help-39

[visual canyon]

shaking in my boots

i can find mean EZ this not so much

[bitter epoch]

As long as you understand the formula intuitively, you don't have to memorise it

The formula is just fancy math symbols for the total sum of each probability times value

[visual canyon]

oh

[bitter epoch]

For example, if you have a 6 sided fair die, the expected value is just 1/6 * 1 + 1/6 * 2 + 1/6 * 3 +...

[visual canyon]

ok one sec im gonna try the problem and see if i can get anything that looks like an answer..

ok i cheated a little and used a calculator

but i understand the formula now and i could probably do it by hand i just dont want to right now....

[bitter epoch]

Ok, if you can solve it then that's good

[visual canyon]

allright its c

.close thank you!

[pearl pond]

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[visual canyon]

how do i find relative frequency for this?

[pearl pond]

@visual canyon Has your question been resolved?

[visual canyon]

<@&286206848099549185>

oh

that was weird

@quiet tendon did you say something and delete it or am i crazy

[quiet tendon]

yeah that’s my bad, i thought i had a good answer but i realized it wasn’t too helpful

[visual canyon]

okok lmao dw

.close

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[silk solar]

Can someone help with this

[ember spear]

yo what did you try?

[silk solar]

after that im not able to solve it

[ember spear]

do the same for 2nd equation

[jolly parrot]

Pro_Hecker

[pearl pond]

@silk solar Has your question been resolved?

[silk solar]

Ok i did for both

Then

@ember spear

[ember spear]

so,

what is the problem then?

can you solve linear equations in two variables?

[silk solar]

I need to find the value of A and B from this how can i proceed after that

[ember spear]

what are the equations you got?

[silk solar]

4a+17=2.6(a+b)
4a+ 17 = 13 (a-b)

[ember spear]

ok nice so

can you simplify them?

[silk solar]

No

I mean i dont know i wanna know

I got the answer but didn't understand the process

[jolly parrot]

Pro_Hecker

[silk solar]

See this solution but there is no explanation about the calculation

[jolly parrot]

Pro_Hecker

[pearl pond]

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[silk solar]

.reopen

[pearl pond]

✅

[silk solar]

@ember spear

[ember spear]

Did you understand til heree?

[silk solar]

Yes i got till there

[ember spear]

then substract and you'll get linear equations in 2 variables

gtg, if you're not able to find an answer ping helpers

[silk solar]

Ok let me try

I got -10.4a-10.4b = 16

[pearl pond]

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[karmic knoll]

how many unordered sets of groups (A, B) such that A, B are subsets of [n] and:

[karmic knoll]

for any k that belongs to [n]

(write the answer as a function of k)

i want to make sure that my answer is correct

so what i did was choose k elements from [n]

and then for the remaining n-k elements, i split them into 2 unlabelled groups (A and B)

[pearl pond]

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[karmic knoll]

<@&286206848099549185>

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[arctic edge]

i dont know if i got it right

[arctic edge]

my equation/answer was

y=9x+18

i used (-2, 0) as my point

[pearl pond]

@arctic edge Has your question been resolved?

[arctic edge]

<@&286206848099549185>

[winter pawn]

yess

not me ig

[arctic edge]

sorry

the equation up top

[pearl pond]

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[pearl pond]

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[trail belfry]

$15 \sin^2(a) - 7 \sin(a) - 2 = 0 \ \sin(2a) = ?$

[jolly parrot]

nashira._.

[trail belfry]

so here i got

$\sin a = \frac 23 \ \ \sin a = - \frac 15$

[jolly parrot]

nashira._.

[trail belfry]

so I know that

$\sin a = \frac{1 - \cos 2a}{2}$

[jolly parrot]

nashira._.

[trail belfry]

the first one becomes

[pliant vessel]

why not use sin2x=2sinxcosx

[trail belfry]

I am trying to simplify to sin(2x)

[pliant vessel]

ah not needed

[trail belfry]

i have no idea how i would rearrange to make it 2sin(x)cos(x)

[pliant vessel]

unless its a formula

sin2x=2sinx cosx

i suppose you know the relation between sin and cos

[trail belfry]

yeah thats why I was going that route 😅

[pliant vessel]

not that one

sin^2x + cos^x=1

[trail belfry]

yeah thats where I was going

kinda

coz

[pliant vessel]

why did you do this

[merry carbon]

(A couple more things:
Do they give you any more context to the question?
Also, your identity, here, should be sin^2(a) rather than sin(a), ofc)

[trail belfry]

to get a cos(2a) square both sides after collecting, then convert cos^2(2a) to sin^2

[pliant vessel]

what im trying to say is

you already have sina

[trail belfry]

ohhh yeah one more thing

[pliant vessel]

cos a would simply be sqrt(1-sin^2a)

[trail belfry]

$a \epsilon (\pi ; \frac{3 \pi}{2})$

[jolly parrot]

nashira._.

[pliant vessel]

just multiply them both and multiply by 2

[merry carbon]

(Could be positive or negative, hence the ask for context)

[pliant vessel]

yes i thought so too

[merry carbon]

But yea, with this, as per above-

[pliant vessel]

so in 3rd quadrant both should be negative

[trail belfry]

yeah could I then finish my thought process then you could please tell me what is wrong with it?

[pliant vessel]

did you get this?

[trail belfry]

one sec

[pliant vessel]

3rd quadrant would mean you take the negative value of sina in this

[merry carbon]

(And, while you can use your way, it’s a bit more work imo than it’s worth)

[pliant vessel]

exactly

[trail belfry]

$\frac{1 - \cos 2a}{2} = \frac 23 \ \ \cos 2a = -\frac{1}{3} \ \ \cos^2(2a) = \frac 19 \ \ 1 - \sin^2 (2a) = \frac 19 \ \ \sin^2(2a) = \frac 89$

[jolly parrot]

nashira._.

