help-13

[warm vector]

I’m saying why did we not set the 2pir = circumference of circle?

That’s what’s confusing me a little

[tropic oxide]

what circle's circumference do you want to set 2πr equal to?

the circle that the original sector came from?

[warm vector]

This sector’s circumference

[tropic oxide]

with all due respect what the actual fuck are you talking about

[warm vector]

I’m saying why didn’t we do that?

[tropic oxide]

you keep changing your story

[warm vector]

So you know this sector

It’s circumference

[tropic oxide]

so why didnt you answer yes here

instead you chose to lead me in fucking circles for 10 minutes straight

i hope it wassnt fuckng deliberate

[warm vector]

I’m confused that’s why

But it wasn’t deliberate

[tropic oxide]

ok heres your sector

now answer me clearly with either a YES or a NO and nothing else:

as far as i understand, you are asking why we set 2πr equal to only the arc in black, and not the black plus the red.
DO I UNDERSTAND YOU CORRECTLY?

[warm vector]

Yes

[tropic oxide]

the sector that we glue into the cone has three sides
two straight sides equal to the slant height
and one curved side
the straight sides are glued together
the curved side becomes the circumference of the base

[warm vector]

Ok

[tropic oxide]

i assume that means "ok fine just stop being mad at me i don't care about the problem anymore i just want to placate you"

[warm vector]

No definitely not

By that I mean that I actually am understanding why

[sweet vapor]

With all due respect, your sorta being toxic Rn. But it’s not my business Oof

[tropic oxide]

But it's not my business

exactly

[warm vector]

Ok I think I get it now, sorry, thanks

.close

[cedar kiln]

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[sweet vapor]

Idk. Your violating the rules. That’s why I pointed it out

[cedar kiln]

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[ripe lance]

7A34B4 can be divided by 36 without any remainders, B is a prime number, how should i find A? (I know tiz is a hw, but at least give me tips on how to solve these kinda problems)

[balmy apex]

is that $7 \cdot A \cdot 34 \cdot B \cdot 4$

[wraith dagger]

illuminator3

[dusty hazel]

No.

I think all those numbers are the digits.

Right?

@ripe lance

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[ripe lance]

@dusty hazel yes... Soooo sry i didnt reply

[ripe lance]

This is my question

[dusty hazel]

Divisible by 36.

So it should of course be divisible by 4 and 9.

Do you realise that?

[ripe lance]

oh

hmm

[dusty hazel]

Do you divisibility of four?

[ripe lance]

wait a min lemme think lol

[dusty hazel]

Of course.

[ripe lance]

so B must be 2

yep

only possible num

[dusty hazel]

Great.

Now the entire number should be divisible by 9.

[ripe lance]

ahhhh

[dusty hazel]

Makes sense?

[ripe lance]

right

gg

[dusty hazel]

👍

[ripe lance]

so... "A" must be 7?

took me long enough lol

[dusty hazel]

Exactly I thought you were done and weren't writing it here.

Anyways,

Yes.

[ripe lance]

yay thx a lot

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[main pagoda]

i know vaguely what i have to do I think i just need someone to help walk me through it. i know i need to get a profit function which is revenue - cost, i know that R(x) = xp(x) but i think i need some help getting the maximum value

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[warm vector]

How would I do this question?

[cosmic steppe]

Make life easier

Small diameter = 2x
Big boy diameter = 4x

Find area in terms of x

[warm vector]

Ok sure

So small area: pi x x^2
Big area: pi x 2x^2

[cosmic steppe]

Yeah

Divide big area by small

[warm vector]

2

Answer says c for some reason

[bold lotus]

yes..

consider 1^2 vs 2^2

you wrote the big area expression incorrectly

2x^2 rather than (2x)^2

@warm vector

[warm vector]

Oh

So 4x^2

[bold lotus]

yes exactly

[warm vector]

OK which makes it 4

I get it

Thks

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[quaint heath]

Can someone make an integral equation with multiple terms like my username where the answer = 7429

[balmy apex]

there's a website for that

let me see if I can find it

https://enjeck.com/num2math/

@quaint heath

[quaint heath]

YES

🙏😋

Aw it only goes to 1000

Do any go higher

I can put the equations as x + y + z and build up to 7k💀

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@quaint heath Has your question been resolved?

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@quaint heath Has your question been resolved?

[dire geode]

.close

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[keen sorrel]

Need help regarding linear transformations

[balmy apex]

just ask your question directly

@keen sorrel

[keen sorrel]

so first we know that T([1,0]) =[5,-6] and that T([0,1)]=[4,2]
but that's as far as I've gone because I'm confused as how to proceed

[balmy apex]

how would the linear transformation look like as a matrix

[keen sorrel]

like this as a 2x2 matrix?

[5 4]
[-6 2]

[balmy apex]

yes

applying a linear transformation to something is nothing other than just multiplying the transformation as a matrix by the thing that you want to transform

[keen sorrel]

so If i'm mapping (-6,-1) to the vector y, then I'm just multiplying that by the 2x2 matrix?

