help-0

[vapid shuttle]

they share the same slope

if they were in the same plane they would never intersect

well

I think they are technically skew

but if they were in the same plane they would be parallel

but since they are not in the same plane they are skew

sorry to backtrack there

[alpine sable]

In three-dimensional geometry, skew lines are two lines that do not intersect and are not parallel
https://en.wikipedia.org/wiki/Skew_lines

they are parallel (and not skew)?

I presume the drawing is a cube

the distance between A and G is the same as the distance between B and H

[vapid shuttle]

parallel is for lines in the same plane

[plucky surge]

Yeah thats what I was confused about

[vapid shuttle]

^

[ripe basin]

it was sufficient i was in a call im so sorry

[alpine sable]

ABGH generates a plane

[plucky surge]

yeah

[vapid shuttle]

ah then nvm again I'll double backtrack

XD

[plucky surge]

hahah

i think the official definition is that same direction vector = parallel

and the lines definitely have the same direction vector

thanks guys!

.close

[lone heart]

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[alpine sable]

Hi can anyone help me?

[ivory pivot]

x=2*55 right?

[alpine sable]

I got that right but idk abotu ahy

about Y*

[ivory pivot]

ADCB its a quadrilateral inscribed in the circle hence y+55 =180 right?

[alpine sable]

yeah

[ivory pivot]

now just solve for y

instead do you now the proof that x=2*55?

[alpine sable]

angles at a center

[ivory pivot]

but do you know the proof?

[alpine sable]

isn't the proof just angles at a circle

or am I wrong g

[ivory pivot]

what says the theorem "angles at a circle"?

[alpine sable]

what does it say?

[ivory pivot]

... i say that x=2*55 because i used this theorem

[alpine sable]

ohh

[alpine sable]

so that'd be the reason?

aigh

[ivory pivot]

do you know the proof of that theorem ?

[ivory pivot]

yes

[alpine sable]

what about this one?

can you help me

[alpine sable]

what about the reason

so it's the same theorm just different positions?

[ivory pivot]

yes for the 2 question

for the first its just the sum of anlges of trianle ...and angle between the ray and the tangent its 90

[chrome vortex]

angles subtended at the centre of the circle is double the angle at the circumference

i think thats how they write it in gcse

[alpine sable]

and for this wuestion it would be related to tangents right?

Angel C would be 90 degrees

or am I wrong

[ivory pivot]

B*

good night

[alpine sable]

Good nihft

I'll try to solve the rest of it alone

I don't think i can but yeah

[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

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[spare fern]

where did the cos^2 go?

[gray isle]

compound angle identity was applied

[spare fern]

which is?

[plain flame]

$\cos (2 x)=2 \cos ^2 x-1$

[ocean seal]

heavy

[gray isle]

also dodgy notation on that page

[spare fern]

how so?

[gray isle]

improper () to clearly indicate arguments of the cos

I dont mid the first two as much
but the
cos2(6x)
is horrid

[spare fern]

yeah It's a screenshot from a yt video

[plain flame]

in my humble opinion

the only time writing cos/sin without brackets is acceptable

is when its just x

so cosx or sinx

[lone heart]

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[alpine sable]

What makes x positive?

[tacit arch]

sqrt(...)<x

Since square root is nonnegative, then x is positive

[alpine sable]

Ohhhhhh

Thanks and sorry 😔

[tacit arch]

All good

[lone heart]

@alpine sable Has your question been resolved?

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[vapid steppe]

yo

[vapid steppe]

,rotate

[vapid steppe]

how do i do this

did i do it correctly so far

[remote heron]

11?

[lone heart]

@vapid steppe Has your question been resolved?

[vapid steppe]

i did (16+6)/2

[tiny fable]

should be right

[lone heart]

@vapid steppe Has your question been resolved?

[vapid steppe]

how do i finish the equation

[tiny fable]

you're already 99% of the way there

what goes after x in there

after y was the major axis length, which you found

so after x is what

[vapid steppe]

4?

[vapid steppe]

how can y be the major axis if it’s 25 while the minor axis is 64

[tiny fable]

oh my bad, i was being dumb

the denominator is the square of the minor and major axes

the stuff after y was the center of the ellipse

so 11 was the y coordinate of the center

and yes

4 is the x coordinate of the center

[lone heart]

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[potent hare]

Hello! I am playing a game and looking for the most mathematically efficient/fastest way to solve the puzzle.