[trail belfry]

and then we would take the negative value of square root of 8/9

do the same for the other one

[merry carbon]

See the points here-

[trail belfry]

the only context is it is in the range I gave and we are finding sin 2a in that range thats all

[merry carbon]

This tells you that you must have sin(a) = -1/5, and the identity is sin^2(a) = [RHS]

[merry carbon]

Negative as per here-

[trail belfry]

aaaaah i get now coz its negative lmao sorry im a dumbass 😂

so when I use -1/5 I get

$\sin^2(2a) = \frac{16}{25}$

[jolly parrot]

nashira._.

[trail belfry]

so then ofc its just

$\sin 2a = \frac 45$

[jolly parrot]

nashira._.

[trail belfry]

so we just get both values of this right?

oh shit and since it is a square root also

$\sin 2a = -\frac 45$

[jolly parrot]

nashira._.

[merry carbon]

Well, I would personally consider the range, 2a is between 2pi and 3pi exclusive, an interval equivalent to 0 to pi, so there you know sin(2a) is positive

[trail belfry]

aaah

[trail belfry]

so we only use the first one here

so it will just be

$\sin 2a = \frac 45$

[jolly parrot]

nashira._.

[merry carbon]

Hmmmmm, one moment-

[merry carbon]

You did note that you're supposed to be using $\sin^{ {\color{green} 2} }(a) = \frac{1 - \cos(2a)}2$, right-

[jolly parrot]

@merry carbon

[trail belfry]

aaah lmao i was supposed to square all the way at the beginning 💀

and u even told me

Istg im just dumb 😂

[merry carbon]

Yeaaaa, of course you could've noted the identity for $\cos(2a) = 1 - 2\sin^2{a}$ was in that form to begin with too haha

[jolly parrot]

@merry carbon

[merry carbon]

And it happens mind you, those were both like afterpoints anyway, should have been in a separate message really

[trail belfry]

nah nah i just cannot read 😢

is there a better way of doing this btw?

Im sure my method is shit anyway

[merry carbon]

Well, the suggested method, as per before [from there, you can figure out that cos(a) = -sqrt{24}/5, from sin^2(a) + cos^2(a) and the quadrant info), from which point the sin(2a) = 2sin(a)cos(a) gets you sin(2a) immediately]

but other than that, not too sure

[trail belfry]

ahaaaaa..... 😅

[merry carbon]

[trail belfry]

wait wait wait i have a question

[trail belfry]

so using quadrant info as here we can "infer" the value of cos?

Since it is in the third quadrant, sin = -cos?

so

$\cos a = \frac 15$

[jolly parrot]

nashira._.

[trail belfry]

is what Im saying making any sense or am I still just dumb

[merry carbon]

In the third quadrant, sin and cos are both negative (whereas tan, of course being the ratio of them, is positive)

[trail belfry]

shit mb. So in this case, sin is negative so is equal to cos?

[merry carbon]

It isn't the case that they're negatives of each other (see as per before, cos(a) is actually -sqrt{24}/5, via Pythag)

[merry carbon]

Not equal, no see what happens when you try and use sin^2(a) + cos^2(a) = 1, even ignoring the signs

[trail belfry]

I cannot understand for the life of me why it is -sqrt(24)/5 😢

[merry carbon]

,w plot y = sin(x), y = cos(x)

[jolly parrot]

Results provided by WolframAlpha

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Requested by chartbit

[merry carbon]

Notice how in between pi and 3pi/2, both sin(x) and cos(x) are negative, and-

[merry carbon]

...you found that sin(a) = - 1/5 (the negative one for the reason just stated), now, using sin^2(a) + cos^2(a) = 1, and noting that cos(a) is negative, what is the value of cos(a)?

[trail belfry]

ohhhh so just substitute?

[merry carbon]

Yep, and cook

[trail belfry]

wait didnt u say this before? I swear I cannot read 💀

[merry carbon]

relatable

[pearl pond]

@trail belfry Has your question been resolved?

[pearl pond]

@trail belfry Has your question been resolved?

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[trim solstice]

Hey all, I'm trying to prove the following $(\lnot p \to p)\to p$ using hilbret's system of axioms, any help? I'm newbie to those systems and having hard time doing this proof.

[jolly parrot]

mtr123

[pearl pond]

@trim solstice Has your question been resolved?

[pearl pond]

@trim solstice Has your question been resolved?

[pearl pond]

@trim solstice Has your question been resolved?

[fathom oasis]

Can you give a reference to the system of axioms?

When I google hilbert's sytem of axioms I get a set of axioms for euclidean geometry

[pearl pond]

@trim solstice Has your question been resolved?

[trim solstice]

Sorry about that, let me find what I’m meaning

[trim solstice]

Here, sorry

actually for the third one we got
$(\lnot B \to \lnot A) \to ((\lnot B \to A)\to B)$

[jolly parrot]

mtr123

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[loud phoenix]

i dont get the wording of this

how can angle PBQ be 70

how is it not 90

[versed remnant]

Pbq?

[loud phoenix]

in the question it says pbq is 70

did I draw it wrong

[pearl pond]

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[polar crest]

can u help me sketch the graph of y=2x+10 but only for x ∈ (-2,3]

[polar crest]

im struggling to understand it

and very urgent

[midnight haven]

Can you plot a straight line ?

[polar crest]

yeah

[midnight haven]

Do it first

[polar crest]

i have

like do u just insert the values of -2,3

into the equation

for x

i mean these values =

x

[midnight haven]

Draw y= 2x +1
Then on the x axis look for the point -2 and point 3
Then the curve in between these two points would be your answer
Rest line beyond these point,you can erase them

[polar crest]

wth

so do u link those points

@midnight haven

like wiht a curved line

please help

this is very urgent

[midnight haven]

After plotting it cut the line

[polar crest]

can u please draw an example i really have no idea what that means 😭

[midnight haven]

I.e remove everything except the part of the line which lies BTW-2 and 3

[polar crest]

so -2,3 are x values right?

