help-10

[teal turret]

oh wow thats interesting

thanks for sharing

[strong forum]

but wouldnt Var(X+X-X) be 4Var(X)?

[wild swallow]

the whole point is that it isn't

[novel knoll]

last time I checked Var(X+X-X)=Var(X)

[strong forum]

why am I being pranked i just want know how to figure out variance

[novel knoll]

what you did is correct

you were told that

[strong forum]

why is Var(X+X-X)=Var(X)

[novel knoll]

X+X-X=X

[strong forum]

[wild swallow]

[strong forum]

X+X-X=X

Oh

My bad guys

@wild swallow@novel knoll thank you for your patience

[teal turret]

quandale dingle here

[wild swallow]

[strong forum]

https://tenor.com/view/quandale-dingle-quandale-dingle-bignleton-gif-25683238

[teal turret]

https://tenor.com/view/anim-mem-move-мем-анимированный-gif-8613665567394800933

[strong forum]

this is quandale dingle signing off

.close

[obtuse pebble]

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[obtuse pebble]

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[tepid remnant]

We look at the wolf population in an area and assume that
The biological sex of the puppies is equally distributed in each pack. Consider
a general pack with n puppies and the events:

A := {the pack has puppies of both biological sexes.}

B := {there is at most one female puppy in the family.}

(i) Formulate a probability space (Ω, P) to describe all of them
possible such packs with n puppies, where n ∈ N, and translate them
Events A and B as subsets of Ω.

(ii) What are the sizes of P(A), P(B) and P(A ∪ B)?

(iii) What is P(A ∩ B)? For which value n are the events A and B un-
dependent?

now i thats all that is given
this is what i have;you can ignore the texts

[tepid remnant]

my previous chat got closed

(witout getting resolved)

[novel knoll]

What about it?

[tepid remnant]

oh sorry, how do i get the intersection P(A n B). now i know that P(A n B) = P(A) * P(B) but that only works if they are independet which we dont know yet

i dont know how to get the intersection of probabilities

[tame narwhal]

I think the formula is P(A n B) = P(A) + P(B) - P(A u B)

[novel knoll]

You haven't quite calculated P(A) and P(B) correct nor do I see where you computed P(A or B)

[tepid remnant]

i thought this formula was used for P(A U B) and not ∩

[novel knoll]

Where did you get P(A and B) from there? And once again, your answers in ii) are wrong.

[tepid remnant]

i just added the probabilities of A and B bc of the union, and substracted it by the entire probability

oh did i need to divide?

[novel knoll]

Feels like you are not reading anything I'm saying.

[tepid remnant]

its really hard to translate the english terms for me because they are so different

i dont understand your point then

[novel knoll]

all answers in ii) are wrong

thats one thing I have said

nothing about english terms

[tepid remnant]

what did i do wrong

i thought for ii) i only need the probability of each case

A, B, A U B

[novel knoll]

what happens if n=1 in your calculations?

[tepid remnant]

we get 0

for P(A) at least

and P(B) = 1

[novel knoll]

okay wait misread that, but still your calculation for

$P(A \cup B)$ makes no sense

[warm shale]

ScapeProf

[novel knoll]

Look your work, you used $P(A \cap B)$, where does that come from?

[warm shale]

ScapeProf

[novel knoll]

And if you knew how to find $P(A \cap B)$ then you could do iii)

[warm shale]

ScapeProf

[novel knoll]

(then the only issue with P(A) and P(B) is you write $P(A), P(B) \subset \Omega$ which is def not true. P(A) and P(B) are numbers)

[warm shale]

ScapeProf

[tepid remnant]

ah

[tepid remnant]

hmmm let me think

is it fine if i write $A \subset \Omega$?

[warm shale]

Gumbutt

[tepid remnant]

instead of P(A) etc

[novel knoll]

I mean no reason to write anything

[tepid remnant]

do i need to find the probability of at least one male puppy and one female?

[novel knoll]

the only thing you should write is for $n \in \bN$

[warm shale]

ScapeProf

[tepid remnant]

aight

[novel knoll]

Yes

[tepid remnant]

aight nice

is 2^n on the right way

[obtuse pebble]

@tepid remnant Has your question been resolved?

[tepid remnant]

its (2^n)-2/2^n

-2 bc of the cases 1 where all male and 2 all female

[obtuse pebble]

@tepid remnant Has your question been resolved?

[obtuse pebble]

@tepid remnant Has your question been resolved?

[tepid remnant]

end

[tepid remnant]

<@&286206848099549185> i am so sorry for bothering you with a ping, i cant close this chat so i cant ask another question in a help chat, could you help me?

[lapis thunder]

whats wrong

do .close if u wanna close it

[tepid remnant]

.close

[obtuse pebble]

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[tepid remnant]

oh thank youu

.close

[obtuse pebble]

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[real vector]

when dealing with vectors in R2, if I vector A, and both vector B and vector C are orthogonal to vector A, does this imply vector B and C are in the same direction and can be made to equal with multiplication of a scalar?

