help-39

[tender siren]

check any identity or anything

[livid bolt]

sin/cos = tan

[next dove]

yes

[livid bolt]

so i can turn top into tan

[tender siren]

trig can snipe

[tender siren]

hm

[livid bolt]

!??!?!

[tender siren]

FIRST SIMPLIFY THE +/- THOUGH

[livid bolt]

oh

[tender siren]

add those sin theta and root 3 cos theta

using identities or some thing

[livid bolt]

pythagorean identity?

[tender siren]

same with the denominator

[tender siren]

yeah, something like that, I am not gonna do that so early lol, try to find it

should be any identity or simplification tactic

[livid bolt]

omg this is a bit too hard

[tender siren]

recall your classes

check notes if you want to though, i aint stopping that

[livid bolt]

idk

this is the furthest for me

[tender siren]

are you gonna give up?

[next dove]

what are you trying to accomplish btw?

[tender siren]

I mean, 2 of the last guys I tried to help, did

[tender siren]

exactly

[livid bolt]

simplify

[next dove]

And you started from?

[livid bolt]

question C

[tender siren]

aw hell no

[next dove]

tf

[livid bolt]

wat

[next dove]

you know the identity for tan(M+N)?

[livid bolt]

um

let me check my formula sheet

[next dove]

kk

[livid bolt]

umm

olh wait

is it the third one

[next dove]

yup

use that

[livid bolt]

so i cant

simplify it further my way?

[tender siren]

wait

I found some stuff

[next dove]

you can

but it'll be a mess

[tender siren]

$$sin(A+B)=sinAcosB+cosAsinB$$

use this

[jolly parrot]

Greydawn Dewer

[tender siren]

for the above numerator

in this

complete this attempt, if it fails, try another approach

[livid bolt]

i thought i did

[tender siren]

damn

[next dove]

my guy will get the same expression he started with

[next dove]

with this

[livid bolt]

wat

[tender siren]

hmm, let me try

[tender siren]

can I say what I found with the simplification?

would that be as giving him the direct answer?

[next dove]

discuss about it

[tender siren]

ok

[livid bolt]

pls give me the answer im in the toilet

[next dove]

@livid bolt you tried that third one?

[next dove]

!noans

[pearl pond]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[tender siren]

exactly thats why I asked

it gives him a sin () simplifier and the denominator gives a root ten times cos () simplifier

that certainly will not give him the the tan question back

[livid bolt]

😖😖😖😔😔😔😭😭😭😭

[tender siren]

let me guide you through

do that exact attempt

continue

if you get errors, tell me, till then, try

[pearl pond]

@livid bolt Has your question been resolved?

[pearl pond]

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[thin sigil]

Please explain what I'm looking at

[frozen dagger]

72

I solved it with three different methods first one i adjusted the power

And last one is I derivated then i got degree is 5

[thin sigil]

Are you sure you're meant to solve it?

[frozen dagger]

what do you mean?

@thin sigil

[thin sigil]

Ah nvm

[pearl pond]

@frozen dagger Has your question been resolved?

[pearl pond]

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[grim phoenix]

Need help so bad on this lol

[slender viper]

8a, your answer should be an equation with variables x, y, p, m

what have you tried so far?

[grim phoenix]

i dont know what to try

[slender viper]

do you know the equation of a line passing through a point

[grim phoenix]

i know answer is y=mx-mp+p^3

no

[slender viper]

it's also $y-y_0=m(x-x_0)$

[jolly parrot]

Element118

[grim phoenix]

oh yeh i remember this

is y0 just p^3

[slender viper]

when you step some amount away from $x_0$, you step a proportional amount away from $y_0$

[jolly parrot]

Element118

[slender viper]

that's the idea

[grim phoenix]

i got b

[slender viper]

alright how about c, what have you tried

did you try factoring the equation in b?

[grim phoenix]

im not sure what to do in c

[grim phoenix]

no i used the fact that A and B lie on both lines, so the equation of both lines substracting would equal 0

[slender viper]

for part c, try factoring the equation in b

you should know one of the solutions, so you can use that to factor

[grim phoenix]

ok

x(x^2-m) - p (p^2-m) = 0

but then what?

[grim phoenix]

What was that one solutuoin

[slender viper]

what point do you know is on both lines?

[grim phoenix]

p

[slender viper]

^

then that should give you one solution

[grim phoenix]

do i sub in p in x?

[slender viper]

substituting should solve the equation

but the problem is you need to find the other solutions

[grim phoenix]

hold up

[grim phoenix]

where do i sub in p to

<@&286206848099549185>

[slender viper]

p is one solution

how do you find the other solutions?

[grim phoenix]

idk

[slender viper]

you have a polynomial and you have one root, how do you find the other roots

can you find a smaller polynomial with just the other roots?

[grim phoenix]

.close

[pearl pond]

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[pearl pond]

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[midnight haven]

linear algebra problem:

Let f:R^3->R^3 with f(x,y,z) = (x+y-z,-x,y-z)
3)Find the basis of R^3 so that the matrix of f with those basis is of this form: (Ir 0 0 0) where r is the rank of f)

I have solved 1) and 2):

The matrix f with normal bases: [[1,1,-1],[-1,0,0],[0,1,-1]]
image(f): span{(1,-1,0),(1,0,1)}
kernel(f):span{(0,1,1)}

for 3) I am not sure how I can solve it, my notes are missing something for sure, I found that rank(f)=2.

GPT said: Doing something like P^-1 * A * P but I don't really get it

[pearl pond]

@midnight haven Has your question been resolved?

