help-10

[feral tide]

is this right

[feral tide]

@everyone

anyone know

is it right

[serene moss]

Question 1 is right

Both are right

[feral tide]

ok ty bro

yo

@serene moss

are these all right

[serene moss]

I'll check.

[feral tide]

ok ty

[serene moss]

Hey @feral tide

[feral tide]

ye

[serene moss]

Look at Q.3

[feral tide]

its wrong?

[serene moss]

I think you are wrong there

[feral tide]

ok

[serene moss]

Say the distance is x

You get tan40=x/45

x=45.tan40=37feets

Q4 is correct

Q5 is correct

[feral tide]

@serene moss

yo i think 3 is right

[serene moss]

one min

Yeah you are right

Sorry, my bad

Made the wrong diagram in my mind lol

[cosmic fog]

Whatever he said is wrong

U cannot do that

U have to convert sinx - cosx purely as sin or purely as cos function

So u have to multiply and divide by the square root of the sum of squares of coefficients of sin and cos

In this case that evaluates to root(1² +( -1²))

So 2 times (sinx - cos x)/root2

2 times sin(x-pi/4)

[obtuse pebble]

@feral tide Has your question been resolved?

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[obtuse pebble]

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[fresh copper]

what are the subgroups of the dihedral group D6 as permutations of the numbers 1-6?

[fresh copper]

<@&286206848099549185>

[rancid shadow]

just ask chatGPT, and it gives you a PERFECT answer.

I just did it.

👍

[fresh copper]

it doesnt, its wrong

[rancid shadow]

then what do you want exactly

[fresh copper]

i want the subgroups of gorup D6

as in C3,C2, V4, etc

[rigid plaza]

Wait so you know the subgroups

But not how to represent them?

Very different problems

@fresh copper

[fresh copper]

i want to know both

ill show u what im making

when i press the subgroup, it should highlight the elements

@rigid plaza

@rancid shadow

[rigid plaza]

What is your question exactly about what you need to do

Where are you stuck

[fresh copper]

i dont know what the subgroups of D6 are and what elements are in them

[rigid plaza]

!status

[obtuse pebble]

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

[fresh copper]

1

[rigid plaza]

Think of rotations and reflections

And how those act as generators of the dihedral group

How can use a subset of them as generators for the subgroups

Once you can identify all the subgroups

The next part is easy

[fresh copper]

i actually dont get this

can you give me an example?

[rigid plaza]

Do you know what a dihedral group is?

[fresh copper]

so D6 is a regular hexagon

and the group of D6 means the ways u can rotate/reflect it to keep it the same

[rigid plaza]

"Is" is a very strong term

[fresh copper]

ok not is, but you can interpret it as one

and i am using a hexagon with vertic as 1 2 3 4 5 6

[rigid plaza]

Yes

What are the generators of D_6 geometrically

[rigid plaza]

You said it here

[fresh copper]

r60, r120, etc, and then a reflection over each vertici

[rigid plaza]

You need a single reflection and a single r60 to generate the whole group

What other rotations will give you sub groups?

[fresh copper]

mupltilpes of 60?

[rigid plaza]

Which ones give you unique subgroups

[fresh copper]

120,

[rigid plaza]

Anyways after you have analyzed the structure in terms of reflections and rotations

Just draw a hexagon with vertices labeled to find permutations in terms of numbers

[fresh copper]

wdym the strcuture

[rigid plaza]

What the subgroups are

gtg good luck

[fresh copper]

https://tenor.com/view/crying-emoji-dies-gif-21956120

[obtuse pebble]

@fresh copper Has your question been resolved?

[fresh copper]

hell naw

[fresh copper]

☠️

[obtuse pebble]

@fresh copper Has your question been resolved?

[obtuse pebble]

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[outer cargo]

i know the answer is 0 i just dont know wtf went wrong

[slate monolith]

!rotate

.rotate

mm

the bot

[outer cargo]

ill rotate it

[slate monolith]

ok thx

.original

!original

[obtuse pebble]

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.

[slate monolith]

is that the original q

[outer cargo]

there

having a hard time with gifs

greatest integer functions

[slate monolith]

^

[outer cargo]

what original?

like the original equation?

[slate monolith]

ye

[outer cargo]

yup

thats the given

[prime pilot]

r u sure the answer is 0?

[slate monolith]

x/(x-1) + x

[outer cargo]

yep, the lecturer told so

what do you guys think

idk what went wrong

[prime pilot]

is that a ceiling function or a floor function

ohh its a floor function

yeah its 0

lemme explain

[outer cargo]

i do not know

[slate monolith]

me neither -_-

[outer cargo]

a floor function involves a positive subscript right?

i think

[prime pilot]

The floor function ⌊x⌋ returns the greatest integer below the input (basically the nearest integer below it)

[thick chasm]

gif is also known as floor function

[slate monolith]

erm is this hs math or uni math or smth else

[prime pilot]

therefore, if x is approaching 1 from beneath it (from the negative side), the floor of all of those values no matter how close they are to 1 will be 0

[outer cargo]

uni

[slate monolith]

oh

[outer cargo]

im trying to study advance because my uni has online lectures uploaded on yt

[prime pilot]

icl theres not much more to it than knowing wht it is

[prime pilot]

u with me @outer cargo

[outer cargo]

yes

one question tho

perhaps it isnt

[eager niche]

!original

[obtuse pebble]

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.