[real vector]

my intuition is ringing alarm bells telling me this is true

[golden oriole]

It is true in R2

[real vector]

perfect, thank you

im doing a homework problem and I needed confirmation this is true for my argument, if this wasnt it wouldve all fallen apart 🤣

in R3 this isnt true is it

[golden oriole]

Nope
In R3 you can have an entire plane of direction vectors perpendicular to one vector

[real vector]

thats what I was thinking too, thank you very much

I feel like a genius right now. been stuck on this problem for days and I finally saw a way through 🤣

@golden oriole thanks for confirmation !

[golden oriole]

Also by same direction i hope you are considering negative scalars and exactly opposite directions aswell

[real vector]

that is a good point, im going to relook at my answer now with this in mind

[obtuse pebble]

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[somber thorn]

How can you prove that every number is either be able to be divided by a prime number or it is a prime number?

[high lily]

that isn't entirely true though

[wise patrol]

Huh

Should apply for natural numbers greater than 1

https://en.m.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic#:~:text=In mathematics%2C the fundamental theorem,the order of the factors.

[somber thorn]

Euclids lemma. Okay thanks!

[kind hawk]

you just need induction for this. not FTA or euclids lemma

[wise patrol]

Yeah or just contradiction

[obtuse pebble]

@somber thorn Has your question been resolved?

[obtuse pebble]

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[sand spruce]

Given these complex numbers $u, v, w$ such that: [
\begin{cases}
|u+4-2i|=2\
|3v-1+i|=|2v+1-i|\
|w|=|\bar{w}+2+2i|
\end{cases}
]
Calculate $|w|$ as $S=|u-w|+|v-w|$ reaches its minimum

[warm shale]

Mildew

[sand spruce]

i let u = a + bi, v = j + ki and w = x + yi, and have found their loci

,,
\begin{cases}
(a+4)^2 + (b-2)^2=4\
(j-1)^2+(k+1)^2=2\
x-y+2=0
\end{cases}

[warm shale]

Mildew

[sand spruce]

but not sure how to plug in these loci into S

[timid silo]

U minus w and minus w

Represent thr shortest distance between the loci

The above loci are 2 circles and a line

Did you get it ?

[sand spruce]

yeah i think so

[timid silo]

You can solve further?

[sand spruce]

ok wait

ok so

for the u-w i got 2sqrt2 - 2

for v-w i got sqrt2

whats next?

[timid silo]

Are those individual minimas ?

[sand spruce]

yes

[timid silo]

But

You have to find a point where where the comb8ned equation is a minima

Try plotting a graph

You find find it

Might*

[timid silo]

Do you have the answer?

[sand spruce]

yeah, but i dont wanna look at it until im done

i do have 4 options to choose though

[timid silo]

Is root 2 an option

?

[sand spruce]

no

[timid silo]

I do not know how to solve it

But the idea is correct :))

[sand spruce]

lol

imma think for a while

might be solvable with geometry

[timid silo]

Is root 10?

[sand spruce]

yes

[timid silo]

:)))))))

[sand spruce]

one of the options is sqrt10 / 2

[timid silo]

Breh

Is there no root 10?

[sand spruce]

no lol

[timid silo]

I made a mistake

1 min

[sand spruce]

i prefer you give me the hints, and ill do the work myself

[timid silo]

I won't give the answer

I am trying to find out if my method is correct

Just try finding thr minimum from the graph

[sand spruce]

sure

[timid silo]

Try

I got it

But I don't know how to explain it

Hint:shortest distance between two circles

@sand spruce

[sand spruce]

hi

[sand spruce]

hm

oh wait

isnt that

distance between 2 centers - radius of circle 1 - radius of circle 2

i can calculate that easily

[timid silo]

Aye

Graph it

[sand spruce]

but isnt that just |u - v|

not exactly |w|

[timid silo]

That is not the a swer

Andwer

Answer

It is a clue

Graph it and you will bet the idea

Get

[sand spruce]

oh wait

it would be a segment connecting 2 circles

and will intersect locus of w, which is a line

[timid silo]

YES

[sand spruce]

and modulus of that intersection is the |w|

ez

ok wait let me

[timid silo]

Ezzz

[sand spruce]

sqrt10 / 2 is the final answer

thanks!

.close

[obtuse pebble]

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[obtuse pebble]

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[sly imp]

ok so idk how to explain it cuz idk anything about math and its not rly a serious question but im curious
but okay so video game example
lets say I have a 0.78% chance to see an item drop from a boss
and I have to roll 1-100 against 20 other people, the one with the highest roll wins.
What would be the full chance of getting that item?

[sly imp]

I don't know how to calculate this becuase I'm not good at math so a simple explenation would actually be nice so i can do it myself next time

[slow dove]

about the question you have there are lot more to it that you have to mention like
1)what if more than one person get 100
2)what if every person get the same number
and more...

