help-13

[maiden venture]

To find a point or do I add t or something?

[burnt vapor]

You want to find the line l

So you need a point and a vector

Q is in l so you already have a point

[maiden venture]

Yes

[burnt vapor]

And l is perpendicular to π, so n is the vector

[maiden venture]

Yes okay

[burnt vapor]

So you can find l

Now, if l is a line that contains Q and is orthogonal to π, what is the orthogonal projection of Q onto π?

[maiden venture]

N - Q?

[burnt vapor]

Why?

[maiden venture]

Well I don’t know I am just guessing

[burnt vapor]

Do you know what the orthogonal projection is?

[maiden venture]

No

[burnt vapor]

Then you should study and understand it before trying to do a problem about it

You should at least know a formula to find it or something

Otherwise you can't do anything

[maiden venture]

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[crimson sedge]

This is so confusing, idk where to start

[sturdy rose]

lets start somewhere

[dusty hazel]

do you know when f(x) is negative? Or when f(x) is positive?

[gilded mesa]

Put the values

[scenic brook]

Should we expand

:?

[crimson sedge]

Positive when x= 0 and 7, negative when x = 4

[gilded mesa]

No expansion

[scenic brook]

Ok

[dusty hazel]

No, in general.

[crimson sedge]

Oh uhh

[scenic brook]

Oh, you can make a graph

🙂

[coral jewel]

i suggest graphing out the equation

[scenic brook]

To find signs

[crimson sedge]

This is a dsat question, I can use desmos

[scenic brook]

Damn...

[coral jewel]

the x^2 coefficient is >0, so the graph should be u-shaped

[gilded mesa]

But there is x

[coral jewel]

and since at x=1 and x=7 f(x) is positive, the minimum of the graph should be between 1-7

that should be your first clue

[scenic brook]

👍

[crimson sedge]

Does that give me anything?

We need a + b

[scenic brook]

Of course

[coral jewel]

a rough sketch of the graph

[scenic brook]

It says possible values

Not exact

Idk

So you just use graph

[gilded mesa]

Graph is difficult

[scenic brook]

I mean table of signs

Method of intervals or whatever

[coral jewel]

heres another hint: the equation is under the form (x-a)(x-b), therefore a and b are the 2 roots of the equation

essentially, you are finding the possible sum of the two roots

[crimson sedge]

Ok can someone tell me steap by step what to do? I'll really confused

Ok I gotta go, I'll just ask my teacher tomorrow

Sorry

Thank you everyone

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[gilded mesa]

I figured out
4<a<7
1<b<4
@crimson sedge

You just have to put value of x inside function

We need just one value a+b not exact value

We can assume a and b by these interval

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[dull plover]

hi

[dull plover]

sorry im a freshman taking calc 1 and im a bit unfamiliar with the general terms of integrals

i know that the integral of any variable ie x^n is x^n+1 / (n+1)

but the symbols here got me confused as im new to them

[chrome flax]

$\int dx = \int 1 dx$

[wraith dagger]

Xotiic

[chrome flax]

1 is just x^0

so add 1 to the power and divide by that and you’d have x

[dull plover]

makes sense but what exactly is dx here

[supple flume]

derivative of x

basically

[dull plover]

im familiar with the derivative notations

[supple flume]

to tell you what you are integrating

[dull plover]

but im completely oblivious to how they work in this context

[chrome flax]

Yh it tells you what you’re integrating with respect to

[dull plover]

so if u were saying

[supple flume]

if it said dk you would be integrating k in this integral

[crimson sedge]

dx is not the derivative of x.

[supple flume]

yeah wrong use of words

[crimson sedge]

dx is a differential, the derivative of x is 1

[dull plover]

so if u were to turn the first note into words

what would it read as

the integral of any constant is x+c?

[supple flume]

no

[supple flume]

integral of 13 dx would be 13x + c

[supple flume]

think of it as k*x^0 and use the power rule

[dull plover]

because its k * x^0

yeah

what about dx alone

isnt it x^1

[supple flume]

well as the other person says its just the integral of 1 dx

or the integral of 1 * x^0 dx

[dull plover]

yes what does that exactly mean

[dull plover]

ohhh

[supple flume]

if you see a constant term just remember theres and x^0 in there

[dull plover]

well what about

x dx

what does this mean

[supple flume]

well whats the power rule?

[dull plover]

is it just talking about the variable x raised to any power

[supple flume]

here

[supple flume]

yes

[dull plover]

yeah i get it now

[supple flume]

k could be anything

well it doesnt work for x^-1

[dull plover]

im familiar with the technicalities but i was just confused with the official formula bc its been a while since i last took integral calc 😭

anyways ty

[supple flume]

but you will learn about that in the future

[dull plover]

wouldnt that be

like

oh i see where this is going

[supple flume]

(u cant divide by 0 so u cant use the power rule for this)

[dull plover]

yeah this is all we took

[supple flume]

oh yeah its just number 5

[dull plover]

figured 😭

i have an exam this saturday with the basics of integration and the fundemental theory of calculus

wish me luck

[supple flume]

good luck lol

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[slender mango]

Hey

[slender mango]

I need to transalte this

for what $a$ is vector u and v paralell if $\vec{u}=(-2,a+1,-4)$ and $v=(a,-1,2)$.