[prime pilot]

well

we’re looking at values really close to 1, but below it

so x-1 is gonna be really close to 0, but below it

meaning that the floor function of all of these numbers will be -1

remember the subscript -, after the 1, just tells you that you’re approaching 1 from values below it

[outer cargo]

even the x-1?

okay i think i get it

i need more exercises

thank you

[prime pilot]

this means the floor function of x - 1 as x tends to 1⁻ is -1

[prime pilot]

nws

[outer cargo]

.close

[obtuse pebble]

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[obtuse pebble]

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[brisk yacht]

kinda confused on what to do

[prime pilot]

in order to combine those fractions, you'll need all of them to have a common denominator

once you've done you can just add/subtract the numerators, while keeping the denominator the same]

[brisk yacht]

whats the common denominator?

[prime pilot]

in this case it's gonna be 12x^2

ik that becasue, the lowest common multiple of 3, 4 and 6 will be 12

and the highest power of x there is x^2

[brisk yacht]

ok

ty

.close

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[onyx sentinel]

I have A^(-1) = [(detA)^2]/detA, can I cancel the determinant in the denominator and the square in the numerator?

[onyx sentinel]

Like we do using numbers?

[gilded needle]

the equation doesn't make sense, the LHS is a matrix and the RHS is a number

[onyx sentinel]

wait

let me show the original problem

@gilded needle

I have done this so far

[timid silo]

You could stop here

Further simplifying would get you |A|^5 =1
Since the power is odd you could say
|A|=1

[onyx sentinel]

Aigh aight

thanks

.close

[obtuse pebble]

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[rough stump]

Prove that { tan ( 135° +θ ) * tan ( 45° - θ ) = sin2θ - 1 / sin2θ + 1 }

[obtuse pebble]

@rough stump Has your question been resolved?

[rough stump]

<@&286206848099549185>

[dreamy fox]

Okay wait I'm doing it on paper

[rough stump]

ok thanks

[dreamy fox]

Is your question correct?

[royal flint]

Exactly what I was thinking

Lhs is just -1

[dreamy fox]

LHS is coming -1

[rough stump]

wait let me re check

[dreamy fox]

Real

[royal flint]

I lying in bed and mentally solving

Rhs is -tan^2 theta-45

[rough stump]

sorry one this is wrong

[royal flint]

Lhs is just -1

I was so confused

[rough stump]

Prove that { tan ( 135° + θ ) * tan ( 45° - θ ) = sin2θ - 1 / sin2θ + 1 }

sorry guys

[royal flint]

Ok now it's right

[rough stump]

u got the answer

so quick

[royal flint]

I did this way too often

[rough stump]

can u teach me

[dreamy fox]

It just takes some practice

[rough stump]

ok

[royal flint]

Exactly

[rough stump]

i did this problem 3 times but not getting answer

Thanks btw

[dreamy fox]

That happens initially to all of us

[royal flint]

Yeah there are some general patterns

[dreamy fox]

True

[royal flint]

Like if lhs has theta and rhs has 2 theta, usually you have to use sin2 theta or cos 2 theta stuff

[rough stump]

ok

Can I ask one more question (short one)

[dreamy fox]

Sure

[rough stump]

The absolute least value of sqrt(2)* sinx - sqrt(2)*cosx is:

Absolute least value mean function smallest value? right

[dreamy fox]

Yes.

[rough stump]

-1 < = sin < = 1 , -1 < = cos < = 1

[dreamy fox]

That approach won't work in this question

[rough stump]

ok

sqrt(2) [ sinx - cosx]

[dreamy fox]

Haven't you heard of the concept of maximum and minimum value of f(x)=Asinx +Bcosx?

There's a general formula

I can show you if you haven't

[rough stump]

hmmm

[dreamy fox]

Hmmm means?

[rough stump]

• sqrt ( a^2 + b^2 )

[dreamy fox]

Yes

[rough stump]

thinking

[dreamy fox]

So in this case you need to use it.

[rough stump]

it is 2

[dreamy fox]

Not really

[rough stump]

2

-2 , 2

[dreamy fox]

Yes.

[rough stump]

but this is not correct option

[dreamy fox]

Huh

[rough stump]

according to my text book

0 is correct option

[dreamy fox]

I'm pretty sure that's wrong.

Does your question

Include

Non negative?

[rough stump]

nope

[dreamy fox]

I believe the given answer is incorrect then.

0 will surely be a vaue at x=π/4

But that isn't the min.

[rough stump]

ok

[dreamy fox]

(a) must be the one

Wait are u a jee aspirant??

[rough stump]

not reallly

but some thing like that

[dreamy fox]

Oh ok

NDA?

[rough stump]

I am from Pakistan

[dreamy fox]

What then?

Oh I see

[rough stump]

You Indian

[dreamy fox]

Yep

[rough stump]

great

Nice to meet you

[dreamy fox]

Same here!

[rough stump]

Can u tell me

[royal flint]

Ooh

[rough stump]

where minus go in above question when we solve tan (135)

[royal flint]

What grade

[rough stump]

12th

[royal flint]

Nice

I just finished 12th

[dreamy fox]

The - sign is multiplied inside the bracket

[royal flint]

Also buddy sleep it's like 1.30 here I am done with school so I can do anything 😂

[dreamy fox]

Same

[royal flint]

Nicee

[dreamy fox]

JEE aspirant??

Bear?

[royal flint]

Nah I am going to us

[dreamy fox]

Oh nice

[royal flint]

Wbu?

[rough stump]

I study late at Night it is 1:00 AM here

[dreamy fox]

Yeah prepared for JEE now going to Jadavpur University EE

[dreamy fox]

Us bro

[royal flint]

Nicee, good luck!

[royal flint]

No worries just take care of your health

12th was so stressful 😭

[rough stump]

Thanks

[royal flint]

Cbse?