[sly imp]

ok well lets say instead

1 in 20 will get it

like random

[slow dove]

still have same problems

for exp you have to mention no one can get the same number

[sly imp]

ah

well lets use my scenario

20 players rolled 1-100

i won with a 94 against the 20

[slow dove]

well if we assume that none can get the same number it will be almost 5% chance of winning

here is what I did
1/100 . (99/100)^19 .100

[sly imp]

ok i see

so how do we calculate the full chance

0.78% then 5%

[slow dove]

you multi it

would be 78/1000 . 5/100

[sly imp]

would it work as

0.0078 x 0,05?

[slow dove]

yes

[sly imp]

which then it means

0.039%

0,04

[slow dove]

yes

I dont know what you playing but that a rough game

[sly imp]

it's world of warcraft classic

I guess I got very very lucky

and now at least I know how to do it next time which is cool, thanks (:

[slow dove]

no problem

[obtuse pebble]

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[tepid path]

Sorry the question is in Japanese but what it is asking for is the length of P-Q
I can’t even start on it.
Help?

[glossy basalt]

Let PQ be h, then we can express QA and QB in terms of h

[tepid path]

I’ll try

Very messy but something like this?

I made pq = s as H gets confusing with hypotenuse

[glossy basalt]

ah i see

you use a, o and h that's why

yea, and then we can have
tan(45°)=QA/s
and
tan(60°)=QB/s

so we can solve those

[tepid path]

Is this right?

[glossy basalt]

reading

correct!

[tepid path]

Thank you so much

[glossy basalt]

Cheers!

[tepid path]

.close

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[frozen charm]

open

[frozen charm]

can someone help me at point C?

[polar fossil]

start by calculating B + 2C

[frozen charm]

-2 8 8 4?

[polar fossil]

sure

so now we have $2XA = \begin{pmatrix}-2 & 8 \ 8 & 4\end{pmatrix}$

[warm shale]

hayley table

[polar fossil]

we want to solve for X

[frozen charm]

so now we need a number for x?

or det for A?

[polar fossil]

X is a matrix

first thing we can do is get rid of that 2

[frozen charm]

ok

[polar fossil]

XA = ?

[frozen charm]

-4 16 16 8?

[polar fossil]

necesitamos dividir

-1 4 4 2

[frozen charm]

ohhhhh

i did 2x everything inside

[polar fossil]

it's just like solving in regular algebra

now we'd like to divide by A but.... we can't do that, A is a matrix

instead we multiply by the inverse

so $XA = \begin{pmatrix} -1 & 4 \ 4 & 2 \end{pmatrix}$

[warm shale]

hayley table

[polar fossil]

$XA\red{A\inv} = \begin{pmatrix} -1 & 4 \ 4 & 2 \end{pmatrix}\red{A\inv}$

[warm shale]

hayley table

[polar fossil]

do you understand?

[frozen charm]

no

[polar fossil]

do you know what A^-1 is?

[frozen charm]

ye

[polar fossil]

we are multiplying both sides of the equation by A^-1

[frozen charm]

thanks for help

im am giving up at math

how do i close?

[polar fossil]

.close

[obtuse pebble]

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[frozen charm]

ok

thanks

[obtuse pebble]

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[ember frost]

How many ways are there to arrange 4 Vietnamese, 5 French and 2 American to sit on a line of 11 chairs such that people of same nationality must sit together?

[ember frost]

not sure how to aprpoach this

[swift stream]

Treat the group of 4 vietnamese, 5 french and 2 americans as one unit each

Now how many ways are there to arrange these three units?

[ember frost]

3P3

6

[swift stream]

Yeah, now we permute the people in the units themselves

For the vietnamese group of people, how many ways are there to arrange them?

[ember frost]

4!

[swift stream]

Do the same for the other two groups

[ember frost]

ok so

4!5!2!

that makes me wonder, wouldnt something like VVFFFAAVVFF work? it doesnt necessarily have to be rigid like VVVVFFFFFAA

the former still satisfies the original question

[swift stream]

I interpreted "so that the same nationality must sit together" as they must sit in one group, like VVVVFFFFFAA

[swift stream]

Whole thing times 6

[ember frost]

yeah just reread the question, this makes more sense

[ember frost]

yes

so its

4!5!2!6

oh wait this is not probabolity

[swift stream]

It doesnt, because the people of the same nationality refers to the whole group of people. Not smaller groups of people of the same nationality like you are suggesting

[ember frost]

right

so the final answer should be 34560

,calc 4!5!2!*6

[warm shale]

Result:

34560

[ember frost]

thanks!

.close

[obtuse pebble]

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[swift stream]

👍

Sorry, wrong emohi

[ember frost]

https://cdn.discordapp.com/emojis/1048660781123780749.gif?v=1&size=48&quality=lossless

technically the pre-edit still fits

[obtuse pebble]

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[silk maple]

hi does anyone know if theres a simple way to find the limit for x approaching infinity of this series function?

[obtuse pebble]

@silk maple Has your question been resolved?

[lapis thunder]

hmmm

ratio test?

An+1 / An

[silk maple]

hmm

.close

[obtuse pebble]

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[lapis thunder]

you got it or u need help?