[balmy apex]

yes but order matters here

$AB \neq BA$ for matrices

[wraith dagger]

illuminator3

[keen sorrel]

yeah so the 2x2 matrix times (-6,-1) itself

[balmy apex]

yes

[keen sorrel]

got it thank you!

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@crimson sedge Has your question been resolved?

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[rugged tusk]

can someone explain me what is happening at line 3 ?

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[rocky sable]

let me show u what i did

what did i do wrong

how

what u mean

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[woeful musk]

can you guys provide an example of a function f that satisfies f(f(x))=f(x)^f(x)

[balmy apex]

$f(x) = 1$

[wraith dagger]

illuminator3

[woeful musk]

kekw yh nvm .close

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[crimson sedge]

if it was the 5th term

then 12288 is the 6th term

[modern compass]

yes

I suspect though that the question is asking you to find r in terms of n

if n=5 then yes, 2n-4 =6... but what if n is 25? or 73? you have a different r

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[crimson sedge]

how to solve this?

[upper abyss]

There's no way

For small angles, cos(y) ≈ 1 and we can grab an approximate solution this way

[crimson sedge]

what is am? :

is that like the error function

[crimson sedge]

why is math broken

[gray condor]

https://mathworld.wolfram.com/JacobiAmplitude.html

[crimson sedge]

fuk

[upper abyss]

Like I said, no way to do it.

A solution doesn't exist in the elementary functions

[crimson sedge]

ah ok

thats depressing

so what are non elementary functions?

[upper abyss]

Error function is one

[crimson sedge]

mr gauss man

[upper abyss]

Error function is common and well documented though, so that would be "nice enough" imo

Then there's whatever that is, the JaCoBi AmPlItUdE fUnCtIoN

[crimson sedge]

what makes non elementary functions spawn tho

[upper abyss]

Lots of integrals don't exist

Like ∫ e^(x²) dx

So we give them a name. That one is the error function

[crimson sedge]

sin(x)/x

its possible with DUIS tho

[cedar kiln]

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[ember python]

I'll have you know I use the Jacobi (Godsent) amplitude function on a daily basis.

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[teal lantern]

^3sqrt24m^3

[teal lantern]

would I need to separate the m^3 from the 24?

disregard actually

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[honest bobcat]

I am struggling

with 8a

[cedar kiln]

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[honest bobcat]

<@&286206848099549185>

this is something i vaguely wrote

but i dont really understand

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@honest bobcat Has your question been resolved?

[honest bobcat]

heol0po

[honest bobcat]

helpo

[south juniper]

@honest bobcat Let |f(x)| = g(x). We know for all M > 0, there exists S > 0 s.t 0 < |x-a| < S ==> g(x) > M.

Now do rough working for the limit we want to show.

We need | 1/g(x) - 0 | < e which is just g(x) > 1/e

Take M = 1/e. (M > 0 implies e > 0). So for all e > 0, we know that there exists S > 0 s.t 0 < | x - a | < S ==> g(x) > 1/e which is equivalent to 0 < | x - a | < S ==> | 1 /g(x) - 0 | < e

[cedar kiln]

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[honest bobcat]

@south juniper are you calling e epsilon here

[south juniper]

Yes

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[spark idol]

can i get help with this
tried finding x through x^2 - x + 1 but discriminate is -ve and it doesnt seem to be solveable through long division

@Helpers

[faint dirge]

,rotate

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[slim parcel]

How would I differentiate the x squared root x

[mental granite]

$x^2 \sqrt(x) = x^{5/2}$

[wraith dagger]

duckduckdo

[plucky pawn]

,rotate

[mental granite]

lol soz

[plucky pawn]

yes convert it to exp form it will be easier to work with

[slim parcel]

Did u add the powers?

[plucky pawn]

yes

[slim parcel]

2 plus 1/2 is 3/2

[plucky pawn]

x^2 is the same as x^4/2

[dusty hazel]

$x^m \cdot x^n = x^{m+n}$

[wraith dagger]

What the hell am I doing here?

[plucky pawn]

but you need common demoniator to do that

[slim parcel]

Omggggggg

Pleaeeee kill meee

Next question

A

I tried to do simplify the fraction by writing it as sin over cos sin over cos

[dusty hazel]

Nah that's a lot of work.

Do you know,

Componendo dividendo?

[slim parcel]

Wot

I am in a level

[dusty hazel]

If $\frac{a}{b} = \frac{c}{d}$\
Then, \
$$\frac{a+b}{a-b}=\frac{c+d}{c-d}$$
This is how it looks.

[wraith dagger]

What the hell am I doing here?

[slim parcel]

I see

[dusty hazel]

Well I don't want you to believe me just like that. Try looking up proof.

For this.

[slim parcel]

I’ll do that

[dusty hazel]

It makes it a lot easier.