[dreamy fox]

Real, lost my socializing skillz fr

[royal flint]

Ikr 💀

[dreamy fox]

Nahh west Bengal board

[royal flint]

Either way it's stressful 💀

[dreamy fox]

Which uni are u going bro

[stoic yacht]

!redir

[obtuse pebble]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[rough stump]

ok

[royal flint]

Carnegie Mellon

[royal flint]

Sorry

[dreamy fox]

Woahh 😭

[stoic yacht]

all good :)

[dreamy fox]

Ok we'll leave now

[rough stump]

Thanks guys

again

[royal flint]

Byee so nice talking to u

[rough stump]

You to bye

.clsoe

.close

[obtuse pebble]

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[onyx sentinel]

I am 🙏

[obtuse pebble]

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[lone bobcat]

wtf does this even mean

thats not even echelon is it

type 2 echelon matrix, is a regular upper triangle

with extra columns

and to say first and third colomns are basic, what?

[stoic yacht]

basic columns are sometimes called pivoted or leading

[lone bobcat]

pivoted

[stoic yacht]

as in, that column has a pivot in RREF

[lone bobcat]

meaning 0s infront

[stoic yacht]

pivoted columns have a nonzero entry in that position in RREF

the first pivot is the 2 in the first column, so the first column is pivoted

then stair-stepping down by moving one down and one to the right, there isn't a 0 there so the second column isn't pivoted

the next nonzero entry at that level is the 5 in the third column, so the third column is pivoted

[lone bobcat]

but shouldnt the 5, be a 1?

for echelon

[stoic yacht]

you could technically further simplify the second row by dividing by 5, yes

[lone bobcat]

so summarising

2 and 5 are the leading elements of their respective rows so those columns are the pivots

or basic columns

[stoic yacht]

you usually refer to the 2 and 5 as being the pivots

[lone bobcat]

hm

[stoic yacht]

and you say those columns are "pivoted" or "leading" or "basic"

[lone bobcat]

hm

k

[stoic yacht]

just a terminology thing

[lone bobcat]

2,5 pivots, basic columns

tyvm

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[tardy heart]

Hi I am confused why there are 2 cos^2

[tardy heart]

Where did this come from?

[latent walrus]

thats just equivalent to 1

theyre making a common denominator

[tardy heart]

How do we know though?

Like why isnt it sin over sin

[latent walrus]

because the denominator of the fraction is cos^2

[tardy heart]

Ohhh

[latent walrus]

you could do sin/sin sure but it wouldnt be useful in any way

[tardy heart]

They just want it to be same?

[latent walrus]

they need a common denom to combine the terms into a single fraction

[stoic yacht]

they have a/b + 1, and they want to write it as one fraction. by writing 1 as b/b, that becomes a/b + b/b = (a+b) / b

[tardy heart]

OHHHH

I SEEE

Thank you so much!

.close

[obtuse pebble]

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[tardy heart]

.reopen

[obtuse pebble]

✅

[tardy heart]

Sorry one more question

Why do we need to simplify this to begin with?

tan^2 looks simplified?

[latent walrus]

i wouldnt say you need to, its just what it wants you to do, having a single term is arguably simpler than multiple if possible

[tardy heart]

Oh ok thank you

I just have a hard time figuirng out if I need to simplify or not

[latent walrus]

unless it asked you to i wouldnt stress about it

[tardy heart]

Kk awesome

Thanks again

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[astral cypress]

I was studying for descrete mathematics(not really sure if it's the right name since im not a native english speaker). It's basically the base of the computer thinking, graphs, combinatorics, relations etc.. so not to prolong it too much the topic im confused about is combinatorics precisely combinations, for example question says you have a 5 digit number then, it has to be palindrom and the sum of all digits has to be 6, i have couple of setups im thinking about. I would split all possible forms of 5 digit number in "categories" so 1) the "shape" of the number should be aabaa 2) abbba etc.. the part im stuck at is the how do i set this up as a combination like n over k for combinations since aabaa can only be 1 number and it's 11211 but that's just using logic i want to know how to "mechanically" get to the answer coz i assume there won't be always logical answers or there might be just too many of them to do it this way. here's also a picture of how i was thinking

[astral cypress]

a

[cunning pebble]

honestly its probably the best way of doing that, its a very simple problem and theres no need to overthink it. Once you get to more difficult ones they will force you to produce elegant and efficient solutions, since if there are for example 10^9 combiantions, you cant just count them.

[astral cypress]

thanks for your time and replying, i'd just correct you i am not overthinking it i just really need to know the usual route for exactly those purposes because i do know for a fact there won't be such easy problems so im trying to set up this particular "easy" problem so i can learn from it

[cunning pebble]

oh and yeah the area of maths is called discrete mathematics

[astral cypress]

amazing

[cunning pebble]

its a good way of learning, but this problem might just be too easy to look for another approaches. I think any other approach here would have to start with setting some constraints on the numbers, which is what you did, and that basically gives the answer

Discrete maths is in my opinion very intuition based class, you need to practice attention to details, this is essential especially when it comes to combinations and stuff like that. Concepts arent that difficult(they get more difficult at higher level), but yeah theres a lot of room for silly mistakes or wrong assumptions.

[astral cypress]

i understand that but i just got to it several days ago because i couldn't follow the lectures from university because of private reasons so any info on anything will be helpful in passing this exam

the only reason im just a bit irritated about this is that i can't see that way, for example counting all palindrom 5 digit numbers i just wouldn't know how to set that up that's what i was asking in the first place since there are a ton of them i can't count them

.close

[obtuse pebble]

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[weary reef]

If you labelled the digits $a_1, a_2, ...a_5$

For a palindrome
$$ a_1 = a_5, a_2 = a_4 $$

The number of non-negative solutions to this equation

$$ 2a_1 + 2a_2 + a_3 = 6 $$

[warm shale]

StrangeQuarkAL

[weary reef]

0 <= a_1, a_2, a_3 <= 6

a_1 > 0

Still gotta use some logic to count it but it's slightly easy I'd say

[astral cypress]

thanks for replying, im a bit confused on how do i use these i did pull them out as seen on picture i just had no idea how to use them so i just left it there

im assuming my labling is fine as well?

