help-23

[teal kite]

Yah, tbh(not being elitist) but math olympiads in my country is not really the focus compared to country like china, us, etc, they basically just give like the simples of simple. So actually its not really part of some curriculums

[dapper star]

but I just got an idea

[teal kite]

Ohhh

[dapper star]

I will be back

[teal kite]

[dapper star]

so

this can maybe work?

wait

I will come back with solution

[teal kite]

No worries

[dapper star]

so I thought of

making three equations

with three variables

using pythagorean theorem

[teal kite]

Woah

[dapper star]

so Im getting IJ ~= 163 smth

163.7

so Put IJ as z
QJ as x and
KJ as y

and then

equations would be

(163+x)^2 = 156^2 + z^2

PIJ ^

260^2 = (163+x)^2 + y^2

PJK^

QJK :- 143^2 = x^2 + y^2

also I gotta sleep

@teal kite

you should ask for answers

[teal kite]

Are you southeast asian?

[dapper star]

yea

[teal kite]

Same

[dapper star]

India

[teal kite]

Oh

I'm much more earlier(PH)

It's 2 here

Okay, i'll just ask

THANKS THOUGH

.close

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[lunar prawn]

Excuse me, how can I convert radians and degrees to points that perhaps may not be on a unit circle?

[fathom adder]

If you have rad you multiply by 180/pi to get degree

[lunar prawn]

ok

but how do i get the points then? like for example (1/square root 2)

[fathom adder]

But the may not be on a circle makes me doubt

[lunar prawn]

hmmm

what to do then?

[fathom adder]

You want what points ?

[lunar prawn]

for example lets say (1/square root 2, - 2/3)

is it possible?

another question I had is that for each angle, lets say 135 degrees, why does it have the points of (- square root 2/2,square root 2/2 )? why not any other point?

sorry im new to radians, etc so I had trouble mainly with point and the standarm position etc

[fathom adder]

Cos(135) and sin(135)?

Ok ic do you know polar ?

Polar coordinates ?

[fathom adder]

It depends of the distance from the center

For the one you cite its distance from the center = 1

You just need to change this distance

[lunar prawn]

hmmm

[fathom adder]

Ila show

[fathom adder]

We call the length from the center
r = sqrt(x^2 + y^2)

[lunar prawn]

cos135 is -0.99608783514 and sin135 is 0.0883686861. or do you perhaps mean archcos and atchsin?

[fathom adder]

No in rad

[lunar prawn]

i see

[fathom adder]

Cos (3pi/4)?

[lunar prawn]

true i forgot

[lunar prawn]

lets see........

[lunar prawn]

do you want the answer in degrees or points?

[fathom adder]

Value

[lunar prawn]

hmm

lets see

i dont know tbh

[fathom adder]

Lets get back to this

So r = sqrt(1/2 + 4/9) = sqrt(9/18 + 8/18) = sqrt (17/18)= sqrt(34)/6

[lunar prawn]

hmm

where did r = sqrt(1/2 + 4/9) come from though? I apologzie if my questions may appear a bit baisc and irritating

[lunar prawn]

did it come from here?

[fathom adder]

It is pythagore in fact

[lunar prawn]

oh my

this will be helpful very much

[fathom adder]

Your question is all about this

[lunar prawn]

ah yes indeed

[lunar prawn]

but even so, my y was 2/3, so w\here did 4/9 come from?

[fathom adder]

Its squared

(2/3)^2 = 4/9

[lunar prawn]

oh true

but in that case

why wasnt 1/2 also squared? although the radical is no more, shouldnt the 2 be a 4?

[fathom adder]

No it was 1/sqrt2

So squared it gives 1/2

[lunar prawn]

ah i see

go on please

[fathom adder]

Then for the angles in rad you take arctan(y/x) the answer is not pretty but its how to it

[lunar prawn]

take your time

[lunar prawn]

i see......

so if I was give 11pi/4 for example, would I use cos or sin for it?

so cos(11pi/4)

[fathom adder]

Both

[lunar prawn]

or sin(11pi/4) correct?

i see

then in what case would I use tangent?

[fathom adder]

In the case where you dont have the angle

[lunar prawn]

i understand

[fathom adder]

To find it

With x and y

[lunar prawn]

but if I were give a degree, would I also use sin or cos?

to find the points

[fathom adder]

Degrees are not used in trig, we use rad so ig convert into radians first

[lunar prawn]

and if I were given the points, would I use sin and cos also?

[lunar prawn]

I see. BUt for example what I mean is that pi/2 = 90 degrees for instance. so if I were given an angle, would I change it into a radian first and then use sin and cos after?

[fathom adder]

Yeah its better to go in radian

[lunar prawn]

i see

[lunar prawn]

the for my last annoying questions what about this: ..

[fathom adder]

What you mean by use cos and sin ?

[lunar prawn]

would i do this: sin(point x, point y) and cos ""? just how i did to the radians?

[fathom adder]

Nah you have the point (1,1) u find that r = sqrt(2) so
1 = sqrt(2)*cos(theta)
1 = sqrt(2)*sin(theta)

[lunar prawn]

so in theta, do i write the points in its place??

