help-17

[candid magnet]

and take x+2 = t

[vocal sleet]

@indigo wave Has your question been resolved?

[meager shoal]

Complete the square?

Then u can split fraction and get them into arctan

[night wave]

no, i didn't ecieved any help.

[vocal sleet]

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[grizzled crest]

If 3x−y=12, what is the value of 8^x/2^y?

A) 2^12

B) 4^4

C) 8^2

D) The value cannot be determined from the information given.

[vocal sleet]

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

[grizzled crest]

I don't know where to begin.

im stupid

[shrewd moth]

for this type of problem its easy]\

just put 8 in terms of 2

[loud walrus]

Start by writing 8 in base 2

[grizzled crest]

mark it

[shrewd moth]

aka 2^3

so then it turns into (2^3)^x/2^y which is the same as 2^3x/2^y

using exponent rules you get 2^(3x-y)

3x-y is given as 12

so the answer is 2^12

[loud walrus]

!nosols

[vocal sleet]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[shrewd moth]

looks like an sat question

[grizzled crest]

k thanks man

.close

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[grizzled crest]

turns out it was one of the hardest sat questions

i just wanted to find math 💀

i just wanted to find math 💀

and understand it

to fill my hollow brain

[shrewd moth]

hardest?????

mf

thats like

the easiest question

on the entire sat

[grizzled crest]

i didnt send an invite

[shrewd moth]

no

goosination probably got hacked i guess

[shrewd moth]

change ur password my man

[plucky trout]

you leaked the invite link

time to join-

[vocal sleet]

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[woven stream]

Hi! I'm actually so confused and I have no idea where to start. Picture was taken from a practice test done a few hours ago!

[woven stream]

log(100000)?

[prisma rain]

remember when base is hidden, it normally means base 10

[woven stream]

10^4?

4?

yipee

my horses are held because i think its e

instead of 10?

i missed like 3 lectures cause i got super sick

😭 i guess the first term? i dont know why the log is exponited

i did not

we arent allowed one

math department said nope

oh yes sorry

im exhausted i apologize

[subtle summit]

log_a(x) is the opposite of exponentiation a^x

[woven stream]

so is it just 7?

cause sqrt4 is 2

5+2 is seven

[subtle summit]

[\log_a(a^x) = a^{\log_a(x)} = x]

[twin meteor]

Invariance

[subtle summit]

That's the entire point of log

[woven stream]

ohh

[subtle summit]

All the other log formulae come from that one

[woven stream]

okay tyyy

.close

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[velvet cove]

how to show that (x,y) is perpendicular to (-y,x)

[sharp rain]

You mean x= y and x= -y

?

[velvet cove]

these are vectors btw

[sharp rain]

If yes then just find slope of respective lines

Multiply to get -1

[sharp rain]

Oh

[full summit]

i was just abt to say points can't be perpendicular

[sharp rain]

Dot profuct = 0

[keen siren]

Dot product -xy+xy=0

Sorry Imma echo

[sharp rain]

That's why I was getting that confirmed at the start lol

[velvet cove]

oh ok\

also how to show that |u cross w| is the area of a paralleogram with sides |u| and |w|

[sharp rain]

You can't show that

Because it isn't true

[velvet cove]

like i gotta prove it

[sharp rain]

That for cross

[velvet cove]

i meant to put magnitude

[sharp rain]

Oh you wrote cross product

Sorry lol

[sharp rain]

This make sense?

Yellow one is u

Purple one is w

Red is |w| sin theta

[sharp rain]

Rectangle ABCD has the same area as the original parallelogram

Which is |u||w|sin theta

Which is exactly what the cross product is

This make sense? @velvet cove

[velvet cove]

hold on i didn’t see it

[velvet cove]

how you get this

oh wait nvm

ohhh

i thought u had to use that one A=1/2 ab sin(c) formula

ty

[sharp rain]

Ur welcome

[velvet cove]

.close

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[glass galleon]

am i right with B here?

<@&286206848099549185>

[warm vine]

yes

[vocal sleet]

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.

[glass galleon]

tysm

.close

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[silk osprey]

!show

[vocal sleet]

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

[keen siren]

Haven’t looked too closely but i’m guessing maybe you didn’t account for the pasture sharing a fenced wall?

[silk osprey]

the perimeter isn’t 2x + 2y

there’s the fence part in the middle that divides the pastures

draw a diagram for these problems

it makes it much easier

@fallen fulcrum

bruh

you asked for help then disappeared

😟

T_T

draw it

it’ll make more sense

two pastures

separated from each other

just draw it please

send it

perfect, now the line in the middle will have a length of what

yea P = 2x + 3y

well, maybe we should label it better

because the larger pasture isn’t x by y

x is the length of the length of the larger pasture + length of smaller pasture

in your diagram

how about you label just the length of the larger one x

and the length of the smaller one idk w or something

you choose

then the perimeter will be 2(x+w) + 3y

@fallen fulcrum do you follow?