[wraith dagger]

Totalani

[slender mango]

what I did was $(-2,a+1,-4)=\lambda(a,-1,2)$

[wraith dagger]

Totalani

[slender mango]

Which gave me $$\begin{cases}
a\lambda=-2 \
-\lambda=a+1 \
2\lambda=-4
\end{cases}$$

[wraith dagger]

Totalani

[slender mango]

Kinda stuck here, not sure if I did it correcly up to now?

nvm think I got it

yup I got it

.closer

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[slender acorn]

Find the square of:
3a-2b-5c

[slender acorn]

Shall I start like this?

[dire geode]

How did -2b become +2b

[slender acorn]

oh sorry

actually 2b-5c should have become 2b+5c

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[wet fossil]

R = {(1,3),(1,5),(2,3),(2,5),(3,5),(4,5)}
Is R^-1 = {(y,x) | (x,y) belongs to R}?

[slate lintel]

yeah

[deep hull]

Are you trying to expand a trinomial?

[wet fossil]

Alright, so
R⁻¹ = {(3,1),(5,1),(3,2),(5,2),(5,3),(5,4)}
R○R⁻¹ = {(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(4,4)??? but this is not even in options

[deep hull]

oh it ended

[slate lintel]

what does ○ mean?

[wet fossil]

Composition

[novel isle]

From a point P on the bisector of angle AÔB, draw a line that forms four congruent angles with and intersects the sides of angle AOB, or their extensions, at points C and D. Prove that:
a. OC = OD;
b. for any point Q on OP, QC = QD

[wet fossil]

Does the O (in ROR⁻¹) mean something else here?

[slate lintel]

hi this isn't your channel

[novel isle]

I’m sorry where I have to go?

[slate lintel]

#❓how-to-get-help

[slate lintel]

i think you did R⁻¹○R

remember in R○R⁻¹, R⁻¹ gets applied first

[wet fossil]

ohh yeah right my bad

thanks!!

[slate lintel]

here's this to help you lol

[wet fossil]

thanks

got it

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[woeful mason]

Hello, I appologies as im French Canadian the terms I use might be different than what ive wrote as ive basically just tried to translate it.

I'm having a real hard time to understand and do "Formal System"

The first image are the rules that I have that allows me to transform things and so on

The second image is an example, but when Im trying to follow it I get blocked at the 2nd step.

[woeful mason]

So basically each line represents one step the result, and the rule used one what

(1) x v z premise
Means that is step one of my formal system, the result is x v z, and its just a premise available in my theorem

while
(2) ~x => v (1), r(9)

means its step two, the result is ~x => v after than I applied the rule 9 on the first result

When I look in the tables of rules above, I dont understand how the rule 9 is being applied and where the negation comes from

as x v z doesnt contain any negation

[upper abyss]

Rule 9 says:
A → B
Is exactly the same as
~A V B

Oh wait nvm they're combining two rules into rule 9

Rule 9 also says:
A → B
Is exactly the same as
~B → ~A

That "second rule 9" is what they're using in step 5.

[woeful mason]

I understand that, but what I dont understand is how my x v z can be applied in the rule nine, there is no negation in it

[upper abyss]

Okay we are talking about rule 2, not rule 5. My bad, I got mixed up

Then we refer back to the "first rule 9" lol

Rule 9 says:
A → B
Is exactly the same as
~A V B

Now, you're saying there's no negation. Note the negation needs not be there

It's the relationship between the symbols that actually matters

If I were to, say, substitute ~A in for A, then this is also rule 9:
~A → B
Is exactly the same as
A V B

[woeful mason]

so A v B can be applied the rule 9 ? you then switch it to ~A => B because you just "note" that there is supposed to have a negation at the start meaning that you have to inverse it at the end?

or yeah as you say you just subsitute it, but at the end you have to put it back?

(A => B) <-> (~B => ~A) <-> (~A v B)
is the same as
(~A => B) <-> (~B => A) <-> (A v B)

if I understand

[woeful mason]

ok thank you!

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[dull plover]

how is this expression valid

[dusty hazel]

Is this photomath?

[dull plover]

sin^2(x)+ cos^2(x) = 1

shouldnt it be 1 - cos^2(x)

[dull plover]

yes

[dusty hazel]

They want to say sin^2(x) or (sin(x))^2

[dusty hazel]

Yes it is that as well.

[dull plover]

how do u get this from that

[dusty hazel]

But cos(2x) = cos^2(x) - sin^2(x)

[dusty hazel]

So from here.

[dull plover]

ohhh

we used that formula instead

[dusty hazel]

Pretty much used both of them, but yeah.

[dull plover]

wait

can u do a step by step

of the substitution

[dusty hazel]

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

Solve for sin^2(x) in terms of cos(2x)

[dull plover]

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

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

yea makes sense now

ty /!

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[livid stratus]

An infinite Geometric series that is going down is given. a1,a2,a3,,,an. its Quotient is Q. and its numbers are positive. using the numbers from the given sequence we set a new.

[livid stratus]

prove if the new sequence is a Geometric series or not.

<@&286206848099549185>

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[latent wolf]

find a set of parametric equations for the line of intersection of the planes 3x+2y-z=7 and x-4y+2z=0

[latent wolf]

Hello 🙂

[lavish quest]

3x+2y-z-7 = x-4y+2z ?

wait

[latent wolf]

Is that how I need to rewrite my equation?