[weary reef]

Yeah that's fine

[astral cypress]

i did get similar solution where it says 2a+2b+c=6 i just had no clue how do i use that

[weary reef]

My answer was more in response to "how to mechanically"

[astral cypress]

yeah and im seeing the pattern, in these im just not really getting what these equations are telling me and i feel like im missing out on a lot

[weary reef]

Well, all the variables are non-negative and definitely less than or equal to 6

[astral cypress]

so let me get this straight, not to bother you all night, is that equation able to tell me what the solutions are and if not what at least how many are there ?

like the exact numbers

[weary reef]

It should be possible but I can't say I'm familiar with the methods unfortunately

[astral cypress]

yeah yeah understandable, doesn't matter imma dig anyways thanks a lot for reaching out you did put me on a right track tho so i appreciate it a lot

[weary reef]

I usually just the equations to make it into a simpler exhaustive process

[obtuse pebble]

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[twilit bay]

theres a second part to this question\

b. Calculate an estimate of the lower boundary of the masses of the heaviest 50% of these animals.

[twilit bay]

a is 159 and b is 636

[twilit bay]

could someone explain to me how and why the answer to this is 23.5kg

like i first did 50% of 2226 which is 1113 so we know we get heaviest starting from number 1114

but afterwards then i calculated cumulative freq which showed me that 13-32 has 1590

but if we bring that in half to 13-22 and 23-32 which is a freq of 530 and 530 each

then we get to now 1060 which is even closer

then finally using that i got 23kg

but how do u get 23.5

[obtuse pebble]

@twilit bay Has your question been resolved?

[obtuse pebble]

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[tardy heart]

How do I solve this if the tangent of 270* is undefined?

[tardy heart]

Nvm it doesnt want me to solve

332

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[bold zinc]

is this saying $s = \sqrt{((R-r)^2}+h^2)$

[warm shale]

pixel

[short spire]

Well that extra parenthesis is unambiguously telling you it’s $s = \sqrt{(R-r)^2+h^2}$

[warm shale]

Aslan

[craggy osprey]

the root is all thw way

[bold zinc]

oh ok

thank you for clearing that up

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[full egret]

whenever i put the matrice into my calculator it doesn't give me an answer, what am i doing wrong?

[full egret]

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[drifting nova]

am i supposed to reject?

[drifting nova]

at 0.05

[obtuse pebble]

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[drifting nova]

.reopen

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✅

@drifting nova Has your question been resolved?

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[desert cipher]

Just making sure I’m right on 2. I’m not exactly sure how to do this so I could be wrong, but the RREF form of the matrix just doesn’t make sense to me if U is a linear combination of those vectors. Do I have this right?

[grizzled shore]

If your rref is correct then you’re right

[desert cipher]

Row 3 shows that the system is inconsistent right?

[grizzled shore]

An inconsistent system means there does not exist a, b, c ∈ ℝ such that v = av₁ + bv₂ + cv₃

[desert cipher]

We’re allowed to use a calculator for RREF so I just did that

[grizzled shore]

So you can’t make v out of a linear combination of v₁, v₂ and v₃

[desert cipher]

Yea

Alright

Just making sure I wasn’t wrong cause I’m still a little shaky

[grizzled shore]

It’s correct!

[desert cipher]

Thanks bro!

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[viral valley]

Hey! Trying to solve this problem regarding subspaces. It's from the "Linear Algebra Done Right" Textbook. I know how to prove if a set U is a subspace of a vector space V. There are 3 rules: 0 element in U, closed under addition, and closed under multiplication.
However, I am really struggling to understand what "set of continuous real-valued functions on the interval [0,1]" means, and what R^[0,1] means. I've been searching the internet to find an explanation and I can't.
In addition, I don't know how a "0 element" could exist in a set of functions. Also how can we be closed under multiplication if there are only functions and no scalars?

[stoic yacht]

iirc, R^[0,1] is notation for the set of all functions f: [0,1] -> R

[viral valley]

[0,1] is the domain (continuous interval)?

[stoic yacht]

yes

[viral valley]

would the function work on any number outside of [0,1]?

[stoic yacht]

its defined domain would be specifically that interval

for example, f(x) = 0 is an element of R^[0,1]

any value in [0,1] is in the domain and gets mapped to a real number

[viral valley]

ah okay awesome

[stoic yacht]

so clearly continuous functions f: [0,1] -> R is a subset of R^[0,1]

just a matter of showing it's a subspace, i think you can prove all three conditions pretty easily

adding continuous functions gives a continuous function and since R is closed under addition, it's still real valued

scaling a continuous function by a constant gives a continuous function and since R is closed under multiplication, it's real valued

[viral valley]

Ah

[stoic yacht]

as for what the zero element looks like in a set of functions

the zero element is the unique "vector" such that if it's added to any other vector, that vector's value is unchanged after the addition

[viral valley]

f(x) = 0?

[stoic yacht]

so my earlier example, f(x) = 0, is exactly your zero element

yeah

[viral valley]

Ohhh my gosh

[viral valley]

are they the same?

[stoic yacht]

are the two spaces the same?

[viral valley]

yes

[stoic yacht]

no

R^[0,1] is the set of all functions that map from [0,1] to R

but not every function is continuous

which means the space of all continuous functions mapping from [0,1] to R is a subset of that space

[viral valley]

oh thats true

oh this makes so much sense now

thank you so much!!!!

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[random grotto]

Hi, I'm studying matrices and I had a question. If the rows of a matrix are linearly independent, does that also menat the determinant of A is 0?

