lim x --> 2^-?

[olive zephyr]

no

cause the discontinuity is at -2 lol

not 2

[vague marsh]

Oh wait yeah

.close

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[chilly cobalt]

ignore ts wrong image

[chilly cobalt]

wrong img wait

blehh

is my proof correct

[midnight plank]

@chilly cobalt Has your question been resolved?

[chilly cobalt]

<@&286206848099549185>

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@chilly cobalt Has your question been resolved?

[trim ether]

your proof is absolutely solid

[midnight plank]

@chilly cobalt Has your question been resolved?

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[neat silo]

how many ways to arrange 1, 2, 3, ..., 7 in a line such that no two adjacent sum is prime

[neat silo]

its impossible isnt it?

[gleaming latch]

2 4 6 3 1 5 7

[runic hamlet]

draw all numbers and then draw connections between them if you can write them next to each other

[neat silo]

oops

[neat silo]

yea i was also thinking about this

[runic hamlet]

it does help to realize that you can just put all odds next to each other and all evens

so you just need one hop between odd and even

[neat silo]

1, 2, 4 has 3 connections
3, 5, 6 has 2 connections
7 has 1

[neat silo]

ooh ok but how do i count the number of ways?

[runic hamlet]

well thats wrong

[neat silo]

oops

[lyric charm]

so ig we are counting the number of hamiltonian cycles in this graph?

[runic hamlet]

you can only switch between odd and even with 2-7, 3-6 and 4-5

some annoying casework should probably give you the result

but eww

[neat silo]

ah ok

so i just put 2-7, 3-6, and 4-5 in the middle and do casework for the rest?

[runic hamlet]

you also gotta do casework on the number of times you switch between odd and even

[neat silo]

ok

.close

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[dusky island]

How would I put this in the calculator? I don’t know how to if there is 2 unknowns

[gleaming latch]

,rccw

[gleaming latch]

what calculator model are you using

[dusky island]

Casio classwiz

would you know how to do it?

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[signal rock]

For part iii), is it possible to just multiply the result yielded from ii) by pi?, or should I do the normal way i.e pi integral y^2 dx way

[pine wave]

yeah you need to do it the normal way

different heights from the x axis contribute different amounts of volume

[signal rock]

ight

ima leave this open coz ill try work out the area and confirm if its correct

[signal rock]

i may be wrong but i got 0.89pi

[pine wave]

maybe show your working?

[signal rock]

yep

only the blue

[pine wave]

hmm, looks good, lemme do a quick and dirty estimate and a calculation

$2^3$

[grand pond]

Element118

[pine wave]

\calc 2^3

,calc 2^3

[grand pond]

Result:

8

[pine wave]

,calc 8/3+2-2^(3/2)/3-2^(3/2)

[grand pond]

Result:

0.89543050033841

[pine wave]

okay that kinda rounds better to 0.90

,calc ln(2)

[grand pond]

The following error occured while calculating:
Error: Undefined function ln

[pine wave]

,calc log(2)

[grand pond]

Result:

0.69314718055995

[pine wave]

,calc 2pisqrt(2)*(1-log(2))

[grand pond]

Result:

2.726622312032

[pine wave]

,calc 0.89543050033841*pi

[grand pond]

Result:

2.8130778816634

[pine wave]

yeah checks out

sanity check worked, integration looks fine, calculation looks fine

[signal rock]

ight

[pine wave]

(this is kinda the multiply by pi trick, but yeah, it can't provide the exact answer, as you seen)

[signal rock]

ight thanks

.close

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[ionic magnet]

The derivative of the linear function y = f(x) is 8, the graph of the function y = f(x) intersects the y-axis at a point with an ordinate of 4. Calculate the value of f(-1,2).

[languid mica]

If the derivative is 8 AxEDf

Then f'(x)=8 => f(x) = 8x +c

By integration

And you use f(0)= 4 to find that c =4

So f(x) =8x+4

But idk what you mean f(-1,2)?

[ionic magnet]

what

what does mean AxEDf

[languid mica]

For every x in Df

[ionic magnet]

f(0)

[languid mica]

it says the line cuts y axis at 4

So f(0) =4 <=> c = 4

[ionic magnet]

substitute into a function

[flat spire]

u mean -1.2?