[vocal sleet]

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[silk osprey]

what

did you not get it?

oh don’t feel embarrassed though

you don’t have to leave just because you can’t figure it out

but you can try again tomorrow ig

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[languid ridge]

Help

[arctic mantle]

yooooo same tag

,rcw

[languid ridge]

Can you guys help to double check

[steep crater]

wait what

oh its not ren its ren

[signal pendant]

3^6 is not 216

[steep crater]

your work looks okay except for (4)

[arctic mantle]

looks good apart from 4

[arctic mantle]

handwriting?

[languid ridge]

It's 5

[arctic mantle]

wtf is 15

it's legit incorrect

[prisma rain]

ah

[arctic mantle]

ik what you mean in 17

but please, neater work

[prisma rain]

looks like a 6

[languid ridge]

Im rushingg

Is the part one good?

I mean A.

[vocal sleet]

@languid ridge Has your question been resolved?

[vocal sleet]

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[grave anvil]

hello I need help...

[grave anvil]

how do I find the distance when they connect?

[vocal sleet]

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[hearty nova]

need help with these I aint getting the correct answer I have done it but I guess it is wrong

[hybrid flicker]

what happens when you subtract row 2 from row 3?

[hybrid flicker]

pinging just in case

[hybrid flicker]

maybe you want to show your work to see why it's wrong before doing this?

[hearty nova]

Oh our teachers here taught us a different method that you might also know

theres a method to calculate a 3x3 determinant,where we take the minor of all the elements in first row

something like this

[hybrid flicker]

yes development into minors

[hearty nova]

yep I am trying this method

the end result I got simplifying this equation is

I just wanted to make sure if it seems correct to you

[hybrid flicker]

my answer is (a-b)(b-c)(2a+b+c)

I would expand it to check

[hearty nova]

and Can you also if you have time,explain this method to me too

[hybrid flicker]

so

do you remember that adding/subtracting a row a certain amount of times to another row doesn't change the determinant?

[hearty nova]

oh I have to take commons i guess then It can be simplified more

[hybrid flicker]

for example

[hearty nova]

oh yep

[hybrid flicker]

if you replaced row 3 by (row 3) - 56(row 2)

the determinant is unchanged

or row 1 replaced by (row 1) + pi * (row 2)

[hearty nova]

I see

[hybrid flicker]

ok

[hearty nova]

okay I got it

[hybrid flicker]

so what happens when you replace row 3

by row 3 - row 2

[hearty nova]

that means I have to substract like (a+c a+b a) - (a+c a+b a) then the answer will be the third row right?

[hybrid flicker]

you mean (a+c a+b a) - (a+c a+b b)

right?

and yeah you replace row 3 by the result

[hearty nova]

oh yeah misplaced a again

oh okay

oh btw we substract the corresponding element with each other like (a+c) - (a+c) or something else

[vocal sleet]

@hearty nova Has your question been resolved?

[hybrid flicker]

why would there be something wrong with this method

[hybrid flicker]

yeah

that's how you add/subtract rows together

entry by entry

[hearty nova]

oh there is nothing wrong

[hearty nova]

okay btw have you tried expand the answer you got I tried it I guess i got something wrong

[hybrid flicker]

(a-b)(b-c)(2a+b+c) = (a-b)(2ab-2ac+b^2-c^2) = 2a^2b - 2a^2c + 2abc -ab^2-b^3-ac^2+bc^2

mmh wait

the abc coefficient should also be 2abc

oh yeah there's some phantom coefficient that appeared

[hybrid flicker]

I get the same thing as you

[vocal sleet]

@hearty nova Has your question been resolved?

[hearty nova]

Phantom coefficient?

[vocal sleet]

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[heavy palm]

Did I do 16 correct

[vocal sleet]

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[pine salmon]

(x - in_min) * (out_max - out_min) / (in_max - in_min) + out_min

this is the formula for mapping a number in a range into other number range

[pine salmon]

i don't understand why we divide it with this (in_max - in_min)

+out_min is offset i guess

x/(in_max-in_min) would be percentage between inputs if we multiply it

with (out_max-out_min)

and offsetting it with out_min would be the same?

oh i got it nvm

.close

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[slim kernel]

X = pi

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[still berry]

f(t) is given graphically

[still berry]

$$F(x) = \int_4^x f(t) dt\ $$

[twin meteor]

bonk

[still berry]

for

$$x \in [0,12]$$

[twin meteor]

bonk

[still berry]

what are these asking for conceptually?

[keen siren]

the evaluations at a particular x of the anti derivative

[still berry]

how could i do this without having either one though?

only graphically?