[midnight haven]

Yes

[polar crest]

so anypart of the line which lies anywhere on thee axis where x =3 or -2 must be erased?\

[midnight haven]

No

Everything else exc3pt that part must be erased

[polar crest]

is my understand coorect 😭

[midnight haven]

✅

[polar crest]

kk

tysm

[midnight haven]

There is one more thing though

X is not equal to -2

So I guess you have to keep a point circle at that point

[polar crest]

why is it not equal

it says draw the line between x values from -2 up to 3

[midnight haven]

Belongs to (-2,3]

That is an open bracket which means X is not rual to -2

Equal

[polar crest]

so what do i draw instead

[midnight haven]

^

[polar crest]

sl just a dot there

[midnight haven]

Not a dot

Dot 2ould mean it includes that value

A circle

[polar crest]

kk

[midnight haven]

I am not sure if it is right though

[pearl pond]

@polar crest Has your question been resolved?

[pearl pond]

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[meager imp]

Someone please help me check this result

<@&286206848099549185>

[cursive wraith]

while the first equality is true, in general you need more justification for interverting infinite sum and integral

[pearl pond]

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[pearl pond]

@meager imp Has your question been resolved?

[dapper path]

Are you sure? Considering the infinte sum only acts upon the n variable, surely this proof holds? I am not sure what other more explicit way you can prove this.

@meager imp Is this proof correct? IMO this is a-ok

[pearl pond]

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[pearl pond]

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[fast lion]

Guys what does the (>) mean?

[cloud wing]

Parallel

[fast lion]

Oh, thank you!

.close

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[dusky scaffold]

why I am getting a different answer with these vectors

[dusky scaffold]

one sec

A motorboat is racing
towards north at 25 km/h and the water
current in that region is 10 km/h in the
direction of 60° east of south. Find the
resultant velocity of the boat.

[steel canopy]

Ok what did u do?

[dusky scaffold]

So if A is the vector representing the motorboat

and C is the current of the river

[steel canopy]

A is a point?

[dusky scaffold]

why is it that the resultant vector is A1 C and not AC1

[steel canopy]

?

[dusky scaffold]

A A1

vector

[steel canopy]

Ok

It should be AC1

Cuz the boat will move up

[dusky scaffold]

why do they take the angle as 120 and not 60

i thought you had to move the bottom vector up

then the angle would be 60

[steel canopy]

Angle u should consider is angle between vectors

Which is thetha

And not 60

[dusky scaffold]

but that doesnt give the magnitude of R does it

[steel canopy]

Thetha will be 120

And it will give the right answer

[dusky scaffold]

so is R the sum of the two vectors

[steel canopy]

Yea

[dusky scaffold]

this vector is the resultant right

[steel canopy]

Yea

[dusky scaffold]

but isnt the angle here 60

the angle between the vertical vector and top one

[steel canopy]

Um i didnt get u?

[steel canopy]

Is A1C1 padallel to that AC?

If yes then yes

[dusky scaffold]

so shouldnt we take cos(60)

and not 120

to get the magnitude

oh i think you get the same answer both cases then

• cos(120) is the same as - cos(60)

aight got it now

.close

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[trail linden]

but in graph jumps:
Closed + closed circle = included
closed + open = not included
open + open circle = not included?

[iron cobalt]

what

[pearl pond]

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[opal patio]

Hey folks, quick combinatorial question:
If I have combine two same size sets of distinct elements what is the chance that all elements that I draw randomly from the combined set come from a specific one of the original sets?
Example: I have a box with 50 red balls and 50 blue balls. I draw 5 balls randomly without replacing them. What is the chance that they're all blue?

[opal patio]

In the example it should be 50/100 * 49/99 * 48/98 * 47/97 * 46/96, right?
Is there a general formula for this kind of situation?

[timid spindle]

You can alternatively do (number of ways to pick 5 blue balls)/(number of ways to pick 5 balls)

[opal patio]

Number of ways to pick 5 balls is just 100 choose 5, right? How do I get number of ways to pick 5 blue balls?

[timid spindle]

How many blue balls are there and how many do you have to choose?

[pearl pond]

@opal patio Has your question been resolved?

[opal patio]

For the example, 50 blue balls of 100 and choose 5

[trail linden]

piecewise functions

[timid spindle]

Right so it's the same logic as your 100 choose 5 here

[opal patio]

Could you just tell me? This is not homework, just a minute detail in an optimizing problem I have 😄

[midnight haven]

Two

[timid spindle]

You quite literally have to choose 5 blue balls out of 50 blue balls to choose 5 blue balls out of 50 blue balls

[pearl pond]

@opal patio Has your question been resolved?

[opal patio]

There seems to be some sort of miscommunication here

At least I'm confused. Maybe I can rephrase my example more clearly:
I have a box with 50 blue balls and 50 red balls. If I draw 5 random balls at once, what are the odds that all of them are blue?

Alright, I got it, nevermind

[opal patio]

A bit of honest feedback, that's not a helpful response

Thanks for trying to help though

.close

[pearl pond]

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[proud reef]

hi

[proud reef]

there isnt a concrete solution to this right?

the final ecuation i get is 5a+b = -19

[hollow cobalt]

There is a unique solution

[sharp vigil]

you should check both potential points of discontinuity

[pearl pond]

@proud reef Has your question been resolved?

[proud reef]

sorry but which is the other one?

apart from 5?