[random grotto]

Because I think its right in saying that if the columns of A are linearly dependent then the det of A is 0

[remote skiff]

Not as stated no

[random grotto]

So I was thinking from another angle

[random grotto]

May I ask why?

[gilded needle]

if the rows are LI (and it's a square matrix) then the det is nonzero

[stoic yacht]

determinant is 0 when there's linear dependence, otherwise nonzero

[random grotto]

Yea sorry I should mention that the matrix A is n * n

[stoic yacht]

this could be column-wise or row-wise linear dependence since det(A) = det(A^T)

[random grotto]

Ah sorry I see now

Okay so that means if the rows of A are linearly independent, then the detmerminant of A is not 0

I see, thank you all

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[obtuse pebble]

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[mint nebula]

Consider the square ABCD and the points K (AB), L (BC) and M (CD) such that the Triangle AKLM is
rectangular isosceles, with the right angle in L. prove that the right AT and DK are perpendicular.

[mint nebula]

I think that the problem is really easy, yet I'm missing on a theorem or lemma.

Here's the drawing.

[obtuse pebble]

@mint nebula Has your question been resolved?

[mint nebula]

<@&286206848099549185>

I've found that LC + AK = KL

If it helps

[obtuse pebble]

@mint nebula Has your question been resolved?

[mint nebula]

<@&286206848099549185>

[obtuse pebble]

@mint nebula Has your question been resolved?

[mint nebula]

<@&286206848099549185>

[warm sparrow]

theres not T in your drawing or in the explenation so AT is the T suppose to be something else?

[mint nebula]

I meant to say AL

not AT

[warm sparrow]

oke well lets see then

oke so to prove that those are pupandicalair you got to prove that DKL~ ALK

and for that you need those variables that you already put in a and x

[mint nebula]

are kbl and lcm congruent?

[warm sparrow]

well yes

but you got to give prove for that

[mint nebula]

we have ML = KL

and i think the agles LMC and KLB are congruent

[mint nebula]

@warm sparrow why do I need to prove that these triangles are similar

i don't think they even are similar

[mint nebula]

<@&286206848099549185>

[rocky galleon]

yes?

[mint nebula]

i need help with this problem Consider the square ABCD and the points K (AB), L (BC) and M (CD) such that the Triangle AKLM is
rectangular isosceles, with the right angle in L. prove that the right AL and DK are perpendicular.

[rocky galleon]

kk wait

[mint nebula]

alr

[tranquil sonnet]

You can just length bash the whole thing

Solve for x

Solve for relevant lengths, And use converse of Pythagorean

[mint nebula]

Okay

And it should end up to knowing the sides of AKLD

[mint nebula]

I dont think I have enough lenghts though

like enough given lenghts

[tranquil sonnet]

assume teh square has side length 1

[mint nebula]

i'd rather go on a synthetic solve

rather than calculus

but idk

[gilded flower]

Can i get help with something in this channel?

[tranquil sonnet]

no

[mint nebula]

no, make ur own

sorry

[gilded flower]

Hooooow

[tranquil sonnet]

yeah i would prefer that but the computation is direct, ill see if theres an elegant way

[mint nebula]

i'll do the same

[mint nebula]

i'm thinking ab what this guy said

yeah those are not similar

[tranquil sonnet]

nvm yeah synthetic is probably better

i dont think computation is good

[mint nebula]

right

and also this is a 7th grade national olympics problem

they probably don't know computations lol

[tranquil sonnet]

ok

so do you see why JK=MC?

[mint nebula]

where's J?

you mean L?

[tranquil sonnet]

sorry AK=MC

[mint nebula]

oh

yeah KB = DM=x

[tranquil sonnet]

yeah

[mint nebula]

so by difference AK = MC

[tranquil sonnet]

ok note that DAK is congruent to ABC

[mint nebula]

Right

[tranquil sonnet]

now you can just calculate the slope of the lines and multiply them to get -1

because if the product of the slopes is -1 they are perpendicular

[mint nebula]

wait one sec

i wanna go sythetic all the way lol

[tranquil sonnet]

theres an angle chase too if you want

ok

[mint nebula]

let me see smth

[tranquil sonnet]

let <AKD=x

then

focus on quadrilateral KBL(intersection)

what letter is labeled at the intersection? sorry i cant see

[mint nebula]

I didn't lable it. Let's say it's T

[tranquil sonnet]

Ok

[mint nebula]

so KBLT

[tranquil sonnet]

yeah look at its angles

<BKD=180-x, <KBL=90, <BLA=<AKD=x

[mint nebula]

B has 90, TKL has 180 - x -

[tranquil sonnet]

and the sum of all angles should be 360

yeah

and this gives <KTL=90

so they are perpendicular

[mint nebula]

amazing

that was a beautiful solve tbh

[tranquil sonnet]

i suggested slope because its also very easy

[mint nebula]

yeah true

[tranquil sonnet]

rise/run = slope

so

AD/AK * -BL/AB

using the congruent triangles it simplifies to -1

[mint nebula]

what's nice about geometry is synthetic solving, at least thats what i like

[tranquil sonnet]

good

[mint nebula]

but when you're absolutely out of ideas, you can do that too lol

[tranquil sonnet]

i love synthetic

when you learn more advanced geometry youll see the beauty of synthetic

its absolutely amazing

[mint nebula]

i'm looking forward to that

anyways, thank you and have a great day

[tranquil sonnet]

np

u2

[mint nebula]

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[young crypt]

can you make a formula for the distance between the point and the equation perhaps

[unkempt lotus]

Yeah d = sqrt((x-3)^2 + (y-0)^2)

[vale pelican]

who is y?

[unkempt lotus]

So y would equal sqrt(x)

[young crypt]

yes

[vale pelican]

we know who y is now lol

[unkempt lotus]

And x is ……. Something

Can u not skep steps

Skip*

[young crypt]

yea sorry mb

[unkempt lotus]

I don’t know how to get x

[young crypt]

thats fine we have an equation for the distance in terms of only x correct?