[ionic magnet]

4= 8*(-1,2) +b

right?

[languid mica]

Oh

-1.2?

[ionic magnet]

f(0)= 8*0 +b

[languid mica]

Alr

Yea substitute

Oh wait it says its linear

Dont need integration

[ionic magnet]

yes

[languid mica]

Yea so b = 4

And 1.2 times 8 is 9.6 so its 13.6

Oh -1.2 wait no

Its -5.6

[ionic magnet]

yea

thank you

i have exam tomorrow

[languid mica]

Good luck

[ionic magnet]

that task was on previous days

[ionic magnet]

thank you man

can u say me any recommendation for derivative

like this

[languid mica]

Wdym

What level are you on?

[midnight plank]

@ionic magnet Has your question been resolved?

[ionic magnet]

how i can check my level

[midnight plank]

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[wanton glacier]

For (b), do i use s=0.0198 or sigma =0.0199 for bar(X)?

[wanton glacier]

Wrong picture sry

.close

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[unique beacon]

how do i do this

[midnight plank]

@unique beacon Has your question been resolved?

[unique beacon]

yeah whatever i got it

.close

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[tawdry laurel]

What is this?

[blissful sigil]

what is bro on about

[midnight snow]

<@&268886789983436800>

[tawdry laurel]

i thought bro was asking a finance question for some reason lmao

[pearl hull]

.close

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[pearl hull]

Not in every channel LOL

[tawdry laurel]

Wait he did?

[midnight snow]

ye

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[latent wadi]

!redir

[midnight plank]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[frosty ice]

<@&268886789983436800>

[midnight plank]

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[fair frigate]

Jennifer and Muhammad are training for a marathon. If Jennifer runs 2 miles per hour faster than Muhammad, and one training day she ran 16 miles in the same time Muhammad ran 12 miles, how fast was each person running? Hint: recall the science formula d=rt.

[fair frigate]

<@&286206848099549185>

[midnight plank]

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

[fair frigate]

bruh

!status

[midnight plank]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[fair frigate]

1

[dire hornet]

rearrange to formula to get speed

since distance=speed x time

speed=distance/time

leave that on the side

calculate the difference in distance

@fair frigate are you getting it?

[fair frigate]

yea

[dire hornet]

do you get why we calculate the difference in distance?

[fair frigate]

becuase we need to find the total distance

in time

like time over distance to figure spped

speed

[dire hornet]

it is to find the time

since jennifer runs 2 miles faster per hour

with that information and the 2 distances we can find the time by finding the difference and dividing it by 2 to find the amount of hours

@fair frigate have you gotten the time now?

with the time (if you have it) find the speed now

[fair frigate]

i mean if we have 16 minles and 12 miles sicne we dont know hours but there is a 2 miles increase per hours so y=x+2 so show differnces

right

[dire hornet]

yes

[fair frigate]

for hours can we just subticue till 4 miles a hour

and go up till 8

just do this with a table

?

[dire hornet]

no

since it is 2 miles faster per hor

hour*

jennifer's per hour speed will be if we take muhammad's speed as x

x+2

with the information of 2 per hour

(y-x)/2=time

[fair frigate]

yea

8 and 6 could work as valuse no

sincr 8-6=2

and 8 and 6 *2 =16 and 12

no

is that the answer

[dire hornet]

yes

i was misunderstanding your work mb

yes

8 and 6 are the answers

but put the units

miles/hr

[fair frigate]

yea 8 miles per hour

and 6 per hour

thanks

[dire hornet]

yes

[fair frigate]

.close

[midnight plank]

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[dire hornet]

np

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[fair frigate]

why is c right

[fair frigate]

this is what my teacher got

<@&286206848099549185>

[fallow scarab]

!15m

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

[tame fox]

What had you done when attempting this problem?
What part of your teacher's answer doesnt make sense?

[fair frigate]

i didnt get how to favotr higher temrs

terms

!status

[midnight plank]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[fair frigate]

1

[tame fox]

Okay
Lets start first with figuring out a factor of the polynomial
Since we dont want to just guess and pray we get lucky on finding f(x)=0
We have this method called rational root theorem
If p are the factors of the coefficient of the highest powered term, and q are the factors of the constant, then + or - q/p will eventually provide a solution to the the function

Are you with me here so far?