[lavish quest]

i guess

What does z represent?

[latent wolf]

I'm supposed to write a parametric equation for the line of intersection

I have the answer, I'm just not sure how they got there

x = 2
y = 1 + t
z = 1 + 2t

[lavish quest]

So the question is what exactly did he want?

if x = 2 3x = 6 btw

[latent wolf]

Right

[lavish quest]

then 3x+2y-z = 7

[latent wolf]

The question is how did he come up with the equation for the line

[lavish quest]

2y-z-1 = 0?

[latent wolf]

I don't understand how he came up with 0i+1j+2k

[lavish quest]

You can produce an equation that depends on y

or z whatever

what did he do

i dont see

[latent wolf]

I don't know all I have is the answer

one sec

[lavish quest]

Does it want intersection point

[lament root]

The first one can use the dot product definiton with cosine?

[latent wolf]

Ya I managed to get the angle easy enough

[lavish quest]

wait

let me look at it

[lament root]

Well lets start by taking z=0

Solve the first two equations for x and y

[lavish quest]

lets look at here

[lament root]

Then, we use that vector to take a cross product with the normals of both planes

[latent wolf]

Lemme see if I can snip my work into here

[lavish quest]

ok

[young bane]

hello is someone willing to help me with analytics geomtry ?

[lament root]

Did you finish the cross product calculation?

That is crucial in finding the normal to the intersection

[latent wolf]

When I did the cross products I ended up wiht 0i-7j-14k

[lavish quest]

hello

[young bane]

hi

[lavish quest]

@latent wolf You will handle it my friend

@young bane You can write to another channel, or dm

[young bane]

okay

[lavish quest]

@young bane But next time you want help, write to another empty channel

[young bane]

oh okay sorry

[lament root]

So then add that to the point you found by taking z=0, finding x and y from the two equations

[latent wolf]

Is it possible for me to share my screen in here?

[lament root]

No but we can have a miro https://miro.com/app/board/uXjVNOH9Mag=/

[latent wolf]

I'll snip

So if I know that x = 2 and I set z=0

or am i missing something

[lament root]

So you have a point. Now you need the normal

The normal is found by cross product of the normals of eq1 and eq2

[latent wolf]

So for the cross product of the normals of eq1 and eq2

I end up with 0i-7j-14k

<@&286206848099549185>

<@&286206848099549185>

Hello?

<@&286206848099549185>

[cedar kiln]

@latent wolf Has your question been resolved?

[latent wolf]

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

[frigid hemlock]

buddy relax

[latent wolf]

just tryna get some help fam

[cedar kiln]

@latent wolf Has your question been resolved?

[latent wolf]

.close\

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[untold holly]

My guy is in middle of the exam

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[vital beacon]

Does anyone know how to get the horizontal asymptote of this equation?

[slate lintel]

limit as x goes to infinity or negative infinity

[vital beacon]

oh

how does that work

OHHH I get it

thanks

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[cedar kiln]

Please don't occupy multiple help channels.

[keen raven]

im not

oh wait

ok i closed my old one

Why do I multiply by conjugate

Does it matter if I multiply by 5-sqrt(7)

Wont that still cancel out

why is it +

[crimson sedge]

,,<cancel>\def\CancelColor{\tc r}\f4{\cancel{5-\s7}} \c b{\cd \f{\cancel{5-\s7}}{5-\s7}} =\f4{5-\s7}

[void sand]

difference of squares

[crimson sedge]

so you are not saying much by what you are doing

[keen raven]

Oh

Oops

ye...

[lavish quest]

i think it is (-10-2√7)/12

multiply by 5+√7

top and bottom

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[last trench]

14B please

[last trench]

how do i find the centre and radius of the circle of intersection

thats the equations

plsss

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[upper abyss]

9 = (10 - 1)

Mind you, that leads heavily into number theory

But binomial-esc

You'll see it every time now lol

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[celest marsh]

Is there any software I can use to double check my tuple relational calculus queries?

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[autumn tusk]

This is cc differential equations. He's explaining ERF.
I don't understand the first line. Shouldn't be (D+a)(D+b) ?

a and b are roots

[upper laurel]

on the second line, p' presumably means dp/dD or if you want ∂p/∂D

[autumn tusk]

He's differentiating the first line. A step which I also don't understand

My first problem is the first line

[silver oxide]

yeah no the video is right

(d-b) means that b is a root

same concept for a

[gilded mesa]

So a and b are root of D^2-(a+b)D+ab

=

0

What you mean by root

It says root case

[autumn tusk]

The roots of the characteristic polinomial.

[gilded mesa]

Oh

[silver oxide]

if (d-b)(d-a) = 0, meaning they are roots then d-b=0 and therefore d=b, same concept for a

[autumn tusk]

Yes this is what I thought, but why having a minutes there

I mean the second term coul also be positive right?

[silver oxide]

why would it be positive?

[gilded mesa]

Ya not postive it could be anything

[autumn tusk]

Why not?