[sharp vigil]

the function switches from quadratic to linear to quadratic

[proud reef]

ohh

let me try

it worked thankss

[pearl pond]

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[lethal estuary]

so i have a quick question

[lethal estuary]

if i were to get asked

to get the basis of vectors given

i would put them in columns into a matrix

get echelon

then pick out the vectors that dont have free variables

correct ?

would those be the basis vectors ?

as opposed if i were to get he basis to a null

id have to actually solve the thing

get the vectors that are the answers to the null a

and those would be the basis vectors

can someone confirm and clear up misconceptions that i have if i have any

[jade sorrel]

what you actually can do is much simpler

you can just put the vectors as rows of a matrix

then when you run gauss jordan elimination on the matrix you get basis vectors that still generate the same space

[jade sorrel]

and not really what you described

[sharp vigil]

well putting them as rows and running gaussian elimination is really not any more or less simpler then putting them in columns and running gaussian elimination

[jade sorrel]

gauss jordan elimination does not change the space generated, so the space is the same but duplicit vectors get eliminated as a byproduct of the elimination

[jade sorrel]

is echelon what gauss elimination is called?

[sharp vigil]

gaussian elimination gives row echelon form, whereas gauss-jordan gives reduced row echelon form (although both work for the purposes of finding bases)

[jade sorrel]

i otherwise agree with you, i just did it always by rows, but it doesnt matter

[sharp vigil]

useful reading: https://smashmath.github.io/blog/rowcolspace/#column-space

[jade sorrel]

oh i never heard the term echelon, sorry, that was my bad

[sharp vigil]

well rows gives you a simple basis, whereas columns gives you a basis of the vectors you already have. both of which can be useful for different purposes

[jade sorrel]

but all vectors that remain after gauss elimination (are not null) are basis vectors

if you want them orthogonal to each other you need to run graham schmidt as well

the second part of the question is i suppose about how to find the matrix kernel (a space that is mapped by the matrix to zero)

[pearl pond]

@lethal estuary Has your question been resolved?

[pearl pond]

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[tidal sonnet]

How many did I get wrong and which ones

[regal herald]

this a test of some form?

[timid spindle]

if you submit your answers you'll find out

[regal herald]

and you really want people to scroll through 25 questions to check them

as fellow said, submit tool will give you that instant feedback

[tidal sonnet]

It’s a pre test and I want to know what ones are wrong and fix them because they don’t tell me how to fix them

[half saddle]

@timid spindle i really need some help if you have the time, pleasee

[regal herald]

submit it, look at whats wrong, if you cant figure it out, then ask again

[tidal sonnet]

100/100

Let’s goo

Guessed on all of them pretty much

.close

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[pearl pond]

Please don't occupy multiple help channels.

[chrome raptor]

im cooked lol

[pearl pond]

@chrome raptor Has your question been resolved?

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[tawny talon]

what unit is this?

[tawny talon]

I have exam which look similar to this paper and ik nothing

[pearl pond]

@tawny talon Has your question been resolved?

[tawny talon]

.close

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[unkempt rose]

hi guys there are a lot of things i dont know

[unkempt rose]

\begin{align*}
\pi \int_{-1}^{3} [8 + 2y - y^2]^2 &= \pi \int_{-1}^{3} 64 + 4y^2 - y^4 \
&= \pi \int_{-1}^{3} 64y + \frac{4}{3}y^3 - \frac{y^5}{5} \
&= \left[ 64(3) + \frac{4}{3}(27) - \frac{243}{5}\right] - \left[ -64 - \frac{4}{3} + \frac{1}{5} \right]
\end{align*}

[jolly parrot]

Juan

[unkempt rose]

first, in the y replacement where i shut put the pi

now, as far as im doing it, its right?

obvioussly i need to put the pi in to multiply both [] i think

btw
$(8 + 2y - y^2)^2$ is equivalent to $(64 + 4y^2 - y^4)$ or i should use the trinomial method

[jolly parrot]

Juan

[unkempt rose]

so

[summer imp]

No, it's generally false that $(a+b)^2 = a^2 + b^2$, let alone three terms

[jolly parrot]

Azyrashacorki

[unkempt rose]

so i started wrong_

?

[summer imp]

Yeah

Also, don't forget dy's if you're submitting this

[unkempt rose]

this is just for you

[summer imp]

Okok

Just in case, it's good practice

[unkempt rose]

.close

[pearl pond]

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[desert solar]

$\sum_{n=2}^{\infty}\left(1 - \frac{\ln(n)}{n}\right)^{n}$

[jolly parrot]

James Price

[desert solar]

I'm stuck

Here

$\frac{1}{(\frac{n-\ln(n)}{n})^{-n}}$

[inland ivy]

do you need to find this series explicitly, or determine convergence/divergence?

[desert solar]

😭

Convergence

[inland ivy]

well, is that not important enough to be mentioned?

[desert solar]

How do I put that denominator to -n

I was first trying to write where I had arrived, actually I had to write it straight away

[jolly parrot]

James Price

[desert solar]

Nice

I came back to this harmonic series

Given that the exponent of the harmonic series is <1, does the series converge?

<@&286206848099549185>

It should be right

@inland ivy

Now I'm here, so I think I did well

$\frac{1}{n^{-n}}$

[jolly parrot]

James Price

[pearl pond]

@desert solar Has your question been resolved?

[cursive wraith]

i think it's actually 1/(n^(1/n))

but you can notice that it's always >= 1/2

or find through log that it converges to 1

[pearl pond]

@desert solar Has your question been resolved?