[mental granite]

$\tan^2(x) + 1 = \sec^2 (x)$

[slim parcel]

Once again I am in a level

As to be exact

[mental granite]

thats a-level stuff

trig identities

[slim parcel]

What’s sec?

[plucky pawn]

what does that equate to in grade>

1/cos^2x

[livid hound]

check your sign

[wraith dagger]

duckduckdo

[mental granite]

$\sec^2(x) = \dfrac{1}{\cos^2(x)}$

[wraith dagger]

duckduckdo

[mental granite]

AS level is year before final, so grade 11 i think

your final grade - 1

[slim parcel]

That is correct

What is sec?

[plucky pawn]

dam derivitives in grade 11 geez

[mental granite]

sec, cosec and cot are ways to write 1/cos, 1/sin and 1/tan

[slim parcel]

I see

That’s very helpful

[mental granite]

so $\dfrac{\cos(x)}{\sin(x)} = \cot(x)$

[wraith dagger]

duckduckdo

[slim parcel]

I see

[mental granite]

do these ring any bells ?

[slim parcel]

Yes

Well I just learnt what it means thanks that’s so helpful

Am I correct in assuming that cotx is the inverse of tan x

[mental granite]

$1 + \cot^2(x) = \dfrac{1}{sin^2(x)} = \csc^2(x)$

noo

[patent flame]

it's \csc

for cosec

you didn't do it the first time

[wraith dagger]

duckduckdo

[mental granite]

was just trying to say csc = cosec

[mental granite]

noo, $ \cot(x) \neq \arctan(x)$

[cosmic steppe]

Cotangent is the reciprocal of tangent, not the inverse

[slim parcel]

I see

[dusty hazel]

$\cot(x)=[\tan(x)]^{-1}$\
$\cot(x) \neq tan^{-1}(x)$

[wraith dagger]

What the hell am I doing here?

[cosmic steppe]

That's why I don't like using tan^-1

I always use arcfunctions

The same thing but

[mental granite]

with AS and A-Level maths just do the all ! past papers and youll be fine

[cosmic steppe]

Visually, it's better for me

[slim parcel]

I get it

[slim parcel]

Thanks

[mental granite]

you solved it yet ?

or need help still

i meant all past papers not the all! past papers btw

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[dire pier]

how did my friend get the highlighted part?

is it the rational roots theorm

[deft forum]

Umm he just multiplied the two brackets bro

…

[eternal comet]

FOIL

well the more general version since one the parentheses has three terms

[dire pier]

wdym

i know foil

with like (x+3)(x-4) for ex

but not 3 times

would i use synthetic?

[eternal comet]

(a+b+c)(d+e) = ad + bd + cd + ae + be + ce

[glad kestrel]

it's just distribution

[eternal comet]

so that's the word

[dire pier]

so i would just kinda memorize this?

[glad kestrel]

??

[dire pier]

coulld i not use synthetic in this situaiton

[deft forum]

Or a better way to write maybe $$(a + b +c)(x+y) = (a + b +c)x + (a + b +c)y$$

[glad kestrel]

synthetic division does not have anything to do with distribution

[wraith dagger]

Deep.

[balmy apex]

$$a \cdot (b + c) = a \cdot b + a \cdot c$$ this is the only thing you need to "memorize"

[wraith dagger]

illuminator3

[balmy apex]

when it comes to distribution

[dire pier]

ohh

how about this?

[dire pier]

this was very easy to understand with this

[dire pier]

would i use synthetic

or like idk

[eternal comet]

sure

[dire pier]

@eternal comet

would this be on the right track

to guess and guess

like i would guess f(1)= to check if equals 0

[deft forum]

Synthetic is what I would use

[dire pier]

im using the rational roots theorm at the start

is that incorrect

[deft forum]

Tbh idk what is it

[eternal comet]

Just use synthetic division

I would use polynomial division since that is what I am used to (maybe I should take the time to get used to synthetic), but it's the same thing.

[dire pier]

ohhh

i got it

[cedar kiln]

@dire pier Has your question been resolved?

[cedar kiln]

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[faint dirge]

u=x-2

[stoic basin]

ohh ok sec

would this be 2/sqrt3

wait lemme just check

damn it's just 2. Okay so I'll say what I did

[upper garnet]

just u sub x-2 and ur done

[stoic basin]

that's what I did

I got -2/sqrtu

[eternal comet]

show your work

[stoic basin]

ohh I forgot I need to input u

dope

[faint dirge]

wdym show work I thought all math should be done mentally

[stoic basin]

every day I wish I was a math god and could do this

[faint dirge]

you're telling me I was allowed to write down my partial fractions when doing laplace transforms??