[gilded mesa]

Function of x and constant

[autumn tusk]

I mean the second term of the diff

[upper laurel]

the minus sign isnt to tell you that the number is positive or negative

[gilded mesa]

Not have function of y there

[upper laurel]

its to have you subtract instead of add if the root is plugged in

(x - 3)(x - 2) has roots 3 and 2 which can be seen by plugging them in
(x + 3)(x + 2) has roots -3 and -2 which can be seen by plugging them in

[gilded mesa]

For second term try uv formula of differentiation

[silver oxide]

P(D) has roots when P(D)=0 therefore (d-b)(d-a) = 0, meaning that d-b is a root and d-a is a root therefore d=b and d=a are roots

[autumn tusk]

I tried but o don't know what I have to differentiate

[gilded mesa]

You have to differentiate p(D) with respect to D

Not respect to x

[upper laurel]

try differentiating (x - 3)(x - 2)

youll see you get (x - 2) + (x - 3)

[autumn tusk]

Ooooo

Ok gotcha

[upper laurel]

bruh

[autumn tusk]

What I'm saying is
I could also have
(D+a)(D +b) right? This is just an example not a general rule

[upper laurel]

it no longer means anything if you do that but p(D) = (D + a)(D + b) sure is a polynomial that has the letters D, a, and b in it

its derivative dp/dD is (D + b) + (D + a)

[autumn tusk]

Why means anything?

[upper laurel]

is that a bad thing?

[autumn tusk]

I don't get it...

I'm lost

[silver oxide]

well what even is the goal?

[upper laurel]

if it takes a single misread sentence to dislodge your understanding then you have some big questions to answer

what did you get before

[autumn tusk]

I was following a video and he jumped on this proof.

I'll rewatch the video after work.

[autumn tusk]

This doesn't help

[upper laurel]

it helps you no longer get lost by restarting back to "what was I stuck on again"

[autumn tusk]

Ok thanks

.close

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[elder elm]

could someone help me with this problem'

[cedar kiln]

@elder elm Has your question been resolved?

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[novel ginkgo]

hello

[novel ginkgo]

how can i solve this

2:3 is the ratio of foods delivered (fries to chips). if they had delivered 4 more fries, the rario would be 3:4, how mang fires n chips they delivered in total?

[quiet plover]

Assume there are 2y fries and 3y chips

Then the ratio would be 2:3

And then solve

[novel ginkgo]

how can i solve it is there a way to find out the total number other than substituting the choices

thank u for your response!

[quiet plover]

Wdym

Ok wait

Say there are f number of fries

And c number of chips

so f/c = 2/3 right?

[novel ginkgo]

yes

[quiet plover]

Now cross multiply

3f=2c

[novel ginkgo]

okok

[quiet plover]

After dividing by 2

c = 3f/2

[novel ginkgo]

okioki

[quiet plover]

Now we just need to find c

So how can we represent 4 more fries

f + 4

[novel ginkgo]

ohhh

c = 3(f+4)/2

?

[quiet plover]

Nono

The ratio is not 2:3

[novel ginkgo]

ohh i see whre i got wrong

[quiet plover]

Look at the questin

[novel ginkgo]

is 3:4 after adding fries

i see now

[quiet plover]

(f+4)/c = 3/4

[quiet plover]

Yes

[novel ginkgo]

ohhh nice

[quiet plover]

You'll get 3c = 4(f+4)

3c = 4f + 16

Now you can substitute the value of c

And solve for f

[novel ginkgo]

THTS coooool

thank u so much

!!!

[quiet plover]

But I think the neater answer would start with the fries and chips being 2y and 3y

[novel ginkgo]

ohhh i see i see

thank u blurple !

[quiet plover]

welcome

You can close the chat using .close now

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[near quiver]

The sum for the real numbers a and b for which the limit (photo) is equal to:

[near quiver]

!status

[cedar kiln]

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

[orchid pilot]

start by calculting/simplifying the given limit

[near quiver]

The limit of the square root would be +infinity I assume.

I never had this a and b unknowns to limits, I am quite new to limits, actually. Just about 2 weeks in.

[orchid pilot]

true but what is the limit for ax

[near quiver]

Minus infinity I assume?

a can be any real number though...

[orchid pilot]

so you have infinity plus infinity right

we have to search for a value a thats gives us a finite limit

try to do that

try to get x^2 out of the sqrt

so get something like |x|sqrt(.....) - ax

[near quiver]

Mhm..I will certainly try and think of a way, I will come once I'd find anything or give up.

[orchid pilot]

yes try and lmk

[near quiver]

You mean a forced factoring?

Of the x^2, such that we would have the x^2 out as x(something)?

[orchid pilot]

yes so factor what is under the sqrt as x^2(......)

and then you can get the x^2 out of the sqrt

[near quiver]

Now. I assume I can't factor x and its absolute value.

[orchid pilot]

perfect

no so first

we have to simplify this

this can be done by getting rid of the sqrt

do you have an idea how?

[near quiver]

Well, the limit of 1/x for either plus or minus infinity will always be 0?

[orchid pilot]

yes

[near quiver]

So, the radical will be 1

[orchid pilot]

yes so what will be left is |x| + ax = b

right?

[near quiver]

Precisely.

I make two scenarios?

[orchid pilot]

no

you know that x will be negative

[near quiver]

Ah, true.

[orchid pilot]

so only the case where x is negative

so what do get?

[near quiver]

x - ax = b?