[pearl pond]

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[glacial sequoia]

I don't know if (b) is super obvious for people who work with two variable functions, but I don't so

For (a) I basically shifted everything on the real line into the square

something like (x, 0) --> (x/10, 1/2)

[rough stream]

You want to construct two real numbers, using the decimal places of one real number

I'm nervous to explore that fully, since infinite decimal expansions are weird, though

[glacial sequoia]

Is that meant to be for (a)

[rough stream]

Oh, yeah you have a

[glacial sequoia]

Because does this work

[glacial sequoia]

yeahh

I think your idea is useful for (b) though

take two real numbers and file them into a single one

and is this meant to be an obvious thing to do

[rough stream]

Definitely not going to call this obvious. Again, I'm nervous. This seems like the kind of thing that causes problems

[glacial sequoia]

Okay if I have like 0.7583276428... and 0.584739248309...
You can just pick alternating digits right so those two map to 0.7558843...

[glacial sequoia]

Well that's easily done too I guess

Yeah that would be for (a)

Now I look like a lunatic

[rough stream]

Better you than me

[glacial sequoia]

@summer imp the idea works?

This works fine for infinite decimal expansions I guess

[summer imp]

Yeah I was trying to point towards that actually

[glacial sequoia]

I mean we wouldn't consider 0.9999999... to be in this set would we?

[cursive wraith]

it's 1

[glacial sequoia]

Yeah

So if I had 0.9, 0.9 I would just map that to 0.99

[glacial sequoia]

Hey I think this works just as well for ginite ones

[cursive wraith]

!occupied

[pearl pond]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[signal finch]

HELP I HAVE AN EXAM

HELP PLSS

[glacial sequoia]

You didn't even pick a top channel why me

[cursive wraith]

read

[glacial sequoia]

[rough stream]

Yeah I'm not nervous about finite decimals, this obviously works

[glacial sequoia]

Open your own channel

[signal finch]

CAN SOMEBODY HELP?

[cursive wraith]

This channel is occupied. Last warning

[rough stream]

But then infinite ones are

[glacial sequoia]

Also we don't help with exams

[signal finch]

Ahh ok how can i open?

[summer imp]

It's an exam either way.

[glacial sequoia]

Is there an outright problem though with the alternating approach

[cursive wraith]

eh, not really
you can show that [0,1) is in bijection with the sequences of {0,1,...,9} that don't station to 9

[glacial sequoia]

Sorry I don't quite know what that means

[cursive wraith]

meaning no 8 3 5 2 5 2 9 9 9 9 9 9 9 9 9 9...

for example

(you can also replace 9 with any base b-1)

[rough stream]

I think your solution is good 😊

[glacial sequoia]

Uh...

[glacial sequoia]

Right, thanks 😭

[summer imp]

In other words, even though every number has a representations with eventually infinitely many repeating 9's, you can always pick a nice one that doesn't have that.

[glacial sequoia]

Ohh

Yeah fair

[cursive wraith]

not just infinitely many, like only 9s after some point

if that's what you meant by repeating I'm sorry for not understanding

[summer imp]

Yeah I meant that but it's true reading through again it wasn't phrased properly haha

[glacial sequoia]

I understand now, thanks all three of you :)

.close

[pearl pond]

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[pearl pond]

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[eager sun]

I forgot how to integrate this

[tacit mauve]

Hello, can anybody help explain a statistics question, because I have been wrecking my brain for the last 15mins trying to figure this out?

[dim dagger]

create a new help channel above, see #❓how-to-get-help

[leaden marsh]

Nah im sorry
Didnt work

[eager sun]

All good

@dim dagger I see u typing pls help 🙏🙏😓

[dim dagger]

I am trying but its taking too long lol

give me a min :S

[eager sun]

IS IT HARD

how they expect me to do this

Brother 💀

[dim dagger]

I don't think so, its because of the latex kjdkasjdkaj

Let $2x+3 = u$, then $du = 2$ and $5x^2 + 6x = x(5x+6) = x(2(2x+3)+x)$. See that $\frac{u-3}{2} = x$. So, we get $\frac{u-3}{2} \cdot (2u + \frac{u-3}{2}) = \frac{u-3}{2} \cdot \frac{5u-3}{2}$. At the end, we get $\int_1^4 \frac{5x^2+6x}{\sqrt{2x+3}} = \int_5^{11} \frac{(5u-3) \cdot (u-3)}{\sqrt(u)} \cdot 1/4 \cdot \frac{du}{2} $.

Now, the final form is much easier to integrate

[eager sun]

Holy

[dim dagger]

Its seems way to confusing haha, let me clear it up a bit, it becomes $1/8 \cdot \int_5^{11} \frac{(5u-3) \cdot (u-3)}{\sqrt(u)} du$

omg it did not write 11 at the top

lord help me with latex, im losing it

[jolly parrot]

banana of rivia

banana of rivia

[leaden marsh]

Mmmmk and then using integration by parts right?

[eager sun]

okay I’m gonna go through this topic again

[dim dagger]

Yea, but im too lazy to write that shit in latex

[eager sun]

Thx guys

[leaden marsh]

lol

[eager sun]

.close

[pearl pond]

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[hybrid haven]

how do i write the integral for this

i dont have a domain for the probability density funciton

is it that lim-->infinity shit

[quartz citrus]

wtf

[hybrid haven]

wat

@quartz citrus

how do i do that

how do i sub in infiinity

i dont rly understand

i literally have no idea what u mean

can u show me

[jolly parrot]

..,,

[timid spindle]

you can also do $\lim_{p \to \infty} \int_0^p f(t), \dd t$, i.e. do the symbollic integration and worry about the limit after

[jolly parrot]

ΣAC

[hybrid haven]

yeah thats what my teacher taught me but i have no idea how it works

[timid spindle]

well just do the integration with upper limit p for a start

[pearl pond]

@hybrid haven Has your question been resolved?

[pearl pond]

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[mental verge]

I dont understand question 14.15

My work so far

How can 1+cosx be equal to 2cos^2(x/2)

[dim dagger]

use the equality cos(x) = 2cos^2(x/2) - 1

If you know how to calculate cos(a+b), you can see cos(x) as (cos(x/2 + x/2)), that would give you the same equality as well.