[silk maple]

ill just give up

[silk maple]

it does converge

but idk to what

[lapis thunder]

i dont think i understood your question

[hybrid gyro]

the question is asking what is the limit of the series as x tends towards infinity, not whether if the series converges

[lapis thunder]

oh

i thought it was about range of x

[hybrid gyro]

no worries lol

[hybrid gyro]

oh wait I may have figured it out. As x tends towards infinity, we can drop the absolute

so we have $\frac{ln(x-n)}{n^2ln(x)}$

[warm shale]

lgkoo

[hybrid gyro]

but $\frac{ln(x-n)}{ln(x)}$ tends to 1

[warm shale]

lgkoo

[hybrid gyro]

so the series tends to $\sum_{n=1}^{\infty} \frac{1}{n^2}$

[warm shale]

lgkoo

[hybrid gyro]

which is famously known as $\frac{\pi^2}{6}$

[warm shale]

lgkoo

[lapis thunder]

most beautiful explaination ive seen

[dark nexus]

Can someone help me to understand if the answer of this problems are correct?

[lapis thunder]

#❓how-to-get-help

[hybrid gyro]

thx, but idk if this is rigorous enough though

pls send in a new channel as this is occupied atm

[dark nexus]

Yes sure! Sorry

[obtuse pebble]

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[graceful pumice]

how to do

[graceful pumice]

i tried ratio test

[lapis thunder]

p-series

so

(e/pi)^n

because the base is < 1

it converges

and you should use the convergent series sum formula

[graceful pumice]

there is a sum formula??

lemme search rq

[lapis thunder]

yep

[graceful pumice]

oooh

the stupid geometric series thingy

[lapis thunder]

yess

[graceful pumice]

i always forget to apply

a1/1-r

that thing

[lapis thunder]

yeah

[graceful pumice]

icic

thx

[lapis thunder]

np

[graceful pumice]

👍

[obtuse pebble]

@graceful pumice Has your question been resolved?

[graceful pumice]

what would be the first step in this?

[hybrid gyro]

by "inspection"

what does e^x-x differentiates to?

[graceful pumice]

e^x-1

ohhhhh

[hybrid gyro]

yup, which is the numerator in this case

[graceful pumice]

omfg

ur literally him

thx

[hybrid gyro]

np

[obtuse pebble]

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[timid silo]

How would I do this?

I've figured that f(x) cuts through -2 and 10

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

<@&286206848099549185>

[oblique grail]

what?

[timid silo]

I need assistance

[stiff glade]

What do you need help with ?

[oblique grail]

with what?

[timid silo]

part a

the equation for f(x)

[stiff glade]

It's a quadratic euatiion

[oblique grail]

k ill take a look at it

[stiff glade]

equation

[timid silo]

ik

turning points are 4,24

[stiff glade]

Okay

[timid silo]

(x-4)^2 + 24

since it's (x-4)^2

[stiff glade]

You use the quadratic formula

Hello

[oblique grail]

(10,0)

[timid silo]

what's (10,0)?

[timid silo]

no u dont

[oblique grail]

the positive x-intercept

answer to part a

u know why?

[timid silo]

i know it's 10

i meant part 2, lmao

yeah

[stiff glade]

@oblique grail You stole my answer

[oblique grail]

oh ok

[timid silo]

quadratics are symmetrical

2 + 4 = 6, so add 6 onto 4 = 10

[oblique grail]

y=-2x-4

for part b

@timid silo i got the answer to part b

[timid silo]

is it?

[oblique grail]

yes

[timid silo]

says something completely different here

[oblique grail]

f(x)=-2/3(x^2-4)+24

[timid silo]

it wants the equation of the quadratic

[oblique grail]

oh that

for the equation i got -2/3(x^2-4)+24

[timid silo]

huh

?

[oblique grail]

wait -2/3(x-4)^2+24

[timid silo]

did u get the answer up there?

[oblique grail]

no i found the quadratic that works

[timid silo]

i need to get the exact quadratic

[oblique grail]

-2/3(x^2-8x+16)+24

[oblique grail]

so -2/3x^2+16/3x+40/3

[timid silo]

really weird explanation

[oblique grail]

-2/3(x^2)+16/3(x)+40/3

[timid silo]

how did u get it?

mind explaining

[oblique grail]

so u know the vertex, which is (4,24) right?

[timid silo]

yes

turning point

[oblique grail]

so to get to the vertex you have a(x-4)^2+24, where a is a constant

do u understand why?

[timid silo]

no

[oblique grail]

ok

[timid silo]

i get this part

just not the a

[oblique grail]

because you haven't been given the x-intercept yet, and the a value determines the x-intercepts

so the x-intercept is -2

you then plug in -2 and get -2=a(-6)^2+24

so then you get a=-2/3 if you solve the quadratic equation

so the solution is -2/3(x-4)^2+24, which is -2/3(x^2)+16/3(x)+40/3

[timid silo]

okay wait so

lemme go through this

a(x-4) + 24

[oblique grail]

a(x-4)^2+24

[timid silo]

a determines x as we haven't been given it yet

[oblique grail]

yes

[timid silo]

ye that

okay what do you do exactly from there

make it

what

[oblique grail]

but then it tells you that it intersects the x-axis at (-2,0)

so you plug in -2 for x

so you have a(-6)^2+24=0

so -36a+24=0

[timid silo]

-6?

a(-6)^2+24
36a+24

oh ye

-6

[oblique grail]

wait 36a+24=0

im stupid

so then you have 36a=-24, so a=-2/3

so -2/3(x-4)^2+24=0

do u get why now?