[keen siren]

You have f(t)

[still berry]

a graph of it, yes

[keen siren]

Can you send a pic of the question

[grizzled vault]

i guess you can only estimate

[keen siren]

Ok yes, so plug in the value from F(x) into your integral and compute

No you can get definite answers

[grizzled vault]

treat them as straight lines

[keen siren]

The integral from 4 to 0 would be the negative area of the triangle

for example

[grizzled vault]

?

the lines aren't even straight

[keen siren]

It doesn’t matter

[grizzled vault]

the best you can do is estimate

[keen siren]

Integrals don’t require the lines to be curved

Just stop talking please

You can use triangles to figure out the area under/above the curve

0 to 4 is a triangle with base 4 and height 7 for example

[still berry]

so for F(0) im integrating "f(t)" with bounds 0 and 4

[keen siren]

And it’s below the curve so it’ll be negative (which can also be seen by switching the bounds and giving it a negative)

[grizzled vault]

charlie... you literally assume that the blue thing is a straight line in the process

[keen siren]

4 as the lower and 0 as the upper

[grizzled vault]

the question is trivial

[prisma rain]

The area below the x-axis is negative, but the bounds are inverted (4 to 0 instead of 0 to 4) so it will invert the sign again, giving a positive overall result

[still berry]

so when im at F(10) am i subracting the part between 8 and 10?

[keen siren]

At F(10) you’re finding the integral from 4 to 10

Which will be the whole triangle above the x axis, and a negative little triangle from 8 to 10

It’ll help you visualize if you draw lines in your head with the bounds at a particular x value to see the resulting triangle

[still berry]

F(8) is 12 and the little triangle is looks to be of area 2.

[keen siren]

Yes so subtract 2 from the integral from 4 to 8

[still berry]

its not accepting 10.. maybe im reading the lines wrong

it could be 2.5 x 2

making a = 2.5

[keen siren]

What’s the area of the triangle in the middle

[still berry]

the large triangle in the middle F(8) = 12

so stupid.. that little triangle is base =2 and height = 2.5

[keen siren]

You can see it by looking, I didn’t notice myself

You can also infer it from the area of the whole triangle @still berry

Look at the area of the triangle on the right and see where it gets split

It’ll be a ratio of the total area

[still berry]

this one?

[keen siren]

Yes

[still berry]

makes sense

[keen siren]

Yes

Very good

[still berry]

intervals were simple enough so i got this question knocked down. I have another one to do so ill see if that goes smoothly.

thanks for your help!

[keen siren]

No problem at all

[still berry]

.close

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[cunning galleon]

could anybody solve this?
I tried but didn't succeed

[vocal sleet]

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

@cunning galleon Has your question been resolved?

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[candid magnet]

try taking

[marble tinsel]

how to find the recursive rule for the sequence 1,1,1/3,1/4,1/15

[candid magnet]

x+1 = 1/t

oh shit the the channel closed im sorry

.close

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[marble tinsel]

.reopen

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[fading marsh]

Why doesn't the rocket have an initial speed?

[heavy yoke]

context?

[fading marsh]

I know it's a stupid question, but I want proof

When does the body have an initial speed and when not?

[spiral turtle]

When the problem specifies that it does.

Or if you're in a practical situation, when the practical situation says it does.

Rockets can absolutely have an initial velocity, if your problem is regarding, for instance, the calculation of a burn for a transfer orbit.

[heavy yoke]

on the other hand we would expect it to be stationary if we started at the launch pad. so very context dependent

[fading marsh]

So you say that if the rocket is in static surface and then launches, it does not have an initial speed, but if it starts from a moving surface, here the surface has given it an initial speed because it is moving?

Is the same with cars and all objects

[empty frigate]

yeah, or more generally just, if it's... moving, at the start

where what counts as the "start" is kind of just dependent on where you start calculating from

if you happen to know that the rocket launched from a static surface ten seconds ago, but what you have are numbers for what it's doing now and you want to predict the rest of its trajectory from that, then it would have an initial velocity, because it's moving

[fading marsh]

Do you mean, according to the pedal, determine the initial speed, for example, we study the movement of a rocket from t=7, then t=12, where the moment t=7 is the first moment here, the rocket has an initial speed

[vocal sleet]

@fading marsh Has your question been resolved?

[vocal sleet]

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[ruby crest]

Is there a way of constructing a surjective ring homomorphism from R to Q? I saw on Stack Exchange that if you treat R as a Q-vector space, you can sort of construct a surjective group homomorphism by taking a basis for R and sending everything except the multiples of one basis vector to 0 (I say sort of, because this requires axiom of choice apparently). Is there a similar thing we can do for a surjective ring homomorphism from R to Q?

[ruby crest]

I guess what I'm asking is, can we vaguely know what a surjective ring homomorphism from R to Q looks like? Because, in the group case, while we do use choice, you can get an idea what the map is actually doing.

Oh wait, I guess I should've asked first whether such a thing even exists, because I assumed that between a ring and a subring, a surjection always exists, but that might be wrong.

[vocal sleet]

@ruby crest Has your question been resolved?

[dim pumice]

say such a homomorphism phi exists. It needs to satisfy phi(1) = 1 and then phi(2)=phi(1+1)=phi(1)+phi(1)=2. Now consider phi(sqrt(2) * sqrt(2))

[ruby crest]

Oh of course, thanks.

.close

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[dim pumice]

another way, using a bit more ring/field theory, there is a theorem that states that any field homomorphism is always injective, so phi would have to be bijective

[vocal sleet]

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[vast shale]

May I get help?

[vast shale]

don't ask to ask

just ask

[vast shale]

I know c is 65 and d 146

[vast shale]

whats your question

did u use a venn diagram

[vast shale]

Yes

can u show it

[silver prairie]

well, it gives you B "82 take only cream"

[vast shale]

a and e?