[orchid pilot]

no you made a mistake

ax will not change

it is |x| that has to be rewritten for negative x

because of the absolute value

so the term ax will remain untouched

[near quiver]

Hm, because we don't touch ax at all and let it remain positive?

I thought by x being negative, it would be a no brainer to write "-ax"

The absolute value makes sense, obviously.

[orchid pilot]

so want to rewrite the absolute value in terms of x and not |x| so we can a common factor with ax

how to do that in case that x is negative

[near quiver]

x(1-a)=b?

[orchid pilot]

not really\

[near quiver]

-x(1+a)=b ?

[orchid pilot]

no

so what is |x| when x is negative

[near quiver]

still x

Positive*

My bad.

[orchid pilot]

yeah so we have to turn the sign

so how can we do that?

[near quiver]

Into negative, by multiplying with (-1)?

[orchid pilot]

yes so we find that...

|x| = -x when x<0

which is true in our case because the limit is negative

so what do we get now?

[near quiver]

ax - x = b?

[orchid pilot]

yes

we can rewrite as

(a-1)x=b

now think of a case where this limit can be true

[near quiver]

I realised, didn't I make a mistake? -b?

Or I am multiplying only the LHS?

[orchid pilot]

we do you think that

yes only lhs

[near quiver]

Ohh because of the specific x requirements.

[orchid pilot]

because we want to get rid of abs of x

yes yes

[near quiver]

Great, now...

[orchid pilot]

this step

in what case is this limit never infinity

[near quiver]

When a-1 and b are 0?

[orchid pilot]

yes

so we now solve for a

and take the sum

[near quiver]

Which a+b is just 1

Bingo. We did it.

Thanks for the help man, appreciate it lots.

Have a great day/evening!

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[dense herald]

.close

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[royal finch]

If you already know how to calculate the maximum likelihood from earlier parts, what part of this alternate version is challenging

I'm not going to read that, but if you can calculate this you should recognize that the computation for an MLE is broken up into a few steps. Namely writing down the likelihood function, then optimizing it.

The optimization step is just calculus, you find a critical point and you exploit the log convexity of the normal to guarantee that the critical point is the optimal value

Now, you have a case where mu must lie in [0,1]

So either the critical point is in [0,1] or it isn't. And you should know from Calculus where the optima lies if there is no critical point in your region of interest

The point is that by drawing a picture of a quadratic on the interval [0,1]

You have your answer

[cedar kiln]

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[bold wyvern]

can you reformulate the question neatly from the beginning ?

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[sturdy rose]

Find the number of ways to distribute 21 identical oranges among 3 children such that all of them get atleast 2

[sturdy rose]

i know i can use ramanujans partions

and i can use multinomial

apart from these two, is there any other way to solve these questions?

[cedar kiln]

@sturdy rose Has your question been resolved?

[tropic slate]

just wondering

is the asnwer 136

[sturdy rose]

i dont know the answer

but using partions, it does come out to be 136

[tropic slate]

im pretty sure i used partitions

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[sturdy rose]

.close

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[bold mauve]

If you draw 3 cards from a standard deck of 52 what is the probability your first card is a space given the second and third card are a spade?

[sturdy rose]

do you know what revised probability is?

[bold mauve]

No

I have setup $E_i$ to be the event that the $i$-th card was a space: $$P(E_1|E_2E_3)$$

[wraith dagger]

o.O

[sturdy rose]

hm

[bold mauve]

Im doing it like this

I expand $$P(E_1|E_2E_3) = \frac{P\left(E_{1}E_{2}E_{3}\right)}{P\left(E_{2}E_{3}\right)}$$

[wraith dagger]

o.O

[bold mauve]

Using basic properties of conditional probability

but I don't understand why my answer is wrong

$$\frac{\frac{4}{52}\cdot\frac{3}{51}\cdot\frac{2}{50}}{\frac{4}{52}\cdot\frac{3}{51}\cdot\frac{2}{50}+\frac{48}{52}\cdot\frac{4}{51}\cdot\frac{3}{50}}$$

[wraith dagger]

o.O

[bold mauve]

For P(E1 E2 E3) I thought we have 4 in 52 cards being a space, 3 in 51 on the second draw, and 2 in 50 on the third draw

then for P(E2 E3) that is the situation P(E1 complement E2 E3) + P(E1 E2 E3)

[sturdy rose]

the cards arent replaced right?

[bold mauve]

They are not replaced

Is revised probability bayes theore

I know what that is

[sturdy rose]

well what youre doing is what i would do, so the name isnt important

[bold mauve]

these are the identities we are given

I used the top one

oh I computed it wrong because I didnt know what a deck of cards was I thought spade was a card denomination like king or queen

$$\frac{\frac{13}{52}\cdot\frac{12}{51}\cdot\frac{11}{50}}{\frac{13}{52}\cdot\frac{12}{51}\cdot\frac{11}{50}+\frac{39}{52}\cdot\frac{13}{51}\cdot\frac{12}{50}}\ $$

[wraith dagger]

o.O

[bold mauve]

this comes out to the right answer

.close

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[alpine rune]

,rotate

[cedar solstice]

Question 85

[alpine rune]

okay

what did you try so far?