[fair frigate]

yes

@tame fox

[tame fox]

Cool
So now that we have a solution, here being x = -2
We factor out x+2 by long division or through synthetic
Which ever is more comfortable
Just remember that synthetic uses the solution, not the factor, when running through the method

Do you understand how to do both?

[fair frigate]

i know long but not with variales

and terms

[tame fox]

It follows the same process as numerical long division. each term is similar to a digit so to speak
You just match whats needed and when you subtract it fixes itself

I can write it out for you if you're able to wait a couple minutes

[fair frigate]

yea sure

[midnight plank]

@fair frigate Has your question been resolved?

[tame fox]

I'm having trouble uploading it from my mobile
Give me a few more minutes to try something else

i think discord is having issues in uploading files. Even from pc i can't and a friend wasn't able to send me a picture either

[sudden yacht]

Yeah it's a problem of the whole Discord

[tame fox]

hopefully it comes back soon

anyways, @fair frigate until discord figures itself out, we can continue with the problem and i'll upload the images later on

[midnight plank]

@fair frigate Has your question been resolved?

[fair frigate]

okay ig

[tame fox]

I dont think theres any voice chats otherwise i'd offer to show you there
And it still wont upload files. I just checked

[fair frigate]

oh okay but can we still go with the problem

?

[tame fox]

Anyways, aside from dividing polynomials, are you okay with the steps taken so far?

[fair frigate]

yea

[tame fox]

Cool
So far then you have factored the expression to (x+2)(6x³+23x²-4x-4)
So we repeat the proccess again again where we pick out factors p and q from 6 and 4
You'll get x = -2 again and you repeat either the long division or synthetic

Which ever of your chosing. You'll get to the same place

[fair frigate]

and then facotr the second term

but how got (x+2)

?

[tame fox]

Ah sending pictures work again finally

[tame fox]

There is no (x - 2)
Where do you see that?

Oh ok

[fair frigate]

is that syenthic or long

is that is sytneic then i am stupid

as hell

[tame fox]

X + 2 is from x = -2
You make it equal to 0

[fair frigate]

nah not that

[tame fox]

Its long division

[fair frigate]

yea thanks

what if we got a remider how to write that

[tame fox]

For factoring you dont need to worry about a remainder. If you get one then it wasnt a solution to beging with

But depending on what the instructions are you can just say r. ___ or write it as a fraction (remainder)/(divisor)

[fair frigate]

oh okay

[tame fox]

Of course
Now if the problem hadnt already told you what each of the factors were, you'd continue factoring the quadratic using methods you should already be familiar with
Since we got to the quadratic expression by this point the problem is finished

[midnight plank]

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[midnight plank]

@fair frigate Has your question been resolved?

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[short hill]

it is not base 😔 can someone explain

[short hill]

ping if respond

[midnight plank]

@short hill Has your question been resolved?

[dawn dagger]

If you rotate around the y-axis then you measure from the origin up until the left edge of the rectangle.

[short hill]

ah i see. so it should be left edge

[dawn dagger]

Yes, that's what I think.

[short hill]

ok. if it is wrong i will enter your dreams and stuff your face until you're as fat as my cat <3

[dawn dagger]

Maybe think about what I wrote, rather than depend on my opinion.

[short hill]

.close thx it was center

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[tidal turret]

need help with 2

[dire hornet]

translate the question

[tidal turret]

just says express all z in C in binomic form

[gaunt nimbus]

cant u

define ${z = a + bi}$

[grand pond]

k

[tidal turret]

?

[gaunt nimbus]

ye

that works

[gaunt nimbus]

define ${z = a + bi}$ for some ${a,b \in \mathbb{R}}$

[grand pond]

k

[gaunt nimbus]

then use that definition

distribute

and compare coefficient

note also that ${z \bar{z} = a^2 + b^2}$

[grand pond]

k

[tidal turret]

whats your point?

