help-36

[twin pivot]

I have to see if this is true or not

[scarlet sequoia]

The implication as you wrote it is incorrect

[twin pivot]

I wrote |x(t)|≤M

[scarlet sequoia]

Think about what happens if x is very close to 0

[twin pivot]

So |y(t)| ≤ |log(M²)| so limited constant?

[scarlet sequoia]

No you can't conclude that

[twin pivot]

Mmm

[scarlet sequoia]

Take your time to write down exactly what the inequality |x(t)| <= M implies

[twin pivot]

I just used the definition

y(t) = x(t)

So |y(t)| =|x(t)| ≤ M

[scarlet sequoia]

y(t) = log(...) you mean

[scarlet sequoia]

?

[twin pivot]

Its an example

[scarlet sequoia]

Ok but in your actual exercise

y(t) = log(x²(t+1))

[twin pivot]

Yes

[scarlet sequoia]

Take your time to go through the inequalities' implications one by one

You start with |x(t)| <= M

What can we do to get closer to the expression of y

[twin pivot]

Yes

|y(t)| = |log(x²(t+1))|

Then what do I have to do?

If I know that |x(t)| ≤ M

s=t+1

[scarlet sequoia]

You can apply functions on each side of the inequality

[scarlet sequoia]

Yes also substituting t by t+1

|x(t+1)| <= M

[twin pivot]

So |y(t)| = |log(x²(s))| ?

[scarlet sequoia]

Take it the other way around

We start with the inequality

And from the inequality, we try to recover y(t)

[scarlet sequoia]

So we start with this

In y(t), there's no |x| but x²

So what should we do?

[twin pivot]

I bring the absolute value into the log

[scarlet sequoia]

Once again, you're looking at it the wrong way. And no, you can't do that

what's inside the log already needs to be positive

How do we go from |x| to x²? What's the link between those two quantities

[twin pivot]

But if I can't get |x|≤M, how can I write x as M? Then it's not stable because I can never get |x|

|x| ≤ M -> |x|² ≤ M²

[scarlet sequoia]

Yeah

And what is |x|²

|||x|² = x²||

[twin pivot]

x • x*

[scarlet sequoia]

x is a complex function now?

[twin pivot]

No

[scarlet sequoia]

So |x|² is x • x

[twin pivot]

How do I know if x is real?

I don't know what function it is

[scarlet sequoia]

You know how to take the log of any complex number?

[twin pivot]

I only know that x:D->R or x:D->C

[scarlet sequoia]

Let's assume x is real for now

If x is complex then you can take x = i and the log makes no sense

[twin pivot]

But is it real because if it were complex it would be x^*?

x²(t) ≤ M²

Right ?

[scarlet sequoia]

Yes

x²(t+1) <= M²

[twin pivot]

if x(t) = 0 the log is not defined => not stable?

[scarlet sequoia]

That's the idea

Log explodes for small values of x

[scarlet sequoia]

From here you get y(t) <= log(M²)

And not |y(t)| <= |log(M²)|

[twin pivot]

I can't apply ABS both sided?

[twin pivot]

.

In the definition it is only written |x(t)|≤M

I Need a lower bound

<@&286206848099549185>

@scarlet sequoia

[scenic pendant]

@scarlet sequoia maybe he can take x(t) = 1/(|t| +1) <= 1 ?

so M = 1

$y(t) = \log\left(\frac1{|t+1|+1}\right) = -\log(|t+1|+1) \rightarrow -\infty$

[soft zealot]

alee

[scenic pendant]

so $|y(t)| \rightarrow \infty$

[soft zealot]

alee

[scenic pendant]

so limited input that does not produce limited output

im studying this

but I'd like to see how you proceeded

[scarlet sequoia]

It's the same, just take x(t) any positive function that goes to 0

[scenic pendant]

did you study this by any chance ?

it should be called signal processing / signal theory

btw before i missed a -2

and it was (....)^2

[final saddle]

@twin pivot Has your question been resolved?

[final saddle]

@twin pivot Has your question been resolved?

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[soft pewter]

Consider an exponential function whose rule is of the form f(x) = acb(x-h) + k, where 0 < c < 1.

What are the possible values of the parameters a and b?

[soft pewter]

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[sturdy grove]

I need help with my geometry work

[craggy plank]

,rccw

[shell condor]

are those two lines parallel?

[craggy plank]

it does not matter honestly

[shell condor]

the 51m and the 68m line

[shell condor]

No but for similarity it would

[lilac moat]

You have three angles that are similar.

That is enough conditions

[craggy plank]

it already fulfills AA similarity

[shell condor]

oh wait I missed the label indicating equal angles

[lilac moat]

@sturdy grove, hopefully that gives you an idea of where to go.
Use the fact that if two triangles are similar, then the ratio of their sides are equivalent

[sturdy grove]

okay

Is it 36?

I got 36, is that correct

I did 68/51 multiplied by 48/z

oh

Yay

Thank you everyone for your help

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[strange sparrow]

you just have to write out what g(f(x)) is

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[prime tusk]

GENUINELY what am i doing wrong

[prime tusk]

i bet its something super stupid

[loud sundial]

there shouldn't be a boundary at 1

recheck your signs

[prime tusk]

oh is it

01

-1

yeah

whopp

whoops

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[prime tusk]

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[tulip panther]

can someone explain the solution? I dont understand the derivations for a > 1, 0 < a < 1

[final saddle]

@tulip panther Has your question been resolved?