[eternal comet]

I'll remember that next time I'm solving an ODE

[faint dirge]

and here I was, solving sixth order system of equations mentally

[eternal comet]

That pales

in comparison to mental arithmetics

hardest task you could ask of a human being

[stoic basin]

okay cool thanks yall

.close

[cedar kiln]

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[faint dirge]

lol have a good day

[cedar kiln]

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[clear glen]

and the equations needed to solve are as follows

[clear glen]

so my question is that i managed to find the rCN value by solving eqn 1 with the info given, then plugging it into eqn 2 and 3 respectively. From there I found that I could use elimination/substitution to get rid of the rHCrCN term and the rHC^2 term and solve for rCN

basically for that part I did (roughly) I2-(2I1)=-168rCN^2

which gave me the answer rCN = 1.158*10^-10. My question is, Is there a possible way for solving for rHC without using the quadratic equation on one of the eqns?

(if it helps I can send a picture of my "simplifed" equations that i got rCN from)

This is the eqns from plugging everything in

the 52 in the 2nd eqn should be squared fyi

<@&286206848099549185> anyone?

[cedar kiln]

@clear glen Has your question been resolved?

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@clear glen Has your question been resolved?

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[south dawn]

Why some math boards are hidden?\

[dusky oracle]

wdym

[south dawn]

what's wdym

[dusky oracle]

what do you mean

do you refer to channels

[south dawn]

When I go to unread in both, it says hidden

[south dawn]

I guess it is channel. I am sort of new

[dusky oracle]

if you right click on a category then you should be able to make them visible again via a checkbox

[south dawn]

Thank you. Let me try.

[south dawn]

I did not see a category. sorry.

[dusky oracle]

oh so the entire Math Help category is hidden

uh

[south dawn]

I don't know why.

[dusky oracle]

hm confuses me as well

[south dawn]

Maybe BOT can help

or @rustic coyote

[rustic coyote]

hmmm

its the open and closed system for the help channels

once the channel is closed it gets hidden and u will not be able to talk in that channel until its made available again automatically later

[south dawn]

I think I got it. Thank you.

.close

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[dire pier]

i was confused what this question is asking me

and how even start it

[south dawn]

Integrate?

[dire pier]

?

I'm not quite sure what u mean

[south juniper]

First try to find where f'(x) = 0

[leaden snow]

@dire pier What happens to f'(x) if you plug in x = -2?

[south juniper]

maybe x=-2 is one possibility

[dire pier]

I was told if f'(x) = 0

and its an extreme

then i got this question

and i didnt get f'(x)=0

and now im a bit lost

anyone

@south juniper

@leaden snow

[south juniper]

f'(-3) isn't 0 so that means u dont have a stationary point

that gets rid of 2 options

Now it is possible that it may be an extremum end point

[dire pier]

Do you mind coming into a call with me

to explain just for 2 mins

[dire pier]

wait im not sure if i remember correctly

but whenefer f'(x)=0

its an extremum

[south juniper]

uh not necessarily

take y = x^3

are you being taught online?

like online classes of college or school

[dire pier]

school

wait why

ive had covid for last 2 weeks so im a bit confused on whats going on ;/

[dire pier]

wait what

oh wait i got it

does anyone know about this one

<@&286206848099549185>

[steel cave]

So you are asking

[dire pier]

for the question above

[dire pier]

.

[steel cave]

Well whats the average rate of change over that interval?

Thats a question for you

[dire pier]

ok

[steel cave]

Hint:

Let me know if you arent understanding

[dire pier]

i am

not understanding

[steel cave]

Aight

So the secant line

Is going to have the average slope of the line it is secant to over that interval

In this case, since -4, and 7 are the x intercepts

the secant line is just y = 0

It is asking at what x does f'(x) = 0

[dire pier]

i dont understand ;/

[steel cave]

Do you know what a derivative is?

[dire pier]

ye

[steel cave]

Do you know what a secant line is?

[dire pier]

nope

[steel cave]

A secant line is just the line between two points on another function

Here i've graphed y = sqrt(x) and its secant line between x = 1, x = 4

Or the secant line for y =x^2 between x = -1, x =2

Understand?

@dire pier

[dire pier]

gimme a min

Yea

I understand what a secant is now

when it passes two points

[steel cave]

Ok

So basically

A secant line will always have the average slope on that interval of the function it is secant to

[crystal pelican]

@dire pier

[cedar kiln]

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[tame shadow]

May I get help with changing order of integration inside of a call please?

[tame shadow]

in mathematics voice

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[forest lark]

Given that f : A → B and g : B → C are injective functions
f is one-to-one
g is one-to-one

Prove : h = g o f : A → C

Proof :
Suppose that x and y are arbitrary elements in A such that x = y. Then, f(x) = f(y) since function f is injective and therefore must be one-to-one. Then, g(f(x)) = g(f(y)) since function g is also injective therefore must also be one-to-one. Hence, g o f is one-to-one and function h is an injective function.

[forest lark]

Does my proof make sense/is the logic valid??

[crimson sedge]

I could be wrong but

isn't the implication the other way?

[strange grove]

It's the other way

[forest lark]

what do you mean by the other way?

[crimson sedge]

you would start out with h(x)=h(y)

then somehow arrive at

x=y

wait hold on

is this question asking you to prove that h is injective?