There are a few rules and some information to help:

• This game has 6 players, split in to two teams of 3 players, those teams are divided between the blue and brown colors on the map. Any player can cross between the two halves of the map any number of times (ideally the fewest times possible).
• The two teams of 3 players will both start at their respective number 1 on each half of the map.
• The numbers must be 'activated' in order to complete the puzzle.
• In order to activate a number, you must first stand at n-2 of the current number you are trying to activate to receive a buff that will allow you to activate a number.**
  Any number of players may stand at a number at one time to recieve this buff. i.e: all three players will stand at number 1 to receive this buff.
• **The numbers may only be 'activated' by a player if you are at n-2 of the current number you are trying to activate. This rule only applies to number 4 and beyond. Prior to number 4, the buff you need to activate a number will always be at number 1. i.e: a player could not skip to number 5 and wait for 4 to be done without first standing at number 3 to recieve the buff necessary to activate the number.
• Any player can activate any of the numbers respective to their colour, so long as they are done in order.
• The goal is to activate all six numbers in order in the fewest moves possible, as fast as you can.

If you'd like to try and help, try just doing the blue side only first! Remember each side only has 3 players, and a number cannot activated without first standing at
n-2 of the number you are trying to activate.

Thank you :D

[lone heart]

@potent hare Has your question been resolved?

[potent hare]

For some more information, the way we solve it right now is very inefficient. Only 1 person out of each team will oversee activating the numbers, in an order that goes like this: 1->2->1->3->1->4->3->5->4->6.

[lone heart]

@potent hare Has your question been resolved?

[lone heart]

@potent hare Has your question been resolved?

[lone heart]

@potent hare Has your question been resolved?

[lone heart]

@potent hare Has your question been resolved?

[shell widget]

Try #discrete-math

[prime badge]

so what stops you from doing this

or it's like, someone else must be at n−2 in the present, not just you in the past

[prime crown]

masmis

[proven solar]

yo

How can I calculate volume in cm^3

[pseudo schooner]

Hello,I need help in my assignment and the submission is after tomorrow, I searched on google but idk if I can trust it cuz its different from the example its central limit theorem,z score btw here is the question

In a study of the life expectancy of 400 people in a certain geographic region, the mean age at death was 70 years, and the standard deviation was 5.1 years.. If a sample of 50 people from this region is selected, what in the probability that the mean life expectancy will be less than 68 years!

I'm having hardtime in labeling and idk what to do with "less than 68 years".
Should I ignore the 400 people? Cuz in google they did ignore it

[lone heart]

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[pseudo schooner]

.reopen

[green gulch]

How does this have a solution?

[granite wing]

!show

[lone heart]

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

[green gulch]

I really don't know where to start

I mean, when x>0 its always increasing

[granite wing]

its not

,w plot x^4*ln(x)

[ocean seal]

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[green gulch]

oh sorry

I did not see lnx

[granite wing]

🙂

[green gulch]

that makes a lot more sense

.close

[lone heart]

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[near crag]

Wtf

[alpine sable]

does that say n! * 2n! in the denominator? and n^3 * 2n! in the numerator?

[near crag]

yes to both

and + (2n-1)*n!

[alpine sable]

Ok, thanks

[near crag]

np, but i've been stuck on this problem for the last 15 minutes

i'm just dying

[lone heart]

@near crag Has your question been resolved?

[lone heart]

@near crag Has your question been resolved?

[alpine sable]

i got to:

$ae + \frac{b}{e} + c = \sum_{n=0}^{\infty}\frac{n^3}{2n!} \quad + \quad \sum_{n=0}^{\infty}\frac{1}{(n-1)!} \quad - \quad \frac{1}{2}e$

[ocean seal]

hannibal

[alpine sable]

before getting stuck

can't help past that

[lone heart]

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[green gulch]

Not getting much progress with this question

[green gulch]

just to make sure, you need to prove IVT applies first right?

[last ether]

Yes

[green gulch]

so I have that x must be > 0

Im sure im doing something wrong trying to figure out IVT

[last ether]

Determine whether the function is continuous on its domain

[green gulch]

but basically I thought that if a function has a point below and above 0 and is continous

a solution exists

but for this case there is none below 0

[last ether]

Are you sure

Your solution curve is this:

$$s(x) = 207e^x + 2023\ln(x) - \pi^{24}$$

[ocean seal]

Umbraleviathan

[last ether]

There are values of x such that x > 0 such that s(x) < 0

[green gulch]

Oh right

s(x)

I guess what im having trouble with then is finding the values

because I cant solve for s(x) = 0

[last ether]

No

But you can prove that s(x) is continuous for all x > 0

And then you can prove that there exists a value of x such that s(x) < 0 and there exists another such that s(x) > 0

Then you need to prove that s'(x) is continuous for all x > 0

[green gulch]

so to get those two points id kinda have to guess and check in a way?