[unkempt lotus]

I kno x = x^1/2

Yeah true

My b

I was trying to get x in terms of y

But I forgot we just need the objective in terms of one variable

[young crypt]

so if we have an equation for d in terms of x, and we want to find the minium value of d what do we do

[unkempt lotus]

So I have sqrt(x^2-5x+9)

[young crypt]

yep

[unkempt lotus]

Which is (x^2-5x+9)^1/2

Then we take the derivative and get critical point

[young crypt]

exactly

[unkempt lotus]

So I need chain rule

[young crypt]

yea

[unkempt lotus]

Is that just bringing the (x) to 1/x and putting. x’ on top?

Like x’/2x?

And x is the (x^2-5x+9)

[icy mirage]

sqrt(x^2 + x - 6sqrt(x) + 9)

[unkempt lotus]

Wut

[young crypt]

i got sqrt(x^2-5x+9)

[unkempt lotus]

They r trolling

<@&268886789983436800>

[young crypt]

you need to decrease the power still

[unkempt lotus]

<@&268886789983436800>

[dapper bloom]

Don't troll here.

[icy mirage]

Sorry, my bad. Read (0,3) instead of (3,0)

[unkempt lotus]

Oh yea

Ok let me try that again

Now solve for d’(x)=0

[young crypt]

yes

[unkempt lotus]

And I can just multiply both side by denominator. So it is just 2x-5 = 0

[young crypt]

yea or you can think of it like you cant divide by zero so the only way for the right hand side to be zero is for the top of the fraction to be zero

[unkempt lotus]

So i get x is 5/2

[young crypt]

correct

[unkempt lotus]

Ok now get d(5/2)

[young crypt]

we dont need the distance just the y coordinate now

[unkempt lotus]

O

So I just plug it into the constraint

Y = sqrt(5/2)

[young crypt]

yes the original equation

yep

[unkempt lotus]

Ok cool

Thx

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[urban topaz]

Hey can someone double check for me. I got 20 for my answer but the website says im wrong

[frosty crypt]

perhaps it's supposed to be 2,000 because x is in 100 dollars

because x=20 is correct

[urban topaz]

it was 2000 😭

thank you

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[karmic hedge]

Where was my sign error?

[karmic hedge]

sin^3(x)cos^2(x)

sin^2(x)sin(x)cos^2(x)

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

(1-u^2)(u^2)

u^2-u^4

[haughty coyote]

you missed the - sign

[karmic hedge]

Where?

[wooden cipher]

what is du?

[high lily]

forgot about the conversion of dx to du

[karmic hedge]

u = cos(x)

du = sin(x)

[wooden cipher]

youre missing a dx

also are you sure about that

derivative of cos

[karmic hedge]

oh

Youre right

Mb I need a break I have been studying for too long 😂

Thank you everyone

❤️

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[glass pilot]

Can someone please help me with this problem?

[smoky vigil]

!status

[obtuse pebble]

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

[smoky vigil]

!showwork

[obtuse pebble]

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

[glass pilot]

1 or 2, I know the formula, but I don't know how to solve the integral from there

I checked calculators online, but they show up as a complex conjugate

[smoky vigil]

can you draw a picture

[obtuse pebble]

@glass pilot Has your question been resolved?

[glass pilot]

What would the picture depict?

[smoky vigil]

just a picture of the area, normal integral

[smoky vigil]

oh

well, show the integral you've set up

[obtuse pebble]

@glass pilot Has your question been resolved?

[glass pilot]

Sorry for the late response I was travelling

[smoky vigil]

you can distribute, and split this into two integrals

both can be solved with trig sub

@glass pilot

[glass pilot]

How do I do that?

[obtuse pebble]

@glass pilot Has your question been resolved?

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[brisk arrow]

My result seems a bit different from the answer?

Is it wrong?

[narrow fiber]

No it's correct

It just needs to be simplified

[brisk arrow]

How

[deft magnet]

(6*(2^-1)*3)^1/6
=(6*1/2*3)^1/6
=9^1/6
=3^2^1/6
=3^1/3

=(thirdroot)3

[brisk arrow]

I see

[brisk arrow]

Go study

[deft magnet]

ty

[obtuse pebble]

@brisk arrow Has your question been resolved?

[obtuse pebble]

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[lilac bloom]

Can someone explain to me why $\sum\limits_{n=1}^{\infty}2\sin(\frac{3}{n})$ diverges? $n^\mathrm{th}$-term test is inconclusive here, I can't apply geometric tests, I can't apply power series tests, and we're not super far into series so I don't know how sine is described as a series. My gut instinct tells me that it does diverge since it oscillates, and i know that it does diverge anyways, but I don't know exactly why or how to prove it.

[warm shale]

lILi

[deft magnet]

sin(3/n) approaches sin(0) = 1?

[lilac bloom]

sin(0)=0

[deft magnet]

o

mb

not thinking

💀

[lilac bloom]

if it were 1 then the nth term test would have told me it diverges and it's the first thing i checked

[bronze mica]

when x is small, sinx is approximately x, so in particular greater than x/2

so for sufficiently large n the terms of the series are greater than the terms of a harmonic series

[wooden cipher]

you can do limit comparison with 3/n

[deft magnet]

m

[lilac bloom]

oh

interesting

[bronze mica]

if you've not seen a formal definition of sine yet, I assume just stating that this is true will be fine in terms of rigour

[lilac bloom]

so, since $2\sin(\frac{3}{n})>\frac{3}{n}$ for sufficiently large values of $n$, the comparison test still applies?