[forest lark]

yeah that is correct

[crimson sedge]

ok yeah then you'd follow that

everything you have is right just flip it around

[forest lark]

ah i see, do i need to do anything with the A --> B and B --> C info?

[crimson sedge]

I dont think so

[forest lark]

ok thank you so much for your help!!

[crimson sedge]

np

[forest lark]

.close

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[cedar kiln]

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[sand ether]

I'm having a hard time putting together in my head what the following functions will look and the their respective integrals

$\int_0^a f^{-1}(x) ~dx$, $\int_0^b f^{-1}(y) ~dy$, $\int_0^c f(y) ~dy$, $\int_0^d f(x) ~ dx$

[wraith dagger]

azeem321

[sand ether]

hard to word this question, but maybe a sketch with all the functions and the areas they represent on the same graph would help?

if this area under f(x) from 0 to d, what would f(y) between 0 and c look like?

[cedar kiln]

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[sand ether]

.close

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[cedar kiln]

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[acoustic frost]

“A college student is preparing a course schedule for the next semester. The student may select one of four mathematic courses, one of five science courses, and one of five courses from the social sciences and humanities. How many schedules are possible”. This question is part of my review and I’m not sure how to solve or what to look up on google. Please and thank you!

[acoustic frost]

& is there a way to get into a vc with a helper so they could tutor the rest of the review?

[ancient nebula]

Hii I am emma

[acoustic frost]

Hi Emma, are you gonna help?

[ancient nebula]

No sorry I have a question

[acoustic frost]

This channel is currently taken; though there are other open help channels

[ancient nebula]

Sure thank you

[acoustic frost]

<@&286206848099549185>

[south dawn]

4 math5science5Social and humanities

[south dawn]

There suppose to be multiplication

[acoustic frost]

are you sure?

[south dawn]

That is how I see it. 4(5)5

[south dawn]

You can list all possible ways if you want

[acoustic frost]

Okay, I see. so for the rest of the similar questions i just multiply to find the possibilities?

[south dawn]

Yes.

[acoustic frost]

Wonderful, thank you so much

[south dawn]

🙂

[acoustic frost]

.close

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[cedar kiln]

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[gaunt glacier]

Not quite sure how to do this. What are the first few steps?

[gaunt glacier]

plz ping

[glass blade]

That means theta lies in 4th quadrant

Now evaluate

tan = sin/cos
sin= - sqrt(1-cos²)

[gaunt glacier]

ok so on the unit circle in the 4th quadrant

[glass blade]

What answers you got putting the values?

[gaunt glacier]

lemme put them in

[glass blade]

Sure

[gaunt glacier]

um

its saying it cant understand it

[glass blade]

No no

Place the values

like sinx = ± sqrt(1-cos²x)

cosx = 2/11 (given)

And since sinx <0 we can conclude that x lies in 4th quadrant of the unit circle
And thus sinx = -sqrt(1-cos²x)

@gaunt glacier

sinx = -sqrt(1- (2/11)²)

Do like this

[gaunt glacier]

ah sorry I was eating

[gaunt glacier]

like this

it said that it couldnt understand it

[glass blade]

Yes

Evaluate it bro lol

[gaunt glacier]

ahh ok

thought I was missing something

[glass blade]

cosx = 2/11 given

I just replaced the theta with x

[gaunt glacier]

-sqrt sin^2(theta)?

[glass blade]

I meant this

• sqrt(117/121)

You know right the basic Trigonometic formulas ?

[gaunt glacier]

yeah

so evaluate -sqrt(117/121)

?

[glass blade]

No need to just write it as -3(√13)/11

Beside sin(theta)

[gaunt glacier]

yeah thats what I meant

[glass blade]

It's saying wrong?

It shouldn't be tho

[gaunt glacier]

oh no it said its right

I forgot to click submit

ok so how did you get -sqrt(117/121) again

cus I understand after that

[glass blade]

.

[gaunt glacier]

I cant really read that ._.

[glass blade]

sin²x + cos²x= 1

Agree ?

@gaunt glacier

sin²x = 1-cos²x
sinx = ±√(1-cos²x)

The sign before the root comes depending on which quadrant the "x" lies

[gaunt glacier]

Ah agree

but where did you get the 117/121 from

.close

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[manic acorn]

How do i evaluate this?

[manic acorn]

Am i missing something obvious 💀

[deft forum]

Take a look at rule3.

[manic acorn]

Then?

[deft forum]

Also idk if its useful or not but $log_b a = \frac {log_x a}{log_x b}$

Oops

[wraith dagger]

Deep.

[manic acorn]

But what do i do from here💀

[deft forum]

Yup this is useful

[manic acorn]

Ok let me see

Yep i got it

[deft forum]

Cool

[manic acorn]

15/2 thanks

.close

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[magic storm]

Hello! I was wondering if anyone could please tell me a quick mnemonic or way to memorize the convergence test for geometric and p series? I have some troubles differentiating between them

[upper garnet]

-1<r<1 for geo and r>1 for p

between and bigger? I mean I don't think it's hard to differentiate these two

[magic storm]

actually yeah I was just over complicating by only trying to remember the divergences and convergences of both series. Also just a lack of practice, thanks for clearing that up.