[mental verge]

how can that be equal

which formula is that

[mental verge]

i know how to calculate cos(a+b) but never seen this equation u shared

<@&286206848099549185>

[rough otter]

Formula of Carnot

[midnight haven]

Cosa*cosb -sin a * sin v

[blazing inlet]

if you want to prove why cos(x) = 2cos^2(x/2) - 1, take cos(x) form it as cos( x/2 + x/2) on applying the formula cos(a+b) you would get cos^2(x/2) - sin^2(x/2), now add and subtract 2cos^2(x/2) you would get, 2cos^2(x/2) - cos^2(x/2) - sin^2(x/2) now on applying sin^2(x) + cos^2(x) = 1 you would get 2cos^2(x/2) - 1 = cos(x)

there you go, now we have cos(x) + 1 = 2cos^2(x/2) proved

[midnight haven]

cos2x = 2cos^2 x - 1

clearly cos x = 2cos^2 (x/2) -1

[blazing inlet]

they asked for an explanation on why these 2 would be equal

[midnight haven]

💀

[pearl pond]

@mental verge Has your question been resolved?

[hot lodge]

It is straightforward

cos(x)=cos(x/2 + x/2)

[mental verge]

oh

[hot lodge]

=cos^2(x/2) - sin^2(x/2) = 2cos^2(x/2) - 1

[mental verge]

okay uhm still

how am i gonna solve the question

it got 2cos^2x/2 but not with -1

what do i do

[dim dagger]

When you apply this equality, I believe that you should get something in the form of $a^2 -a - 1 = 0$, where $a = cosx$. Then you find the roots of this equation.

[jolly parrot]

banana of rivia

[dim dagger]

This gives you the value of cosx. Finding what x should be is a different matter, I feel like the solution on the photo you posted uses some kind of a calculator.

I do not think it can immediately be found, additional work is required.

[mental verge]

but how do i apply?

like as i said

we got cos^2x and 2cos^2(x/2)

no cosx or 2cos^2(x/2)-1

i dont get it like i get the formula but idk what should i be doing atm

[dim dagger]

because we know that $2cos^2(x/2) -1 = cos(x)$, we can say $2cos^2(x/2) = 1 + cos(x)$. So, you can substitute this value to the left side of the equation.

[jolly parrot]

banana of rivia

[dim dagger]

That gives you $cos(x) + 1 = cos^2(x)$, and then we try to solve this equation. Is that clear?

[jolly parrot]

banana of rivia

[dim dagger]

I guess what I told you is the first solution, are you trying the second one ?

[mental verge]

oh okay

okay alr lemme try

[mental verge]

i was doin some other questions

[mental verge]

no im not trying the second solution tbh i dont understand from their solutions

the ppl in this channel explain it better than the book and im grateful for everyone that helps me here

[dim dagger]

the solutions are a bit confusing yeah, it does not include many middle steps to get there

[mental verge]

i got cos^2x-cosx-1

[mental verge]

cant perform quadratic thing here
b^2-4ac should i do this?

[dim dagger]

Yeah, that is your discriminant, the roots are defined as $\frac{-b \pm 2\sqrt{\triangle}}{2a}$, where $\triangle$ is your discriminant, that is $b^2 - 4ac$.

[jolly parrot]

banana of rivia

[dim dagger]

You will find two values as the root of this equation.

one of which, in this case, is the golden ratio.

[pearl pond]

@mental verge Has your question been resolved?

[mental verge]

dude the solution finds a number but i cant

look i've done the thing u said

its just cos things

no specific number

[blazing inlet]

you took the values of variables, you were supposed to take the values of the coefficients

in this case, the equation being cos^2(x) -cos(x) - 1

so in this case we have the coefficients, a=1 b= -1 and c = -1

by applying the quadratic formula we would get two values for cos(x)

applying the quadratic formula here, would give us two roots of the equation that is (1 ± √5)/2

so we have cos(x) = (1 ± √5)/2

so x = cos^-1 {(1 ± √5)/2 }

cos has a value between -1 and 1 so we cannot have (1 + √5)/2 as one of the possible answers

Therefore, x = cos^-1 {(1 - √5)/2}

[pearl pond]

@mental verge Has your question been resolved?

[ember spear]

what is your doubt?

[mental verge]

i dont see a relation of the question and its solution and the solution shared here

[mental verge]

values of coefficients?

[blazing inlet]

general form of any equation, quadratic in this case can be written in the form ax^2 + bx + c = 0, over here a, b and c are the coefficients so and x is the variable, in the equation cos^2(x) - cos(x) -1, cos(x) is the variable and 1, -1 and -1 are the values of its coefficients in this case

[pearl pond]

@mental verge Has your question been resolved?

[mental verge]

i still dont get how can we get 1 +- √5 over 2 though

do we simply remove cosx?

[mental verge]

how can we be sure cos x = 1?

[pearl pond]

@mental verge Has your question been resolved?

[pearl pond]

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[waxen condor]

Hi

[midnight haven]

hi

[waxen condor]

!status

[pearl pond]

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

[midnight haven]

1

[waxen condor]

Ok

So Given student doesnt knows answer and guesses it correct is 1/2

In the second part it talks of bayes theorem

Do you know it ?

[midnight haven]

yeah i konw that

[waxen condor]

So you want me to

Do the solution

Or try some more ?

[midnight haven]

do the solution, cause i am bad at probability

[waxen condor]

Ok

So see the probability the student guesses be x

Then the probability he doesnt guess is (1-x)

[midnight haven]

yeah

[waxen condor]

Now

The conditional probability that he guesses and its correct is 1/2

C = correct

[midnight haven]

mhm cause it is i true/flase type question

[waxen condor]

P(C/x) = 1/2

So

Now

In the second part its says

P(x/C) = 1/6

So now

Expand it

[midnight haven]

oh damn

[waxen condor]

Can you ?