[warm shale]

lILi

[wooden cipher]

in calc, its common for people to be given lim x->0 sin(x)/x = 1

[lilac bloom]

also I realise now that $2\sin(\frac{3}{n})$ doesn't oscillate on $n\geq1$

[warm shale]

lILi

[bronze mica]

it changes direction once actually

it doesn't oscillate for n > 1

[lilac bloom]

I see

One more question

How would i prove $2\sin(\frac{3}{n})>0,;1<n<\infty$

[warm shale]

lILi

[bronze mica]

also yeah comparison test applies, because you can rewrite the series as (finitely many terms where the inequality doesn't work) + (still an infinite series where it does which is what you actually apply the test to)

[bronze mica]

0 < 3/n < pi?

[lilac bloom]

Oh

fair

Alright, thanks for clearing that up for me

That makes sense now

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[wraith barn]

6a

[wraith barn]

i thought itd be like AB = AO - BO, but ig not

nvm

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[timid silo]

What do i do with the root

[shut lagoon]

Where did the bottom factor go?

[timid silo]

? Wdym

[shut lagoon]

You just multiplied the numerator by 1+sqrt(sin^2(x))

$\frac{a}{b}\cdot \frac{c}{c} = \frac{ac}{bc}$

[warm shale]

Azyrashacorki

[timid silo]

Oh..

Right

[shut lagoon]

That should help, along with the identity $\sin^2(x) + \cos^2(x) = 1$

[warm shale]

Azyrashacorki

[timid silo]

Hm, what do i do next?

[shut lagoon]

Well the point isn't to cancel them out, otherwise you just did..nothing

Leave the factors there and expand the denominator.

[timid silo]

Ohh

[minor furnace]

I think the right answer isn't shown in the answer choices..?

[timid silo]

Might be

[high lily]

as is, you can just direct sub
(but that doesn't lead to any of the answer options)

do you have the intended answer?

[timid silo]

No..

[shut lagoon]

Oop didn't even think of plugging in 😅

[minor furnace]

it's supposed to be 1 + sqrt(3) / 2

it's not on there

[high lily]

I suspect they may have intended
x to pi/2 (instead of pi/3)

[timid silo]

Hmm

[minor furnace]

I would take that up with the instructor if applicable

[timid silo]

Tysmm yall

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[rose ferry]

hey could someone show me the equation to this?

[stoic yacht]

they've actually given you the expression for the mass of the isotope over time

0.9174^t

just a matter of seeing when that equals 0.2

[obtuse pebble]

@rose ferry Has your question been resolved?

[obtuse pebble]

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[near cairn]

Does Jordan’s Lemma hold true for any $f(z)$or is it only for functions of the form $f(z)= e^{izt}g(z)$? I’m seeing conflicting results online\
I’m looking within a textbook and saw a video which both stated $f(z) = e^{izt}g(z)$ but in a different video I’m looking at, the person just writes a generic $f(z)$\
Did the person make a mistake here or is the $e^{izt}$ just kinda implied here in a $\frac{a}{a}$ kinda way?

[warm shale]

KySquared

[brazen viper]

So you generally use Jordan's Lemma to find the values of a path integral around a particular semicircular arc. The fact that it goes around the arc is where the exponential comes from, it's part of the parametrization of the curve.

So your function f(z) doesn't, in the general case, contain e^(izt), it's just that it picks up the e^(izt) because that's how we walk around a circular arc on the complex plane.

[near cairn]

Wait how?
I understand rewriting z as Re^(it) but its not like that can get factored out…?

Wait

Are you considering the e^(izt) as getting factored out?

[brazen viper]

It's not that it gets factored out

Ok, so starting from the beginning. Let's take an ancient problem I solved once.

https://math.stackexchange.com/questions/3735801/easier-approach-to-int-0-infty-frac-mathrme-x-cosh2x-51-math

So this contains a detailed work up of a particular integral

The complex analysis started at "our contour"

So the idea is, I want to evaluate the integral on 0 to infinity. But I only know how to evaluate it on a closed contour because of the residue theorem, right?

[near cairn]

Mhmm

So make a contour which includes the region you want

That much I get

[brazen viper]

So we set up a contour from -R to R along the real axis, and then a semicircular part with radius R

Our goal is to show that the semicircular part vanishes

[near cairn]

Agreed

Everytime I tried so far I would manually show it goes to 0 by expanding and pulling out Re^(i*theta)’s

[brazen viper]

$$
\begin{align*}
\oint_C \frac{z^{-2/5}}{1 + z^2} , \mathrm{d}z &= \int_{-R}^0 \frac{z^{-2/5}}{1 + z^2} , \mathrm{d}z + \int_0^{R} \frac{z^{-2/5}}{1 + z^2} , \mathrm{d}z + \int_0^{\pi} \frac{{\left(R \mathrm{e}^{i \phi}\right)}^{-2/5}}{1 + {\left(R \mathrm{e}^{i \phi}\right)}^2} , iR\mathrm{e}^{i \phi} , \mathrm{d}\phi \
\lim_{R \rightarrow \infty} \int_0^{\pi} \frac{{\left(R \mathrm{e}^{i \phi}\right)}^{-2/5}}{1 + {\left(R \mathrm{e}^{i \phi}\right)}^2} , iR\mathrm{e}^{i \phi} , \mathrm{d}\phi &= 0 \
\oint_{C, R \rightarrow \infty} \frac{z^{-2/5}}{1 + z^2} , \mathrm{d}z &= \int_{-\infty}^0 \frac{z^{-2/5}}{1 + z^2} , \mathrm{d}z + \int_0^{\infty} \frac{z^{-2/5}}{1 + z^2} , \mathrm{d}z \
\int_{-\infty}^0 \frac{z^{-2/5}}{1 + z^2} , \mathrm{d}z &= - \int_0^{\infty} \frac{(-z)^{-2/5}}{1 + (-z)^2} , \mathrm{d}(-z) \
\oint_{C, R \rightarrow \infty} \frac{z^{-2/5}}{1 + z^2} , \mathrm{d}z &= \left(1 + \mathrm{e}^{-2\pi i/5}\right) \int_0^{\infty} \frac{z^{-2/5}}{1 + z^2} , \mathrm{d}z \
\end{align*}
$$