[opal plinth]

When a > 1, a^{sin(x)} <= a

When 0 < a < 1, a^{sin(x)} >= a

(when a = 1, a^{sin(x)} = a, obviously)

[tulip panther]

ah

[tulip panther]

it took me a while to see it but i do now

why do we select the greater value of x for a > 1 and vice versa for 0 < a < 1

and are there any resources online so i can practice considering cases for exponentials (and maybe logs)

[opal plinth]

Wdym "the greater value of x"?

[tulip panther]

so for the case $a > 1$, we say that $a \ge \frac{3 + \sqrt{5}}{2}$, and for $0<a<1$ we select the lower value (the value with the - instead of the +)

[soft zealot]

keqae

[opal plinth]

Oh that

how do I explain

I think it's kind of implicit but they're taking sin(x) > 0 because the equation has both sin(x) and -sin(x)

Like it's a^{sin(x)} + a^{-sin(x)} and one of these terms is > 1 while the other is < 1

a^0 = 1 and a^u as a function of u is monotonic, so a^u and a^-u are on either side of the (0,1) point, both horizontally and vertically

So just pick u to be the one on the right, so u > 0, and if a > 1 that gives you a^u > 1

(3+sqrt(5))/2 > 1

On the other hand, if 0 < a < 1, that gives you a^u < 1

(3-sqrt(5))/2 < 1

[tulip panther]

ive been staring at this for a while

its not going into my head

[opal plinth]

Just graph it

[tulip panther]

what do i graph? a^sin(x)?

[opal plinth]

No we don't really care about sin(x)

It's just there to restrict the exponent to [-1, 1]

Think of the question as a^x + a^-x = 3, but with the added restriction that x must be in [-1, 1]

So graph a^x and play around with a

Maybe graph both a^x and a^-x at the same time, though really you should be able to imagine a^-x as the reflection about the y axis

[tulip panther]

hmm okay

[opal plinth]

Also graph y=a so you can visualize when a^x is above or below a

[tulip panther]

alright

i got it graphed

what am i looking for?

[tulip panther]

i can definitely see that a = 1 => y = 1 for all x

although trivial

[tulip panther]

yeah can see that

[opal plinth]

Sorry I have to go

I made you this: https://www.desmos.com/calculator/3rxz87lm0k

When the vertical green lines show up, that means there's a solution

Should be enough to understand what the question is really about

Good luck

[tulip panther]

oh alrighty thank you so much

i understand that it works now

i just dont understand how

[final saddle]

@tulip panther Has your question been resolved?

[final saddle]

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[bold flume]

Given a circle with center O and diameter CE.
From point C, draw the tangent Cx to the circle with center O (C is the point of tangency).
On ray Cx, take a point A.
From A, draw the tangent AB to the circle with center O (B is the point of tangency, B ≠ C).
Let H be the intersection of OA and BC.
Draw AE, which intersects the circle at point D (D ≠ E).

Prove that HB is the angle bisector of ∠DHE.

[bold flume]

I’ve been thinking about it for two hours and still can’t solve it.

[thin cloud]

Have you made a diagram?

What have you tried?

[bold flume]

i made

[bold flume]

ye

do you need it

i will send

[thin cloud]

ofc

[bold flume]

ok , wait for 5 mins

[final saddle]

@bold flume Has your question been resolved?

[bold flume]

@thin cloud

[thin cloud]

Here's a better diagram I made, sorry I had things to do, I'm free now

Also you made me recall the reason I hate Geo

If we show $\angle DHA =\angle OHE$, then we can deduce the given

[bold flume]

hmm

i think not

[thin cloud]

typo sorry

no

[soft zealot]

Alexis_Fx

[bold flume]

why DHA = OHE ?

[thin cloud]

Similar triangles

[bold flume]

What conditions are needed for those two triangles to be similar?

[thin cloud]

Since we need the angles of those triangle

We might want to look for the ratios

It's a good idea to try showing Side-Angle-Side

[bold flume]

now we have a new problem

Which sides are proportional to which sides, and which angles are equal to which angles?

[thin cloud]

Why don't you try first

[bold flume]

alr

i will think

but I find your method a bit hard to understand.

[thin cloud]

My solution is pretty short tbh, take like 5 lines

[bold flume]

I feel completely stuck now.

[thin cloud]

well I can give you a hint: triangle ADH and triangle AOE are similar

oh lmao

[bold flume]

are u sure , angle ADC is 90o , angle AOE is not 90o

[thin cloud]

There's a typo

should be good now

[loud sundial]

hi

[thin cloud]

gm csp

[bold flume]

oh i see

now we have AHD = AEO

um

still hard

can u gimme the next hint

[thin cloud]

no, focus one the ratios

We need those

[bold flume]

wait , we need to show triangle HDA similar to triangle HOE , dont we?

[thin cloud]

yeah

[bold flume]

now , we have angle ADH = angle HOE

[thin cloud]

well yeah

That's the pair of angles we need

[bold flume]

the ratios are hard now

[thin cloud]

AD/DH=OE/OH, right?

From the two similar triangles I said

[bold flume]

ye , showing it is a problem

[thin cloud]

why is that a problem

triangle ADH and triangle AOE are similar

That should directly deduce AD/DH=OE/OH

[bold flume]

what

I thought this couldn’t be deduced directly, because my teacher taught me that it can’t be inferred directly.

[thin cloud]

Oh wait

My dumbass

I meant

AD/DH=OA/OE

[bold flume]

This doesn’t relate to the two main triangles we need to prove, because it doesn’t involve OH.

[thin cloud]

It does

[glossy zephyr]

Also, i hope this helps somewhat declutter the drawings,

[thin cloud]

not directly tho

[bold flume]

So can OA be replaced by OH, or what?