[timid silo]

y = -2/3x

[oblique grail]

y=-2/3(x-4)^2+24

because it is a(x-4)^2+24, but a is -2/3

[timid silo]

yes

[oblique grail]

ok i gtg now

bye

[timid silo]

oh shoot

bet bet

lemme do it now

[oblique grail]

ok

good luck

[timid silo]

thank you

why did i get sum different

i got

[obtuse pebble]

@timid silo Has your question been resolved?

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[latent frost]

Hello! So, will try to translate this question to my best of abilities but let me know if there's something unclear

[latent frost]

A is a 42 by 40 unknown matrix where 0 is the 42 by 1 matrix with only zeros

Figure out which of the following statements are true and motivate why so

If the equation Ax = 0 has one particular solution, then the equation Ax = y has one solution for every 42x1 matrix y

[versed stratus]

$$42 by 40$$?

[warm shale]

ƒ(Why am. I here)=I don't Know

[latent frost]

So this one I would say is false, since the last two could just be zeros

yep

[modest ingot]

what does "where 0 is the 42x1" even mean?

[latent frost]

Ax = 0

[versed stratus]

you mean $O$

[warm shale]

ƒ(Why am. I here)=I don't Know

[latent frost]

uhhh

[versed stratus]

right, null matrix of order 42 by 1

[latent frost]

i mean it does look like a 0 ngl

yes

so that one I would say is false because the last two rows could just be 0 0 0 0 ...

then for the next one

the null dimension / zero dimension for A is >= 2

and this I would say is false, since the dimensionformula(???) says that the

man this is not easy to translate

but linearly independent + linearly dependent columns = n in R^n

[versed stratus]

use google translate?

[latent frost]

it aint good with math stuff :/

ill try

rank + zero dimension = R^n

so to me the first one and second one shouldn't be true

but for the third one

If the vectors v 1, v 2 are bases for the zero room to a, there is linearly
independent y 1, y 2, y 3, so that to each of the equations
Ax = y1, ax = y2, ax = y3
there's no solution

My original thought here is that the zero dimension / null dimension is given by the amount of parameters we get from gaussian elimination

but then I think that those vectors are what we would get the null matrix if we put in right?

so what does it mean to.. not have a solution?

the answer to the question is that C is true, if v1 and v2 are bases for the null space then the null dimension is 2. Then the Rank–nullity theorem gives that the "value room" is of dimension 38, there is therefore 4 linearly independent y

but I am so confused by this explanation :C

welp

essentially what im asking is what does it mean, in terms of matrixes, that for example Ax = y1 doesn't have a solution and how would I calculate the amount of unsolveable equations?

[modest ingot]

ngl the translation sucks

[modest ingot]

from the rank nullity theorem, you have that the columns of A = rank(A) + nullity(A)

[latent frost]

yep

that's my thought too

[modest ingot]

since you did affirm that v1 and v2 are basis for the null space then you have that nullity(A) = 2 then rank(A) = 38 right?

[latent frost]

but what would it mean that Ax = y doesn't have a solution?

yes

so my understanding is that if I multiply A with v1

[modest ingot]

if rank(A) = 38 that means there are 38/42 rows that are linearly independent and 4 are linearly dependent

[latent frost]

i would get the null matrix

is that correct?

[modest ingot]

it's a basis so any linear combination of those basis would give you Ax = 0

but yes sure

[latent frost]

so what would it then mean that Ax = y doesn't have a solution?

That if I do gaussian elimination I get for example that b = 2 and at the same time in another equation b = 6?

[modest ingot]

you have 38 rows that are linearly independent which means when you do gaussian elimination

those 38 "variables" can be solved for

[latent frost]

wait

[modest ingot]

but the other 4 can't really be

[latent frost]

is it rows or columns? or both?

[modest ingot]

because you'd have rows of 0s on doing gaussian elimination

[modest ingot]

number of linearly independent rows = number of linearly independent columns

[latent frost]

oh

so I have 2 linearly dependent columns

which means 4 linearly dependent rows

thank you

you are an angel

[modest ingot]

rows makes sense since we're talking about gaussian elimination

that's why i switched to rows

[latent frost]

yeah but that just made me realize where we get the number from

[modest ingot]

[latent frost]

awesome

hope you have an amazing day !!