[silver prairie]

374 coffee drinkers, only 65 do not add sugar or cream

374-65=309

309 use at least one of the addatives

so e is 309

and a is 309-64-82=163

[vocal sleet]

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[visual forge]

1. The trolley can carry 293 kg safely.
2. Each block has a mass of 10 kg.

Hence,

$\text{Maximum Number of Blocks} = \frac{293 , \text{kg}}{10 , \text{kg/block}} = 29.3$

[twin meteor]

DW1

[visual forge]

but since the number of blocks have to be a whole number

we round down

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[velvet cove]

how can i fix this

[flat whale]

download more ram

https://mathematica.stackexchange.com/questions/194878/limit-the-amount-of-ram-mathematica-may-access

anyway, probably not a math question

[grand vortex]

more wam*

[dim pumice]

dedotated wam

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[frank plover]

How would i translate what I circled in green into words

?

[olive igloo]

every subset $A_i$ is contained in $A$

[twin meteor]

rain

[frank plover]

Ah ic

ty

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[young imp]

3. Suppose there are two urns. Urn 1 contains 100 chips: 30 are labelled 1, 40 are labelled
   2 and 30 are labelled 3. Urn 2 contains 100 chips: 20 are labelled 1, 50 are labelled 2,
   and 30 are labelled 3. A coin is tossed and if a head is observed, then a chip is randomly
   drawn from urn 1, otherwise a chip is randomly drawn from urn 2. The value Y on the
   chip is recorded. If an occurrence on a head on the coin is denoted by X = 1, a tail by
   X = 0, and X ∼ Bernoulli(3/4). Compute the following:
   (a) E(Y | X = 0), E(Y | X = 1), E(Y | X)
   (b) Check that E(E(Y | X)) = E(Y )
   (c) E(X | Y = 1), E(X | Y = 2), E(X | Y = 3), E(X | Y )
   (d) Check that E(E(X | Y )) = E(X)

I've found E(Y|X=0) and E(Y|X=1), how do i find E(Y|X)

[olive igloo]

dawg

[young imp]

I accidentally typed enter too early mb

[olive igloo]

E(Y|X) is not a number, its a function of X

[young imp]

I'm aware of that

[olive igloo]

so the function has to return what you got for X = 0 and 1

[young imp]

Yeah but how am I supposed to show as summation to find it? Like it's intuitively 2.1 - 0.1x (E[Y|X=0] is 2.1 and E[Y|X=1] is 1)

But I'm not sure how to sure the work of how I got there

I just kinda saw it

I'm sorry if I'm coming off as rude

Not at all my intention

[languid sparrow]

so E(y | x. = 0) is expected value of y when tails
so from urn 2
is $0.2*1 + 0.5 * 2 + 0.3 * 3=2.1$
i think?

oops

[twin meteor]

reaver

[languid sparrow]

using expected value we weight each by their probability of occuring

[young imp]

That is correct

[languid sparrow]

isnt that it?

[young imp]

I'm struggling to find E[Y|X] in general terms

As in a function of X that describes it

[languid sparrow]

oh

[young imp]

Well, I have it

I just don't know how I got there

It's whatever, I think I got it

Thank you

[languid sparrow]

wouldnt you just weight each one or am i tweaking

[young imp]

Like E[Y|X=0] * P(X = 0) + E[Y|X=1] * P(X=1)?

Wouldn't that give you E[Y]?

[languid sparrow]

well

[grand vortex]

That is the law of total expectation

[languid sparrow]

isnt that right or am i tweakingnginginggngn

[grand vortex]

https://en.m.wikipedia.org/wiki/Law_of_total_expectation

[languid sparrow]

or is that just E(Y) then?

[young imp]

Yeah that's E[Y] not E[Y|X]

[languid sparrow]

is it this $0.50.21 + 0.50.52 + 0.50.33 + 0.50.31 + 0.50.42 + 0.50.33$

bruh

[twin meteor]

reaver

[grand vortex]

Can I ask what part specifically you are trying to solve?

[languid sparrow]

E(Y|X)

[true matrix]

been watching for a while, what topic is this

[young imp]

Probability theory and statistics

[grand vortex]

Well E(Y|X) is a random variable. It doesn’t have a determined value

[young imp]

I am aware of that

I know it's 2.1 - 0.1X just from looking

I just don't know how to show my work for that

I know that E[Y|X=0] is 2.1 and E[Y|X=1] is 2

[languid sparrow]

i think it should be 2 maybe?

[young imp]

It's not a numerical value

[grand vortex]

I see, one way to state the equivalency would be to show that the pmf of E(Y|X) and 2.1-.1X are the same

In this case that would be computing the outcome table for both

[young imp]

I figured it out

I greatly appreciate the help from all of you!

?close

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[tropic brook]

could someone please help me on how to do this

what i did was squaring the bottom equation

then subtracting

3y^2-4xy=9-k^2

is there someone significant about the difference of squares?