[fathom adder]

You solve for theta and you have the angle

So theta = pi/4

[fathom adder]

In a graph, that lead to take the angle (pi/4, 45°) and you use a ruler to draw a segment of length r = sqrt(2) , the end of the segment will be your points

[lunar prawn]

so for points (1,1) shown above, if the points were the 1/2,1/3, would I use 1/2 = sqrt(2)sin(theta) and 1/3 = sqrt(2)sin(theta)

?

[lunar prawn]

i see

[fathom adder]

Just change sqrt2 by r since its not the same lentgh from the center as (1,1)

[lunar prawn]

ok

[fathom adder]

All good ?

[lunar prawn]

all good!

thank you Yaku-senpai or should I say

[fathom adder]

Polar really interesting and im sure there is many video about this subject

[fathom adder]

Nah you can just say yaku

[lunar prawn]

sure Mr. Yaku

[lunar prawn]

indeed. However alot of videos dont answer my questions at all

thank you very much Mr. Yaku for all your help

have a nice day, evening, morning, or night in which is your case!

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[hardy vapor]

Let f be a function from R->C of period 2pi, such that f can be integrated over [0,2pi].
We then define the complex Fourier coefficients as

[hardy vapor]

and the nth Fourier polynomial of f

then our lecture goes on to explain that we can express these as

which is supposedly always two sequences of real numbers for any of real-valued functions of the given properties

... why?

I mean I can see the right part there obviously is real, it's a pretty simple integral... but why does the equation c_k+c_{-k} = 1/pi integral... hold (or the one below? why is i(c_k-c_{-k})=... ?)

and why, for that matter, can I write the sum over c_ke_k (ie. the Fourier polynomial) in this form?

Also what does any of this even describe, considering that

[flat frigate]

Bob Goldham

[hardy vapor]

is not guaranteed to exist

and if it exists, it does not necessarily approach f

so what are we even doing with any of this?

[frank horizon]

well essentially all of this is saying that periodic functions can be written as infinite sums of sins and cosins

[hardy vapor]

but we're not writing the function as a sum

because this doesn't even approach the original function in the first place, apparently

[frank horizon]

some functions don't have valid fourier decompositions

but other times, functions are uniformly approximated by their fourier expansions

[hardy vapor]

so this is basically fortune telling and hoping that you were right?

I can define a series expansion that's sometimes right. Why is this particular one any more useful?

[frank horizon]

no, there are conditions under which the fourier series converges uniformly, namely if f is continuous and the sum of fourier coefficients converges absolutely

you can think of the fourier expansion as another taylor expansion

[hardy vapor]

okay so if (|c_k|)_{k in N} converges and f is continuous then

[hardy vapor]

this=f(x)?

[frank horizon]

yeah

or rather the sum of the fourier coefficients i believe

so $\sum |c_k| < \infty$

[flat frigate]

Bair

[hardy vapor]

is that the same as saying the series converges absolutely?

should be

okay I can see how that's useful

[frank horizon]

err do you mean sequence? yeah this is the saying that the series converging absolutely

[hardy vapor]

that answers 1 question, leaving the 2 up here

[frank horizon]

well it's really just because $e^{i\theta}=\cos\theta + i\sin\theta$

[flat frigate]

Bair

[frank horizon]

and a little algebra

[hardy vapor]

oh so we're just applying Eulers identity to the e^ikx part and then splitting it into two different factors in a way that has the construction working as we want it to? Makes sense

[frank horizon]

yeah in essence

[hardy vapor]

That leaves the question of why a_k and b_k are always real numbers for real-valued f specifically?

[frank horizon]

well $a_k = \frac{1}{\pi}\int^{2\pi}_0 f(x)\cos kx dx$ which is necessarily real

same for b_k

[flat frigate]

Bair

[frank horizon]

there's a 1/pi there i forgot

[hardy vapor]

why can't this hold for f that have non-real values?

Or is it a case of "it can, but it won't always"?

[frank horizon]

yes

[hardy vapor]

Alright... I think that answers my questions for now. Thanks for helping!

.close

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[dark bobcat]

hello

[dark bobcat]

can someone tell me if i did these questions right

these are the questions

<@&286206848099549185>

[safe radish]

@dark bobcat Has your question been resolved?

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[unkempt thicket]

What did I do wrong?

[inland flicker]

is the purpose to show that the trig identity is true?

[unkempt thicket]

Yes

To prove it’s true

And CSC squared is equal to 1 over sin squared so I replaced these with the appropriate replacements

Yet one side of the equation is addition and the other is multiplication so it’s not true

[inland flicker]

if I recall, when I learned this, our teacher said that over time we will begin to recognize pathways to solving questions like these

with that being said, personally I would try to manipulate the left hand side of the equation

[unkempt thicket]

We just got a chart lol

[inland flicker]

not saying that starting with the right side is wrong, but I think it takes a much larger gap in understanding

I suggest trying to combine the fractions on the left side by giving them the same denominator

[unkempt thicket]

How’d you manipulate the left side? Wouldn’t that be the same? Just changing it to CSC squared plus SEC squared

[unkempt thicket]

Ohh! That’s probably what we needed to do I’ll try that

[inland flicker]

come back if it doesnt immediately pop out at you

[lime dust]

Remember that sin^2x + cos^2x = 1

[safe radish]

@unkempt thicket Has your question been resolved?