[last ether]

And prove that s'(x) > 0 for all x > 0

So there's a lot

[green gulch]

right

MVT

But to get those two points where s(x) is < and > 0

is it kinda like guess and check

[last ether]

I mean let's be real

Let x = 0.001

That's definitely gonna be below 0

And then let x = 10000000000000000

That's definitely gonna be above 0 lol

[green gulch]

okay lol I see I see

I just wanted to make sure

[last ether]

And then you just need to show that s'(x) is either monotonically decreasing or increasing

[green gulch]

just to make sure, I have to make sure s'(x) is dec/inc or s(x) is dec/inc?

I have to study more on MVT

.close

[lone heart]

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[prime vessel]

can someone help me get the answer to part b? my brain isnt braining

[prime vessel]

i know that the answer to part a is 156cm

[last ether]

Do you know what integers are

[prime vessel]

yes

whole number

[last ether]

Well whole numbers and integers are different sets but sure

And I'm assuming they want a nondegenerate rectangle so the side lengths will end up being natural numbers

You know that to get the perimeter you need to add up 4 numbers, 2 of which are the same and 2 are also same

[prime vessel]

yep

[last ether]

It's also possible that all sides are congruent

Start with the lowest natural number

[prime vessel]

in this case would it be 3

[alpine sable]

bro how to start help

mine is not starting

[last ether]

#❓how-to-get-help

[last ether]

3 is not the lowest natural number

[prime vessel]

it cant be 1 can it

[last ether]

Oh wait I'm dumb

[prime vessel]

so 2?

[last ether]

I didn't read the text above it

[prime vessel]

oh

[last ether]

Dude

#❓how-to-get-help

Did you read it

I don't think you did

[alpine sable]

yes

it said to send in help 0

and this probably help 0

[median oar]

Send it in 15

[last ether]

No, it said to choose any vacant channel

Not just help-0

[median oar]

#help-15

[alpine sable]

oh ok sorry very very sorry

pls

[prime vessel]

👍🏻

[last ether]

Well hm now I'm thinking

Because I didn't read the excerpt at the top

Do you know calculus optimization?

[prime vessel]

i dont

[near crag]

how did you get 1/2 e?

[prime vessel]

bro?

[last ether]

Ffs

[near crag]

1/2n! =/= 1/2*n!

[last ether]

Dude

#❓how-to-get-help

[last ether]

Maybe list out different combinations of integers that multiply to 1521

[prime vessel]

wouldnt that take ages

[last ether]

Uh

Maybe

Not really

,w all integer factors of 1521

Oh apparently 9 and 169 are the only integers LOL

Well

No

[prime vessel]

oh lol

[last ether]

That's false

There's 1 and 1521

[ocean seal]

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[last ether]

Stupid machine

1 • 1521
9 • 169
3 • 507
17 • 13
39 • 39

That's it lol

[prime vessel]

oh

[last ether]

Wait no

There

[prime vessel]

oh so

all i need to do

[last ether]

Hopefully I didn't do a dumb

[alpine sable]

it doesn't?

[prime vessel]

is 17+17+13+13?

[last ether]

Alright I got every one

[last ether]

Should be

Yeah

[prime vessel]

wait no

17x13 is 221

[last ether]

It shouldn't

Did I do a dumb again

Oh my god

I'm failing as a helper lol

Let me

[prime vessel]

thats okay lol

(1, 1521) (3, 507) (9, 169) (13, 117) (39, 39)

here

[last ether]

It's prime factorization is:

3 • 3 • 13 • 13

Any product combination of those

[last ether]

Yeah this

I meant 117 I guess lol

[prime vessel]

oh lol

and it cant be 39x39

since that would be a square

the 13x117 combination is 260 if im not mistaken

which should be the answer

bro

[last ether]

#❓how-to-get-help

You are the third guy to interrupt yznx

[prime vessel]

lol

[last ether]

I'm sure you can read, can you?

I'm sure your ability to navigate is not reduced down to red arrows and not making connections

[pseudo schooner]

Oh sorry

[last ether]

It should

[prime vessel]

ill try 260 as an answer for b

[last ether]

Because 39(5) = 156

39(4)*

Equals 156

Iirc the theorem is that the square has the smallest perimeter given area

I'll have to check

[prime vessel]

lets go i got it

great

but i had that answer b4 and it said it was wrong lol

weird

thats why i came here

[pseudo schooner]

They did not explain well cuz the tutorial said claim help 0

[prime vessel]

oh well, thank you for your help @last ether

.close

[lone heart]

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[last ether]

Lemme see if I can prove it for a sec

.reppen

[prime vessel]

.reopen

[last ether]

.reopen

[lone heart]

✅

[last ether]

Let $A$ be a fixed area for some quadrilateral. The area and perimeter of this quadrilateral can be expressed as:

$$A = xy$$
$$P(x,y) = 2x + 2y$$

Since $A = xy \implies y = \frac A x$, we can rewrite $P(x,y):$

$$P(x) = 2x + \frac{2A}x$$

Taking the derivative and optimizing:

$$\begin{align*}
P'(x) &= 2 - \frac{2A}{x^2} &= 0 \
&\implies x &= \sqrt{A} \
&\implies y &= \sqrt{A}
\end{align*}$$

Therefore, the quadrilateral of area $A$ that minimizes perimeter is a square of side lengths $\sqrt{A}$

[ocean seal]

Umbraleviathan
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[last ether]

Alright so I ain't tripping balls lol

@prime vessel if you don't know derivatives, it's fine, derivatives are basically slope of a function.