[pearl pond]

@midnight haven Has your question been resolved?

[plucky python]

when you say (Ir 0 0 0) i assume you mean like
1 0 0
0 1 0
0 0 0

if so i'm a little confused cus that would imply that the matrix representation of f has an e-val of 1, which it doesn't

[midnight haven]

what is e-val?

but yes (lr 0 0 0)
is
1 0 0
0 1 0
0 0 0

[pearl pond]

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[fossil needle]

Sup everyone, I have 10 excercises about Geometry, I have reached ratios of area in math just so you know what level I am i will need help in all those 10 excercises since this is a really hard subject for me. Here is the first excercise:

The quadrangle ABCD is a parallelogram.

1. Proof that BF/FA = AD/AE
   2A. Proof that S△adf/S△aef = AD/AE
   2B. Use questions number 1 and 2A and prove that S△adf= S△bef

[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

[fossil needle]

1

[brisk steeple]

hi

[fossil needle]

Hi

[brisk steeple]

what is given?

[fossil needle]

Lemme send

[brisk steeple]

Okay

[fossil needle]

The only thing given is that ABCD is a parallelogram

thats it

[brisk steeple]

ohh

AEF ad BFC are similar

as the lines are parallel

[fossil needle]

Yeah

[brisk steeple]

Okay so

What will the ratio of BF and AF be equal to?

[fossil needle]

I have no idea

[brisk steeple]

similar triangle have the ratio of corresponding sides equal

So BF/FA = FC/EF

Do you understand?

[fossil needle]

Oh yeah

i do

Oh we also have a tales here

[brisk steeple]

do you understand that AF and DC are parallel?

I mean

AD and DC

[fossil needle]

Uhhhhh

[brisk steeple]

because it's a parallelogram

[fossil needle]

I forgot what parallel means

Im not a english speaker

[brisk steeple]

lines which never meet

[fossil needle]

Ok yeah i understand that

[brisk steeple]

As AF is parallel to DC, AE/AD = EF/FC

[fossil needle]

Yes

[brisk steeple]

or AD/AE = FC/EF

[brisk steeple]

It was shown FC/EF = BF/FA

so AD/AE = BF/FA

[fossil needle]

Ohhh yeah

i get it

[brisk steeple]

Great. I won't be able to help you now. Sorry I have got to go.

Bye

[fossil needle]

Oh

Bye

<@&286206848099549185> yall can help? Look at pinned + i solved 1

[pearl pond]

@fossil needle Has your question been resolved?

[pearl pond]

@fossil needle Has your question been resolved?

[midnight haven]

@fossil needle what does S triangle adf mean?

[fossil needle]

Area of triangle adf

[midnight haven]

@fossil needle , how are u expected to solve these questions, using similarity of triangles or some other concepts??

[fossil needle]

tangens too

[midnight haven]

Tangents?

[fossil needle]

yes

tangents theorem

[midnight haven]

Oh okk so do u know some basic trigonometry?

[fossil needle]

Yes but these question dont include trigonometry

[midnight haven]

Ohkk

Are we allowed to make constructions in the figures??

[fossil needle]

yes

[fossil needle]

Yes

[midnight haven]

Shall I send the soln??

For like 2A)

@fossil needle

[fossil needle]

Yeah

for 2B too if you can

[midnight haven]

. rcacw

[fossil needle]

thank you so much

[midnight haven]

Is it right...?

Or there is some other method expected?

[fossil needle]

yes it is right but i need more explanation on how you got 1/2 x fg x ae

[midnight haven]

See, dude area of any triangle can be written as 1/2*height *base...

I constructed FG as the height...

[fossil needle]

Yeah

And fp

[midnight haven]

Since both triangles are between the same parallels they have the same height...

FP is constructed parallel to AD...

[fossil needle]

Ohhh

i get it

[midnight haven]

Good

[fossil needle]

Fp isnt a height right?

[midnight haven]

Nope

[fossil needle]

Yeah thanks I will try to solve 2B but im not sure i know how

If you can help it would be appreciated

[midnight haven]

2B) requires u to use the results of 1 and 2A) and some manipulations..

[fossil needle]

Yeah

[midnight haven]

Now triangles AEF and BEF have the same height...

[fossil needle]

oh you formed a new triangle

[midnight haven]

Yep

[fossil needle]

Im very bad when it comes to forming new stuff inside the drawings

[fossil needle]

Well you solved it

[midnight haven]

See it actually is simple since the question needs triangle BEF...

[fossil needle]

Yeah

[midnight haven]

Are u in like grade 9 or something...

[fossil needle]

10

[midnight haven]

Wow

[fossil needle]

Im awful at geometry tho

[midnight haven]

U must be in Europe I guess

[fossil needle]

I barely passed this year

[fossil needle]

Middle east

[midnight haven]

Ohh that's interesting..

I'm from india

Okk if u have any other questions feel free to dm or ping me....

[fossil needle]

So, although i have 9 more question i need help with i dont really want to you to do all of them since i feel very bad about it

Like i dont want you to do questions for a random person you dont know i feel bad about it

[midnight haven]

Okk no issues

Im bored asf anyways at home

[fossil needle]

Oh, well do you mind doing 9 more questions?

In this type

[midnight haven]

See geometry is easy if u think in a systematic way...

[fossil needle]

Yeah im very bad at that

[midnight haven]

Sure send all the questions at once..

[fossil needle]

im more a fan of algebra

[fossil needle]

Ok this will take me 2 minutes im sending

[midnight haven]

Sure I'm waiting

[fossil needle]

Given ABCD is a square.
From vertex B drew a line TR.