[unkempt thicket]

Working on it lol

[unkempt thicket]

Ohh right!

Do I just replace this entire thing by 1 over

[inland flicker]

Yes

[flat frigate]

Savage_Cat

[inland flicker]

So uhhh

[unkempt thicket]

Now I’m stuck🤣

[inland flicker]

Yes

Precisely

Oh

Wait

[inland flicker]

These two are equivalent

[unkempt thicket]

Yes I replaced it with 1

That was the left side of the equation

[inland flicker]

Oh yes i see I was confused by which step followed which

I understand now

[unkempt thicket]

Sorry haha

Oo wait if I just flip the fraction won’t I get the reciprocal functions??

What’s L

[inland flicker]

Where tf did that L

[unkempt thicket]

🤣🤣

[inland flicker]

It’s supposed to be a multiplication sign

[flat frigate]

Savage_Cat

[unkempt thicket]

Wait but can’t I just flip the fraction

[inland flicker]

You can, whichever you prefer

[unkempt thicket]

And flipping it would cause cos to become sec and sin to become csc right?

[inland flicker]

Yes

[unkempt thicket]

Let’s goo, thank you!

[flat frigate]

Savage_Cat

[inland flicker]

As desired.

[unkempt thicket]

Why do you not get cos 1/2x when taking the square root?

[lime dust]

Sqrt(cos^2x) is not cosx

[unkempt thicket]

It’s not?? But the answer came out correct

[lime dust]

No

[unkempt thicket]

That’s how he taught us to do it lol

[lime dust]

It is |cosx|

[unkempt thicket]

Absolute value?

[lime dust]

Yes

Because it is squared

For example sqrt(x^2) = 1

You can say here that x can be 1 or -1 because

[unkempt thicket]

Ohh

He never told us that, he just said that they cancel each other out

And you are left with x

[lime dust]

Sqrt((1)^2) is 1 but sqrt((-1)^2) is also 1

Not true

There are two different things

[unkempt thicket]

Wait so that applies to functions as well?

[lime dust]

First

(Sqrt(x))^2

This is x

But

Sqrt(x^2) is |x|

You can see with numbers

(Sqrt(4))^2

This would be 2^2

But now lets say

Sqrt(2^2) or sqrt((-2)^2)

Both are 4

But in the first case u cant

(Sqrt(-4))^2

This would be (2i)^2

Which is -4

[unkempt thicket]

Ohhh I was gonna ask what if it’s a negative haha

So they do kinda cancel out no?

Technically it more in depth

But end result is the same

[lime dust]

So when u have

Sqrt(cos^2x)

The sqrt is outside and the squared inside

So that would be |cosx|

And from there u get two solutions

Cosx = sqrt3/2 and cosx = -sqrt3/2

[unkempt thicket]

Hmm.. he told us that this would be the case only if the x is squared

We have that as a separate topic

Quadradic function equations

[lime dust]

Cosine is a function applied to x

Cos(x)

Is something that depends on x

His value is a number

[unkempt thicket]

So wait why is the cos squared and not the x

[lime dust]

This is another way of writing it so u can see clearly

Cos^2(x) = (cosx)^2

[unkempt thicket]

So cos^2x=cosx^2

Oh nice haha

[lime dust]

But not the one u said

The one u said cos(x^2) is different

[unkempt thicket]

How so?

Wait so the exponent does apply to both??

[lime dust]

Ok so take cosine like the function ok?

We insert a number to that function

For example 0

Cos(0)

What is this?

[unkempt thicket]

0 I assume

Oh 1

Apparently

[lime dust]

Ok so

If we put another number

[unkempt thicket]

What does cos do

[lime dust]

We will get a different answer right?

[unkempt thicket]

Like what does the actual function do

Maybe that’ll help me understand

[lime dust]

So u have to have in mind that cos is a function where

If u insert a number

U get an outcome

[unkempt thicket]

But what’s the process from the input to the output?

[lime dust]

For 0 is 1 for others is a different outcome

[unkempt thicket]

Why does 0 become 1

[lime dust]

That you learnt for sure in trigonometry

Right?

[unkempt thicket]

Nope

[lime dust]

With triangles

[unkempt thicket]

Right side triangle?

Wait it has to do with the sides of the triangle?

How can a side even equal to 0 in the first place

Length can’t be 0

We only learned right side triangles idk about other ones

Leg leg and hyp

So opp adj and hyp

[lime dust]

Aha

And if angle is 0 grades

[unkempt thicket]

How does that make sense

[lime dust]

That would mean they are equal

[unkempt thicket]

An angle can be 0??

[lime dust]

It cannot in a triangle, you have to imagine how you make the angle

Smaller and smaller

And u will see how adjacent side

Is more and more equal to hypotenuse size

So when u reach 0

They are equal

[unkempt thicket]

So cos doesn’t only apply to right side triangle?