But yeah I can confirm the 39x39 square

[prime vessel]

alright

thank you very much

should i close it?

[last ether]

Yeah

I got it

.close

[lone heart]

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[mystic orchid]

hi

[long axle]

hel

lo

[mystic orchid]

I have a question ..
y=(1/x)^x , find its second derivative at x=e and then find e^e (y")

I've tried it many times, it becomes messy and everytime i do it , i get a different answer , i dont know what the right answer is

[long axle]

Show ur work

[mystic orchid]

And for all times i tried i got a different answer

Am i just dumb

Just were am i going wrong??

[tacit arch]

,w d/dx(d/dx(x^(-x)))

[ocean seal]

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Requested by riemann#2529

[mystic orchid]

I tried using the partial derivate for the first order derivative then differentiated it again to get the second derivative

And then i put x=e and multiplied by e^e

<@&286206848099549185>

[mystic orchid]

By substituting x=e in the equation given by wolfram alpha and then multiplying by e^e gives me 4-1/e , is it right

[analog tangle]

yes?

[mystic orchid]

I dont know im confused

[analog tangle]

?

show me

[mystic orchid]

Wait give me a sec let me write it down neatly

[analog tangle]

k\

[mystic orchid]

Sorry thats the neatest i could write

[analog tangle]

its alright

[mystic orchid]

You sure?

I dont know what the answer is

Ty for your help

.close

[lone heart]

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[lone heart]

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[lost relic]

Hello, can someone help me understand some questions on exponents im struggling with

[frozen crown]

ye ok

what are they

[lost relic]

how do i write math with a keyboard

[past hornet]

use bot

[robust vapor]

you can sen a picture?

[lost relic]

let me send a pic wait one sec

actually the question is (3squared+4squared+10squared) to the power of 1/3

does that make sense

that

[past hornet]

9+16+100=125 the cube root is 5

[lost relic]

ok so let me send the full question

right so the answer for k would be 5?

[past hornet]

yep

[lost relic]

thank you

.close

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[alpine sable]

i think its c because the det b is 0

[alpine sable]

.close

[lone heart]

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[neon sierra]

I need help understanding the question

pls bruh

@past hornet

@lost relic

@marsh rapids

[marsh rapids]

don't ping random people

[neon sierra]

i need help

i need 1 more to finish this

[lone heart]

@neon sierra Has your question been resolved?

[lone heart]

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[worthy agate]

i need to calculate whole triangle area

.close

[lone heart]

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[alpine sable]

i have not taken matrix algebra yet but i need to solve something like this by hand

i understand how the multiplication part works, but how do i account for the 36 and 0? how do i compute all of this?

[vapid shuttle]

if you multiply that matrix by that column vector

it will be the case that you get the top row

with its first entry times v1

plus its second entry times v2

will be equal to 36

and the bottom row with its first entry times v1

plus its second entry times v2

will be equal to 0

[alpine sable]

alright

[vapid shuttle]

this is just an ordinary system of equations

[alpine sable]

then its just like solving- ye

tyty

.close

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[dim matrix]

I'm trying to get values of a and b, while solving I'm getting b =0, which seems wrong.

[wary stream]

Fyi your C is wrong

That is probably why your A and B are wrong

[dim matrix]

yeah it is 1/2

[wary stream]

Yeah C = 1/2

[wary stream]

Because you used C = 1 here, it lead to A = 3/8 and B = 0, but if you used C = 1/2, you'll get different values

[dim matrix]

I'm solving again. Many thanks 🙂

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[high wolf]

Can someone help me understand how they simplified this:

[shrewd egret]

Well, everything is square rooted (they have the same exponent), so you can multiply the two arguments

so you can take (4-x^2) and multiply it w/ what is inside the other root

and then take the square root of all of htat

from there, it's simple algebra

distributive property and simplifying

it's like √(2) * √(3) = √(6)

[high wolf]

Yes, but I'm not sure how that works..we'd get sqrt((4-x^2)(x^2/4-x^2)+1)..