1. prove that AR+CT=TR
2. Express the area of the square with TR

First question

Im sending them like this

Given:
ABCD is a trapezoid with AB parallel to DC (AB||DC).
DC is twice the length of AB (DC = 2AB).
Point E lies on side BC such that BC is three times the length of BE (BC = 3BE).
Point F lies on diagonal AC such that FE is parallel to DC (FE || DC).
Diagonals AC and DE intersect at point M.

To find:

1. Find the ratios FE/DC and FE/AB.
2. Prove that MC is three times FM (MC = 3FM).
3. Find the ratio AM/MC.

Given: In trapezoid ABCD, where BC is parallel to AD (BC || AD), the midpoints of the bases are N and M, and the segments CN and DM intersect at point O. The segments BN and AM intersect at point Q, and the segments CO and DO intersect at point P (see figure).

1. Prove: PQ is parallel to AD

b. Given: AD = 2a, BC = a
express with A how much PQ is

Dude what is going on

[midnight haven]

I'm trying...

[fossil needle]

for some reason its sending as spoilers

[midnight haven]

Questions are not very straightforward lol

[fossil needle]

You can talk to me if you want there might be a problem with translation

[midnight haven]

Sure

In the first question use congruency...

Like triangle abr is congruent to traingle bct

[fossil needle]

Okay

[midnight haven]

By Angle-side-angle congruence

[fossil needle]

Ok im sending the rest of the questions.... if you have any problems make sure to tell me them

[midnight haven]

Sure

Did u understand the first part of the first question...?

[fossil needle]

Yes

[midnight haven]

Nice

Feel free to interrupt me if u don't understand anything...

[fossil needle]

Sure

In triangle ABC, BD is the height to AC. Point E is located on the extension of height BD so that AB bisects angle EAC (see figure).

Given: ∠BCA = 2∠BAC

Prove:

1. BC * ED = BD * EA
2. Using the given information and 1, prove: BC * ED = AD * BE

[midnight haven]

When do u have to submit these 10 problems...?

[fossil needle]

By the end of the month

september 1st

[midnight haven]

So send the other questions later to me...

These are enough for today...

[fossil needle]

Should i dm you all the questions?

[midnight haven]

Yep that would be better

[fossil needle]

Ok i sent you friend request

@midnight haven you here?

.close

[pearl pond]

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[pearl pond]

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[lavish basin]

@indigo pewter @warm current @next dove

[pearl pond]

Please don't occupy multiple help channels.

[lavish basin]

guys

[warm current]

!noping

[pearl pond]

Please do not ping individual helpers unprompted.

[next dove]

?

[lavish basin]

i figured it out thanks

[next dove]

figured out what?

[lavish basin]

for this

my stupid ahh wasnt counting n = 0

💀

[lavish basin]

sorry but thanks

.close

[pearl pond]

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[frail root]

btw you should count negative values also!

there are 9 possible values for n

[pearl pond]

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[arctic roost]

can someone please help me with question 4

[pearl pond]

@arctic roost Has your question been resolved?

[tough depot]

hello, I was wondering what you set your x and y variables to represent?

[arctic roost]

length(x)

width (y)

[tough depot]

I meant like

which length and width specifically

like this?

[arctic roost]

L= 60ft W=40ft

[pearl pond]

@arctic roost Has your question been resolved?

[tough depot]

@arctic roost ?

[pearl pond]

@arctic roost Has your question been resolved?

[arctic roost]

what is not clear? @tough depot

[tough depot]

are those what you have down as x and y?

or another measurement

[arctic roost]

yes

that is what i have down

im not sure what my next steps are after what i have written down

[tough depot]

ok last question are we assuming there's also fencing here

hence the 2x + 3y = L?

[arctic roost]

no

[tough depot]

oh so in that case it's like this?

just getting clarification cause the phrasing of the question is a bit confusing to me

[arctic roost]

i think that is correct

[tough depot]

ok cool

can you explain to me how you got the original cost function $C = 3(2x+3y) + 2(3y)$ ?

[jolly parrot]

AwesomeRat

[arctic roost]

I just plugged the $ amount into the equation I got in #3

[tough depot]

well if this is how you represent the x and y values, shouldn't the resultant cost function be C = 3(2x + 2y) + 2(y) where 2x+2y is the length of the outer fence and y is the length of the inner fence

correct me if I'm wrong

[arctic roost]

i think that is correct

[tough depot]

ok well in that case it's gonna be very similar to the first problem

because you're trying to minimize the cost

[arctic roost]

wait why not c= 3(2x+2y)+2(2y)?

[tough depot]

where'd you get the 2y

in the second term

[arctic roost]

Outside fencing: 2x+2y (the two lengths and the two outer widths)
Inside fencing: 2y (the shared width in the middle)

[tough depot]

well there's only one inside fancing though right

fencing*

[arctic roost]

ok

C= 6x+8y

[tough depot]

ok now you can't differentiate to find the minimum quite yet though

because since you have three variables that would require implicit differentiation which isn't really applicable in this scenario

so think of a way to replace the x variable in terms of y or vice versa

[arctic roost]

is this correct?

[pearl pond]

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[warm charm]

yo

[warm charm]

C = 10w^2 + 90/w

C' = 20w-90w^-2

zero of C' at 1.65

plug in zero to C

i get that answer

and its wrong?

[tough depot]

how did you get the 90/w in this equation?