[cedar solstice]

I choosed one seat for the child that come first

6c1

And

Then

For worst case

Another child have to choose another seat

So 5c1

+1 for the the possibility that the second child get the seat he needed

And other three got their respective seats

Divided by total cases

I am not able to find total cases

And idk if what I did is correct or not

[alpine rune]

hi

sorry im here

[cedar solstice]

Did i choosed the cases correct

[alpine rune]

hold up

[cedar solstice]

K

[alpine rune]

ooh aakash byju's

[cedar solstice]

Lol

[alpine rune]

jee? 😭

[cedar solstice]

Answer the question

Yup

[alpine rune]

yea doing that

[alpine rune]

cool

well from what I could deduce, we know that the first student must select a seat out of any of the 3 seats that are not of the last 3 people. so the probability he does that hs 1/2 (3/6). the next student must also select a seat randomly that isn't of the last 3 people, so 2/5, and then 1/4. I'm giving it a second thought though because im not really sure if that would be the answer because I think im missing something

so like (3/6 * 2/5 * 1/4)*120 🤷‍♂️

[cedar solstice]

Might be the answer

I couldn't find the answer online anywhere to this question

[alpine rune]

ok

well check if this is right then, I think that should be it

[cedar solstice]

It seems right to me

Thanks for the help

[alpine rune]

alright

[cedar solstice]

Can you please check 84 also I got the answer 6

[alpine rune]

yeah that's right

[cedar solstice]

K

Thanks

Bye

[alpine rune]

no worries

ok

[cedar solstice]

.close

[cedar kiln]

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[cedar solstice]

.reopen

[cedar kiln]

✅

[cedar solstice]

Wait I have one more question

[alpine rune]

shoot

[cedar solstice]

Question no 82

[alpine rune]

ok

hmm

you know that omega = 1/omega^2

[cedar solstice]

Yup

[alpine rune]

nevermind I don't think that's taking us anywhere

hmm

back

[cedar solstice]

Any progress??

[alpine rune]

well all I could think of is using brute force and expanding it

but im sure there's some other way out

[cedar solstice]

It would take years

[alpine rune]

yeah it would

is it 2

[cedar solstice]

Idk

[autumn kindle]

im getting 4

no wait

let me do it again

[autumn kindle]

bro

i got it

what u can do is

u can create a polynomial equation

1/(a+x) + 1/(b+x) + 1/(c+x) + 1/(d+x) = 2/x

now roots of this are w, w^2

make it a 4 degree equation

and use sum of roots

product of roots

the answer is 2

[alpine rune]

or you could divide both sides in the first equation by omega, then compare the denominators or something (aw^2 + 1 = a + w^2) and I I'd like to think that'd get you somewhere

[autumn kindle]

u got the answer?

[alpine rune]

yeah

[autumn kindle]

kk

[alpine rune]

this would make it easier

[autumn kindle]

yeah

[alpine rune]

because that gives you a = b = c = d = 1

[autumn kindle]

ohh

i see

[alpine rune]

so its 1/2 * 4 = 2

[autumn kindle]

right

faster

lol

no wait what

how did u get

a = b = c = d = 1

[alpine rune]

aw^2 + 1 = a + w^2

solve this, same for bw^2 + 1 = b + w^2

[autumn kindle]

oh im dumb

nvm

the thing is

here this method is allowed

cuz there is symmetry

[alpine rune]

yeah

that's why I used this

[autumn kindle]

yeah

[alpine rune]

its not the best method in other cases but in other cases im sure there's some other way

[autumn kindle]

yeahhh

[cedar solstice]

Thanks guys

[alpine rune]

ok

[cedar solstice]

.close

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[autumn kindle]

k

[cedar kiln]

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[eternal pollen]

What is derivatives of functions?

[sturdy rose]

?

"functions"?

a derivative is the differentiation of a function wrt to x (mostly)

[eternal pollen]

wrt?

[sturdy rose]

with respect to

[eternal pollen]

ahh okay

also

since the average rate of change is written as Δy/Δx, why is instantaneous rate of change written as dy/dx?

is it just to make them visually different?

[sturdy rose]

some people even interchange the two

but theres a big difference

like literally

one of them indicates a very minute change

and one indicates a considerably big chance

[eternal pollen]

so which is which?

[sturdy rose]

this will help better

[eternal pollen]

okay thank you^^

.close

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[random grove]

Trying to better understand the proof behind this proposition in my lecture notes.

I think the idea behind the whole proof is that if you can find a bounded subsequence, then by Bolzano-Weirestrass theorem, there exists a convergent subsequence of the original bounded subsequence. If there isn't a bounded subsequence , meaning no convergent subsequence, then by the definition of a limitpoint, then the limitpoint set is empty. i've probably worded it really poorly but I hope it makes sense. pictures include the proposition with its proof and then the definition of a limitpoint set

[random grove]

I just want some clarification is my reasoning is correct or not in order to better my understanding

ah the definition didnt send. here it is:

[cedar kiln]

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[spare mountain]

Is anyone here familiar with Radix Conversion in Hash values?

[spare mountain]

Here's one of the given:
Key Value: 24680
Radix Conversion (Base 9 to Base 10): 56

But when I converted 24680 from base 9 to base 10, I got 16596 so the Radix Conversion value should have been 96.

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[warm spear]

Idk how to start... Do I rationalize the denominator? Do I open tan into sin and cos?