[shrewd egret]

yep

so you would distribute the (4-x^2) to ((x^2/4-x^2) +1) using the distributive property

resulting in

sqrt(x^2+(4-x^2))

simplifying results in sqrt(4) = 2

[high wolf]

How did that work? How did the distributive process worked here..sorry, this all looks visually confusing.

[shrewd egret]

ok hold up

[wary stream]

FOIL it out

[shrewd egret]

u wouldn't need to

[wary stream]

Or do a substitution, and let a = 4-x^2

And then you have a * ((x^2/4-x^2) +1)

When you multiply it, you get (a * (x^2/4-x^2) + a * 1)

Plug back in a

[high wolf]

a u substitution?

[wary stream]

Just a substitution in general

[high wolf]

How does the distributive process work here, the one o7 recommended..

that seems easier..I just don't understand how it works on these terms..

[wary stream]

$(4-x^2) * \frac{x^2}{4-x^2} + (4-x^2) * 1$

[ocean seal]

dldh06

[wary stream]

It's literally what I showed

[wary stream]

.

[high wolf]

Sorry..makes alot of sense. I'm just shit at math. Thanks for putting up with me..

[lone heart]

@high wolf Has your question been resolved?

[vapid shuttle]

Sonny I've helped you before, you aren't shit at math. we all get stuck sometimes

keep it up

[lone heart]

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[quasi crater]

where did the (7n)(7n+1) come from

[quasi crater]

original

[jagged cobalt]

in $a_n$, the numerator is $7\cdot 14\cdot 21\cdot...\cdot 7(n-1) \cdot (7n)$ so the numerator of $a_{n+1}$ is $$7\cdot 14\cdot 21\cdot...\cdot 7(n) \cdot (7n+7)$

[quasi crater]

uhhh

wtf

[ocean seal]

AℤØ
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[quasi crater]

thats not the numeratore

[jagged cobalt]

i extended it back one bit

you would have that 7(n-1) before the 7(n)

which is why when you replace n with n+1, you get 7n(7n+7)

[lone heart]

@quasi crater Has your question been resolved?

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[tough zephyr]

Help

[tough zephyr]

I wanna start studying

[pine kettle]

what do you need help with

[tough zephyr]

math

for state testing

And I don’t wanna fail

How can I start

@pine kettle

[pine kettle]

take practice tests

review questions you have no idea how to do and also the ones you take a guess on

[tough zephyr]

But how tho

[pine kettle]

research online

there's HUNDREDS of resoucres

[lone heart]

@tough zephyr Has your question been resolved?

[lone heart]

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[hard mountain]

Let $F = \mathbb{Z}_2[x] /(x^2 + x + \bar{1})$. How many elements does F have?

[alpine sable]

perhaps the top should be Z_2[x]?

[hard mountain]

yes sry

[ocean seal]

OOKK

[hard mountain]

F should be the polynomials with degrees less than two right?

Is this right? $F = {x+1, x, 1} \implies |F| = 3$

[ocean seal]

OOKK

[alpine sable]

what about the identity class?

[hard mountain]

makes sense

so F = {x+1, x, 1, 0}, |F| = 4

thank u so mch!

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[summer trail]

Ok so I have that since f(x)>epsilon we know that the set will contain the x's that result in f(x)=1/n, or that we will have a subset of the rationals

And then by the archimedean property, if you give me some eps>0 I will be able to produce a 1/n that is smaller than that epsilon

so we can only use rationals with n<c where c is some real number

Is that enough to say that this is a finite set?

Please ping me if you respond I am cooking right now

[lone heart]

@summer trail Has your question been resolved?

[summer trail]

<@&286206848099549185>

[lone heart]

@summer trail Has your question been resolved?

[lone heart]

@summer trail Has your question been resolved?

[lone heart]

@summer trail Has your question been resolved?

[raven hill]

I guess so, we all can guess the idea. I would just say for some c>0, f(x)>=c.
Let n be the largest integer such that 1/n >= c, then for some natural k with n>=k, we have f(x) >=c for x being a rational with denominator k. Let Sk denote the set of rationals in [0,1] with denominator k, (lowest terms) , notice Sk is finite as you have only finite amount of naturals m, such that m/k =< 1. Then you notice the set that you want is just the union of Sk for 1=<k=<n, which is certainly finite.

[open linden]

hi katherine!!

ignored for 4 weeks 😔

ok hmm

m

m

sorry i suck at stats

[lone heart]

@summer trail Has your question been resolved?

[timber grotto]

women ☕

[lone heart]

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[median oar]

is sample variance just defined as this

[median oar]

is there anything special about it because my material only talks about it twice and only briefly mentions that it changes with sample (same as sample mean) and that it's not normally distributed

i understand that a rv can be normally distributed but i dont get why my slides say the sample mean is normally distributed but the sample variance is not

[livid widget]

As far as I understand it, you can use the sample variance to calculate the variance of the distribution of the mean

although I may be mistaken and it would me the population variance

let me check rq

[keen plinth]

[median oar]

so the sample variance is talking about what the sample mean can take on?