[warm charm]

30/w+60/w

[light helm]

where did that come from

can you show full work

[warm charm]

C = c1(2w^2) + c2(2w*10/2w^2 + 40w/2w^2)

= 10w^2+30/w+60/w

=10w^2+90/w

c1 and c2 are given in feedback

i mean in question

[tough depot]

ok looks good

lemme see

[light helm]

rounding seems to be off

[warm charm]

OH

i read that as decimal not cent

thanks!

.close

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[midnight haven]

Please help???

[pearl pond]

@midnight haven Has your question been resolved?

[last summit]

Which part are you confused on

[naive hemlock]

Ok

Oh

Yo Austin

[pearl pond]

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[untold dust]

I have the function $G(\omega)= \frac{-1}{\omega^2+2\gamma\omega i - \omega_0^2}$ where $\omega , \gamma,\omega_0 \in \mathbb{R}$.
I need to find $\int_{-\infty}^{\infty} G(\omega)e^{-i\omega t} d\omega$. I managed to evaluate the integral by instead $\omega \in \mathbb{C}$ and using Cauchy’s and Residue Theorem. However, I’m curious as to the intuition as to why changing fields worked

[untold dust]

My initial thought was that its allowed since $G$ has complex roots, so allowing $\omega$ to be complex naturally makes sense. But is that the actual reasoning as to why thats okay?

[jolly parrot]

KySquared

[prime rock]

brother i think you might be waiting a while to get an answer 💀

[untold dust]

[prime rock]

thoughts on its a typo and omega and gamma are supposed to be in the complex field originally?

oh wait nvm

[untold dust]

Oh I should have specified omega naught to be real as well

Im bad lol

[jolly parrot]

KySquared

[untold dust]

No, gamma isn’t complex

[prime rock]

damn changing the field feels so illegal lmao

[untold dust]

It does that’s why I’m looking for a justification as to why its okay

Instead of just “oh the roots are complex so lets just pretend omega is complex as well”

That feels sloppy

[prime rock]

ok wait, how did you even conclude G has complex roots? does that shit even have roots??

[untold dust]

Quadratic formula

[prime rock]

ok but the quadratic is in the denominator

[untold dust]

1 sec
I’ll send the full derivation

[prime rock]

ok

[untold dust]

@prime rock

[prime rock]

ok but you found the roots of the denominator of G(w) no?

idk ive never seen someone find the roots of a hyperbola

[untold dust]

Right, and the roots are complex
I’m trying to figure out if the roots being complex is enough justification for allowing omega be complex?

[untold dust]

Its a method of solving ODE’s

[prime rock]

i see

[prime rock]

lowkey i dont think it does, but honestly i wouldnt be surprised if it was

sorry idk, probably you should repost your problem here again, got a bit burried

although one small thing, i remembered cauchy/residue theorem involving contour integrals but idk if it changes anything tbh

[untold dust]

Yep I changed it to being contour integrals over the entire upper & lower planes and got an answer that made sense

I’ll go and ask again later today then
Thanks for your time

.close

[pearl pond]

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[pallid mural]

24 part i
Help plz

[acoustic path]

hint: try to find ways to force the sin double angle identity using the fact that theta = pi/15

[pallid mural]

Can ya write and show?

[acoustic path]

sin(2x) = 2sin(x)cos(x)

[pallid mural]

Can't ta just show step by step?

[acoustic path]

the entire solution? no

i am simply telling you the tools you should use to do such problem

[pallid mural]

I used the cos A.cos B formula

Reached till half way

[acoustic path]

i believe that will make things more difficult for you here

[pallid mural]

Even if i force that idk what to do next

Like force which one to be sin

[acoustic path]

it helps to be familiar with shift/reflection identities in trig functions

i.e. cos(theta + pi) = cos(theta - pi) = -cos(theta)

[pallid mural]

So what? Like use sq.(1 - sin sq. A)

[acoustic path]

no

that is not a shift/reflection identity

[pallid mural]

Oh that one

[acoustic path]

that is the pythagorean identity which will not help you much here

[pallid mural]

I understand that

So... Cos (theta + pi) one... Oof that's tougher for me to do but okay I'll see

[acoustic path]

have you done 24 part ii?

[pallid mural]

Nope... If i knew part ii I'd do part i easy

[acoustic path]

i see

i recommend doing part ii first as it is an easier problem than part i

(i do not know why they are written in that order)

[acoustic path]

for part ii, trying to force an application of this identity is the way to start

[pearl pond]

@pallid mural Has your question been resolved?

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[sage jay]

a, b, c are single digit positive integers. Find a, b, c respectively. There is more than one solution. Find all solutions.

[sage jay]

I did trial and error and found 2 solutions

[dapper kraken]

is a,b,c diffrent?

[sage jay]

Yes

[dapper kraken]

should say that :p

err

what was the solutions you found?

[sage jay]

164 and 195

[dapper kraken]

yeah im pretty sure you can deduce a must be 1

you can narrow the list for the last digit

[sage jay]

How many solutions did you find

[dapper kraken]

still working on it

[sage jay]

Ok

[dapper kraken]

(3,5) (6,4) (5,5) (7,5) (5,3) (5,7) (9,5) (5,9) (6,6) (4,6) for the last digits

[acoustic path]

do you need to prove these are the only ones?

[sage jay]

No those are the ones I found

[sage jay]

Those don’t all work tho

[acoustic path]

but are you required to provide a proof?

[sage jay]

No

[acoustic path]

then ok i tell you those are the only two solutions

[sage jay]

My friend found a third though apparently

[dapper kraken]

yes

[sage jay]

But I couldn’t ask him what it was

[dapper kraken]

trust mqnic not me

[acoustic path]

i think they made a mistake somewhere

[sage jay]

Ohh ok

Yay

How did you find it tho

@acoustic path

[acoustic path]

i just tested every possibility and checked my work

[acoustic path]

something along the lines of this

[sage jay]

Oh same

[acoustic path]

is this what the "summary" is supposed to be?