[gaunt nimbus]

like

${z(\bar{z} - 2)} = z\bar{z} - 2z = (a^2 + b^2) - 2(a + bi)$

[grand pond]

k

[gaunt nimbus]

then add up like terms

and compare coefficient

find a and b

to find z

[tidal turret]

zz* = |z|^2

[gaunt nimbus]

yup

and ${|z|^2 = |\sqrt{a^2 + b^2}|^2 = a^2 + b^2}$

[grand pond]

k

[tidal turret]

so?

[gaunt nimbus]

we use that in here

[tidal turret]

what is the benefit of it

[gaunt nimbus]

so that its easy to compare coefficients

[tidal turret]

maybe is better if we use polar?

[gaunt nimbus]

polar doesnt work well with additions and subtractions imo..

[tidal turret]

ok but I dont follow

[gaunt nimbus]

which part are u confused

abt

[tidal turret]

why

[tidal turret]

the why

[gaunt nimbus]

we are rewriting the left hand side

so that it is similar to the right hand side

[tidal turret]

ok

ok

ok

${z(\bar{z} - 2)} = (a^2 + b^2) - 2(a + bi) = (a^2 + b^2 -2a - 2bi$

[grand pond]

Renato

[gaunt nimbus]

yup

now separate into real and imaginary parts

[tidal turret]

(a^2 -2a) + (b^2 -2bi)

[gaunt nimbus]

b^2 is real

[tidal turret]

(a,b)=(Re(z),Im(z))

[gaunt nimbus]

not imaginary

[tidal turret]

(a^2 -2a+b^2) + (-2bi)

a^2 -2a + b^2 = 7
-2b = 4

@gaunt nimbus

[gaunt nimbus]

yup

[tidal turret]

@gaunt nimbus

b = -2

[gaunt nimbus]

yup

[tidal turret]

a^2 -2a = 7-4 = 3

a^2 -2a - 3 = 0

,w solve a^2 -2a - 3 = 0

[grand pond]

Results provided by WolframAlpha

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

Requested by renatoxbz

[tidal turret]

z1 = -1 -2i, z2 = 3 - 2i

@gaunt nimbus

[gaunt nimbus]

,w (-1 -2i)(-1 + 2i - 2)

[tidal turret]

?

[grand pond]

Results provided by WolframAlpha

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

Requested by nuclearinteraction

[gaunt nimbus]

,w (3 -2i)(3 + 2i - 2)

[tidal turret]

?

[grand pond]

Results provided by WolframAlpha

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

Requested by nuclearinteraction

[gaunt nimbus]

yup

correcto

[tidal turret]

how do I check

[gaunt nimbus]

expand back

[tidal turret]

im lazy

is there a quick way?

[gaunt nimbus]

nah i checked already

[tidal turret]

thanks i trust you my life k

❤️ 😈

[tidal turret]

thanks ^^

.close

[midnight plank]

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[neat silo]

how do i use holders inequality here?

[lyric charm]

hölder's inequality?

do you have a statement of it in front of you

[neat silo]

yea too lazy to type that o

[lyric charm]

were you instructed to use hölder here somehow?

[neat silo]

yea my teacher gave me a hint to use holder

[lyric charm]

like ok we want to minimize p(a)p(b)p(c)/(ab+bc+ca) over all (a,b,c) such that the denom is positive

hmmm

ok yeah no sorry i dont see it

but for real maybe you could at least show us what form the inequality itself was given to you or shown in class

[neat silo]

alr

$\left(a_{1} + a_{2} + \cdots + a_{n}\right)^{\lambda_{a}} \cdots \left(z_{1} + z_{2} + \cdots + z_{n}\right)^{\lambda_{z}} \geqslant a_{1}^{\lambda_{a}} b_{1}^{\lambda{b}} \cdots z_{1}^{\lambda_{z}} + \cdots + a_{n}^{\lambda_{a}} b_{n}^{\lambda{b}} \cdots z_{n}^{\lambda_{z}}$

[grand pond]

acgn

[neat silo]

holds for $\lambda_{a} + \lambda_{b} + \cdots + \lambda_{z} = 1$

[grand pond]

acgn

[lyric charm]

so your version of hölder's inequality requires a product of exactly 26 brackets?

[neat silo]

not really

[lyric charm]

why's it indexed with the entire latin alphabet from a to z then 😛

[neat silo]

doesnt have to be 26 brackets

[neat silo]

just to give the form lol

[lyric charm]

well presumably you'll want to have p(a), p(b) and p(c) act as the brackets on the left hand side.

somehow.

don't know with what exponents.