[unkempt thicket]

So wouldn’t that make hyp=0 as well

Why is it 1 then

[lime dust]

No

[gritty kindle]

its not what happens when the angle is zero, but what happens as you approach zero!

[lime dust]

The sides have a length

[unkempt thicket]

So it can never be zero only approach it,

[lime dust]

I think i said that

[unkempt thicket]

?

So how does the calculator not show an error message when you do cos(0)

[lime dust]

But lets focus first in our initial problem

[unkempt thicket]

If it can’t be that

My initial question btw was why do you take the square root of the cos but not of the x,
But this is interesting as well 🤣

[lime dust]

Ok so the thing is you have a function where u insert a value and this function gives you an answer

[gritty kindle]

you can plug zero, but usually, thinking about what happens as you approach zero can be useful, maybe not in this one but in many other areas

[lime dust]

We called that function cosine

And we can input numbers there

When u write cos^2(x)

This means (cos(x))^2

You square the whole function

Not x

You square the output not the input

So when you square root that output, because it is squared, you get the absolute value

[unkempt thicket]

So technically you could ignore the square until you get the output and then square the output?

[lime dust]

So sqrt(cos^2(x)) = |cos(x)|

[lime dust]

You can say like that

Is like you say f(x) = x^2 and u ignore the ^2 until u input the x

[unkempt thicket]

Hmm.. so if I had cos^2(60) would that be 1/4

[lime dust]

Yes

Assuming 60 degrees

[unkempt thicket]

I see that’s way easier to look at it this way

Just taking the square afterwards

Instead of of the function itself

So that applies to square root as well?
You take that of the output?

[lime dust]

What would be cos^2(120)?

[unkempt thicket]

Also 1/4?

[lime dust]

Yes

[unkempt thicket]

I don’t like that🤣

[lime dust]

[unkempt thicket]

Wait so what does cos do to 120 to become 1/2

Is it division

Multiplication

Like what’s the process the calculator does

[lime dust]

No cos(120) is -1/2

[unkempt thicket]

Negative yes

My bad

[lime dust]

Anyways

I recommend you to study the definition of absolute value

The classic and the one i told u

[unkempt thicket]

We studied it but we only touched it when we did absolute value graphs

[lime dust]

About sqrt(x^2)

[unkempt thicket]

Y=|x|

[unkempt thicket]

I’m still interested to know that too tho lol

[lime dust]

|x| and sqrt(x^2) is the same

[unkempt thicket]

If you get |x| afterwards how come
Y=|x|
And
Y=x^2
Don’t look the same

[lime dust]

I did not write that

[unkempt thicket]

Cuz you’d have the take the square root of x^2 giving you Y=|x|

So aren’t they supposed to look the same

[lime dust]

I think I didn’t follow your last question

[unkempt thicket]

The graphs of these two looks different

But when you solve the x squared one you said you get the absolute value of x

So how come the graphs look different if the answer is the same

[lime dust]

Not x^2

Sqrt(x^2)

,w plot sqrt(x^2)

[flat frigate]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by _arkimond_

[lime dust]

,w plot |x|

[unkempt thicket]

OHH!

[flat frigate]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by _arkimond_

[lime dust]

They are exactly the same

[unkempt thicket]

Okay my brain can relax now 🤣

Thank you!

[safe radish]

@unkempt thicket Has your question been resolved?

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[quaint aspen]

Is this correct was ask to find maclaurin series?

[quaint aspen]

.close

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[quaint aspen]

.reopen

[safe radish]

✅

[quaint aspen]

this

[safe radish]

@quaint aspen Has your question been resolved?

[quaint aspen]

<@&286206848099549185>

.close

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[strong cosmos]

Hey, what does the numerator (x + 2) mean here

the denominator is the symptote

[safe radish]

@strong cosmos Has your question been resolved?

[strong cosmos]

also 1 more question here

[strong cosmos]

if x is 1, is y 1? if x is 10, what is y, how did y get all the way up there

or is the line drawns after the asymptotes are found? The Vertical-Asymptote is the denominator and the Y Asymptote is the answer of the fraction?

[strong cosmos]

And if x = 2 ( vertical asymptote ) and y = 1 ( horizontal asymptote ) where did the bottom left come from , the line there, theres no asymptote at -1 xand - 2 y

Thats all my questions

[strong cosmos]

1st is here

[safe radish]

@strong cosmos Has your question been resolved?

[safe radish]

@strong cosmos Has your question been resolved?

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[wispy vortex]

what does it mean when set B is shaded and the universe is shaded

but set A isn't

[gusty trench]

try visualizing each of the six options

make sure you know the definition of complement

[wispy vortex]

my teacher didn't teach

[wispy vortex]

but is it A' meaning "Not A"

I searched up the notations and stuff

[gusty trench]

https://en.wikipedia.org/wiki/Complement_(set_theory)

[wispy vortex]

I think I've figured this problem out

are you able to help me on another one

[gusty trench]

maybe not me specifically but someone eventually will

[wispy vortex]

ok

.close

[safe radish]

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[crude peak]

Can someone help i kidna forgot how to estimate with midpoints

[crude peak]

please ping 🙏

[granite flower]

So you have no problem with the Left or the right ?