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[keen mulch]

Find the coefficient of $t^{47}$ in the expansion of $(t + 2)^{50}$

[wraith dagger]

Jef Costello

[crimson sedge]

use binomial theorem find general term and set the variables such that power of t is 47

[keen mulch]

is it $50 \choose 3.{2}^3$?

[crimson sedge]

yeah

if $\binom{50}{3} 2^3$ is what you mean

[wraith dagger]

Shunon

[keen mulch]

yes

[crimson sedge]

its correct

[keen mulch]

thanks :')

[crimson sedge]

np (:

[keen mulch]

.close

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[magic oxide]

Hi i need some help with this. I was absent when this was being discussed and I checked the PowerPoint presentation they have sent but i still couldn't answer it. Please help me understand how to answer it thanks

[cedar kiln]

@magic oxide Has your question been resolved?

[magic oxide]

I believe this is theorem on parallelogram

[cedar kiln]

@magic oxide Has your question been resolved?

[cedar kiln]

@magic oxide Has your question been resolved?

[dire geode]

.close

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[quiet hazel]

If you flip a coin twice, what is the game's expected value?

[quiet hazel]

i got 2 for this one

[bold hinge]

What

[quiet hazel]

thats my answer

[bold hinge]

What game

[reef venture]

what is the game

[quiet hazel]

it didnt specify

[bold hinge]

And coins don't have numerical values either

[quiet hazel]

this is a bad q

[quiet hazel]

wdym

[bold hinge]

Like it's just heads and tails

Are you supposed to assign different numbers

[quiet hazel]

no

[reef venture]

? there must be some info like you get 2$ when its head and lose 1$ on tails or something

[bold hinge]

To each of those outcomes

[livid hound]

there's got to be some set of rules

[bold hinge]

What i think this is getting at is just linearity of expectation

[quiet hazel]

no

yeah this q is incomplete i feel

but it should be

[bold hinge]

Let's say you have a 10% chance of successfully making a free throw in basketball

Then the expected score on the nth shot would be n*10%

So after 10 shots you would have made a single basket

[quiet hazel]

ye

its a bad q

ill give constraints

[bold hinge]

Do you understand this tho

[quiet hazel]

yeah

[bold hinge]

Thats what the question is really trying to ask

Ig

[quiet hazel]

when you say n shot

its the numerical value of shots the prob gives righ?

[bold hinge]

Yes

Your expected score is 1 after 10 shots

10 after 100 shots. Because you make 1 out of every 10 shots

Yes

[quiet hazel]

since you have a 10% of making a shot successfully right?

[quiet hazel]

u mind deleting and sending again?

so if the problem states that if you have 15% chance of making a shot, and you can only make 20 total shots

the expected value of this wiuld be (.15)(20) right?

[bold hinge]

Yes

[quiet hazel]

thats what the question is asking

If you flip a coin three times, you get 1$ if you flip heads and $1.50 if you flip tails. What is the game's expected value?

EV = 3 [(.5* 1) + (.5*1.50)] this would be the ev of this right?

[bold hinge]

Yeah

Since the outcomes are equally likely you can use the arithmetic mean

(1+1.5)/2 = 1.25

[quiet hazel]

is 2 still right for this?

[bold hinge]

I'm not sure

We don't know what the game is in that scenario

[quiet hazel]

tru

You have 4 coins. You throw them in the air. For every H that lends, you get 1 dollar. How much would you pay to play this game?

this would be $2 right?^

[dire geode]

what's the expected number of heads that appears?

[quiet hazel]

2

[dire geode]

how did you calculate this ?

[quiet hazel]

no its logic

if there is 50% chance of you flipping heads

and you have 4 coins

your gonna flip heads half the time

EV = 4[(.5 * 1) + (.5 * 0)] = $2

[dire geode]

sounds right

[rustic siren]

i wouldnt pay 2 dollars

i want profit

[cedar kiln]

@quiet hazel Has your question been resolved?

[cedar kiln]

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[crimson sedge]

Hi, could someone explain me this exercise?

I have the solution and i dont get it tbh.

[upper garnet]

,rotate

[crimson sedge]

My initial idea:
We can find the vector n through the plane equation. So n = 5 , -1, 1

Then we know that n must be perpendicular to our line r, and also the line g must be perpendicular

So I guess I could do the dor product between r and g to see whether the result is 0

And then, I also need to do the same thing, so the dot product between n and r which must result 0

The solution simply does: (6 3 -2) CROSS PRODUCT n

I have no idea why though

,rotate

[crimson sedge]

And this is the solution. No idea why ur is parallel to (nw cross ug)

[cedar kiln]

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@crimson sedge Has your question been resolved?