[sage jay]

‘Summary of all of the solutions’

Nvm dw

Thanks

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[midnight haven]

anyone?

[midnight haven]

i have no idea

lol

anyone?

[spare lark]

You can factor a polynomial by his roots

Which one is a root ?

[midnight haven]

i have no idea

you explain teacher?

3x-2

[spare lark]

3x-2 ?

Oh ok ic

[midnight haven]

this one?

[spare lark]

Well yes a root is x when g(x) = 0

[midnight haven]

yeah

so 3x-2

[spare lark]

I was confused cuz you ask explain and just after put the answer xd

Anyway

[spare lark]

What is g(2) ?

[midnight haven]

im an idiot bro

[midnight haven]

2 = f(x)

[spare lark]

What is g(2) ?

[midnight haven]

idk

2,-3

[spare lark]

Yeah -3 so what what is f(-3) ?

[midnight haven]

3

[spare lark]

So f(g(2)) = 3

[midnight haven]

can you explain again lol?

i mean yes it seems to be correct but

[spare lark]

we want to find f(g(2))

So we search g(2) first

We found -3

So f(g(2)) = f(-3) = ?

3 !

[midnight haven]

hmm

pointing

got it

thx

[spare lark]

Any Questions left ?

[pearl pond]

If you are done with this channel, please mark your problem as solved by typing .close

[midnight haven]

here

[spare lark]

You have a graph ?

Or something else with ?

[midnight haven]

no

just this

[midnight haven]

bro you wanna do voice chat or something? it will be quicker and easier

[spare lark]

I cant

[midnight haven]

oh

[spare lark]

Well it provides information on the vertex, direction of opening

And intersect of the parabolic graph

[midnight haven]

so?

[spare lark]

Oh yeah youre asked to distinct

So

The - infront tell the direction of opening

It goes down

The constant 1 and 3 are the coordinates of the vertex (1,3)

And intersect are given by constants, letting y=0 firstly and x=0 secondly

[midnight haven]

i don't know man

i don't understand this

[pearl pond]

@midnight haven Has your question been resolved?

[pearl pond]

@midnight haven Has your question been resolved?

[pearl pond]

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[vernal grove]

would this be correct

or will the nfa for regular expression epsilon be the same as the regular expression a

like this but instead epsilon

I think the first image would work but im just not sure

[scenic aurora]

<@&268886789983436800>

[sweet bone]

replacing a with epsilon also works, but your first solution is fine

[vernal grove]

ai

rah my second mssg

can you explain this

whats the dotted line for

[sweet bone]

what is this supposed to accept

it doesn't have an accept state

[vernal grove]

this should be for alpha^* right

oh shit true

lemme find a diff pic

[sweet bone]

I mean, if you connect like this

[vernal grove]

thats what I was thinking

[sweet bone]

then this consumes a* yes

[vernal grove]

alr it makes sense

oh and

for emptyset

it will just be a self loop

[sweet bone]

you can just have one state

[vernal grove]

and no final state

right

[sweet bone]

initial and not accept

yeah

the loop is extraneous

[vernal grove]

bet tysm

oh not needed?

[sweet bone]

yeah

the NFA branches just immediately fail; there are no valid transitions if there are no transitions in the first place

[vernal grove]

bet tysm

.close

[pearl pond]

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[jolly parrot]

Retrieving the previous image failed.

Aug 7, 2024, 9:22 AM

[next dove]

, rotate 15 degrees acc

[cosmic charm]

you can do this normal way or smart way

[next dove]

what's that

[cosmic charm]

normal way is just use the quotient rule

smart way is notice a certain relation between the numerator and denominator

[next dove]

😭 why degrees thing doesn't exist

[cosmic charm]

idk just sacrifice your own neck i suppose

[marble bear]

can u send solutions thru

[cosmic charm]

no

[marble bear]

need explanation

[cosmic charm]

,rccw

[cosmic charm]

if P(X) * P(Y) = P(X∩Y), then X and Y are independent

[marble bear]

why are they doing it 2 ways

[cosmic charm]

because there are two ways to find P(X∩Y) from the given information

if they are independent, then it should be the same

[pearl pond]

@marble bear Has your question been resolved?

[pearl pond]

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[fluid schooner]

.open

hlo anyone

[fluid schooner]

<@&286206848099549185>

[robust kraken]

hello

[sharp smelt]

!15m

[pearl pond]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[next dove]

!15min

[pearl pond]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[fluid schooner]

i have a doubt'

[marble bear]

.reopen

[fluid schooner]

5cosA=7cosA
find 7sinA+5cosA/5sinA+7cosA

can you guys resolve it

[next dove]

5cosA=7cosA?
are you sure that it is the given info?

[robust kraken]

7/5?

[fluid schooner]

yea you have to do it like that

but i cant reach the solution

[robust kraken]

according to wolfram its 7/5 lol

[next dove]

!nosols

[pearl pond]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[robust kraken]

mb

[next dove]

maybe guide him to the solution

[fluid schooner]

bro let me see the answer atleast

[cosmic charm]

!xy

[pearl pond]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[thick galleon]

.open

hlo

[fluid schooner]

Hlo

[fluid schooner]

!xy

[pearl pond]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[cosmic charm]

,rccw

[cosmic charm]

.