@crude peak

[crude peak]

nope

[granite flower]

$$\sum_{n=1}^{6} 2\cdot f(X_ {middle})$$

[flat frigate]

Sherif Player

[crude peak]

wait shouldnt n=0?

[granite flower]

Where Xmiddle here is just the values in the middle of the rectangles
1, 3, 5, 7, 9, and 11

[granite flower]

If it is zero then it would be 7 sum operation because 6 is inclusive in the sum function

[crude peak]

ah i see

[crude peak]

right thats what i got

i drew the rectangles and all

my problem is im not sure how to find the area for each

i rememeber vaguly that Mid point was a lil different from left and right

bbut idk anything else from there

[granite flower]

The base of the rectangle times there heights.
The base of all rectangles are equal which is 12/6 = 2
The height of each rectangle is the value of the function from the place you choose the maximum height to be like right or middle or left

[crude peak]

okay so we do still do 2*f(x)?

[granite flower]

Choosing right means that you will start each rectangle from the right, like draw from 12 until it hit the curve then just complete the rectangle to 10, then start from 10 until hit the curve then complete to 2, etc until you complete to 0
The left is the opposite you start from the left of each rectangle, continue to get up until hitting the curve then completing the rectangle to the right

Middle is choosing to start at the middle of each rectangle until hitting the curve and completing each one to the right and left together

[granite flower]

Yeah

[crude peak]

so like for the first sub interval's area it would be 2*f(1) and f(1) is around 8.9 or smtg so itd be 2x8.9 = 17.2

[granite flower]

Yeah

[crude peak]

alr alr ty

[granite flower]

The second one is 2*f(3) and so on

[crude peak]

ok ty for ur help!

[granite flower]

No problem

[crude peak]

have a good day/night

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[tired kernel]

Prove that √2 is an irrational number.

[tired kernel]

how will we prove it???🤔

[main mural]

by contradiction

[tired kernel]

meaning??

[main mural]

suppose that it is a rational number, then do some steps to show that it leads to something absurd

which would mean that it cannot be a rational number => hence irrational

start by supposing that sqrt(2) is, rational, meaning:\
$\sqrt{2} = \frac{a}{b}, a,b\in \bZ$ and $\gcd(a,b)=1$

[flat frigate]

artemetra

[tired kernel]

ok\

√2=a/b after that i did not understood

[main mural]

a and b are integers, and they don't share any factors

[tired kernel]

ok

only 1

is a factor

[main mural]

in other words a/b is in its simplest form: like 1/2, 5/6, but not 2/6, 5/20 etc.

[main mural]

precisely

[tired kernel]

ok

after that how will we prove it irrational

we will divide it but which till place then???

[main mural]

there's thousands of proofs of these online, i'll recommend you to read some of these first and come back if you have any questions

short text proof: https://www.homeschoolmath.net/teaching/proof_square_root_2_irrational.php
khan academy video: https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:irrational-numbers/x2f8bb11595b61c86:proofs-concerning-irrational-numbers/v/proof-that-square-root-of-2-is-irrational

i don't feel like repeating everything they say there lol

[tired kernel]

ok thx

have a good day

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[dusky tundra]

ABCDEFG is a square number such that 8ABC = DEFG, find the square root of the number

[dusky tundra]

I'm so clueless, i'm thinking about starting to ABCDEFG as 1000000A + 100000B... etc but got nowhere

[round shore]

just a clarification, are ABC and DEFG also three-digit numbers and four-digit numbers respectively in the the 2nd equation?

[dusky tundra]

Yes

[round shore]

and is it 8*ABC or 8000+ABC?

[dusky tundra]

8*(digit ABC)

[desert swallow]

ABCDEFG = 10000(ABC) + DEFG might be a good place to start idk

[dusky tundra]

I did try that but got nothing

[main mural]

10000ABC = 8ABC*1250 = DEFG*1250

so ABCDEFG = DEFG*1251

,w factor 1251

[flat frigate]

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[main mural]

awesome

this means that DEFG must be of the form 139*N^2 where N is some other integer

@dusky tundra are you following this?

[dusky tundra]

Ohhhhhhh

[desert swallow]

ABC is also of the form 279 x M, M is a square number

Wait no yes

[main mural]

additionally, since DEFG is 4 digits, 3 <= N <= 8

i think you can further narrow it but yeah this is the gist

yeah you can because DEFG must be divisible by 8

[dusky tundra]

Okay, how do you continue?

[main mural]

so $N \in {4, 8}$

[flat frigate]

artemetra

[main mural]

and N can't be 8 because

,calc (139*8^2)/8

[flat frigate]

Result:

1112

[main mural]

that's 4 digits, not 3

so N=4

hence ABCDEFG = 3^2 * 139^2 * 4^2, and the square root is 3*139*4

,calc 31394

[flat frigate]

Result:

1668

[main mural]

that's the answer

@dusky tundra ta-da

[dusky tundra]

OHHHHHH

That's make a whole lot of sense

Thank you!!