[boreal remnant]

think geometrically

this means you want the line to be tangent to the circle

[tropic brook]

oh yeah

im not really sure where to go from there though

[boreal remnant]

can you use calculus?

[tropic brook]

sorry not really

im in my first semester of alg 2

[boreal remnant]

alright then the simplest solution is this

[frosty thorn]

???

[boreal remnant]

take the second eqn, solve for x

[silk osprey]

make it a quadratic

[tropic brook]

idk

[silk osprey]

then set the discriminant equal to 0

[frosty thorn]

what would you have suggested?

[tropic brook]

sorry, make what a quadratic?

the second equation?

[boreal remnant]

x = 2y + 3

(2y+3)^2 + y^2 = k^2

[silk osprey]

substitute either x or y in terms of the other into the first equation

[boreal remnant]

expand

[bright niche]

Sup knief 👋

[silk osprey]

hello sir

how is studying going

[tropic brook]

oh okay

[silk osprey]

what are you up to

[olive igloo]

the line is tangent to the circle

[bright niche]

?

[olive igloo]

im guessing thats what he meant

[silk osprey]

yes that’s what he said

[bright niche]

What’s the problem

[silk osprey]

^

simple

you can solve it goat

this is algebra 2

[frosty thorn]

find the derivative of the first equation at a point in the second equation?

[bright niche]

Oh algebra 2

Just solve it

Just go to the answer key

[silk osprey]

@tropic brook did you understand what we proposed

[tropic brook]

yeah i did

[silk osprey]

what town hall are you

[tropic brook]

sorry idk why i didnt realize i coulve just substituted earlier

uh 14 i think

alright thanks y'all

[frosty thorn]

is this right

[silk osprey]

you’re welcome

[frosty thorn]

oop alr

[tropic brook]

cya

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[sick wave]

△ABC is an equilateral triangle. D is a point on AC, and E is a point on BD. Let P
and Q be the circumcenters of △ABD and △AED, respectively. Prove that △EPQ is an
equilateral triangle if and only if AB ⊥ CE

[sick wave]

can anyone please help me solve this problem

prove the statement

i tried but my teacher said its incorrect

[rocky compass]

Could you draw an accurate figure
and send it

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@sick wave Has your question been resolved?

[sick wave]

i tried but the figure is so confusing 😭

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@sick wave Has your question been resolved?

[split zephyr]

this question is from the ongoing usamts competition, which you aren't allowed help on

<@&268886789983436800>

[lost yarrow]

do you have a link?

[strong isle]

https://files.usamts.org/Problems_36_2.pdf

[lost yarrow]

thanks

dealt with

.close

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[lost yarrow]

(we just like to double-check that the competition is indeed ongoing, since sometimes users erroneously report sample problems as the actual competition)

(but in this case the report was correct)

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[indigo rose]

When doing Sum and Differences and I reach the part where i must convert the sin and cos into the coordinates of points of the unit circle which one do I pick for which one. X or Y?

[olive igloo]

y is sin

x is cos

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[indigo rose]

I understand it now

We passing tran class

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[olive lintel]

can someone explain this solution to me

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[regal agate]

anyone know how i can write this as a general solution of a linear combination

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[elfin pendant]

Hey guys, can anyone explain to me how to calculate the pullback?

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[elfin pendant]

<@&286206848099549185>

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[civic flume]

Right so I don't fully understand this question

I am tempted to understand "closed subspace" here as closure under addition/scalar multiplication

But this exercise is in the context of a chapter about continuity of functions from one metric space to another

So is this asking me to show closure in the topological sense? Is that equivalent to closure under addition/scalar multiplication?

is this theorem relevant?

[wraith venture]

yeah topologically closed

more general theorem: all finite dimensional subspaces are closed

[civic flume]

yeah thats the next part

[civic flume]

good to know, any hint as to how I can approach this?

[potent beacon]

Orthogonal projections

[wraith venture]

you can use the probably known fact that linear maps are continuous (in finite dim, otherwise bounded linear maps)

[potent beacon]

I think could do it

[potent beacon]

Oh yes, thats better

[civic flume]

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[ember bloom]

err how to solve

[unique elm]

best bet when you see a sqare root is to square everything

[ember bloom]

okay

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[jade stone]

tran???

tf is that

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[bold scarab]

I don’t understand how the teacher did this

[bold scarab]

Nvm .close

.close

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[winged vine]

i got confused

please help

i dont know whag to put for the next stepp

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[valid birch]

I need a lot of help in this

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[tulip jewel]

hi im learning about proofs
i'm currently working on a proof and i want to use pigeonhole principle to supplement my proof

my problem is, pigeonhole principle is typically stated as follows:
if n items are put into m containers, with n > m, then at least one container must contain more than one item

but in my case, i want to show that if n items are put into m containers, with m > n, and strictly 1 item per container, then at least one container must be empty

is this still pigeonhole principle? and is it applicable in my case?

[fast ledge]

Pigeonhole is great only for finite element proof

Depending what you want to proof though

You still have to do an induction like argument

[fossil dawn]

n and m are finite?