<333

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[latent frost]

Hilfe ;D

[latent frost]

So this is in swedish, ill translate asap

The linear transformation F transforms the vector (0,1,0) to (1,1,0) and transforms the vector (0,0,1) to (1,-1,2), later (1,1,1) is an eigenvector with the eigenvalue 1 to F

Find the A for F

so my idea was

the columns in a matrix should be the place the x, y and z ends up

so I did

a1 1 1
a2 1 -1
a3 0 2

as the column

and then as for X, well we know that (1,1,1) becomes (1,1,1)

so I set up the equation system

a1 + 1 + 1 = 1
a2 + 1 - 1 = 1
a3 + 2 = 1


which gives

a1 = -1
a2 = 1
a3 = -1

plug the values back in

-1 1 1
1 1 -1
-1 0 2

and that's our A matrix, right?

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[teal granite]

I am having trouble with this exercise I used the example to do it but im off and I think its because there are 3 dice so I dont know how it will affect the probabilities

[obtuse pebble]

@teal granite Has your question been resolved?

[obtuse pebble]

@teal granite Has your question been resolved?

[obtuse pebble]

@teal granite Has your question been resolved?

[obtuse pebble]

@teal granite Has your question been resolved?

[obtuse pebble]

@teal granite Has your question been resolved?

[obtuse pebble]

@teal granite Has your question been resolved?

[teal granite]

<@&286206848099549185>

[teal granite]

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[dreamy hull]

If set A has cardinality m and B has cardinality n then what is the cardinality of A^B?

[fossil crag]

what's your guess?

[obtuse pebble]

@dreamy hull Has your question been resolved?

[dreamy hull]

Yes they have but I can't figure out how

Please help to derive

[obtuse pebble]

@dreamy hull Has your question been resolved?

[molten night]

Do you know how A^B is defined?

(Set theory context)

[obtuse pebble]

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[candid charm]

Which row and column is 2025 in?

[obtuse pebble]

@candid charm Has your question been resolved?

[timid silo]

What?

[candid charm]

What ?

[timid silo]

Wait, let me try to think

I see the pattern but I can’t remember the shortcut I learned before 😭

[polar fossil]

look at just the first row or the first column. how do the numbers go up?

[candid charm]

Uhh idk

<@&286206848099549185>

[obtuse pebble]

@candid charm Has your question been resolved?

[obtuse pebble]

@candid charm Has your question been resolved?

[candid charm]

.close

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[neon iron]

hi so theres this question and I am not sure what I'm doing wrong

[neon iron]

the highlighted part

I keep getting sin(theta)+4cos(theta)

this is after expanding the second bracket on the top line

[stark thorn]

can you explain where u r getting stuck exacly?

[neon iron]

so you see the highlighted part

[stark thorn]

yup

[neon iron]

I expanded the sin(theta) + 2cos(theta) squared and I keep getting something different than what that mark scheme says

[stark thorn]

what r u getting

[neon iron]

I am getting sin(theta)4cos(theta)

for the highlighted part

[stark thorn]

isn't that the same as this?

[neon iron]

the terms around it I got

[neon iron]

no, notice where the 4 is

[stark thorn]

by commutativity of multiplication

[neon iron]

what is that?

[stark thorn]

how does that affect it

[neon iron]

it wont cancel out then

[stark thorn]

ab = ba for real numbers a and b

so 4 (sin x) (cos x) = (sin x) 4 (cos x)

isn't it?

[neon iron]

so this is a rule? cause I did not know that until now.

[stark thorn]

5 times 2 = 2 times 5 right?

[neon iron]

yes

[stark thorn]

that's what this 'rule' is saying

[neon iron]

so basically the thing I worked out is correct because ab=ba?

[stark thorn]

u can swap the numbers for the multiplication operator

[neon iron]

so I can cancel it

[stark thorn]

yeah

[neon iron]

alright thanks

I get it now

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[obtuse pebble]

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[solemn cave]

the exponent is a sqrt

and the number is a negative

[winged estuary]

Do I flip it

[solemn cave]

yea but thats not the main concern

this number is negative too

[winged estuary]

Flip both

[solemn cave]

nah..the thing is

you cant really take a sqrt of a negative number

[winged estuary]

So what do I do

[solemn cave]

nothing basically

theres no whole number

[winged estuary]

So it's 999

[solemn cave]

pretty much

[winged estuary]

One sec, so if a number is negative and has a negative exponent it just doesn't exist right

[solemn cave]

no no no

negative exponent flips the number

i was talking about the 1/2

x^1/2 is sqrt(x)

[winged estuary]

Oh I think I get it now thank u

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[ruby sequoia]

huh

[heady dock]

helpp

[ruby sequoia]

put the values in the formula

little

fish

[pastel wren]

lol

[ruby sequoia]

fish snake!!!!!!!!!!!

[heady dock]

sawwy

[pastel wren]

yeha put in 2b instead of r and 5b+3 instead of h

and simplify

!nosols

[obtuse pebble]

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.

[ruby sequoia]

meh

what is this question anyways

[pastel wren]

its that easy question its a crime to tell the answer

[heady dock]

what youtube vid can i watch to help with this like topic in particular

[ruby sequoia]

[pastel wren]

do you know how to solve this problem?

[ruby sequoia]

@pastel wren bro

[heady dock]

no

not at all

[ruby sequoia]

im gonna cry

[heady dock]

thats why im here

LOL

[pastel wren]

that doesnt help you?