[viral dagger]

im betting its 3 of 1/3 so each of them are equal and still sum to 1

[midnight plank]

@neat silo Has your question been resolved?

[midnight plank]

@neat silo Has your question been resolved?

[pine wave]

maybe we should try to find the equality case first

knowing this equality case can eliminate a lot of steps which are too weak because they don't respect the equality case

if it were up to me i would basically take the partial derivative with respect to a, b, c and set it to 0 to see what candidate solutions i could find

That would be $\frac{p'(a)p(b)p(c)}{ab+bc+ca}-(b+c)\frac{p(a)p(b)p(c)}{(ab+bc+ca)^2}=0$

Simplifying, I get $p'(a)(ab+bc+ca)=(b+c)p(a)$ and... it's probably time to substitute $p$ in

$(4a^3-12a^2+14a-8)(ab+bc+ca)=(b+c)(a^4-4a^3+7a^2-8a+16)$

Of course, one way is hope $a=b=c>0$, in which case, we get one of the solutions as $a=b=c=2$, so if the questions is not broken, chances are the problem-setter set it as that.

[grand pond]

Element118

[neat silo]

it is a = b = c = 2 and k = 144

[neat silo]

is this like lagrange?

[midnight plank]

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[tidal turret]

can someone explain to me why the determinant is the volume or area of a square

[topaz epoch]

https://images.app.goo.gl/6rD6fmRSvv5821Mr8

[tidal turret]

wdym?

[topaz epoch]

just search up any determinant area geometric proof

[tidal turret]

?

can u explain

[barren trail]

this might help https://www.youtube.com/watch?v=Ip3X9LOh2dk

[midnight plank]

@tidal turret Has your question been resolved?

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[midnight plank]

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[neat silo]

Given a rectangle ABCD, then points P and Q are respectively on segment AB and segment BC. If the area of ​​triangles DCQ, BPQ, and ADP are respectively 6, 8, and 10, then the area of ​​triangle APQ

[neat silo]

Do i use the sine area formula?

like 1/2 a b sin theta

[lyric charm]

diagram?

@neat silo

[neat silo]

cant make an exact one

[runic hamlet]

no one cares about exact diagrams

[midnight plank]

@neat silo Has your question been resolved?

[rough geyser]

?

Can't you get the height through expanding the areas?

[neat silo]

nope dont think so

[rough geyser]

The bases of cq and bq would equate to da

[neat silo]

lemme try

[rough geyser]

Similarly ap and pb would be dc

[neat silo]

no progress

.close

[midnight plank]

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[midnight plank]

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[vague eagle]

i think ur confusing

implicit and partial differentiation

oh wait

im silly

[safe onyx]

it's both

[vague eagle]

im silly ahahha

i didnt see the question properl

y

[safe onyx]

I gotta take derivative with respect to x, z is function and y is constant

[vague eagle]

yep yep

well

actually

it seems

[dreamy lichen]

shouldnt the y here be as well?

[desert siren]

.

[safe onyx]

well y is constant and it's plus y right?

[vague eagle]

yea i think thats it

[safe onyx]

d/dx ( y) where y is a constant is 0

[vague eagle]

i cant see anything else

[safe onyx]

(partial not d)

[vague eagle]

negatives are all okay

8xyz

-> 8xy

w.r.t. z

d/dz (x^5 + y^7 + z^5 - 8xyz = 0) -> 5z^4 - 8xy

[fallow scarab]

where did 8y go

from here

[safe onyx]

oooh

Thank you

[midnight plank]

@safe onyx Has your question been resolved?

[desert siren]

Gollum.

[midnight plank]

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[safe onyx]

that's me

[midnight plank]

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[mint ravine]

Can someone please help me with this question

[tough stag]

try 599

[mint ravine]

ok

wrong bruh

please dont give random answers, it will take away the credits i get from the question

[round parcel]

Factor 600.

[mint ravine]

yeah i alr did that

[tough stag]

oh wait its 180

[desert siren]

What.