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[waxen hinge]

i am doing a homework problem and im god awful and probability, i believe i need to use the chain rule but don't really think im using it properly

[waxen hinge]

don't know how to solve the second term

I think I need P(~M) and P(M) but I don't know how to find those, I think I need P(A) to find them, and I believe that is given from P(A | ~B,~E) but im not sure if this is a correct assumption

[upper abyss]

Why are they using , when they mean ∩ yikes

[waxen hinge]

old book, i was confused for a whilee about that too

although my professor uses it too -_-

i also found this in my notes but it doesn't explain how this got simplified

[upper abyss]

P(A | B) = P(A,B)/P(B)

That's the definition for P(A|B)

[waxen hinge]

i know most of the general formulas i just don't think i know how to apply them when there are joint probabilities

[upper abyss]

There's really only two options:

• Break all of the joint probabilities using the definition and work with intersections instead, which is messy, but certain
• Law of total probability, which can be clever and clean

[waxen hinge]

not sure i entierly know how to do either :/

I think I neeed to figure out P(~M | B , E)

which would be P( B , E | ~M) / P(B | E) but i still don't really know how to solve the numerator

i am really bad at probability so far, have trouble wrapping my head around it

[upper abyss]

That doesn't follow, as far as I can tell?

P(-M | BE)
= P(-MBE) / P(BE)

Using the normal people's "just write the sets together to indicate intersection" and not using commas

[waxen hinge]

found this online as an eexample with joint probaility so i was just going off this

[upper abyss]

Probabilities with - in them can always be written without

[waxen hinge]

?

[cedar kiln]

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[waxen hinge]

.close

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[unique wind]

all integer solutions of the inequality?

[cedar kiln]

@unique wind Has your question been resolved?

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[crimson sedge]

Start by expanding out both sides

[cedar kiln]

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[keen birch]

can somebody explain why the derivative of something like 5e is 0? online calculators say that it is a constant but im not sure I completely understand why

[upper abyss]

Depends what your variable is

If you are taking the derivative in terms of x, then 5e is, of course, a constant.

[keen birch]

so yeah what about something exponential like 5e

im not sure i understand

cuz my knowledge of a constant is limited to a number in front of a variable

[upper abyss]

So 5e^x is an exponential function

[keen birch]

its just given as 5e

[upper abyss]

But 5e is just a constant

[keen birch]

is that the same as 5e^x

[dire geode]

,calc 5e

[wraith dagger]

Result:

13.591409142295

[upper abyss]

Nope. Never.

[keen birch]

ok j making sure

so how exactly is the rule applied to 5e

or what specific rule

[upper abyss]

5e is a constant. The derivative of a constant is 0.

As calculated above, this is just 13.5 in disguise.

[keen birch]

so the value of e itself is different to e^x?

[dire geode]

,calc e

[wraith dagger]

Result:

2.718281828459

[keen birch]

ahh i see

wait

isnt that the same as e^x

[upper abyss]

Maybe take a quick look into what e is, haha.

e is a constant, like pi

e^x is (2.71828)^x

[keen birch]

ohhh right ok eulers number that makes sense

thank u, that makes much more sense

.close

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[gleaming harbor]

I keep getting 1.923 instead of 1.762. I have no idea what I did wrong. I checked my work 100 times.

[gleaming harbor]

Sorry for it being incredibly messy.

<@&286206848099549185>

[cedar kiln]

@gleaming harbor Has your question been resolved?

[gleaming harbor]

No

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[fair narwhal]

hey, does anyone have an idea on this?

[cedar kiln]

@fair narwhal Has your question been resolved?

[upper abyss]

Where does the matrix take (1, 2, 1)? How do you represent that answer in terms of B?

That's the first column

[fair narwhal]

so i first need a function phi for the endomorphism

and then write that as a matrix or something like that?

[cedar kiln]

@fair narwhal Has your question been resolved?

[fair narwhal]

@upper abyss ([1,2,1], [2,0,-2], [1,1,0]) is the transformation matrix right?

[upper abyss]

If you want to call it that!

[fair narwhal]

so is it for the first column: [1,2,1] = A * [1,2,1] ?

[upper abyss]

So you are given the function. φ takes an input, multiplies it by that matrix, and that's the output

[fair narwhal]

i dont need the canonical basis here right?

like if i wanted to go from B to E i just do φ([1,2,1]) = ([1,0,0])

[upper abyss]

The matrix is stated in terms of that basis, so yes.

[fair narwhal]

but i dont know how i get that third matrix involved

[upper abyss]

Is φ the endomorphism or a change of basis matrix? I think I got confused

[fair narwhal]

its the endomorphism

from a linear space V -> V

[upper abyss]

Then this isn't true

[fair narwhal]

what

[upper abyss]

Where does the matrix take (1, 2, 1)? How do you represent that answer in terms of B?

[fair narwhal]

i did that to get the matrix when going from B to E

[upper abyss]

Did what?