[ruby sequoia]

are ufor sure?

if i give u r =5 and h =1

what would be the volume

[heady dock]

yes like i see that u plugged it in but how did u get 20?

[pastel wren]

plug this and show us

if you know what we arew even talking about

[heady dock]

;-;

[ruby sequoia]

write it

[heady dock]

okay

i think i did it wrong

[pastel wren]

dont write pi as 3.14

just keep it as pi

[heady dock]

oh ok

[pastel wren]

its correct

[heady dock]

but idk how u got 20

[pastel wren]

do you know how to simplify this?

[heady dock]

no

not at all

[pastel wren]

start with the square

whats (2b)^2?

[heady dock]

4b?

[pastel wren]

close

[heady dock]

;-;

[pastel wren]

its 4b^2

you square the 2 and the b

[heady dock]

ohh

does the parenthesis make a difference

like with the parenthesis are they two completely different things

[pastel wren]

so we got 4pib^2(5b+3) right?

[pastel wren]

it does

make a diffrence

[heady dock]

ok

hmmmm

[pastel wren]

do you know how to multiply things like this: a*(b+c)?

[heady dock]

lemme try

[pastel wren]

ok

[heady dock]

no

nvm

i do with normal numbers

but with exponents no

i should just look at a youtube video

;-;

[pastel wren]

those numbers are completely normal ;_;

[ruby sequoia]

little fish

[heady dock]

do i do 2(5b+3)?

but isnt that an exponent

[pastel wren]

what have you got so far?

just write it on paper and show

[heady dock]

theres a ^ in front of it

okay

[pastel wren]

keep the (5b+3) for now

[ruby sequoia]

snake teaching fish

[pastel wren]

yeah

[ruby sequoia]

how to swim

[pastel wren]

evolution

lol

[ruby sequoia]

yea

[heady dock]

I only have this

Idk what to do next

[pastel wren]

its correct thats good

[ruby sequoia]

@pastel wren can i scare lil fish?

[pastel wren]

yeah sure lol

[heady dock]

Noooo

Noooo

[ruby sequoia]

solve this @heady dock

[heady dock]

Waittt so now I see how u got 20

😭 nooooo

[pastel wren]

i love my diffrential equations

[heady dock]

I don’t wanna look at that

[pastel wren]

imagine this a single number (in practice it is lol)

[heady dock]

Is that right?

[pastel wren]

no

[heady dock]

Aw

[pastel wren]

how would you multiply 3*(5b+3)?

[heady dock]

hold on guys i need to help my friend stick mail under the door

can i close this?

thank you so much for ur time i need to go

byebye

[pastel wren]

yeah sure you can open a new one anytime

bye

[ruby sequoia]

aw

bye lil fish

[obtuse pebble]

@heady dock Has your question been resolved?

[heady dock]

20pib^3+3?

no?

im back

is anyone still here?

https://tenor.com/view/judgement-cat-cat-angry-cat-cat-face-cat-forehead-gif-4731341898773190051

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[harsh goblet]

a regular convex polygon has 10 times as many diagonals as its sides. Determine the sum of the interior angles

[harsh goblet]

I wrote up this and got that the poligon is has 23 or 0 sides:

but now I don't know what do

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[topaz adder]

How does (6)2-3(6)-18 = -24? When I am calculating it, I am getting 12 or 16, a positive number but I know that is incorrect. I know it reduces down to 12-18-18 but I am unable to obtain -24. What am I doing wrong?

[deep skiff]

6 * 2 - 3 * 6 - 18

think of it as

(6 * 2) - (3 * 6) - 18

12 - 18 - 18

[topaz adder]

That's where I am stuck at.

I have already reached that part of the equation.

-6?

OOO

[deep skiff]

lmao

[topaz adder]

Ok, ok. I think i got it now

Thakn you

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[robust current]

I'm working on a personal project and this trig problem came up. I'm using FreeCad drawer to give the visuals.
My knowns are in pink and I want to solve for θ.
The second image are values that FreeCad gives for the "answer" but I cannot come up with it mathematically.

This is as far as I got:
c1 + b2 = 50
c1 = a1/sin(θ)
b2 = a2/tan(θ)
a1/sin(θ) + a2/tan(θ) = 50
10/sin(θ) + 15/tan(θ) = 50 ...stuck here, I don't know how to reduce it to θ = ___ ?
Plugging in the the θ that FreeCad gives, returns 50, so the form must be correct so far.

[deep skiff]

I feel like we're missing some info here, no?

[robust current]

FreeCad considers it a constrained sketch given the 3 sides and the 90 deg angles. So I would think it could be solvable...maybe a quadratic equation?

[deep skiff]

Oh wait

c1 = 10 / tan(theta)

b2 = 15 / tan(theta)

c1 + b2 = 50

solve for theta

[robust current]

I'll try on paper.