[mint ravine]

35^22^3

[tough stag]

lol idk what i was thinking

[mint ravine]

3 times 5^2 times 2^3

[round parcel]

OK, and factor the six possible rolls.

(1 - 6)

[mint ravine]

1, 2, 3, 2^2, 5, 2*3

[round parcel]

OK, so we need one 3.

[thorn sierra]

2 + 10 is 12

[round parcel]

So, we can choose between which two numbers for that?

[mint ravine]

3 and 1

[round parcel]

Well 1 won't give us a 3 factor.

[thorn sierra]

12+2 is 14

[mint ravine]

factor of 3 is 3 and 1?

[thorn sierra]

Factor of 3 is 2 and 1

[mint ravine]

can you please stop @thorn sierra

[round parcel]

We got the factorizations of 1 through 6, right?

[mint ravine]

Yes

[round parcel]

We factored 600 as well, right?

[mint ravine]

yes

[round parcel]

OK, so 600 is 2^3 * 3^1 * 5^2, right?

[mint ravine]

yeah

[round parcel]

OK, so there's one 3 in 600.

Which die rolls give you a 3 factor?

[lyric charm]

<@&268886789983436800> spam

[round parcel]

<@&268886789983436800> Was asked to stop helping, keeps giving nonsense.

[carmine sigil]

He's been lobbying to get himself banned for a while now

Sorry about that

[mint ravine]

all 6 dies have a 3

[round parcel]

No, in their factorizations.

Like 2 doesn't have a 3 factor in it. It's 2^1.

[mint ravine]

oh 3 and 6

[round parcel]

Right.

So, let's look at 3.

If we pick that, we can't pick 6.

So, we had 600 = 2^3 * 3^1 * 5^2.

We got rid of one of the 3s, so we have 2^3 * 5^2 left.

Let's look at the 5s.

Which die rolls have a 5 factor?

[mint ravine]

just 5

[round parcel]

OK, so we need two 5 rolls.

[mint ravine]

Yes

[round parcel]

So, 3, 5, and 5.

That gives us 2^3 left.

Which die roles give us 2?

[mint ravine]

2, 4, 6 but we can't use 6 if we already use a 3

[round parcel]

Right.

So, we can pick 2, 2, 2.

We can pick what other combinations to take care of our 2^3?

[mint ravine]

4 and 2

but then we have to add a 1 to make sure that there are 6 rolls

[round parcel]

OK, so 4, 2, 1.

Right.

Any other combos?

[mint ravine]

no

[round parcel]

OK, so:

• 3, 5, 5, 2, 2, 2
• 3, 5, 5, 4, 2, 1

What about 6 instead of 3?

[mint ravine]

then we have 1, 2, 2, 5, 5, 6

[round parcel]

What else?

[mint ravine]

1, 1, 4, 5, 5, 6

[round parcel]

OK, anything else?

[mint ravine]

I don't think so

i forgot to include (1, 1, 4, 5, 5, 6) in my original calculations

[round parcel]

OK, so we have:

• 1, 1, 4, 5, 5, 6
• 1, 2, 2, 5, 5, 6
• 1, 2, 3, 4, 5, 5
• 2, 2, 2, 3, 5, 5

[mint ravine]

yeah

[round parcel]

OK, now order matters, so we need to get the number of orders for each of those.

Have you dealt with choose like 5C2?

[mint ravine]

yes

[round parcel]

OK, there's a way to do these like that.

[mint ravine]

6!/(2!2!) for 1, 1, 4, 5, 5, 6,

[round parcel]

5C2 is like when you have groups of 2 and 3 out of the 5.

So, you do (\frac{5!}{2! 3!}).

[grand pond]

Chai T. Rex

[mint ravine]

yeah i know im learning that rn

[round parcel]

Now for ours, we have 6! on the top, then we have one factorial for each die roll in the answer in the denominator.

So, for 1, 1, 4, 5, 5, 6, note that there are two 1s, one 4, two 5s, and one 6.

So, (\frac{6!}{2! 1! 2! 1!}).

[grand pond]

Chai T. Rex

[round parcel]

Note that the numbers on bottom sum to the top.

[mint ravine]

I think I can go on from here

[round parcel]

This is called a multinomial coefficient.

OK.

[mint ravine]

so the answer is 780?

ok thank you for your help!