[livid hound]

i'd recommend starting with a sub like u=sqrt(x)

[crimson sedge]

same

then it will become easier to work your head around it.

[mighty shuttle]

try $u=\sqrt(x)$ as others here suggested

[wraith dagger]

Why am. I here

[warm spear]

I don't know what sub is...

[sturdy rose]

uh

you really should

[warm spear]

Okay I think I get what sub means

So far correct?

[sage forge]

And sqrt(x) = u

[warm spear]

?

But that one is from dx=du?

[sage forge]

Yes, but you need to rewrite it in terms of u as well

[warm spear]

Really?

[sage forge]

Yes

[warm spear]

Huh that's new

Good?

[sage forge]

Perfect!

[warm spear]

So now I should claim integral and open (u) and my question is solved?

It's correct,Awesome! Thanks!

Now I know how to use "u substitution"

Can this also be used in limits?

Or it's for integrals only

[sage forge]

Do you mean improper integrals?

[warm spear]

Limits and integrals HAVE A CONNECTION?!?!

[warm spear]

Well just regular limits, but I guess not.

What.,...

How and where do I even begin?

[sage forge]

The sin(4x) is bothering me a bit. Can you use a trig identity to rewrite it?

Maybe in terms of something depending on 2x

[sage forge]

Nah, in regular limits there is no chain rule.

[warm spear]

For the cos²(3x) yes but for the sin i don't think so

Can I claim a 2 in any way from the sin's degree (4x)? To make it a 2x or that's mathatically impossible

My phone battery is dying, I'll be back after a few minutes

[sage forge]

You can use double angle identity for sin(4x)

[warm spear]

Making it (2sin(2x)cos(2x))?

[sage forge]

Exactly

[sage forge]

The reason why you want to do this is to substitute.

Because now you only have sin(2x) and cos(2x). That is advantageous for a substitution

[warm spear]

I wonder can i also pull out a (minus) from the denominator and separate the minus with the (1) to make it (1-cos^2(2x)) to tun it into -sin^2(2x)? Or I am now making things up?

[slow kiln]

do u need the steps?

[warm spear]

I think I have to claim the sub for cos(2x)?

[slow kiln]

1+cos2(2x) actually

[warm spear]

All of it?

Oh I see

[slow kiln]

if u take it, differentiate it with respect to x, you'll get the numerator which is 2sin(2x)cos(2x) with an extra -2

[slow kiln]

u already made the numerator like this

so its easy from there

[warm spear]

Got it, thanks!

[slow kiln]

no problem 😄

[warm spear]

Also this was correct?

[slow kiln]

yep

[warm spear]

I had doubts about it

[slow kiln]

ignore the messy numbers its a bit rushed my bad

[slow kiln]

nah you were on the right track

[sage forge]

Please do not give out full solutions. The point of a help channel is to guide the helpees and not do the exercises for them.

[warm spear]

Just a question off that problem, I can also do the same for let’s say sin(3x), right? As long as I can pull 2 out of its “degree” that identity can be used?

[slow kiln]

oh shit alright ill delete it

[slow kiln]

how would u do the same for sin(3x)?

[warm spear]

Well it helped me understand but yeah, thanks for confirming my answer xd

[sage forge]

Don't worry 🙂 It's nice that you are taking your time to help others!

[warm spear]

Ohhh you can’t? Oooh i thought It’s done by like this (3x -> 2(x))

[slow kiln]

as long as you get an even number, but i doubt sin(3x) can be made into an identity because those are double angled identities, you would need an even angle for sin to pull of that

[warm spear]

Is multiplication

Yes yes got it, my brain was in an other world

[slow kiln]

its alright we all go through that lmao

[warm spear]

Also sorry for not using mathematical terms I didn’t learn it in English to recognize them

[slow kiln]

nah even i dont know enough and appropriate mathematical terms, im literally gambling with the words im using here

[warm spear]

Lol

[sage forge]

From which country are you if I may ask?

[warm spear]

Ok that’s all for now, I will go back to study and might…well…will come back later because I’m sure I will have more questions

Thank you all again!

[sage forge]

So the channel can be closed?

[warm spear]

Yep

I will just open another when I come back

[slow kiln]

im from india gang 💯

[slow kiln]

i should do that too

[slow kiln]

and what country might you be from

[sage forge]

Germany

[slow kiln]

oh wow

i plan to do my higher studies there

[warm spear]

You are The true hero 🙏

[slow kiln]

hopefully

[slow kiln]

thank you little soldier

im assuming you're from india too

[warm spear]

Neighbors xd

[sage forge]

Have you already decided for a university here?

[slow kiln]

not yet, but the uni i plan to go this year advances in letting their students go abroad to take their studies further so uh yeah

all depends on that one uni

besides i would need to learn german

[sage forge]

Would be funny if I was tutoring one of your exercises But ofc unlikely

[slow kiln]

would be really funny

[sage forge]

.close

[cedar kiln]

Channel closed

Closed by @sage forge

Use .reopen if this was a mistake.

[slow kiln]

oh uh

[cedar kiln]

Available help channel!

Send your question here to claim the channel.