So according to that I get :
10/tanθ + 15/tanθ = 50
10 + 15 = 50tanθ
θ = atan((10+15)/50)
θ = 26.565051177 ...bit off

[deep skiff]

ye, weird

[humble valve]

c1 isnt equal to 10/tantheta

[deep skiff]

Oh it's the hypotenuse

cursed be my inability to perceive upside down triangles

[humble valve]

so true

[robust current]

Plugging 27.7434 into 10/sin(θ) + 15/tan(θ) = 50 holds true...just don't know how to reduce the equation :/

[humble valve]

i mean you can change 15/tantheta into 15costheta/sintheta

idk that it helps much though

[deep skiff]

it breaks down to 50 sin(t) + 15 cos(t) + 10 = 0

this is managable but it's ugly

isolate 50 sin(t) on one side

apply trig identitiy to turn it into cos as sqrt(1-cos^2(t))

square both sides

solve quadratic

[humble valve]

-50sint

but yeah

[deep skiff]

How dare you correct me for a second time

😭

I had a rough day okay

[humble valve]

mb bro you good

it happens

fr

[robust current]

lol ...so what's the quadratic look like?

[deep skiff]

hell on earth

[robust current]

trig hell

[humble valve]

100 + 300cost + 225cos^2 t = 50 - 50cos^2 t

something like that

fr

[deep skiff]

ya didn't square the 50

[humble valve]

you right fr

rhs should be 2500 - 2500cos^2 t

fr

[robust current]

hmm, it's getting uglier by the minute

[humble valve]

indeed

[deep skiff]

And then you solve that for the sharp angle

and you get it

[robust current]

So I have to plug that last line in a quadratic formula?

[deep skiff]

Here it is

ugly as hell

[humble valve]

never catch me doing that

fr

[robust current]

Nice, that's what I was looking for. Thanks.
I had a deep feeling there was a quadratic involved...

How do I close the topic?

[humble valve]

.close

[robust current]

Sweet, thanks.

.close

[obtuse pebble]

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[obtuse pebble]

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[strange stag]

In this definition of the invariant space-time interval we see this weird product of matrices, i think im confused about the einstein notation, why does the metric tensor (Eta,mu,nu) has its indices in the bottom? why arent them on the top? what would the difference be? or even worse, what if mu was a superscript but nu a subscript (and vice-versa) im confused :(

[obtuse pebble]

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@strange stag Has your question been resolved?

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@strange stag Has your question been resolved?

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[atomic cape]

Need help for the 2929 time

[atomic cape]

We have to make a function for f(x) and g(x)

The thing is I don't get why we need growth factor and how to use it here

[woeful void]

What is the question?

[atomic cape]

Idk the exact translation in English but "find the function expression for f(x) and g(x)"

[woeful void]

Could we use something like e^x?

[atomic cape]

Nope

[woeful void]

Well g is very simple

Think of some kind of bases like 2 or 10

[atomic cape]

I'm still confused

This is what my textbook said

[woeful void]

g=2^x

Well I mean I just did some logical guessing lol

[atomic cape]

I just don't get how they find the growth factor and why😭

.close

[obtuse pebble]

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[wide oracle]

how does this work?

[crude coral]

definition of current?

@wide oracle u there

[wide oracle]

i dont get the n part

what is n

[crude coral]

n is a integer

nq is number of charges passing through

current is define as number of charging passing through a conductor in a unit time

[wide oracle]

oh okay

ok thanks

.close

[obtuse pebble]

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[timid silo]

how do i do this? i have no working out idk how to do it

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

@analog vault

[glossy basalt]

Hint: Consider from 1° to 89°
sin(theta)=cos(90°-theta)
sin²(1°)=cos²(89°)=1-sin²(89°)

so,
sin²(1°)+sin²(89°)
=1-sin²(89°)+sin(89°)
=1

then,
sin²(1°)+sin²(2°)+...+sin²(88°)+sin²(89°)
=1+1+...+1+(1/√2)²
=44+(1/2)

[timid silo]

wait im confused as to why u consider from 1 to 89

[glossy basalt]

that's the hint for the first quadrant

[timid silo]

oh so for like the 359 do i have to consider the fourth quadrant how would i do that?

[obtuse pebble]

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[mellow epoch]

Hint: what is the value at sin 1 degree? and what is the value of sin 359 degrees?

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[limpid saffron]

What is the upper time complexity of Euclid’s algorithm for computing GCD in terms of the number of bits representing the two numbers?
Group of answer choices

n1/2

log2 n

n3

n

[limpid saffron]

log_2(n)?

[obtuse pebble]

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[timid silo]

I need to find if it is bounded or not

[heavy shore]

root test

[timid silo]

What is that

[woeful void]

Heyyy math man nice to see u again hahaha

Saw u yesterday

[timid silo]

Yes

I remember you

Funny cat

[woeful void]

Lol

[heavy shore]

um you haven't done root test?

[timid silo]

I think no

I will check it on youtube

[heavy shore]

would be better

[timid silo]

Ok

[wise patrol]

when applying the root test, it might be easier to convert to sigma notation first $a_n= \sum_{n=1}^{\infty} \frac{1}{n^n}$

[timid silo]

I think i understand

[warm shale]

Cyphercrit

[timid silo]

I see

Done