[latent quail]

cosA = 0

[fluid schooner]

yea

[next dove]

it is sine,

[cosmic charm]

is that supposed to be a 4 or 7

[fluid schooner]

its 5 and 7

[latent quail]

wdym, they stated cosine

[next dove]

here

[latent quail]

oh, alr

[cosmic charm]

5cos = 7sin

divide sin by cos

for reasons you should know

[fluid schooner]

bruh

[cosmic charm]

whats sin / cos?

[fluid schooner]

tan

[cosmic charm]

so how can you make 5cosθ = 7sinθ into something with only tan

[fluid schooner]

ok divide sine by cos

then

5/7=sin/cos if iam not worng

[cosmic charm]

yep

[fluid schooner]

but how can that help in solving the equation

[fluid schooner]

i have to evaluate it

[cosmic charm]

now in the equation, how can you make it into having tan

[fluid schooner]

how?

[cosmic charm]

you can divide the top and bottom by cos

or multiply top and bottom by 1/cos

[fluid schooner]

ok then

[cosmic charm]

what does that do?

[fluid schooner]

then the value iwll become tan

sin / cos will happen

[cosmic charm]

can you write it out?

[fluid schooner]

yea lemme try

till here its correct ig

But how can i take 7 and 5 common

[cosmic charm]

?

[fluid schooner]

just a second i am framing the equstion

hey actually i am unable to send a picture rn but my answer is coming to be 10/12

is that right idts

<@&286206848099549185>

[cosmic charm]

how'd you get it

[fluid schooner]

i just subsitited th evalue by seperating the solutions

like sepearting the denomiantor and numerator

[cosmic charm]

how did you seperate it

[fluid schooner]

like 5 x 7/5 + 5 cos/cos

so it got cancelled

[cosmic charm]

sorry to say but you cant seperate fractions like that

[fluid schooner]

yea i thik the same

can you share me the solutuions

[cosmic charm]

well ok lets look at the numerator for now

its 7sin(θ) + 5cos(θ)

what happens if you divide this whole thing by cos

[fluid schooner]

7 sin + 5 cos /cos

[cosmic charm]

sin also gets divided by cos

[chrome river]

Has the answer been solved yet?

[fluid schooner]

nah

ye i was wirint like whole divided by the cos

[cosmic charm]

you can think of it as im saying (7sinθ + 5cosθ) * 1/cosθ

[fluid schooner]

yea

[cosmic charm]

so whats this then

[fluid schooner]

7*tan+5

[cosmic charm]

and what was tan?

[fluid schooner]

5/7

[cosmic charm]

so 7tan + 5 is?

[fluid schooner]

12

[cosmic charm]

no..

[fluid schooner]

why

[cosmic charm]

7(5/7) + 5 is?

[fluid schooner]

ohh

mb

so sorru

10

[cosmic charm]

good

now the denominator is 5sinθ + 7cosθ

also divide that by cos

what do you get

[fluid schooner]

74/7

[cosmic charm]

good

now whats 10 / (74/7)?

[fluid schooner]

70/74

[cosmic charm]

you can reduce that

[fluid schooner]

yea

[cosmic charm]

thats 35/37

[fluid schooner]

yea

i forgot to write

[cosmic charm]

and thats your answer

[fluid schooner]

yea thanks

[chrome river]

$5cos(A) = 7cos(A)$ is never equal unless $cos(A) = 0$
thus:
$\frac{7sinA+5cosA}{5sinA+7cosA} = \frac{7sinA+0}{5sinA+0} = 7/5$

[jolly parrot]

woomy

[chrome river]

note that $5x = 7x$ is only true for $x=0$ in $\mathbb{R}$ (and $\mathbb{C}$)

[jolly parrot]

woomy

[fluid schooner]

ok buddy

i will ask mu teacher

.close

[pearl pond]

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[pearl pond]

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[red trail]

Can anyone help me prove this:
I used cauchy shwartz inequality and proved it but my lecturer does not allow it, if anyone can find another way then please help

If sum an^2 converges then also sum (|an|/n) converges

[pearl pond]

@red trail Has your question been resolved?

[pearl pond]

@red trail Has your question been resolved?

[red trail]

<@&286206848099549185>

[gleaming comet]

Kya hua Bhai??

[red trail]

?

[stone portal]

There's the condensation test for sums that says a sum converges iff sum(2ⁿa(2ⁿ)) also converges, applying to both sums we get-
(I'll mark convergence by C() to make the argument shorter)

1. C(sum(an²))<=> C(sum(2ⁿa²(2ⁿ))<=>C(sum((2ⁿ|a(2ⁿ)|•|a(2ⁿ)|)
   2)C(sum(|an|/n)<=>C(sum(2ⁿ|a(2ⁿ)|/2ⁿ)<=>C(sum(|a(2ⁿ)|)

Assuming (1), we break into cases:
2ⁿ|a(2ⁿ)|<1 for n>N (for some N):
=> |a(2ⁿ)|<2^(-n) => sum{n=N→∞}(a(2ⁿ))≤sum{n=N→∞}(2^(-n))≤2
Proving (2) and the statement.
if 2ⁿ|a(2ⁿ)|≥1 from some N onward:
2ⁿ|a(2ⁿ)|•|a(2ⁿ)|≥|a(2ⁿ)|
=> sum{n=N→∞}(2ⁿ|a(2ⁿ)|²)≥sum{n=N→∞}(|a(2ⁿ)|)
Since (1) implies C(sum{n=N→∞}(2ⁿ|a(2ⁿ)|²)) we have shown C(sum{n=N→∞}(|a(2ⁿ)|)) which implies (2) proving the case.