[thin cloud]

Dude no rush, have you tried to relate OE/OH to other ratios

[bold flume]

no , i didnt

[thin cloud]

Then try that out

[glossy zephyr]

ACO as a triangle itself doesnt tells us jackshit, right?

its just part of the construction

[thin cloud]

yeah, but it's important to keep in mind since we are not sure what related and what not

While solving

[glossy zephyr]

tbh i havent read much of the convo so im just brainstorming

but AOC and OHC are similar

therefore ABH and BOH also are

[bold flume]

then we have AH x HO = BH2
AH x HO = HC2

[bold flume]

i tried and saw nothing

[thin cloud]

AO/OE=AO/OB

[glossy zephyr]

okay i found a way to reduce the problem a lot

[thin cloud]

Both OE=OB=r

[glossy zephyr]

we can literally get rid of B

[thin cloud]

What now

[glossy zephyr]

Extend EH to AC

DH to CE

[thin cloud]

AO/OB=HO/OB=HO/OE

[glossy zephyr]

HC has to be a bisector of these lines too

[thin cloud]

basically we are done

[thin cloud]

Hmm well that could be an approach

[glossy zephyr]

In this new definition, H is just the midpoint of the radius towards A.

[thin cloud]

Have you found the sol with that construction

[glossy zephyr]

still looking at it, but i think it should be relatively simple from here

[thin cloud]

I already sent my method above, so ig Imma head out if OP doesn't have any question

[bold flume]

im asleep , maybe i will think about this tomorrow

[thin cloud]

close this ticket then

[bold flume]

wait

how did u draw

[thin cloud]

Using geogebra

[bold flume]

.close

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[unreal scaffold]

I am unsure how to complete these questions (the text reads “solve the equations”)

[unreal scaffold]

Starting with a)

The correct answer is 6. I got two answers, one of which is 6.18, pretty close but I’m unsure how the correct answer is ONLY 6, given I’m pretty sure the only way to solve is to use the pq formula

Which inherently gives two answers

[thin cloud]

This might be the reason you didn't get the right answer

[unreal scaffold]

Hm

However, when fixing by that mistake I get x^2 = -9x

?

[thin cloud]

wait

[unreal scaffold]

Yeah, the 16 and -16 cancel each other

[thin cloud]

What did you do here

[unreal scaffold]

I multiplied 1 by x+4

[thin cloud]

You can't just do that

You have to multiply both sides

That means you also have to multiply x/(x+4) and -16/(x+4)

[pliant shore]

you have to multiply everything on both sides

[unreal scaffold]

Oh, now I’m getting x = 6

Ok

For b)

It seems pretty clear cut to me, but the answer is that it’s not solveable

The sticking point that I could see being in error is multiplying both sides by t-2, but it seems correct to me

[pliant shore]

you were so close

8 = 4t

that does not imply t = 0.5

everything before that step is correct

[unreal scaffold]

Oh, t= 2

[pliant shore]

yeah

[unreal scaffold]

Right, but according to the textbook it should still not be solveable?

[pliant shore]

indeed

cause you're dividing by t - 2 which would equal 0, in the original equation

[unreal scaffold]

Dividing(

?

Ohhh

If t = 2 then the equation equals 0

Ok

[pliant shore]

actually if you go back to (t - 1)/(t - 2) - 3/(t - 2) = 5

[pliant shore]

the denominator

[unreal scaffold]

Yeah

[pliant shore]

combine the fractions and (t - 2)/(t - 2) = 5

that's impossible

[unreal scaffold]

Ok, so I just did a mistake at the very end, that’s not too bad

Now, for c

I’m unsure how to ensure the denominator is the same across the board given there’s a y^2

Like this

But as there’s a y^2 the denominator is still different

So I am unsure how to rectify this

<@&286206848099549185>

(Pinging due to inactivity)

[rare bridge]

SHAW!

[unreal scaffold]

Hegale!

[wary juniper]

Mutiply y^2 on both sides

[unreal scaffold]

Like this?

[amber narwhal]

yesz

[unreal scaffold]

Ok so using pq it’s either 3 or -2

According to the textbook it’s only -2

However, neither 3 or -2 make sense if you put them into the original equation?

Did I make a mistake in using pq?

[wary juniper]

Its either -3 or 2 so ye i think you made a mistake

[unreal scaffold]

The textbook says the answer is -2? I don’t see any mistake in my working

[wary juniper]

Aint it supposed to be -1/2

[unreal scaffold]

Oh wait

No. That’s wrong, in the answer section it says either 2 or -3.

[unreal scaffold]

?

You’re right

In pq it’s - x/2

Right I’ve made that mistake before

Ok ok

Now I got 2 or -3

Onto the final part

For d

[final saddle]

@unreal scaffold Has your question been resolved?

[unreal scaffold]

This is what I got so far

But the answer is -2 according to the textbook so I must’ve made a mistake somewhere, I just don’t know where

Idk

Ok got it

Thanks for the help!

[pliant shore]

.close

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[uneven token]

@stiff lynx

[hasty mist]

Do you have a maths question?

Does Jai have a maths question?

[final saddle]

@uneven token Has your question been resolved?

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[old pilot]

exercise2: second equation

[steep nest]

you know l'hopital?