.close

[midnight plank]

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[round parcel]

Yes, that's right.

You're welcome.

[midnight plank]

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[mint ravine]

Can someone please help me on this question?

[mint ravine]

I originally got 12 but it was wrong

Here's what I did:
3 options for the left-most box {1, 2, 3}
2 options for the box after that {4, 5}
2 options for the middle box {excludes the number used in the left-most box}
1 option for the box after that {excludes the number after the left-most box}
1 option for the right-most box {the number that hasn't been used yet out of 1, 2, or 3}

Then I multiplied everything out so:
3 times 2 times 2 times 1 times 1 =12

I don't know if I'm overcounting or undercounting, or what I am doing wrong

[dreamy lichen]

1, 3, 2, 5, 4

is valid but not counted in yours

undercounting i think

the two < > boxes dont have to be 4 or 5

[mint ravine]

this doesn't work tho right?

[dreamy lichen]

Which inequality fails?

1 < 3 > 2 < 5 > 4

[mint ravine]

oh nevermind

Oh so the left-most box actually has 4 options {1, 2, 3, 4}

[dreamy lichen]

the options are dependent on each other

the number of options in the 1st box depends on what number is in the 2nd one

[mint ravine]

can we work backwards?

nvm that's just the same thing

[dreamy lichen]

Wdym? I honestly dont know how to do this in a normal-ish way. I'd probably do sth similar like you did, but include some casework. So next I'd deal with the case where 3 and 5 are in the < > boxes

[mint ravine]

I did casework too but I guess I forgot to include some cases

[dreamy lichen]

the 2 cases would be 4, 5 and 3, 5 in the <> boxes

[mint ravine]

please give me a min to add in the extra cases

[mint ravine]

ok im done with the cases

let me take a picture of it first

Sorry about the lighting but I got 16 total cases as my answer

if you want me to take another picture i can

[mint ravine]

so would 16 be right?

nvm it is right

thanks for your help!

.close

[midnight plank]

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[midnight plank]

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[safe onyx]

$dv = \frac{\partial v}{\partial x} dx + \frac{\partial v}{\partial y} dy$

[grand pond]

smeagol

[safe onyx]

partial/partial x (5y cos(xy) )
(5y)(-sin(xy))(y) = -5y^2 sin(xy)

i'm confused where I messed up

[crystal plover]

$$
\frac{\partial}{\partial x} 5y\cos(xy) = (5y)(-\sin(xy))(y) = -5y^2\sin(xy)
$$

Is this what you meant?

[grand pond]

Luigi

[safe onyx]

yes

that's what I had put in, but it says wrong

so I'm confused

[crystal plover]

what's the answer they give you?

[safe onyx]

it doesn't

what

I got it now

I guess I didn't hit submit or the website broke

It never processed my corrected answers

[crystal plover]

shit happens with these websites

[safe onyx]

thanks mate have a nice day :D

.close

[midnight plank]

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[midnight plank]

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[mint ravine]

I have tried to do complementary counting but Idk where i went wrong

[mint ravine]

Here's what i did:

12 choose 3 - (11 choose 2)2= 165

and 12 choose 3 - (11 choose 2) - (10 choose 1) = 155 but that's also wrong

[desert siren]

What's 11 choose 2? There's 12 possible triplets.

Oh nm

It's doubles not triplets

[mint ravine]

yeah

[desert siren]

Ah theres still only 12 lol

[crystal plover]

We walked through the steps already

[desert siren]

Oh, multipost?

[mint ravine]

so 12 choose 2?

[desert siren]

What? I'm saying there's 12 possible pairs next to each other. Why are you choosing pairs from pairs?

[mint ravine]

I don't know what to do subtract by

i'll work on this problem later

[crystal plover]

what didn't you like about the previous explanation?
You agreed you can pick the first in any of the 12 ways
after that you have 9 choices for the 2nd pick
then you have either 6 or 7 depending on the previous choice
You just have to properly multiply

[mint ravine]

that doesn't work cause either way it's going to be more than 12 choose 3

[crystal plover]

by doing this I got 672

Is this the correct answer or not?

[mint ravine]

No, it's not

It's greater than 12 choose 3

max is 12 choose 3 without restrictions