[keen plinth]

are the xs normally distributed

[median oar]

because i also understand that the sample mean itself is a random variable as well

we assume so

[keen plinth]

it is

[livid widget]

yea

[median oar]

this is the slide im on

been procrastinating a lot this weekend

[livid widget]

heres what I was talking about, its known at the central limit theorem: https://en.wikipedia.org/wiki/Central_limit_theorem

[median oar]

i swear i hate CLT everyone just says CLT something something it works

[keen plinth]

well

if the sample is normally distributed

then any sum of them is also normally distributed

[median oar]

yep

[livid widget]

lmao its a very powerful tool in statistics

[median oar]

that was discussed in an earlier slide

something something too hard to prove

[keen plinth]

too hard?

hm

[median oar]

anyway im guessing the sum/difference of 2 independent RV is also normally distributed

[keen plinth]

it's a fundamental property of normal distributions

[median oar]

i'll assume it to be true for now

[keen plinth]

iirc the sample variance is chi^2 distributed

[median oar]

i dont take stat2372 but i go to the lectures in hopes of learning more

anyhow not the point

i dont get why my slides say the sample mean is normally distributed but the sample variance is not

what tells me the sample mean is normally distributed but the sample variance is not

[keen plinth]

well the sample follows a certain distribution

[median oar]

well i suppose if the sum is normally distributed then so is the sample mean that makes sense, kinda

[keen plinth]

like

[median oar]

i just dont understand what makes the sample variance not normally distributed

is sample variance also a RV?

[keen plinth]

X1, ..., Xn are iid normal RVs

yes

[median oar]

oh

[keen plinth]

so like

[median oar]

they should say that

[keen plinth]

what happens is

X1 + X2 + ... + Xn is also a normal RV

[median oar]

yep

[keen plinth]

because each individual Xi is normal

but like

[median oar]

(if they are iid which they are)

[keen plinth]

you wouldn't expect something like X1^2 to be normal

[median oar]

why not

[keen plinth]

well just think about the values

a normal RV is symmetrically distributed on the real line

when you square it

you definitely lose all the symmetry

half the real line is no longer realisable

[median oar]

because smaller numbers get squared less?

[keen plinth]

like

if you have a discrete RV

say binomial

and you square it

you destroy the original distribution

[median oar]

so like if 6 7 8 is symmetric, 36, 49, 64 is further stretched on the 49 - 64 compared to 36-49?

[keen plinth]

the square of a binomial RV is no longer binomial

[median oar]

so adding, subtracting, multiplying, dividing all fine

all gives you back a normally distributed RV

but once u square it it stops being normally distributed?

[keen plinth]

by constants

[median oar]

right constants

[keen plinth]

but not by other RVs

[median oar]

yeah so if i do X1 + Y1 that's fine

[keen plinth]

in fact like

[median oar]

if i do X1 * Y1 that's not fine

[keen plinth]

if you have a binomial RV

and you multiply it by 2

it's no longer binomial

[median oar]

oh what

[keen plinth]

distributions are very specific

[median oar]

but are we not saying all of these are "approximately" normally distributed?

[keen plinth]

the sample mean of an iid sample will be approximately normally distributed

but that's a consequence of CLT

[median oar]

right

[keen plinth]

even then it's not always true if you don't have the right conditions

[median oar]

and sample variance is not approximately normally distributed....

because CLT doesn't apply?

[keen plinth]

it's not in the same form

[median oar]

in what form

[keen plinth]

sum of iid RVs

Xi - X bar are not iid

[median oar]

wait what

i thought it was??

oh wait Xbar depends on each Xi

[keen plinth]

well actually it might work

but not for the reason you expect

[median oar]

well idk the slides seem to also say that even if it's a little bit dependent it still works

[keen plinth]

yeah

but in this case it's chi^2

[median oar]

honestly this week has been quite a jumbled mess i dont really understand why CLT applies in some cases and not others

[keen plinth]

so it is actually the sum of independent squared normals

[median oar]

and what exactly clt is doing to make things approximately normally distributed

[keen plinth]

it might be out of the scope of what you're doing

[median oar]