[old pilot]

yeah

hospital in french

[steep nest]

Yes, true

but l'hopitals rule?

it's a sortof trick in these types of questions

[old pilot]

nah

sadly

[steep nest]

well I'd love to know how you solved the first thing of exercise 2

[stiff lynx]

yo i gotta go now i have some work

[old pilot]

i wrote

and yk the rest

since

the opposite is -4

[old pilot]

wya

me need help

[steep nest]

how do you know this

[old pilot]

the exercise says f'(-2)=4

[steep nest]

very true

[old pilot]

f'(-2) is

[steep nest]

ahhh that's what they want alright

[old pilot]

so

how do i find the second one

[steep nest]

try to write the bottom of the expression as the derivative part

[old pilot]

ok

i couldnt

i should write f(x)-1 as what

[steep nest]

factor out a 3 on the top

we get 3(x+2) right

[old pilot]

yeah

[steep nest]

now devide the top AND bottom by x+2

[old pilot]

there is no x+2 in the bottom

[steep nest]

see what's happening here?

[old pilot]

yes

[steep nest]

Need more help or do you got this

[old pilot]

50%

wait

how did u find the

i mean it does equals to

[old pilot]

but how did u go from this

[old pilot]

to this

[steep nest]

we devide the numerator and the denominator by x + 2

so on top we have 3 (x + 2) devided by (x + 2) and in the denominator we have (f(x) - 1)/(x+2)

in the numerator, x+2 devided by x + 2 is just 1, so we are left with 3

[old pilot]

eh

like this?

understood

ty

[steep nest]

Yeah and I think you get the rest already

<@&268886789983436800>

[old pilot]

.close

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[sinful compass]

I am re-learning some math and I am really struggling with bounds of polar curves. I think it's because they are drawn in cartesian instead of polar coordinates.

Here is a question which doesn't make much sense to me indentifying the boundaries.

[sinful compass]

To provide some of my logical-assumptions.
So the function appears to be in Qiv, so it must be bound by 3pi/2 and 2pi because that's outside the bounds of 0< theta < pi
Obviously that's not right.

Doing some math, I set r(theta) to 0 to find points of theta where the radius is 0.
At that point I get solutions that are 0, pi/2 and pi.
I am not entirely sure what is the upper/lower limits of the integral.

Looking at the R area's loop shape and I think that, well the bottom left loop has a 'negative slope/angle' but so does the upper right loop. But none of my theta values represent negative values; as 0 to pi/2 is Qi that's positive values, and pi/2 to pi is Qii which is again positive values but my R is a region with negative values.

Then I read about 'following the loop' method. But that didn't help me either.

[celest crane]

Are you sure that pi/2 to pi is in Q2?

At θ=3pi/4, sin(2(3pi/4)) = sin(3pi/2) = -1.

[sinful compass]

using this unit circle as an example. yes
https://lh3.googleusercontent.com/proxy/o2BQKpBGTtOGBT7zrPRZC1HJWEqqzfOGISEOPxe23hrrpY-XBfMQ9zyfW8fdGAV9IhK5QTJNTohUQ7mhaZtYCdACgpjnb32wlyicetIF40JAOnozyKfS-SGsSvuk

[celest crane]

The 2 in 2θ changes the periodicity though.

[sinful compass]

I understand the point you're making.

in that case the pi/2 to pi would fall into Qiii and qiv which is both negative

[celest crane]

Correct.

However ... that is only relavent in suggesting that sin(2θ) is negative. It does not necessarily force the curve into Q3 and Q4.

Imagine an arrow pointing in the direction of θ. Along that arrow, there is a negative and positive direction with respect to the origin. If the radius is negative, then the point will be in the opposite direction of θ.

[sinful compass]

I am mainly using khan academy to re-learn. I tried using chatgpt but they aren't as helpful without visual assistance. I tried reading a textbook from Stewart - Calculus - Early Transcedentals 6e and they just skip 'bounds' entirely and the problems themselves are in r, theta plane so the bounds are just readable.
I tried looking up youtube videos but they seem not helpful to me.
I am struggling to re-learn this concept.
My brain feels like it's melting.

[celest crane]

One moment, let me make a graph that may help out.

[sinful compass]

here is another example I am doing it. I already 'solved it' and the website is just re-cycling the question. I can also 'feel' that the correct bounds should be 0 to pi because xy is in Qi and Qii but also i know that isn't a good assumption to make translating to r(theta)

i found theta solutions for when radius is 0 at theta = 0, pi, and 2pi
so i think like this

i am making these assumptions based on the region R having a positive slope and then a negative slope

It would be alot simplier if i found a 'formal' guide explaining how to find these bounds but i've tried using 7 different sources and none of them were helpful.

[celest crane]

https://www.geogebra.org/classic/xxwgzcps

In that graph, the green line is the direction of theta. The distance away from the origin is r(theta).

If r(theta) is negative, the point will go in the opposite direction into the red line.

The whole concept of a negative radius tends to confuse a lot of people.

[celest crane]

Still there?

Finding the bounds in polar integrals can be tricky. Even when you have intersecting curves, theta is not always the same for both curves.

[sinful compass]

i feel like i am going in a feedback loop because i can't grasp the basics

Because in one problem i think i understand the idea but then the next question gets me more confused.
In the second problem

i know radius is zero at theta = 0, pi, 2pi.
So I i think, for what theta value do i have an area R that is above the x-axis.
Well that means, i have positive y values.
But i intuitively equate that, without thinking, that it also means r(theta) must be positive.
But since's sin^2, r(theta) is positive for all theta's then pi to 2pi should also be valid bounds? Obviously not, but i don't grasp it fully.

It just really confusing being given a function a (r, theta) but plotted in (x,y) and then go back and forth making these analytical deductions with a single visualization... it's so confusing because i am using incorrect assumptions.