[main mural]

no problem!

this was a very nice question

[dusky tundra]

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[red delta]

maybe x+2y² = 1

,w plot x+2y² = 1

[flat frigate]

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[red delta]

Substitute x and y.

And if it is equal to one

[cedar void]

I don't think there's any other way besides that double angle formula for cosine

what?//

[safe radish]

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[halcyon light]

How do I know

to draw e^x or e^-x ere

[main mural]

carbon-14 decays over time

[halcyon light]

how am I meant to know that 😭

[main mural]

so clearly N must be decreasing

[halcyon light]

is that meant to just be

common sense

[main mural]

also wait

[halcyon light]

carbon 14 decays over time

[main mural]

why are you talking about e here at all

you are given lambda^t

[halcyon light]

because N = lamda ^ t

[main mural]

and you are given that 0 < lambda < 1

[halcyon light]

tbh I don't get why but I thought e^x would be the graph

but it was e^-x

that was the grap

in the mark

scheme

[main mural]

bruh

my guy

$e^{-x} = (1/e)^x$

[flat frigate]

artemetra

[halcyon light]

well y = e^-x

[main mural]

0 < 1/e < 1 just like lambda

[halcyon light]

as the graph

[main mural]

you are given that lambda is less than 1, this means it's a decreasing function

,w (0.5)^2

[flat frigate]

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[halcyon light]

ah right

[main mural]

,w (0.5)^20

[flat frigate]

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[halcyon light]

thanks

[main mural]

no problem

[halcyon light]

.close

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[obtuse pagoda]

i don't know where to approach this from?
i was instructed not to use derivatives and only trigonometry rules

[obsidian oracle]

you can technically write the thing inside the limit as a function that depends on only one trig function

like only sin(2x) for example

then you can just call y that trig function

and do limit as y->0 of ...

[safe radish]

@obtuse pagoda Has your question been resolved?

[obtuse pagoda]

i'm confused 😅

[summer mountain]

I think it's going to be a kinda long calculation but you can use mathdf.com/lim/ to see steps for it (Input this lim((tan2x+tan8x-tan10x)/(sin2x-tan2x) and change the limit from infinity to 0)

[obtuse pagoda]

i didn't know there's a website like such

thanks!

i have run the calc and yes it is a long calculation

i think i can do it

thankss all

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[summer mountain]

Ye no problem

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[halcyon kindle]

Hi

[halcyon kindle]

I'm a bit confused on what to do

it's incredibly easy but I just want to make sure I do this correctly

for 2. it'll be 600 + 2x

I believe?

[tiny canyon]

I think it is 3(200+2x)

[halcyon kindle]

wasnt that the answer for 1.

oh wait you added a 3

but is that simplified

[tiny canyon]

Yeah because there are two of you so each day your spending is 200+2x

Which is the answer for question 1

Now there are 3 days

[halcyon kindle]

I see

is it fine if I keep this open for longer

I might have more questions

[tiny canyon]

Sure but don't keep it too long, tho 😂

[halcyon kindle]

so for 2

would the answers be 3 and 600

or do I have to do like 3x or something since its asking for variables

[halcyon kindle]

?

[tiny canyon]

I think it is asking you to define the variables that will be used for question 3

[halcyon kindle]

so cant I make it like anything

[tiny canyon]

You can choose some arbitrary character for that

[halcyon kindle]

so like

[tiny canyon]

For example for your "number of nights stayed", you can choose x

[halcyon kindle]

oh

ic

so I can make it

x and f

[tiny canyon]

sure but I think x and y are more convenient

[halcyon kindle]

ill do that then

and for 3. cant I just do 600y + 2x

[tiny canyon]

Is this still the same question as this?

[halcyon kindle]

no

this

for 3. I did

600y + 3x

[tiny canyon]

No, that is wrong

And 3(200 + 2x) = 600 + 6x

[halcyon kindle]

is that my equation?

so why would it be 6x

if they only stayed 3 nights

[tiny canyon]

Wait

[tiny canyon]

What is the original question for this?

[halcyon kindle]

at the top

i think I cropped it out when I posted it

but they're staying for 3 days

[tiny canyon]

Ok, there are a lot of issues here. I thought we were still using the same problems as the above

[halcyon kindle]

nope

if I should change anything from my asnwers let me know

thats my current asnwers

answers

[tiny canyon]

Nope

Because x is the number of nights. And the hotel will charge 200$ each day

So total cost must be 200x

Where does 600 even come from

[halcyon kindle]

i put my total cost variable as y

[halcyon kindle]

I thought since they're staying 3 nights

its 600

[tiny canyon]

They are asking you to write an equation that represents the total cost to the number of nights stayed

That means when x = 3 (3 nights stayed), your total cost must be 600

[halcyon kindle]

oh i see

[tiny canyon]

So if y is the total cost you will have the equation y = 200x that represents the total cost to the number of nights stayed if the hotel charge you 200$ each day

[halcyon kindle]

then for this I follow the y = mx + b

format

right

[tiny canyon]

Yeah

[halcyon kindle]

y = 200x + b

thats the slope

just double checking

[tiny canyon]

That is the whole equation

The slope is just a number in this case

Do you know which number represents the slope in this case, tho?