😦

[keen plinth]

usually i've seen CLT proven using characteristic functions

[median oar]

hard to wrap my head around concepts that they just go: haha too hard to prove just memorise it hope u understand

i tried looking on youtube but they lost me on taylor series

i better learn how to do that it seems extremely useful

[keen plinth]

it takes a bit of theory to develop

[median oar]

ok so reading wikipedia, chi^2 is just some special gamma distribution

i've seen gamma distribution before

[keen plinth]

it's the sum of squared normals

that's more related to what you're doing

[median oar]

and the sum of squares of RV gets you a chi^2 distributed variable

what's normals

[keen plinth]

since Xis are normal

[median oar]

oh normal rv

[keen plinth]

and you square them and sum them in the sample variance

[median oar]

right

[keen plinth]

so it's chi^2

[median oar]

so doing Xi-Xbar is still normally distributed

but when you square it it becomes chi^2 distributed

[keen plinth]

something like that

iirc it's a bit more nuanced

[median oar]

i'll write that down as fact for now i suppose

[keen plinth]

but intuition wise that's fine

[median oar]

otherwise it's a bit too hard to remember or convince myself

aight thanks snow

.close

[lone heart]

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[summer stratus]

q3a

[quasi vector]

what can you add to both sides to complete the square

[median oar]

what does Q2 look like

[summer stratus]

(x+2)²-5?

[summer stratus]

oh hold on

[median oar]

that's correct, ish, you've turned an equation into an expression somehow

[median oar]

expressions have no equality sign (= < > ≤ ≥)

equations do

[summer stratus]

so we're solving the equation for q3?

[median oar]

yyup

[summer stratus]

but it says complete the square tho

[median oar]

yep

that's a method of solving the equation

[summer stratus]

do i continue with this?

[median oar]

you've just turned something that looks like 2x = 4

into 2x

the equal sign just disappeared into thin air

(you should not do that)

(x+2)²-5=0 is correct

[summer stratus]

sorry, (x+2)²-5=0

[median oar]

yep, so solve that

(dont expand)

[summer stratus]

x=sqrt5+2?

[median oar]

not quite

show each step

[summer stratus]

sorry*

sqrt5-2

[median oar]

ok we start here

(x+2)²-5=0

add 5 to both sides

(x+2)² = 5

what's the next step

[summer stratus]

x+2=sqrt5?

[median oar]

ok

if i say x^2 = 9

what is x

[summer stratus]

3

[median oar]

is that the only answer for x

[summer stratus]

-3

[median oar]

right

we say x = ±3

[median oar]

so what should this be instead?

[summer stratus]

btw why does sqrt always have + and - tho

[median oar]

because, when you square root something

you dont know if it originally was positive or negative

it could be either

[median oar]

like this

if x^2 = 9, then x = ± 3

[summer stratus]

x+2=+-sqrt5?

[median oar]

if (x+2)^2 = 5
then (x+2) = ±√5

yep exactly

now what's the next step

[summer stratus]

so x=±sqrt5 -2

[median oar]

fantastic

[median oar]

also this only applies if the thing you're looking at is already squared

so let's take x = 5

√x ≠ ±√5

it's ONLY if it's taken to an even power that we do the ± thing

[summer stratus]

oh ok

[median oar]

so x^4 = 16, x^2 = ±4, x = ± 2

[summer stratus]

btw is it necessary to put it in order like -2±sqrt5?

[median oar]

no

BUT

i would say use brackets

if u wanna put the -2 at the end you want to write +-sqrt(5) - 2

just to make sure everyone knows the -2 is not inside the sqrt

to make it as unambiguous as possible

[summer stratus]

alr

ty:)

[median oar]

if i was writing on paper i would add the hook at the end of the hat of the sqrt

to show that's the end of the sqrt, everything after this hook is NOT inside the sqrt

[summer stratus]

ok

.close

[lone heart]

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[civic karma]

d

d

d

d

d

d

d

d

d

d

d

dd

d

[alpine sable]

Bro

[civic karma]

d

d

d

dd

d

d

d

dd

d

d

d

dd

d

d

d

d

d

d
d

d

d

d

d
d

d

d

d

[alpine sable]

@fading citrus

[civic karma]

d

d

d

d
d

d

d
d
d

d
d

[fathom trout]

Yeah that’s good

You need to be more descriptive with the math question

Like what are the degrees on the angles

Sorry I meant the illustration..

You need to fill out the rest of it bro

You’re missing half of the angles

[lone heart]

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[alpine sable]

Very confused about how to solve functions and finding range.

[winter tangle]

all range is, it's the solution you get from putting the values of x in

[echo socket]

The range of S under f would be the set of outputs that the elements in S have

I.e., simply plug in x = -1, x = 0 and x = 1 and put what you get together in a set

[alpine sable]

I got it, thank you.

.close

[lone heart]

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[lone heart]

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[sage epoch]

is there a formula to be used on the 1st problem?

[pine kettle]

charge formula

?

sorry i mean coulomb's law

[lone heart]

@sage epoch Has your question been resolved?