[fast ledge]

And argue at one point that you have some empty container

Yes

Classic pigeon is for finite

[tulip jewel]

yea

[fast ledge]

Don’t know if you can do infinite with it

I heard you can but not sure

[tulip jewel]

so i cant apply pigeonhole here?

[fast ledge]

For your specific problem you have to create an inductive step that you argue you have m-n empty container

You can

Your prob is classic

[tulip jewel]

ok i dont know mathematical induction yet

[fast ledge]

Then study basic first

Without induction it’s very limited usage

[tulip jewel]

yea pigeonhole principle

[fast ledge]

I feel like induction is more basic..

Because if you want to prove your proposed question

You have to induct on n

[tulip jewel]

can you show that my question is true using only pigeonhole?

or u have 2 use induction

[fast ledge]

It’s definitely true

Because of pigeonhole principle

But I can’t prove it without induction like proof and truth is kinda different

Proof is formalism

Maybe there are other ways I don’t know

[tulip jewel]

so i can just say something like "by pigeonhole principle, at least one container must be empty" ?

without showing induction

[fast ledge]

Yes you can, just don’t do it in exam though

Because you don’t have an argument that there’s an empty container

Even its obvious

[tribal pumice]

I don't think you need induction

[fast ledge]

At least m-n is empty

Maybe it’s enough actually

[tribal pumice]

If you learned stuffs about bijections over finite sets, you can use this

[fast ledge]

If he knows then he would defo know induction

[tulip jewel]

ye

[tribal pumice]

If A and B are finite and there exists a bijection between A and B, then |A| = |B|

This result is sufficient to prove what you are asking

[fast ledge]

It’s actually rigorous to just write by induction it’s clearly that at least m-n container are empty

[fast ledge]

It’s actually rigorous

By induction it’s clearly

Depending the context

[tribal pumice]

Using the word clearly twice won't convince me here! What is your proof by induction?

[fast ledge]

Omitted… I am just joking

My professors do that all the times 🫠

[tulip jewel]

i'll just put the statement to prove here again:
if n items are put into m containers, with m > n, and strictly 1 item per container, then at least one container must be empty

tbh i just want to apply pigeonhole principle so bad, because it literally sounds like pigeonhole principle

[tribal pumice]

Bijection between which sets?

[fast ledge]

Can’t do it 🫠😭brain fog

[tulip jewel]

idk bijection, i jus dont wanna go too abstract u feel me

[tribal pumice]

If you want a rigorous proof, this is a particular case of the following result:

If A and B are finite and there exists a bijection between A and B, then |A| = |B|

Now, if you absolutely want to apply the pigeonhole principle, you can apply it the reverse way: You can associate containers with the items

Since there are more containers, it means at least one item has been associated with two or more containers

You can deduct from this that at least one container will be empty

I edited the more general result, that was badly written, sorry 🙃

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@tulip jewel Has your question been resolved?

[tulip jewel]

i tried proof by contradiction:
suppose that all containers contain strictly 1 or more items,
that means there's a total of w items
where w >= m
but we know that m > n
so there can't be 1 or more item per container
in other words, some containers contain 0 items (as there can't be negative items)

is this correct

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[feral wyvern]

I dont know how they do to know B

<@&286206848099549185>

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[manic axle]

hello i need help with a math problem

[ivory cedar]

this is what the server is for

what's your question

[manic axle]

can u please rotate it

[ivory cedar]

wait i'm not sure what the command is

[ivory cedar]

,rotate 270

[manic axle]

basically

we are supposed to show that 2 < (a^2+b^2) / 16 < 5

ik the left part but i got stuck on the right part

.close

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[upper shell]

but u did it right

[manic axle]

.reopen

[vocal sleet]

✅

[upper shell]

last step was that 2<9/4 + 4/4 = 13/4 = 3 1/4

[manic axle]

i struggle to read that

[upper shell]

ok

so in ur last step

u got

[manic axle]

9/4 + 1/4

[upper shell]

yeah

[manic axle]

= 10/4

=! 2

[upper shell]

and 2 is obviously lesser than 10/4

so 2 < 10/4 < (a^2+b^2)/16 < 5

[manic axle]

ohhhh

[upper shell]

so 2< (a^2+b^2)/16 < 5

[manic axle]

damn isee

well thx for clearing that for me man

tysm

legend

[upper shell]

welcome

[manic axle]

i legit thought u got banned when u deleted ur texts XD

[upper shell]

no i realised my working was wrong

lol

and that u were righ

[manic axle]

well i guess imma close

[upper shell]

yeah cya

[manic axle]

.close

[vocal sleet]

Channel closed

Closed by @manic axle

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[tawny hollow]

Wrong

[manic axle]

why

[upper shell]

hm?