[halcyon kindle]

600

?

[tiny canyon]

For this we can assure b = 0 so the equation is y = 200x + 0

[tiny canyon]

Nope

[halcyon kindle]

I was thinking of something else

nvm

[halcyon kindle]

this goes on 4 right

not 3

[tiny canyon]

I mean do you know which is the slope in the equation: y = mx + b?

[halcyon kindle]

m is the slope

y is the y intercept

b is the intercept

[tiny canyon]

So, y = 200x means m = 200

[halcyon kindle]

oh so question 4 is m = 200

[tiny canyon]

It is just 200 to be precise

[halcyon kindle]

and what does the slope represent

[tiny canyon]

That depends on your problem. In your case, the hotel charges you 200$ each day and you have the equation y = 200x. Do you see the connection here? 🧐

[halcyon kindle]

kind of

im just figuring this out for this

[tiny canyon]

It is more fun if you figure it out yourself, tho. I gave you a hint. What do you think the slope represents in this case?

[halcyon kindle]

my assignment is due soon 😭

ill try and figure it out tho

[tiny canyon]

Ok, then it is the amount of money the hotel will charge you each night stayed

[halcyon kindle]

oh I was thinking nights

[halcyon kindle]

btw

this is the final part

[tiny canyon]

each day or each night

[halcyon kindle]

i dont think you can help with this

since I have to go onto websites

[tiny canyon]

Ok, so they are asking you to use real data 😂

[halcyon kindle]

found this hotel for 213

213 x 3 = 639

average price

i think

[tiny canyon]

Yeah.

You can do this yourself in this problem. It is just some addition and subtraction

[halcyon kindle]

ty

for everything

close.

.close

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[slim dagger]

Why is AD the shortest distance here?

[safe radish]

@slim dagger Has your question been resolved?

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@slim dagger Has your question been resolved?

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@slim dagger Has your question been resolved?

[still thunder]

AD is not the shortest distance... They have found AD like AB and CD and After that substituted the values in the formula

if you want the derivation can search for "Shortest distance between two skew lines"

[slim dagger]

Oh Sorry, I meant like why is AD used here?
The equation usually goes like [(a2-a1)x(n)]/magnitude of n

where a2-a1 represents the difference between two points

and where n is the commom perpendicular

[slim dagger]

So why is that AD specifically?

[still thunder]

Hmm it is not necessary... You can use AC too

and will get the same answer

...

[slim dagger]

Ah I see, thanks

[still thunder]

welcome

[slim dagger]

Will I get the same answer with BD too?

[still thunder]

You can see the derivation of this formula to know how it works

[still thunder]

AD, AC, BD, BC all will give same answer

[slim dagger]

Oh ok thanks!

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[hard crest]

i wouldn't call it a trick question

but it is fiddly

what does "L(E) = ø" mean?

[scarlet comet]

It doesn't print anything?

Theres nothing in the language basically

[hard crest]

yeah this

[scarlet comet]

But thennn

Te empty string is also nothing

no

""

[hard crest]

{e} is not the same thing as {}

remember how on the last problem you talked about the language {e, cat, catcat, catcatcat, ...}

[scarlet comet]

oh true

i was mixing it up since an empty string could technically be like those blank characters so if its in the brakets it would look like the empty set {}

so then its false?

[hard crest]

so "E prints e" is saying that e is an element of L(E) right?

[scarlet comet]

i guess?

[hard crest]

right that's why we don't use "" but we use e instead

[hard crest]

correct it's false, E would not print e because e is not an element of L(E)

(because there are no elements of L(E))

[scarlet comet]

.close

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[slate quail]

Fellas help me with my conjugation here

[slate quail]

Did he just punch in 0 for h at the last two lines after simplifying

I am not sure how it was simplified from that point on

[violet zodiac]

so, whats your question again?

[royal kiln]

Clever girl...

[median vigil]

the h was canceled by being common to numerator & denominator in second to last line, then they plugged in h = 0 for the limit for the last line

[violet zodiac]

basically

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[drifting knot]

@fathom adder

[drifting knot]

@main mural

@gusty trench

Guyz

Anyone help to find the 5th derivate of zeta

Derivative*

[rustic goblet]

don't ping individual helpers

unless they agreed?

[drifting knot]

Sorry lord but they my buddies

They come when I call

[rustic goblet]

fair enough

[drifting knot]

Yes

Me waiting for them

,w zeta'(1/2)

[flat frigate]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by oppailol.

[drifting knot]

,w derivate zeta(s) at s=1/2

[flat frigate]

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[safe radish]

@drifting knot Has your question been resolved?