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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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[subtle hinge]

noticee how every alternate term leaves remainder 0

and the other ones 1

[cedar kiln]

Channel closed

Closed by @upper ocean

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[ocean basalt]

Question:
Jessica stretches her arms out 0.60 m from the center of her body while holding a 2.0 kg mass in each hand. She then spins around on an ice rink at 1.1 m/s.
What is the combined angular momentum of the masses?
If she pulls her arms in to 0.15 m, what is her new linear speed?

My work so far:
r = 0.6 m
m = 2 kg
v = 1.1 m/s

The combined angular momentum of the masses is given as
L = mvr + mvr
L = 2mvr
Inserting the values:
L = 2(2)(1.1)(0.6)
L = 2.64 kg m^2/s

I'm stuck on how to calculate the new linear speed. If anybody could help out, it would be appreciated.

[subtle hinge]

bro mathematics server for a reason

[ocean basalt]

huh

could u help out

[subtle hinge]

i ssuck at phy

[near quiver]

He means that this is a physics problem in a maths server

[subtle hinge]

altho i shall try

[ocean basalt]

oh my bad

[subtle hinge]

actually it doesnt matter

[near quiver]

No worries, well you are lucky someone already tries this

[ocean basalt]

really?

[near quiver]

MagnusOP said he'd try?

[crimson sedge]

https://discord.gg/physics

[ocean basalt]

nvm i thought somebody had already asked this question

[subtle hinge]

but u need phone verificaation and shit to chat in that server

[near quiver]

Ohhh

[subtle hinge]

so idt anyone shud mind phy questions in this server

[dusky peak]

uhm

She then spins around on an ice rink at 1.1 m/s.

how do you spin with a velocity?

or do they mean this as the velocity of the balls

[ocean basalt]

masses

[dusky peak]

kk

ok so

quick tip: dont calculate stuff

in this case, you correctly got the value for L

you also got a formula

L=2mvr

if she pulls her arms in, angular momentum will be conserved

[subtle hinge]

same question

[dusky peak]

if we call that new momentum L', then we get:
L=L'

the new angular momentum will have the same formula:
L'=2m' v' r'
note however, that m didnt change, so m'=m

so now we can set L equal to L' and rearrange to v'

[ocean basalt]

ok

so how would this look with the values inserted in?

[dusky peak]

we dont wanna look at values right now
putting in values before the end just makes it messy

we have L=2mvr, L'=2mv'r'
with L=L', we get
2mvr=2mv'r' => vr=v'r'

that way we also avoid rounding errors

[ocean basalt]

alright

.solved

[cedar kiln]

Channel closed

Closed by @ocean basalt

Use .reopen if this was a mistake.

[ocean basalt]

Ciara is swinging a 0.015 kg ball tied to a string around her head in a flat, horizontal circle. The radius of the circle is 0.70 m. It takes the ball 0.60 seconds to complete one full circle. Calculate the tension in the string and its direction that provides the centripetal force acting on the ball to keep it in the circular path. (3 points)

0.0077 N, toward the center of the circle

1.2 N, toward the center of the circle

0.0077 N, along the line tangent to the circle

1.2 N, along the line tangent to the circle

[dusky peak]

.reopen

[cedar kiln]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[dusky peak]

.

.

open

[dusky peak]

astro said:
Ciara is swinging a 0.015 kg ball tied to a string around her head in a flat, horizontal circle. The radius of the circle is 0.70 m. It takes the ball 0.60 seconds to complete one full circle. Calculate the tension in the string and its direction that provides the centripetal force acting on the ball to keep it in the circular path. (3 points)

0.0077 N, toward the center of the circle

1.2 N, toward the center of the circle

0.0077 N, along the line tangent to the circle

1.2 N, along the line tangent to the circle

ok so

i get something completely different than the 4 choices

but the direction should be easy

do you see what it should be?

@ocean basalt

[ocean basalt]

i think so

[dusky peak]

centripetal forces are perpendicular to the movement

so in this case they point towards the center of the circle

but is the force 1.2 N or 0.0077 N?

given m, r, T
we can first derive the angular velocity w=2 pi / T

and then we can use the formula for the centripetal force
F=m v^2/r = m w^2 r

the thing is that when i do the math here, i dont get 1.2 or 0.0077

oh well, they might have rounded

yeah thats probably it

did you try to calculate it @ocean basalt ?

[cedar kiln]

@dusky peak Has your question been resolved?

Channel closed

Closed by @dusky peak

Use .reopen if this was a mistake.

[cedar kiln]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[green mulch]

Need to find the perimeter of the triangle ABC , the ratio of the sides are not common ratios and the ar. Of the inner triangle is 60cm²

[spiral fog]

So like the ration that some dot split AB is 1 : 3?

[green mulch]

Yeah

This is what I did

Can't figure anything else

[spiral fog]

Yeah thats good

[green mulch]

Anything u can think of doing next?

[spiral fog]

Fun problem i will think about it

[green mulch]

Yeah thanks I have been thinking about it for 2 days

[spiral fog]

Any progress?

[green mulch]

Nothing

[jaunty pumice]

U can use coordinate

Geo

[green mulch]

Came up with something but my teacher proved it wrong

[spiral fog]

I hardly believe this can be solve analytically

[green mulch]

Can u show me how?

[spiral fog]

But i mean everything can be solved analytically so...

[jaunty pumice]

Like let point R be origin