[tawny hollow]

10/4 is 2.5

[manic axle]

which is greater than 2

[tawny hollow]

Yes and that’s a problem

[manic axle]

yet smaller than (a^2 + b^2)/16

[upper shell]

wait i dont get it

why is that a problem

[manic axle]

ms paint goes hard

[upper shell]

yeah

did u know that if u hold spacebar u can draw with the mouse by moving it with arrow keys

anyways thats off topic

[manic axle]

i knew

[upper shell]

oh cool

[tawny hollow]

Oh I’m stupid lmao

[upper shell]

lol

we all make mistakes

[manic axle]

its okey man

[tawny hollow]

Read it off

[manic axle]

i made stupider mistakes

[tawny hollow]

Wrong

[manic axle]

like 3+2 = 6

[upper shell]

yeah 3+2 is so similar to 3*2

[manic axle]

i dont wanna get banned for spam closing can someone close it please

[upper shell]

its already closed

[manic axle]

isee

[upper shell]

lets just leave the channel

[manic axle]

well good bye

[vocal sleet]

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[inland stump]

am i on the right track here?
stats practice question, have a final coming up so really want to understand this

[inland stump]

im not quite sure if im finding these values correctly, so would appreciate help please

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@inland stump Has your question been resolved?

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[proper cosmos]

Alright so I'm trying to understand how linear regression works but its kinda hard for me to understand

[proper cosmos]

Basically I understand using the center of the data to start out with

so basically we caluculate the x and y mean

Once we do that,

We find the slope using this point and each point in the data sest

[vocal sleet]

@proper cosmos Has your question been resolved?

[proper cosmos]

and then

for some reason subtract the x and y mean off each point and multiply the x and y answers together then divide by the variance of x

but i dont understnad why

<@&286206848099549185>

hello!

[valid birch]

hii

[proper cosmos]

plz help me

Ok so basically to solve for line of best fit in linear regression

[valid birch]

i need help too

[proper cosmos]

O

ig imma write something for anyone who sees it

Ok so basically to solve for line of best fit in linear regression

You take the mean and center the data around the mean

because the mean will be in the line of best fit

so we find the mean x and mean y of all data points

then we center the data around the mean by subtracting it by the mean

so we can represent that as

$x_i - m(x)$

[twin meteor]

Sinister

[proper cosmos]

x_i representing the point and m(x) represent the mean that we found

we also do this with the y values aswell

$y_i - m(y)$

[twin meteor]

Sinister

[proper cosmos]

My question is

Why do we multiply the result we get for each points x and y and sum them up

and then put that over the sum of $(x_i - m(x))^2$

[twin meteor]

Sinister

[proper cosmos]

so the equation ends up looking like

$slope = sum (x_i - m(x))(y_i - m(y))/ sum (x_i-m(x)^2$

[twin meteor]

Sinister

[proper cosmos]

First why do we multiply the centered points y and x together and also why do we square the bottom expression

Why isnt it just sum y centered / sum x centered

<@&286206848099549185>

[inland stump]

<@&286206848099549185>

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@proper cosmos Has your question been resolved?

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[safe harness]

I don't know if anyone can exactly help with this but currently I have solved a system of equations, found the respective solutions, and plugged them back into my equation to get the equation in the first image. I know it is correct up until this point as it has been double triple checked. but now, I need to go from this, to the solution in the second image, but I keep getting completely stuck. I have tried writing the e^theta terms as cosh(theta)+sinh(theta), and then using sum to product identities on them (as I was pointed to do) but I keep getting a completely different solution than the one in the second image

[olive igloo]

so you want to go from the first image to the second?

[safe harness]

yes exactly

[olive igloo]

what is $\omega$

[twin meteor]

rain

[safe harness]

sorry their omega is my theta

[olive igloo]

oki

🙂

lemme think abt it

[safe harness]

and I have been able to get to similar to that, with the correct angles in cosh and sinh, but with my terms i keep getting like, cosh(theta/2)(something)+sinh(theta/2)(something) so I cant exactly factor out the cosh(theta/2) like they did, etc

[safe harness]

thank you!

[olive igloo]

is the $\frac{2n-3}{2}$ inside the sin and cos or no

[twin meteor]

rain

[safe harness]

yes it should be, LaTeX error

[olive igloo]

oki

[dull bear]

We're back with another one

[safe harness]

yeah I've been here a good few times with the overall problem 😭

[dull bear]

Awwww it does look like it's one of those painful ones again

[olive igloo]

im workin on this

and i got something completely diff

:3

[safe harness]

I have a notebook full of schizophrenic looking scribbles and 2 weeks worth of headaches ahha

[olive igloo]

my brain is schizophrenic scribbles so

i will continue

[safe harness]

as have I! hence I am majorly lost 😭

that is so real

tysm!

[olive igloo]

i will let u know if i get anything normal

im off my meds so

idk if it will work

🙂

[safe harness]

"normal" 🤣

Putting e^θ=coshθ+sinhθ I get the following

But then its applying the sum to product on these that I absolutely do not get the “correct” final solution

[olive igloo]

my head hurts looking at that

😦

[safe harness]

So I fear I went very wrong at that step^

[olive igloo]

im still tryin

[safe harness]

me too😭

[olive igloo]

😭

I might have it

gimme a sec

I got it

🙂

[safe harness]

HOWWW

[olive igloo]

lemme try to explain

[safe harness]

ur a genius omg

[olive igloo]

🙂

in your second image, I am going to ignore the factor of $\frac{-2i^n}{\sinh(\omega)}$