I know that it looks like the area R is the positive part of the function, so the bounds are probably 0 to pi, but r(theta) is still positive for pi to 2pi bounds using the r(theta) equation.

if the graph was in r, theta coordiantes i wouldn't be struggling at all.

i was watching a youtube video of a slightly more complicated problem, and they made assumptions which i don't understand how they made.
Here it is:

I don't understand how they figured out the shaded region is bound by pi/6 and pi/2 .
I understand mathematically how they found the various points at which radius is 0
But to tell what interval denotes what area makes no sense to me.

[bold zenith]

Are you trying to find which interval of theta R is?

[celest crane]

He is having difficulty understanding how to determine the bounds for regions.

[sinful compass]

yes my issue is only bounds

[celest crane]

The first thing I would do is determine the periodicity of the radius. You can do that by using the identity 2pi/P = B.

For r = 3 cos(3θ), 2pi/P = 3 means that the periodicity is P = 2pi/3.

At that point, barring any external variables, the curve should overlap itself.

[bold zenith]

So you know that at the bounds r(theta)=0

[bold zenith]

And to work out which ones you want, you use this, and if r<0 then it's the opposite

[sinful compass]

but in this question the r(theta) is never 0, so how do you do bounds here...

[bold zenith]

If r(theta) isn't 0 then it's either always r positive or r negative, and the theta would just be #help-36 message (or flipped if negative)

I can't tell you how to do every question because I have no idea what sort of question they'll be but I think these are the two cases

So in this one it starts at theta=0 and ends at theta=3*pi/2

If it's a loop that doesn't appear to have boundaries, (Like this one but in all quadrants), then the bounds are 0 to 2pi or anything to anything + 2pi

Try making them on desmos https://www.desmos.com/calculator

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[left trail]

can someone explain how he goes from these two steps

[dire glacier]

wdym

[barren hound]

the negative from -lim moves into the denominator, and lim(h->0) is the same thing as lim(-h -> 0)

it's two separate steps

[left trail]

I see. Ok thx

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[runic phoenix]

I do not understand example a

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[blissful meadow]

If b is nonzero then the identity is not in the set

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[runic phoenix]

🙁

[vital crag]

Your channel timed out so you need to repost your question

[signal vector]

Just read azyrashacorki's explanation

Identity is a synonym for the 0 vector if that's what was confusing you

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[dusk charm]

Help pls qs1

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[serene panther]

Could you perhaps translate this?

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[soft pewter]

The population P(t) of a village is given by the rule

P(t) = 6530(3)^0.1t

where t is the number of years that have passed since 2011.

By how many inhabitants did the population of this village increase between 2016 and 2023?

[small grail]

what have you tried

[soft pewter]

nothing so far this was b) i did a)

but thsi one im having trouble

[small grail]

you are looking for P(2023-2011)-P(2016-2011 )

[soft pewter]

yup

so i gotta replace f(x) by 2023?

i gotta make 2 equations?

[desert mantle]

not quite, t is the number of years since 2011

[small grail]

right mb

[soft pewter]

how can i make 2 equations

[desert mantle]

you dont need 2 equations

you need two numbers

the population in the year 2016 and in the year 2023

[soft pewter]

oh alright

[desert mantle]

the formula tells you how to compute those

[soft pewter]

its f(x) = 6530(3)^0.1(12)

@small grail

[small grail]

for

2023?

[soft pewter]

yeah

[small grail]

yes

[soft pewter]

alr

and 4 for 2016?

[small grail]

2016 - 2011 = 5

[soft pewter]

oh 5 yeah

btw is the population of 2023 24 403?

[small grail]

,calc 6530(3^((0.1)(12)))

[soft zealot]

Result:

24403.869107068

[small grail]

yes

[soft pewter]

alr

[soft pewter]

btw

its 11 310 for 5 years

but what do i do about the decimals

since its peopel

[small grail]

round to greatest integer less than your number

[soft pewter]

oh so 24 403

[small grail]

24403 for 24403.869107068

[soft pewter]

alr i see

[small grail]

yes

[soft pewter]

now i have to subtract ?

[small grail]

yes

[soft pewter]

alr done

tysm again

[small grail]

np

[soft pewter]

.close

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[twin pivot]

But does the Fourier transform of 1 equal infinity?

[twin pivot]

$\int_{-\infty}^{+\infty} e^{-2\pi iy\xi} dy$

[soft zealot]

Goofy Joe

[twin pivot]

This

,w $\int_{-\infty}^{+\infty} e^{-2\pi iy\xi} dy$

[soft zealot]

Results provided by WolframAlpha

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

Requested by lightningmcqueen04978

[twin pivot]

Help !!!

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[errant oasis]

You can use the sine-cosine form instead of exponential

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[lime crest]

hi! can someone help me with this angle chasing problem my friend sent me

[lime crest]

the only idea i have was to reflect D across AC so that AD'BC are cyclic but i dont think thats getting me anywhere..

i might be kinda stupid guys

[leaden moon]

!status

[final saddle]

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

[jagged wave]

the triangle OBC with that middle intersection point being O is isoceles, to be a little cpt. obvi

since you've worked out that BOC = 6x, then that gives you something to work with

[leaden moon]

im just going to try and angle chase all of the angles in this diagram

with O as the intersection of BD and AC

[echo meteor]

The remaining angle of the isosceles triangle will be 180-6x and you try to find <BCA. After finding <BCA, use sum of angles of triangles is 180 theorem in triangle BCD.

[lime crest]

im probably stuck somewhere midway

[lime crest]

oh, is that it?

[jagged wave]

ayyy

[leaden moon]

or you worked out all 3 angles in this triangle

[echo meteor]

Calculate these angles in order as shown, then you can find the value of x from triangle BCD.