That proves the thing with calc1.

[red trail]

tysm ❤️

[pearl pond]

@red trail Has your question been resolved?

[pearl pond]

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[main briar]

Can anyone help me

<@&286206848099549185>

[last moth]

Please don't ping Helpers unless your question has been unanswered for 15 minutes

[chrome river]

!status @main briar

[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

[chrome river]

also with what do you need help with?

[main briar]

Ok

[main briar]

<@&286206848099549185>

[calm crest]

what Exactly do you need help with

[pearl pond]

@main briar Has your question been resolved?

[pearl pond]

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[nova adder]

someone explain how i do this question?

[latent quail]

Have you tried anything so far?

[naive hemlock]

Do u know the equation of tangent to a curve at a point?

!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

[nova adder]

u find the derivative first but am i supposed to find derivative for y = 2ax - a^2 + 3?

[naive hemlock]

See

I will recommend you to read the question again

show that y=2ax-a^2+3 is the equation of the tangent

Floppy do u know how we find tangent at a point to a curve

[nova adder]

find derivative, plug points in get gradient then use y-y1 = m(x-x1)

is the derivative 0?

[naive hemlock]

See

Derivative of a curve at a point

[midnight haven]

You have the equation y = x²+3. To prove that the TANGENT at given coordinates, you can find the FD. Which y'(x) = 2x. Pluck a into it you get 2a.

So the gradient of the tangent line is 2a. y =mx+c,
m is 2a.
x is well x.

[naive hemlock]

Gives the gradient aka slope of the tangent to the curve at that point

[midnight haven]

This is part of your tangent line already. Y = 2ax.

You can figure out the back:)

[nova adder]

fd?

[midnight haven]

first derivative

[nova adder]

wait so ur trying to prove that the tangent of y=x^2 + 3 at the point (a, a^2+3) equals to y=2ax-a^2 + 3?

[midnight haven]

is it not the question?

To prove that the equation of the tangent line to the curve

[nova adder]

i thought u were supposed to find the derivative of the y = 2ax-a^2 + 3

[midnight haven]

at (a, a2 + 3) is y = 2ax - a^2 + 3

[nova adder]

ok this makes more sense now, thanks

.close

[pearl pond]

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[prime coyote]

why can't i solve it like that?

[prime coyote]

ex - 3

[upper drum]

you cannot as even if your method was correct you would have got

sin x = 3/4 = tan x/sec x

with your logic tan x would be 3 and which is incorrect

[bright kindle]

the value of cos cant be more than 1

[upper drum]

• this

[bright kindle]

sin is perpendicular/hypot.

[west bluff]

*if cos is real

[bright kindle]

which it obvio is

[west bluff]

But yeah in this case probably right, if cos were to be greater than one, then one of the sides of the triangle would be grater than the hypothenuse

Which directly contradicts the Pythagorean theorem

[upper drum]

and theorems by definition are true

[prime coyote]

thanks math gods 🙏🙏

.close

[pearl pond]

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[thin sigil]

Ummm... if you already have tables for ages by gender, to get a genderless one you'd just add them, no?

Or am i missing something here?

Mind sending an example or sample of what you have?

If they're percentage based then I guess you're missing data to allow for such combining

[pearl pond]

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[red falcon]

let's roll a die until we get a 6 and denote with $E_n$ the event that n rolls took place. What would then $(\bigcup_{i=1}^n E_i)^C$ represent? My answer: using DeMorgan $\bigcap\limits_{i=1}^n E_i^C$ where $E_i^C$ is the event where 6 either didn't occur at all, or it didn't occur at the i-th roll. that is, the set of i-tuples of elementary events in which 6 isn't at the i-th place or it isn't in the i-tuple at all. this makes sense?

[jolly parrot]

atif

[red falcon]

hm not sure what the intersection of those $E_i^C$ should be

[jolly parrot]

atif

[red falcon]

ok a perhaps a correction: $E_i^C$ actually represents an event where less or more than i rolls took place

[jolly parrot]

atif

[pearl pond]

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[wooden ermine]

what are the possible things you can use the expected value formula for a discrete random variable?
as in $\sum_{i=1}^{n} x_i p(x_i)$?

[jolly parrot]

AggressiveNooby

[wooden ermine]

is it sufficient to say "this can be used to compute the average trials per success for anything with a consistent probability"

consistent probability
being the key phrase for "possible things" you can use it on

[tall flint]

average trials per success
i'm not sure this parses

[pearl pond]

@wooden ermine Has your question been resolved?

[wooden ermine]

wdym

[tall flint]

i ask you the same thing

what does "average trials per success" mean

[terse tide]

computing the expected value of some "gambling" game, for example, can indicate you whether or not the game is a pure scam lol

[pearl pond]

@wooden ermine Has your question been resolved?

[unique kelp]

Could u help me solve seiberg-wittens equations? In string theory??

[wooden ermine]

uhh this channel is occupied

hmmm?

i don't know anything about that

im a highschooler

all good

[pearl pond]

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[junior vapor]

for b)ii), the answer was tan112.5 but I don't get whr that came from

[sly plover]

How did you solve the first part? Because we can get it easily from the first part

[junior vapor]

in short, I used part a (which I showed) and I just did quadratic formula

[sly plover]

Huh?

[junior vapor]

like from part a u get 2t=1-t^2

[sly plover]

You got an equation from thetha right

[junior vapor]

uh do u mean t?

[sly plover]

Oh

[sly plover]

I'm talking about question b i) lol

[junior vapor]

oh mbmb

[tardy reef]

what is so special about the number 22.5 degrees?