[drifting knot]

,w 4th derivative of π cosec(πs)

[flat frigate]

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[drifting knot]

What

@fathom adder

Yi

Yo

Yo anyone wanna help

Guyz

,w 4th derivative of gamma(s).gamma(1-s)

[flat frigate]

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[drifting knot]

Lol

No bruh

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[drifting knot]

No

Aaj

Aah*

.reopen

[safe radish]

✅

[drifting knot]

Anyone help me

Anyone

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@drifting knot Has your question been resolved?

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@drifting knot Has your question been resolved?

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@drifting knot Has your question been resolved?

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[slim dagger]

Have to find out the area under a polar curve, having difficulty integratin this

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[dull sequoia]

can you show the original question

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[rocky sedge]

how to find the critical points of $f(x,y)=-3-5\cdot x+y$ within the domain $D={(x,y): x^{2}+ y^{2}\leq1.5$}?
i've found out that the function itself doesnt have any so i used the lagrange multipliers method which gave me the following points $\left(\pm\frac{5}{4},\mp\frac{1}{4}\right)$ which are incorrect

[flat frigate]

horizon2.0

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@rocky sedge Has your question been resolved?

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[lost ice]

how do i solve this

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[celest musk]

is it not 3 and 2?

[marsh jungle]

What is the correct ans?

[fringe ingot]

AH should be 4 and CH should be 1

[dry yacht]

AB=2sqrt(5)

BC=sqrt(5)

==> AH,CH

AB^2=AH.AC

BC^2=CH.AC

[fringe ingot]

im using th quadratic formula

the

[dry yacht]

Let AB=x, BC=y

xy=BH.AC=2x5=10

x^2+y^2=5^2=25

==> (x+y)^2=x^2+y^2+2xy=25+2x10=45

x+y=3sqrt(5)

[dry yacht]

In addition, x and y are two solutions of the equation: X^2-3sqrt(5)X+10

[fringe ingot]

Excuse the camera quality

[dry yacht]

X_1=x=2sqrt(5); X_2=y=sqrt(5)

[marsh jungle]

If AH=5-x then CH should be x

[fringe ingot]

yea

just correct that part if you write the solution ig

i did it to simplify it for myself

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@celest musk Has your question been resolved?

[celest musk]

this was one of the solutions i got

apparently its wrong tho

sorry for taking a while to respond

[marsh jungle]

What's the right one?

[celest musk]

i dont know

ive inputted two solutions

neither have been right

[marsh jungle]

Idk but my ans is 2.5 & 2.5 🥹

[celest musk]

interesting let me try this approach

[celest musk]

howd you get that out of curiosity?

[celest musk]

2sqrt5 and sqrt5 isnt right apparently idk bro

[marsh jungle]

Ah no it's wrong

[fringe ingot]

try 2.5 and 2.5

[celest musk]

sure

¯_(ツ)_/¯

[marsh jungle]

Wait ig it was right I'll resend 😂

[celest musk]

nope

nah it doesnt work

[fringe ingot]

wth

[celest musk]

idk

[marsh jungle]

It doesn’t? Check my soln once to see what’s wrong

[celest musk]

im just inputting random numbers at this point

im actually confused as shit

[marsh jungle]

Just check it

Once

[celest musk]

ok

its saying no

[dry yacht]

oh c'mon

[marsh jungle]

It?

[dry yacht]

tan(theta) \neq tan(90-theta)

[celest musk]

its a homework application

for a summer class im taking

[dry yacht]

.

[celest musk]

im gslowly going insane

[dry yacht]

.

[marsh jungle]

Ah okay

[dry yacht]

.

[dry yacht]

.

[dry yacht]

.

[celest musk]

?

what?

[dry yacht]

Your assignment can be solved in many ways

and the way I do it is to find the two sides of a right triangle by applying knowledge of quadratic equations

[celest musk]

ok

im just trying to find a method that works for me right now

.close

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[jovial sun]

can you turn $y=sin(x)$ into $2\pi n \pm arcsin(y) = x$?

[flat frigate]

sealpup321

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[opal oar]

x1 = 0
x2 = x3
x3 = x3 (unbound)

If i have these 3 solutions to a linear system of equations how do i turn them into a "basis solution" and a "general solution"

[marsh jungle]

You write it in the form of vectors. Here it'll be x3(0,1,1)

[opal oar]

is that basis or general

and why x3 and not x2 or x1

[marsh jungle]

It'll be basis too. For basis the criteria is that the vectors must be linearly independent. And since here there is only 1 vector, it is ofc independent

[opal oar]

so the answer is

x3[0]
  [1]
  [1]``` because x3 is the only unbound?

[marsh jungle]

See x1=0 so it is fixed, x2=x3 means it depends on x3, x3 is unbound which means it is free to take any value. So we write the vector in terms of x3

[marsh jungle]

Correct

[opal oar]

is it x = x3... or just x3

[marsh jungle]

No just x3

[opal oar]

thanks you\

!close

.close

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[visual swift]

,rotate

[visual swift]

Doing question 13

,rotate

[visual swift]

So I thought I got the proof

But I didn’t actually get DM and CK in terms of AD and BC

But I did get DM = DK/2, CN = CK/2

but not sure how to convert that into AD and BC

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@visual swift Has your question been resolved?