[fair narwhal]

phi(1,2,1) = (1,0,0); phi(2,0,0) = (0,1,0); phi(1,1,0) = (0,0,1)

[upper abyss]

Okay, so phi is not the endomorphism, but is a change of basis matrix

[fair narwhal]

i did that for the previous exercise: "Determine the matrices for the transition from B to E and back."

[upper abyss]

That's good!

[fair narwhal]

in my script it says that "A linear mapping φ: V → V is called an endomorphism."

[upper abyss]

Right. The change of basis matrix is technically an endomorphism

[fair narwhal]

oh

yes thats true i guess

[upper abyss]

However, I'm referring to the one in your work referred to as "the endomorphism"

[fair narwhal]

i just think they mean the matrix is the endomorphism

[upper abyss]

Multiplying by that matrix is, yeah

[fair narwhal]

but ive tried to write it as a function φ

oh

[upper abyss]

You're using φ as a different change of basis matrix already, and that's confusing you

I'll call that matrix R

[fair narwhal]

v1 * R = v2 no?

[upper abyss]

And "the endomorphism", I'll call that T

[fair narwhal]

Ok

[upper abyss]

T(v) = Rv

That is, to compute T acting on some vector, we really just multiply R by that vector and that's our output

[fair narwhal]

for one column?

no wait wtf

im so confuuuuuuused

[upper abyss]

So let's compute T(1,2,1)

[fair narwhal]

i thought that R was our linear transformation map

[upper abyss]

Kind of. Matrices and the functions they represent blend together a bit

,w {{1,2,1},{2,0,1},{1,-2,0}}{{1},{2},{1}}

[wraith dagger]

Results provided by WolframAlpha

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Requested by Kaynex#7520

[upper abyss]

There we go lol much better

[fair narwhal]

so thats the first column of our matrix

[upper abyss]

So that's T(1,2,1)

[fair narwhal]

oh ok

[upper abyss]

Expressed in E

[fair narwhal]

ok

and then i do that 3 times

and then again with * (1,2,1), * (2,0,0), * (1,1,0) to do that with respect to B?

[cedar kiln]

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[cedar kiln]

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[wicked remnant]

hi

.reopened

,reopen

.reopen

uhhhhhhhhhhhhhhhhhhhhhh

[cedar kiln]

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[wicked remnant]

how do i find lcm and hcf

[wicked remnant]

here are an example number: 34 and 85

<@&286206848099549185>

[celest ledge]

By prime factorization

The index of gcd is min of two indexes, index of lcm is max of two indexes , for any prime factor

[cedar kiln]

@wicked remnant Has your question been resolved?

[wicked remnant]

np

no

bruh nvm someone answered

bruh how to get a funcutional brain

[wicked remnant]

can u show me way to do it in an sheet of paper

calculate it for me

i want to see how do i calculate it

<@&286206848099549185>

[celest ledge]

Let $x=\prod_{1 \leq j \leq r}p_{j}^{r_{j}}$, $y=\prod_{1 \leq j \leq r}p_{j}^{s_{j}}$,
Then $gcd(x,y)=\prod_{1 \leq j \leq r}p_{j}^{min(r_{j},s_{j})}$, and $lcm(x,y)=\prod_{1 \leq j \leq r}p_{j}^{max(r_{j},s_{j})}$,

[wraith dagger]

Cogwheels of the mind

[celest ledge]

So you prime factorize 34 and 85 first

[cedar kiln]

@wicked remnant Has your question been resolved?

[wicked remnant]

what the fuck

[wicked remnant]

i am just grade 7

dude

what is hat

that is not what i teach

that is like grade 12 or something

r u joking

pls tell me fr

okspslpls

pls

plspls

[celest ledge]

Then why your school teaches you gcd and lcm without telling you what prime factorization is…

34=(2)(17),85=(5)(17)

So 34=(2^1)(5^0)(17^1),85=(2^0)(5^1)(17^1)
Min(1,0)=0,min(0,1)=0, min(1,1)=1 so gcd(34,85)=(2^0)(5^0)(17^1)=17. Max(1,0)=1,max(0,1)=0, max(1,1)=1 so gcd(34,85)=(2^1)(5^1)(17^1)=170

[wicked remnant]

its

lcm and hcm

lowest common factor and highest common factor of 34 and 85

qswdefrgetrwedqr34ft5y46htwre4q2rt5yjyrthewreytehrwgtgyhrjkjethwetgryjetwqt5yetjy

4htrnybgersearfgtytukfhngdfsfgrteyrumhndgbxfthr

[celest ledge]

Hcm is gcd

Just different names

[cedar kiln]

@wicked remnant Has your question been resolved?

[grizzled birch]

ok so

[grizzled birch]

lmao same

[wicked remnant]

hiii

[grizzled birch]

ok so

you know what

prime factorization is right?

[wicked remnant]

What

Wys

Yes

[grizzled birch]

okay well

prime factorize 34 and 85

34 would be 2*17

[wicked remnant]

Wait