[leaden moon]

sum of angle in this

[echo meteor]

Ohh wait...

[lime crest]

i mean, <BCA = 180-17x + 3x = 180 - 14x but that doesnt simplify...

[leaden moon]

you worked out this far, try sum of angles in a triangle in this

you should be able to solve for $x$ with some algebra

[soft zealot]

1 divided by 0 equals Infinity

[leaden moon]

or you can do what he said here

[lime crest]

i c

ew why did i get the answer as 90/11

[echo meteor]

Hmm...my idea does not work. We might need some construction

[lime crest]

i just used <AOD' = 90 and then sum of angles

[echo meteor]

Where is O?

[lime crest]

DD' intersection with AB

oh wait

.close

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[arctic sand]

Given 2015 distinct positive integers not exceeding 3019. Prove that among these 2015 numbers, there exist a, b, c, d such that a + b + c = d

[arctic sand]

i will show my work right here

i think maybe i can use this but have no idea how to proceed

:<

[leaden moon]

,rccw

[leaden moon]

,rotate 180

[arctic sand]

it seems like u are a wise guy, pls give me some guidance im so lost 😭

[leaden moon]

diriclet exercise?

[arctic sand]

yeah it is

<@&286206848099549185>

[final saddle]

@arctic sand Has your question been resolved?

[vital crag]

Wow like 80% black

Cropping is your friend

[arctic sand]

isnt this channel occupied?

nvm

[arctic sand]

pls help

!close

.close

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[copper roost]

can someone help me here im confused a bit

[copper roost]

so do i just do 10000 * 100

to get the fillings of the 10000 bottles

[glossy zephyr]

Assuming normal distribution, you are to find the area that satisfies < 98ml and > 105ml

[copper roost]

yeah

but

[glossy zephyr]

You use your z-table for these

[copper roost]

the bottles

theyre irreleant?

[glossy zephyr]

Nope

You find the prob of an individual bottle not fitting the criteria

and multiply by the amount of bottles

[copper roost]

but

how do i know what number of bottles are outside the tolerance

wait

shir

omg

dont i just do 1-p(98<x<105)

[glossy zephyr]

yea

[copper roost]

also its not <= right

[glossy zephyr]

Not really

Continuous Distributions dont differentiate between > and >=

[copper roost]

oh shit

[glossy zephyr]

Since the probability of a single point is 0

[copper roost]

ohhh i remember

my teacher did the = in (=)

man 😭 its 9 pm i gotta learn inverse normal distribution and hypothesis tests

i got a big exam tomorrow

[glossy zephyr]

For any question of this kind, you work with the logic that the expected proportion of valid/unvalid is always equal to the theorical probability of an individual product being valid or not

And in reality, by the law of big numbers, the more products (bigger sample), the closer you get to that expected proportion

[copper roost]

yeah i was wondering like

the mean is the same as 100 ml

[glossy zephyr]

The mean is 100ml, yeah

[copper roost]

is rhat what u mean

[glossy zephyr]

Not really, gimme a sec

[copper roost]

okay

[glossy zephyr]

When you have any distribution you use to find the probability of an event

Here we have the normal

[copper roost]

yeah

[glossy zephyr]

and we want to find the probability of a random value being below 98 or above 105

And once you find that prob., if youre asked "how many products are above/below that threshold /aka invalid"

Then you multiply the probability by the amount of products

lets say here we have p% * 10000

And thats how many products you expect to be invalid

[copper roost]

ohhhhhhhhhh

yess

ty

[glossy zephyr]

Its the same basic idea as coin flipping but with extra steps

[copper roost]

ill do this exercise really quick then move on to the next omfg im gonna br fucked

[copper roost]

yeah

[copper roost]

btw in the example we did its basically "or" right

how do we include that or

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[toxic tree]

Apparently the answers are the 2nd and 3rd

I got the 3rd cause the degree of x in the numerator is 1 higher than the degree of x in the denominator

[toxic tree]

But for the 2nd I got this

Which isnt 1 higher (I think)

Cause you can divide all terms by x, right?

Or does the 1/x mean I cant?

Or is my method just wrong?

[loud sundial]

$\sqrt{x^2}=|x|$, not just $x$.

[soft zealot]

Civil Service Pigeon

[loud sundial]

So $\frac{\sqrt{x^2+x}}{x}=\sqrt{\frac{x^2+x}{x^2}}$ for $x>0$

[soft zealot]

Civil Service Pigeon

[loud sundial]

but not for $x<0$

[soft zealot]

Civil Service Pigeon

[loud sundial]

[https://www.desmos.com/calculator/wiouhs9h0s]

also I'd lowk just say

[soft zealot]

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

[toxic tree]

Wait what

What does ~ mean sry

[loud sundial]

asymptotically equal to

[toxic tree]

Ahh I see

Thank you!

❤️

.cclose

[loud sundial]

[toxic tree]

.close

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[tiny gorge]

looks fine, what's the question?

[toxic tree]

In the solution they gave how come the absolute values just dissappeared?

[tiny gorge]

when x is near zero, x+1 and 1-x are both positive

so you can remove those abs vals

for the one in the denominator, |x| becomes x for the limit from the + side, and it becomes -x for the limit from the - side

[glossy zephyr]

just in case, remember the definition of |x| as a piecewise function

[toxic tree]

Ahhhh ok I see

Thank you!

❤️

.close

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[tiny gorge]

yw

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[latent gazelle]

hi! is parametrisation if i can describe a geometric object with an amount of variables equal to the dimensions of that object?