[midnight haven]

independent or not ?

[waxen condor]

p(x/c) = (P(x) × p(c/x))/(...)

Use the bayes theorem definition

(guesses × correct)/((gusses×correct) + (already known× correct))

Understood ?

[midnight haven]

? is this correct

[waxen condor]

Yeah

But here we need this definition

This question is basically a reverse bayes

[midnight haven]

3 P(x) = P(C)

[waxen condor]

P(B) = answer correct

Which can happen in two ways

[waxen condor]

Here i mentioned

[midnight haven]

is this bayes thm only ?

i think i need to revise

[waxen condor]

$$P\left(\frac{\text { guessed }}{\text { correct answer }}\right)=\frac{P(\text { guessed }) P\left(\frac{\text { correct ans }}{\text { guessed }}\right)}{P(\text { guessed }) P\left(\frac{\text { correct ans }}{\text { guessed }}\right)+P(\text { knows }) P\left(\frac{\text { correct ans }}{\text { knows }}\right)}$$

[jolly parrot]

Cairo

[waxen condor]

P (guessed/correct ans) = 1/6 given

P(correct ans/guessed) = 1/2 given

And we have consider P(guessed) = x

So P(Known) = 1-x

Just put in the formulae

[midnight haven]

oh oh

i see

[waxen condor]

So answer will be 1-x

[midnight haven]

yeah

[waxen condor]

listen

In this type of questions

Always read what they are asking to find

And consider it the variable

It will make it easier for you

[midnight haven]

yup, i made mistake of not considering variables and couldnt find out that
P (guessed/correct ans) = 1/6
P(correct ans/guessed) = 1/2

thanks dude!

[waxen condor]

Welc !

[midnight haven]

have a great day

❤️

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[pearl pond]

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[waxen condor]

Same to you !

[pearl pond]

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[remote ledge]

how do I do this? its pretty simple, but then they intersect at 90 degrees, im confused on that part

[remote ledge]

does that mean I use dot product?

[summer imp]

Can you give a vector that describes each line?

[remote ledge]

what do you mean

[summer imp]

The lines you're given are in vector form, which means that you're given a point and a direction, which generates the line.

So from those equations, can you tell the vector which gives direction in the first line?

[remote ledge]

doesnt the 4, 8, -4 give direction?

[summer imp]

Yes. And similarly (-1,2,3) gives direction on the second line.

How can you tell if two vectors are perpendicular?

[remote ledge]

idk

you use dot product?

[summer imp]

yess

Two vectors are perpendicular iff their dot product is 0.

[remote ledge]

I think i did something wrong

@summer imp

[summer imp]

Well the dot product is 0, so they are perpendicular aren't they

[remote ledge]

yeah

[summer imp]

That's what you wanted to prove for the first part.

For the intersection, you should set the coordinates equal (in terms of s and t) and solve for s,t. This will then give you the point at which they cross eachother.

So for instance, for the first coordinate, we want 4+4t = 1 - s

[pearl pond]

@remote ledge Has your question been resolved?

[remote ledge]

@summer imp ok thank you

for this question do we use cross product

then we use the vector and point to create the vector equation

[summer imp]

Exactly.

[pearl pond]

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[magic knot]

Not sure what happened here, in the underlined parts. Where did the extra h go? 😭 Like 1+h*sqrt(1+h) would make sense to me, but ??

[magic knot]

wait, are the +1 and sqrt stuff multiplied separately? 🤔

[cinder flower]

note that 1 + h - 1 = h

and in the denominator you can factor out an h

[remote ledge]

factored out the h

[magic knot]

I know we factored out the h, but there's 2 D:?

[west sapphire]

the comments in the right column are chatgpt-level of quality

[magic knot]

what?

oh

yeahh xD

they're sometimes better, but not here :')

[latent quail]

Are they generated by gpt?

[cinder flower]

We multiplied.

[magic knot]

like I see why "1" ended up on top, and I know we cancelled out an h, but the other just kinda dipped

[magic knot]

nah this is Quizlet

well

I guess they could be generated

I technically don't know, but I didn't do it at least xD

[magic knot]

but where? :<

this is probably obvious, but I'm very tired and struggling lmao

[latent quail]

of course

[cinder flower]

o i was just making fun of how useless the comments are

[magic knot]

oh lmao yeah, not at all helpful

what did y'all multiply?? 😭 I need details

[latent quail]

1/h

That’s what we multiplied

From the second step to the third, which is also where you circled

[magic knot]

but at that point there was just an h on top

[latent quail]

yeah

So you divide it with h

Which is equivalent to multiply 1/h

[magic knot]

oh I know how a 1 popped up, it's purely the bottom I'm lost on

although they do correlate, so I might be misunderstanding you

[latent quail]

Are you referring to the third to the fourth step?

Just replace h with zero:(

[magic knot]

Idk if this helps, but like. I know how h on the top becomes 1 (something crosses out on top and bottom) but I'm not sure which h crosses out, and I'm not sure why two cross out. Or where the 1 on the bottom came from

[magic knot]

I know that part luckily :D

[latent quail]

Here

Same idea

[cinder flower]

why wouldn’t you make h the thing factored out and cancelled

[magic knot]

I'm so very lost

I might just be too sleep deprived for this problem. 😭 I can try again tomorrow @~@

[latent quail]

I understand your feelings lol

I used to had no idea about how it works as well

cuz it’s weird

[magic knot]

I appreciate your patience, haha

I've decided to cram a term's worth of homework into roughly 3 days, so I am very frazzled :')

kinda. More complicated than that.

Anywho's, I'll just move on for now. Thank you guys! :D

[latent quail]

Have a good one

[magic knot]

you too!