[warm shale]

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

[near cairn]

And then explicitly showing it goes to 0 like that

[brazen viper]

Oh it was cut off, how sad

So you notice that I have a few copies of this integral with minor changes

[near cairn]

I don’t wanna do that each and every time and would prefer to just say “its 0 by jordan’s lemma”

[brazen viper]

But only one copy has the Re^(iφ)

[near cairn]

Mhmm

[brazen viper]

Right, and that's fine

[near cairn]

You switched back to writing it in terms of z for sone reason

[brazen viper]

Because it's pretty well understood that's what you're doing, if you're just doing applications

In the linked post, I had the linear integrals written as a function of z and the semicircular arc as a function of φ.

[near cairn]

Is that the f(z)e^(ikz) part?

[brazen viper]

Of course, it doesn't matter what variables you use

But the intention was there

[near cairn]

Mhmm

[brazen viper]

The point is that Jordan's Lemma only applies to arcs that limit to infinity

[near cairn]

Right, im not worried about the limits

[brazen viper]

And these will have a term of e^(iφ)

[near cairn]

Its specifically the functional part

[brazen viper]

Because they are circular arcs

[near cairn]

Okay I think I get it

[brazen viper]

So your original function can be anything, f(z), but the circular arc part of the integral will be Re^(iφ) f(z) because that's the nature of arc integrals

[near cairn]

Because $\int_c f(z) dz$ can be represented as $\int_{0}^{\pi} f(Re^{i\theta}) iRe^{i\theta} d\theta$ that is the functional representation that is satisfied for Jordan’s lemma because of the arc?

[brazen viper]

Only the semicircular arc portion of the contour integral

It's not the entire contour normally

[near cairn]

Oh I should’ve just wrote it as a curve not a closed loop

But you get the idea

[warm shale]

KySquared

[brazen viper]

Yeah, I see what you're trying to say now

[near cairn]

There we go

[brazen viper]

Nice grookey btw

[near cairn]

Thank you
Nice to see a fellow pokemon fan

Do you like competitive?

[brazen viper]

I... haven't actually played any game since Red Blue. >_>

[near cairn]

[brazen viper]

I pay a little bit of attention to competitive

Like YouTube stuff

I mostly just play poketwo, the discord game

[near cairn]

I played alot of comp a few years back but mostly stopped bc of school

I was kinda losing interest regardless so I doubt I’ll get back to it

Who knows

[near cairn]

Gotcha

[brazen viper]

Kinda wanted to try to make a showdown AI

[near cairn]

Most the time I spend w/ pokemon nowadays is shiny hunting

[near cairn]

Ooo that sounds cool

[brazen viper]

But haven't really sat down and figured out how to make that happen yet.

Like, a random battles NN AI would be sick

[near cairn]

I know the very very basics of ML but if you’d like any sorta input I could try and throw my hat in the ring

Wouldn’t mind learning more ML
Especially since I considered going into data science lol

[brazen viper]

Anyway this is off topic. If you want discuss this we can jump over to #discussion or something. But I don't have anything smart to say about it really lol. I have done ML before, but not specifically reinforcement learning before.

[obtuse pebble]

@near cairn Has your question been resolved?

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[glossy pike]

i dont get why the correct answer is A not B. Is there flaws in my works?

[glossy pike]

<@&286206848099549185>

[rancid shadow]

if you get g(x) from f(x), you must move fx along x-axis -2, multiply 5 along and move -1 along y

that's the reason correct one is A

[glossy pike]

that sounds kinda counter intuitive, does it have to do with the "operation in parenthesis is always opposite of what you see"

[rancid shadow]

just in the case of x axis

[glossy pike]

yeah

[rancid shadow]

let's assume that fx = 1/x and gx = 1/(x-1)

the corresponding answer is (x+1, y)

[glossy pike]

got it, thank you!

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[marble robin]

how do i do this question im really confused !?

[marble robin]

<@&286206848099549185>

please help

[young crypt]

well what’s the formula for speed

[icy oar]

Well, the distance he ran is 450m

[marble robin]

speed = distance by time

[icy oar]

Hmm

[young crypt]

exactly so how far has he travelled

[marble robin]

but theres no time

[marble robin]

450

m

[young crypt]

you can work out the time

[marble robin]

righ

right

ok got it thanks

[young crypt]

cause you have the speed for some sections and the distance travelled

[marble robin]

yes

[obtuse pebble]

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[quaint bluff]

hey, how do i find the intersection points between these 2 functions?

[gilded needle]

the most straightforward way is probably to consider the various cases, where 2x-4 >= 0 vs 2x-4 < 0, and similarly for 6-x

[quaint bluff]

and how do i proceed?

[gilded needle]

well 2x-4 changes sign at x=2, and 6-x changes sign at x=6
so you want to consider these intervals:
(-infty, 2)
(2,6)
(6,infty)

in each of those intervals, you can get rid of the abs vals and apply the appropriate sign

for example:

in (-infty,2) you have 2x-4 < 0 and 6-x > 0, so |2x-4| becomes -(2x-4) and |6-x| becomes 6-x

and similarly for the other two intervals

[quaint bluff]

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[sonic bolt]

https://imgur.com/HBJasyl

[sonic bolt]

does anyone have any idea what the answer would be to this solution?

nothing on the graph matches, quadratic doesn't give me a real number