[ivory vessel]

yes

[latent gazelle]

okay, i have a followup

[ivory vessel]

more specifically, the number of parameters is actually the degree of freedom of the object

which is formalized to define dimensionality

[latent gazelle]

okay thanks for telling me that, i also have a followup question: if the amount of parameters matches the amount of dimensions, but i change the parameters, but still only use the same amount, is that also a parametrisation?

sry for such a confusing way of phrasing it

or am i just changing coordinate systems

or maybe something else

[ivory vessel]

parametrization is not unique, e.g. changing coordinate system

i'm actually not sure if parametrization is unique within one coordinate system

but my bet is yes silly me it's not

[inland birch]

i don't think so

for example (x,y) = (cost,sint) and (x,y) = (cos2t,sin2t)

[ivory vessel]

oh true

[latent gazelle]

sorry in a little slow

is that not a parametrisation?

the example

[ivory vessel]

that is parametrisation

[inland birch]

we were discussing whether a parametrisation is unique or not

[ivory vessel]

if you can express your variables in terms of functions of that variable, then it's a parametrization

[latent gazelle]

oh sry ye i didnt follow that

what does it mean for something to be unique

you can only do it one way?

[latent gazelle]

okay this confuses me

because to me it sounds like

[inland birch]

"Unique" means that if two elements satisfy the property in question, then those two elements are in fact the same element.

copy pasted straight from reddit lol

[latent gazelle]

i mean i could express a variable as something that depends on 1 variable, or multiple

but i thought i wanted to match the amount of parameters to the dimensions

to verify if it is a parametrisation

so if i have a surface, it an object that can be described in R^2 even if it lives in R^3, so i would want to go from some amount of parameters to 2

for it to be a parametrisation

or did i misunderstand –

it is okay if i did!

my goal is just understanding

i feel like those two things contradict eachother

[inland birch]

i don't think this is necessary

i will get an example

[latent gazelle]

damn a guy convinced me of that being how i should think of parametrisation

and i liked it

i feel like i have moved between 3 or 4 definitions and i dont feel closer to understanding it 😦

[inland birch]

it's very likely that i'm the one who doesn't understand, i will illustrate one example but i think that's all i'm good for

not too well versed in this

but let's say you have a 5 variable matrice and you notice that there's 2 non leading columns

[latent gazelle]

none leading means? sry language barrier

[inland birch]

then we have that the 3 leading variables will be parametrised by the 2 non-leading ones

[inland birch]

like uh

it's not the first entry in the row after the matrix has been reduced to row echelon form

wait that might not be a complete description

yeah idk i'll leave this question up to someone else

[latent gazelle]

are they like pivots?

[inland birch]

maybe just re post it so people know what they're coming into

[inland birch]

pivots are what leading columns come off of

[latent gazelle]

should i be picturing some form of staircade but maybe the steps are uneven

staircase*

or does none leading mean that there is no real triangle to be seen

or anything triangle adjacent

[inland birch]

ignore the circles, the 4th and 5th columns are non leading and hence x4 and x5 are non-leading variables

we set them as lambda1 and lambda2

[latent gazelle]

oh i see!

[inland birch]

tho i'm probably not hitting your question, repost your original question below since it's been clogged now

hopefully someone else helps

[latent gazelle]

ok

hi! is parametrisation if i can describe a geometric object with an amount of variables equal to the dimensions of that object? im looking for a way to look at it so i can determine what it is, i really havent understood any definition so far or way to think about it.

[final saddle]

@latent gazelle Has your question been resolved?

[latent gazelle]

basically is this way of thinking about what is and isnt a parametrisation correct, or does it need more context – like is it not all-encompassing for what a parametrisation is

[final saddle]

@latent gazelle Has your question been resolved?

[cerulean epoch]

well you can have more than your dimension

but yes it is a continuous map from some open subset of ℝⁿ to your geometric object that you might want to be surjective or additionally differentiable or smooth or whatever

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[static oak]

Could somebody check my answers and answer the questions I wrote down on 3 of the problems

[steady locust]

the negative angles are not in the interval

because the interval is 0 < theta < 360

which is just positive

the other answers are correct

[static oak]

This is confusing me why would sec not equal 2/sr of 3

Sec is the radius/y

so where did the square root of 3 go

[steady locust]

what is cosine?

the adjacent side right

[static oak]

Oh

I mixed up sin and cosine

[flint sleet]

sine is vertical
sin0 = 0

two ways of remembering it

a 3rd one: sin1 is very low, draw the triangle and see how the sides are oriented

the same idea can be done with sin30, which is also relatively small

[final saddle]

@static oak Has your question been resolved?

[static oak]

Forgot to also send the first page and I have one question at the bottom

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[vapid haven]

shdnt i consider the kinetic energy provided by Fsintheta too

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@vapid haven Has your question been resolved?

[vapid haven]

!helpers

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[vapid haven]

<@&286206848099549185>

[sly forum]

consider work-energy relationship

[small grail]

or Integrate accelereation

[sly forum]

you mean integrate work?

[small grail]

vdv/dx

ma = Fcoskx

[sly forum]

you are not given m

[chilly pollen]

The first line...

[small grail]

take m constant

[chilly pollen]

Well you do not 'take' it as constant. It is given to be constant at m for this problem. The mass may not necessarily always be constant

[small grail]

but work energy is better

[sly forum]

but if you take as m

n would be in terms of m

[chilly pollen]

Read what I wrote

[small grail]

in 1/2 mv^2 m cancels out

[vapid haven]

i asked my doubt

wont Fsintheta also contribute to KE.