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[steel pendant]

help can someone help me understand electric potential

[pearl pond]

@steel pendant Has your question been resolved?

[steel pendant]

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[sharp smelt]

so {0} is an element

next it is closed under addition as $v+u \in \R^2$

[jolly parrot]

ƒ(Why am. I here)=lin-alg

[sharp smelt]

as is $\Xi v$

[jolly parrot]

ƒ(Why am. I here)=lin-alg

[tropic saddle]

no. 0 is an element. {0} is a subset

also, why

[sharp smelt]

ah

[sharp smelt]

why what?

[tropic saddle]

why are you using random letters like Xi

[sharp smelt]

for fun

sorry

[tropic saddle]

why is 0 in U

you need to explain what your letters mean

especially if you arent using the standard letters for which you can guess

[sharp smelt]

They have defined it to be closed under addition

so $v-v=0$

[jolly parrot]

ƒ(Why am. I here)=lin-alg

[tropic saddle]

what is v

[sharp smelt]

$\Xi$ is a constant

[jolly parrot]

ƒ(Why am. I here)=lin-alg

[sharp smelt]

a vector in $R^2$

[jolly parrot]

ƒ(Why am. I here)=lin-alg

[tropic saddle]

ok why do I care about vectors in R^2. I care about U

[sharp smelt]

U is a subset of R^2 that is closed under addition

oo

considering vectors in U

[tropic saddle]

{} is also closed under addition

[sharp smelt]

but U is non-empty

[tropic saddle]

thats what I wanted to hear

because U is nonempty you can take some vector v in U. then -v is also in U and therefore v+(-v)=0 is in U

[sharp smelt]

So my way of working lin alg is still too informal is what you're trying to say ?

now to verify closure under multiplication

[tropic saddle]

yes

[sharp smelt]

let $v$ be. a vector in $U$

[jolly parrot]

ƒ(Why am. I here)=lin-alg

[sharp smelt]

now let $\lambda$ be a scalar multiple

[jolly parrot]

ƒ(Why am. I here)=lin-alg

[sharp smelt]

so we have $\lambda v \in U$

[jolly parrot]

ƒ(Why am. I here)=lin-alg

[sharp smelt]

not sure of how to formally express this though

[tropic saddle]

then express it informally first

[sharp smelt]

wait, it doesn't have to be closed under multiplication, I could define the subset to only be the integral points on the axes

[tropic saddle]

can you?

[sharp smelt]

why can't I? it would still be closed under addition

[tropic saddle]

well I am just asking

demonstrate it formally

[sharp smelt]

define$U =[(x,y): x,y \in \Z]$

[jolly parrot]

ƒ(Why am. I here)=lin-alg

[sharp smelt]

U containes the 0 vector

and is closed under addition

but not under multiplication

[tropic saddle]

thats not a set

closed under inverses?

[sharp smelt]

yes

[hollow cobalt]

||{}, also you could just write Z^2 instead||

[tropic saddle]

prove it. then you are done

[sharp smelt]

if v is the point (x,y) ; then -v =(-x,-y)

then v-v =(0,0)

which is the 0 vector

[tropic saddle]

which is still in U why?

[sharp smelt]

I've defined U to be that

it's closed under addition

and taking additive inverses

[tropic saddle]

no thats not how you defined U

you defined U to be some set

now you have to show that this set is actually closed under addition and inverses

[sharp smelt]

the sum of two integers , is always an integer , so if v=(a,b) where (a,b) are integrers

and u=(e,f) where, e,f are in Z

then v+u= (a+e,b+f)

which proves it's closed under addition

[tropic saddle]

yes

[sharp smelt]

as for proving closure under additive inverses , we have v=(a,b)

then u=(-a,-b)

then we have v+u=0

[tropic saddle]

and u is in U because?

[sharp smelt]

becuase 0 and 0 are both integers

wait

no

negative numbers are integers too

-1,-2...

[tropic saddle]

ok asked a different way, why did you choose integers and not for example the natural numbers

[sharp smelt]

so that it would be closed under additive inverses

[tropic saddle]

ok so you are using some property of the integers which you havent stated yet

[sharp smelt]

the sum of an integer and its additive inverse is zero?

[tropic saddle]

I asked about why u is in U

not about why v+u=0

[sharp smelt]

I don't really know, other than intuition

[tropic saddle]

what I wanted to hear that Z is closed under inverses, so -a and -b are again integers and therefore (-a,-b) is in U

[tropic saddle]

just like here you said that the sum of two integers is again an integer

[sharp smelt]

ah

okay

as for why it's not closed under multiplication , take lambda=1/5 and a=b=1

then (1.5,1.5) doesn't belong to U

is that it?

[tropic saddle]

yes

well, (1/5, 1/5)

[sharp smelt]

yeah

thank you so much !

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[stoic imp]

$a, b \in \mathbb{R}$ show $||a|-|b|| \leq |a-b| \leq |a| + |b|$

[jolly parrot]

calc_and_real_anal

[stoic imp]

please ping

[stoic imp]

there is no c wdym

[latent quail]

I don’t find any better idea other than this lol

square would explain your idea vividly

Since all of them are either positive or zero

ye, squaring is a way to present your idea

[stoic imp]

can do You do more handholding? if I set b = -c and check b>=0 and b<=0 cases? wtf?

wdym squaring, squaring what?

[latent quail]

Square both sides to determine which one is greater

[stoic imp]

aren’t both same?

[latent quail]

wdym

[stoic imp]

squared

[latent quail]

I don’t get it

[stoic imp]

,, ||a|-|b||^2 = |a-b|^2

[jolly parrot]

calc_and_real_anal

[stoic imp]

this?

[latent quail]

yes

[latent quail]

You just need to prove if this is correct

And you’ll be all good

Also, it’s an inequality