[lone heart]

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[upbeat moth]

hello

[upbeat moth]

hii

id no idea

about

maths and i ddint get .

good marks in maths

what i should do

??

[snow ore]

You should get someone to teach you

And practice more

[upbeat moth]

but my teacher teaches me

in school

[snow ore]

Or ask more help from her

Because maths isnt that hard its mostly formulas

Where you are having more problems

What lessons

[upbeat moth]

in circle

[snow ore]

Geometry you mean

[upbeat moth]

yes

and..i will not be able to solve questions

after the chapter is finished

..

[snow ore]

You should watch videos in youtube

[upbeat moth]

i watch

but i did not understand

[snow ore]

What grade are you in if i may ask

[upbeat moth]

wait

,..

few sec

[versed hornet]

k

[snow ore]

Ok

[upbeat moth]

c1

grade

C1

[pine kettle]

wjat is c1

[upbeat moth]

grade

[trail cape]

Which year are you in?

[pine kettle]

grades are numbers not letters

[upbeat moth]

15

[trail cape]

Grade 15?

[upbeat moth]

and..you

no age 15

[trail cape]

So you are in grade 9 right? 9th school year

[upbeat moth]

yes

yes .

in 9th class

but i enterd in class 10th

this year

[versed hornet]

congrats

[upbeat moth]

and..u

[versed hornet]

me?

[upbeat moth]

yes

..

[versed hornet]

8

grade 8

[upbeat moth]

good

[versed hornet]

ok

[upbeat moth]

but i am very weak in maths

you...???

[versed hornet]

not so weak

[upbeat moth]

good

[versed hornet]

k

[last ether]

Patterns more so

Have you done multivariable and vector calculus

It's patterns and bullshittery

[snow ore]

I actually said that on first thought

Because i didnt assume what grade they were

[snow ore]

But now im learning calculus limits

English is not my first language so i thought "formulas" was the right term to use

[lone heart]

@upbeat moth Has your question been resolved?

[lone heart]

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[winter ivy]

Wait why does this not hold? What rule am I breaking?

[lone heart]

Please don't occupy multiple help channels.

[vale wigeon]

1/(a+b) != 1/a + 1/b

[winter ivy]

1/a + 1/b= a-1 + b^-1 right?

Can you please elaborate a bit more? @vale wigeon

[upbeat moth]

it is easy

[vale wigeon]

correct but neither side is equal to 1/(a+b)

[upbeat moth]

minus is turn into 1/

[winter ivy]

how would be the correct way to write everything without a denominator

.close

[lone heart]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[outer lark]

!15m

[lone heart]

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.

[median dirge]

You pinged a user btw

Not Helpers

[outer lark]

[stable plank]

lmfao

x^3 - 8x^2 + 11x + 7

[upbeat moth]

hey

[lone heart]

@alpine sable Has your question been resolved?

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[fluid hornet]

Hey

[fluid hornet]

I need help with exponential equation

When i do logarythms here

i just get into a loop

and i can never solve it

and i cant make the bases the same

[echo socket]

Consider rewriting 3^(x + 1) as 3 * 3^x and factor 3^x out of the left hand side

[fluid hornet]

hmm

i dont understand

how i should do it

[wooden brook]

3^x+3*3^x=4

[echo socket]

Right, now add the similar terms

[fluid hornet]

Could you maybe write it out? i cant follow at all

i know the solution is 0

just out of looking at it

but the way to get to it is fcked up

[echo socket]

$1 + 3 = 4 \$
So
$3^x + 3\cdot3^x = 4\cdot3^x$

[ocean seal]

A Lonely Bean

[echo socket]

Thus you have $4\cdot3^x = 4$ or $3^x = 1$

[ocean seal]

A Lonely Bean

[upper hull]

man i love u

[echo socket]

What?

[fluid hornet]

still dont get it

getting it explained by a friend rn too still not understanding it 💀

[upper hull]

just lovesir bean

[alpine sable]

How I am understanding it (feel free to correct me if I am mistaken):
3^x + 3 * 3^x can be also written as: 3^x + 3 * 3^x * 1. The one does not impact the value of 3^x.
Since 3^x appears twice, you isolate that out so it becomes: 3^x(1+3)
Which then becomes 3^x * 4, or 4 * 3^x

[lone heart]

@fluid hornet Has your question been resolved?

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[spark isle]

Show that (n^2)a^n converges to zero for 0<a<1

[limpid spade]

"Convergence is men's nemesis." Ghashi Balazar

Forgot to say, what have u tried.

[spark isle]

got the proof

.close

[lone heart]

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[tidal gust]

I have been trying to solve this forever.

[tidal gust]

I can't make progress.

The problem is at the top

Below that is what I have tried to come up with

I understand the concept and I can visualize it but I can't come up with a proof/equation