[chilly pollen]

Do you know the WE theorem?

[vapid haven]

yuh

[sly forum]

how/

[chilly pollen]

Okay for your question, the vertical component is adjusted in the normal reaction with the surface

Try to think about this physically, things dont levitate like that

[small grail]

||int vdv = 1/m int
you get v^2/2, multiply m m cancels out||

[chilly pollen]

And in the context of the WE theorem, no work is done in the vertical direction @vapid haven so no K contribution

Sorry Intergalactic

[sly forum]

alg

[vapid haven]

why

[vapid haven]

bruh how.

[chilly pollen]

For reasons I explained above

[vapid haven]

it could overcome normal

reaction

[small grail]

linear motion

[vapid haven]

is given in q?

[small grail]

yes

[vapid haven]

oh ok its given linear motion

k mb

thanks

[sly forum]

👍

[small grail]

80% phy can be solved by energy conserv in jee

[chilly pollen]

It shouldve still said F < mg. Youre right

[vapid haven]

ill keep that in mind fo sho.

i hope i finish WPE by today ngl

[vapid haven]

they gv linear motion so ig we assume, but thanks

[sly forum]

what is WPE/

[vapid haven]

work power energy, short form i use.

[small grail]

cannot be directly you have given forward acceleration

[vapid haven]

true.

[small grail]

like it can be 2D if Fsintheta> mg

[chilly pollen]

edit:mb

[small grail]

elaborate

yes elaborate

[chilly pollen]

Wait nvm

[small grail]

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[vapid haven]

can i close

thanks yall once again

.close

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[sand escarp]

please help

[sand escarp]

i found this answer key online

but i dont get how they got yxy/2 = 1

how do they know that the height of ABE and base of ABE is the same

[keen chasm]

How do we know which square is smaller

[bitter spindle]

DEFG is smaller because Pythagoras

[keen chasm]

👍

[pliant shore]

that relies on showing triangles ABH and ADE are congruent

if you do a bit of angle-chasing

say angle BAH = x

then angle EAD = 90 - x

so angle ADE = ...

or alternatively RHS congruence might be easier if you know it

[sand escarp]

i know the whole reason why its congruent

but after its congruent what do i do next

[thin cloud]

ABH and AED are congruent => AE=HB

[sand escarp]

what does that give you

[thin cloud]

AE and BH

[pliant shore]

because AE and BH are both the shorter sides of the triangles

[keen chasm]

It gives , AE =BH , then we know that
(BH ×AE)/2 = 1, so AE =√2

[keen chasm]

.

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@sand escarp Has your question been resolved?

[final saddle]

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[rain sentinel]

can someone clarify what i've wrote thanks

[small grail]

again same

what did you not understand ?

[final saddle]

@rain sentinel Has your question been resolved?

[rain sentinel]

the reciporcal bit

[small grail]

this ?

if you write T/Tos = a, 1/a = Tos/T

right @rain sentinel

[rain sentinel]

yh]

[small grail]

so what you have is 1/a *(a^g/Rlamda)

then just apply exponent law

a^b/a^c = a^(b-c)

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[merry lodge]

My prof just introduced my class to complex numbers n said there's 3 ways to represent them:
Trigonometric form
Polar form
Exponential form
The class ended before she could explain any of them
Can someone explain em to me

[summer sedge]

Ok

So

Trigonometric form is using trigonometry

Obviously

[void valley]

Just look for some materials I'd say, it's been already explained better than we are able to write down here I guess, there are plenty (videos, websites, tutorials etc.)

[summer sedge]

Actually yeah this

But

Trigonometric and Polar are the same

I think

[void valley]

And then, if you get stuck somewhere, come back and ask

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[sterile willow]

How do I do this? My main issue is that I don't know if I'm supposed to add al lthe steel parts together before doing the equation or not. But I also still just don't know how to do the actual question itself. I get the modular ratio, but other than that, I don't.

[final saddle]

@sterile willow Has your question been resolved?

[void valley]

Now it would be useful to know the position of the neutral axis

then you need to calculate the moments of inertia and the equivalent moment

[sterile willow]

How do I do that?

[void valley]

start with the cross-sectional areas of the individual components (you can divide the steel channel into 3 rectangles)

[sterile willow]

DO I need to find the area of the transformed areas?

I basically looked throughout many websites on how to do this question, but I don't understand this part

[void valley]

If you're using such a method - yes (and I see you do, because you told us about the modular ratio)

[sterile willow]

Okai, would it be right to add the widths up together and then multiply with n? or to do all of those separately

[void valley]

I wouldn't calculate transformed dimensions, rather transformed areas and then transformed second moments of area

[void valley]

Separately, at the end you can apply your modular ratio (n)

This is why I suggested to split it into components

[sterile willow]

So 375 x 16.667 +12x 16.667 (x2)

??

[void valley]

n should be applied to areas, not dimensions

and only to the non-reference material (wood)

[sterile willow]

I'm so sorry, I really don't understand anything-

[void valley]

Okay, do you understand why we need position of the neutral axis to calculate the stress?

[sterile willow]

Not really

[void valley]

Or do you know the formula for the stress (i.e. flexure formula)?

[sterile willow]

this?

[void valley]

Yes

What does "c" letter stand for here?

[sterile willow]

I think the furthest point from the neutral axis

[void valley]

Yes and this is why you have to know its position

[sterile willow]

the y bar is neutral axis, right?

this thing

[void valley]

Yeah, you'll need it

[sterile willow]

Ok then, the later parts make more sense to me now, this is the only part where I'm HIGHLY confused