[twin meteor]

rain

[olive igloo]

so that means your first image will be

$X_n = \frac{(1+i)i}{2i \sinh(\theta)} \cdot [e^{n\theta}(-e^{-2\theta} - ie^{-\theta}) + e^{-n\theta}(ie^\theta + e^{2\theta}]$

[twin meteor]

rain

[olive igloo]

then

$X_n = \frac{-2i^n}{\sinh(\theta)} \cdot \frac{i-1}{4} \cdot [e^{n\theta}(-e^{-2\theta} - ie^{-\theta}) + e^{-n\theta}(ie^\theta + e^{2\theta}]$

[twin meteor]

rain

[olive igloo]

so ignore that first fraction b/c i don't wanna deal with it

and then u can put it back later

then

$\frac{1}{4} \cdot (i-1) \cdot [e^{n\theta}(-e^{-2\theta} - ie^{-\theta}) + e^{-n\theta}(ie^\theta + e^{2\theta})]$

[twin meteor]

rain

[olive igloo]

the easiest way is to

multiply this out

then you get something that looks like

$\sinh( \theta(n-1)) + \sinh(\theta(n-2)) + i[\sinh(\theta(2-n)) + \sinh(\theta(n-1))]$

[twin meteor]

rain

[olive igloo]

maybe i missed a constant 1/4 idk

and then

you use the identity

$\sinh(x)-\sinh(y) = 2 \cosh\left(\frac{x+y}{2}\right) \sinh\left(\frac{x-y}{2}\right)$

[twin meteor]

rain

[olive igloo]

and then u get the answer

🙂

[olive igloo]

ofc, you can group terms to get the $\cosh() \sinh()$ term before converting to this, but what i said is just the way i did it

[twin meteor]

rain

[olive igloo]

its more annoying to factor than to jsut use an identity

make sure to keep track of teh 1/4

should disappear by thee nd

🙂

[safe harness]

sorry I'm at work while doing all of this I'm gonna look at this in a moment but omg you cooked on this

[olive igloo]

ty girlie

[safe harness]

yeah I shouldve just worked with it in the e^theta form rather than keep putting into the huge mess with the hyperbolic trig right away and I think it would've fallen out clearer

[olive igloo]

ya

[vocal sleet]

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[wraith quartz]

I don't know where to start

May someone guide me on how to prove it?

[vocal sleet]

@wraith quartz Has your question been resolved?

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[zinc topaz]

5 ³√64+³√14-6

[zinc topaz]

That cant be possible Right

[heady vine]

It can be

[zinc topaz]

14 cant be simplified

[heady vine]

Looks like it’s written wrong
or it really doesn’t simplify

[zinc topaz]

Try writing it based on what you think i wrote

[spice vortex]

yes so not possible

unless they want you to write 14+ cubed root of 14

[zinc topaz]

What do i write then?

[spice vortex]

do you have a calc

[zinc topaz]

Yeah

[spice vortex]

as in are you allowed one

[zinc topaz]

No

[spice vortex]

then this I guess

[zinc topaz]

Wdym

[empty frigate]

yeah it's just $14 + \sqrt[3]{14}$

[twin meteor]

bee [it/its]

[zinc topaz]

Im confused so bad

Why not

√(7)2

[vocal sleet]

@zinc topaz Has your question been resolved?

[spice vortex]

why would it be that

[zinc topaz]

Idk

How can you even add a whole number to a radical?

They arent the same

[spice vortex]

put it in your calculator and see

ones just a number with decimals on the end

[zinc topaz]

Wut

[spice vortex]

the cube root of 14 = 2.41014226 etc

its still a number like any else

so you can add to other numbers

[gentle sleet]

the 14 there is from 5 (³√64) - 6 btw

[zinc topaz]

Yea

Its 20 but -6

What abt 3√18+√8-5√48

[gentle sleet]

what about it?

[zinc topaz]

Is it possible?

I don think so

[gentle sleet]

it won't be a single term but yea, it's possible to simplify it

it'd be something like a√2 + b√3

[zinc topaz]

So copying it?

[gentle sleet]

wym copying it?

[zinc topaz]

Your essentially just gonna copy it if not possible?

[gentle sleet]

not sure what you mean by copying it

[gentle sleet]

you can simplify your expression until you get to that form

[zinc topaz]

Alright

[vocal sleet]

@zinc topaz Has your question been resolved?

[vocal sleet]

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[prisma rain]

if CH is in the same place, you only need to find how many ways you can arrange AIR

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@vast shale Has your question been resolved?

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[open sage]

Hey folks

So for those who have been helping me for the past few weeks I passed my second linear alg exam

so thanks for that

now i just need to make up for the failed first exam

which somehow i tend to have more trouble with

1a:

Let A,B in R3X3 with det(3A) = 216 and B, with alpha in R

Find for which values of alpha the following is correct:

$det(A^{-1}(B+I))=-5$

[halcyon valley]

Multiply both sides by A

Ull get

(B+I) = -5A

Take det on both sides

@open sage

[open sage]

yep sorry

look like i showed you the exercise incorrectly

[halcyon valley]

Oo