[junior vapor]

when u multiply it by 2 it turns into 45?

[tardy reef]

Yep

[sly plover]

Yea

[tardy reef]

and do you know how the tan of an angle and the tan of an angle thats twice the other relate?

[junior vapor]

not really

[tardy reef]

so look up the formula for tan 2 theta and use it here

[junior vapor]

my problem is b) part 2

idk why u take positive root

and not the negative one

[tardy reef]

because the tan of an angle in 1st quadrant is always +ve?

[junior vapor]

oh...

alr ty

.close

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[mighty ice]

I get how sin and cos refer to the x and y values respectively

how does tan and csc work exactly

[midnight haven]

tangent of any angle in a right angled triangle is the side opposite to the angle divided by adjacent side(not hypotenuse)
same thing here

csc is just 1/sin

[mighty ice]

wdym exactly

like here

[halcyon pawn]

Your question is how tan and csc are related to the sides of a triangle right?

[mighty ice]

more so here

[halcyon pawn]

Yes every point on a circle can be thought of as a vertice of a right triangle with the right angle at the origin.

let the hypotenuse or the distance from the origin to any point on the circle be r. then its projection on the x-axis will be x and it's height will be y. And the hypotenuse makes an angle of theta with the x-axis. so sin(theta) = x / r and you can solve it for x. As for tan it is the ratio of the opposite and the adjacent so the y / x.

[pearl pond]

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[mighty ice]

I see

thanks

.close

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[nova adder]

idk where to begin for this question. am i supposed to get two simulatenous equations with either H or R?

[tardy reef]

get h in terms of r and substitute in volume formula

minimize wrt the variable

[nova adder]

minimize with the variable?

@tardy reef ?

[fickle schooner]

"wrt" => "With respect to"

[nova adder]

how do i do that exactly?

[sacred gate]

derivative

[midnight haven]

You can perform the first derivative test to find the Maximum volume.

$V = \pi r^2 h$ For simplicity sake, lets make r the subject. h = 12 - r. $V = \pi r^2 \times (12-r) = 12 \pi r^2 - \pi r^3$

[jolly parrot]

Sukiyaki

[midnight haven]

Find your first derivative, $24 \pi r - 3 \pi r^2$ (Not sure if i did it correctly, i just did it in my head at 1 am. $\frac{dv}{dr} = 0$

[jolly parrot]

Sukiyaki

[midnight haven]

r = 8 or 0. substitute it to find ur ogiginal.

h = 4, r = 8, $\pi (8)^2 \times 4$ = 256 pi

[jolly parrot]

Sukiyaki

[midnight haven]

that's how I would do it, nto sure about you maybe my method is wrong but ill let other helpers dispute it. Try give it a go and lmk how it goes:)

[nova adder]

so to clarify, in these questions what is the derivative supposed to represent? the volume in the given domain (0<=r<=8) ?

[midnight haven]

Dy/dx is the rate of change, specifically the gradient of the curve.

It just shows how the curve changes at different points. When I did y'(x) [first derivatve], im essentially looking for a maxima or minima

Of course, I just presumed it to be a maxima, you could perform the second derivative test to prove that it is the local maximum.

[nova adder]

so the maxima,minima are the values where the derivative intercepts the x-axis right ?

[midnight haven]

No.

The maximum point is simply where the graph plateaus.

[nova adder]

then what about this part?

[midnight haven]

This is errr whats' it called..

Er the quadratic.

Because i already found that 24pir - 3 pir2 = 0

So i can simply just find r using the quadratic equation

r will return 8 or 0 which are your maximum or minimum radiuses. I'm going to reject 0 because well.. you should know the reason.

[nova adder]

wait so is maximum the same as maxima and minimum the same as minima or those different things

[midnight haven]

I believe theyre simply the same thing

Maxima is just the plural form of maximum

Less about that!

Your question asked you to find the maximum Volume given that you have r+h = 12.

You can reform the equation two ways if you'd like;

$\frac{dv}{dr}$ or $\frac{dv}{dh}$

[jolly parrot]

Sukiyaki

[midnight haven]

It just describes how does the volume of yoru shape changes with the change in radius. But it will reach this maximum point where it is 0, because it no longer changes.

Thus, dv/dr is also the GRADIENT curve at certain points.

Alright, i over yapped. Just ask if you're in doubt:)

[nova adder]

the derivative is the rate of change so as it gets closer to the maximum point the slower the volume is increasing by correct?

[midnight haven]

Err........... you could put it that way.. sure...?

[nova adder]

as in the maximum point is a midpoint on like a parabola or something

[midnight haven]

If given it's just a simple x^2 curve

then sure, you can say that it's the middle.

but if you notice, the volume equation is a x^3

For simplicity of explanation, this would look SOMETHING (its not the same as ur qn) liek the one in your question.

It's not really symmetrical. The maximum point is just the highest point before it goes down agn

[nova adder]

so to reiterate the derivative shows how the volume of the shape changes with the change in radius, and where 0<=r<=8 is the domain of the derivative in the question?

[midnight haven]

That sounds right.

but the part where 0 <= r <= 8

I belive not.

Because they're just the maximum and minimum points

The domain spans beyond that

[nova adder]

so would it be the range? if its maximum and minimum points?

[midnight haven]

nevermind, ignore whatever i said for the domain part.

Im not SUPER sure about the domain part, so i'll prefer not to talk on that. but the derivative part in basic terms soundsright.

[nova adder]

hmm alright, i got what i came for anyways

thanks

.close

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