help-4

[stiff fossil]

It seems to be around 50-51 degrees in magnitude

[ebon glade]

eww its not even a nice angle?

[rough talon]

btw the answer is 54.8 degrees, i feel like theres something amiss?

[stiff fossil]

If P, B and C are colinear then it should be 50-51

[stark wedge]

??????

[ebon glade]

ewww

[stiff fossil]

Is it though?

[stark wedge]

they are, definitionally

P lies on BC by construction

[stiff fossil]

Hmm

[rough talon]

eh wait mv

[stiff fossil]

Then it should be around 50 to 51 degrees in magnituude

But Idk how to formally prove that

[stark wedge]

let me recheck

[rough talon]

mb its 54

i think i accidentally nudged B

[stiff fossil]

ooh

[stark wedge]

original bg

translation:

In parallelogram ABCD, AB is twice a big as BC. Point M is the midpoint of CD, while P is the foot of the perp from A onto line CB. If angle DAB : angle ABC = 1 : 4 (whence they are 36 and 144 degrees resp), find the size of angle DMP in degrees.

[stark wedge]

54 you say? by measurement?

or can we cook up a proof somehow

[rough talon]

by measurement

[lapis agate]

Ok I got a proof give me a sec

First connect AM and BM. Proof that angle AMB is 90 degrees

You need to use the fact that ADM and MCB are isosceles

[stark wedge]

ok, that much i get.

[rough talon]

oh then use the fact that APBM is concyclic?

[ebon glade]

ok I also finally have a proof I think

[lapis agate]

Yes

[stark wedge]

so it's 54 then?

[lapis agate]

Yes

[stark wedge]

right

dear LORD

ok hold on lemme write that shit down

[rough talon]

how is this an 8th grade question D:

is this comp or sum lmao

[ebon glade]

takes me back to my olympiad days. ptsd from geometry

[stiff fossil]

i understand integrals more than this

and i'm not even in my senior high years yet

[stark wedge]

no it's an entry level diagnostic for 8th graders

[rough talon]

cooked

[stark wedge]

hang on

lemme repro the proof

so we are here right

all angles in degrees

[stiff fossil]

Oh wait

[stark wedge]

i don't see how to continue from here

[ebon glade]

||Let Q be the midpoint of AB. it is clear that the angles DMA, AMQ are 18 degrees. it suffices to show angle QMP 18 degrees. AMBP is cyclic. Clearly QM has length a, so by symmetry also QP=a. then QPA=54. from this BQP=108 and then finally in the triangle MQP we find that the angle QMP is 18||

[stark wedge]

wait why is AMQ=18°

oh wait

AQMD is a parallelogram

[lapis agate]

See that AMBP is cyclic because angle AMB = angle APB = 90 degrees or that AB is the diameter of a circle. Then use the fact that angle ABP is the same as angle AMP

[rough talon]

i was kinda going for this but i couldnt get QMP=18

[ebon glade]

oh yeah those two angles are the same

thats so much easier

[stark wedge]

ugh

that was fucking HELLISH

unbelievable

but thanks everyone @ebon glade @rough talon @lapis agate @stiff fossil

[ebon glade]

but Ann, you are not alone

with struggling with that shit

[stark wedge]

weh

[stiff fossil]

No need for credits to me, I'm just listening to yalls explanation lol

[ebon glade]

that was fucking hard

[stiff fossil]

I was tryna help but at some point I gave up

[ebon glade]

fucking 8th grade lul

[stiff fossil]

I can confirm this should NOT be 8th grade math

[ebon glade]

fine for a competition but not normal school

[stiff fossil]

Idk bout your curriculum tho

[stark wedge]

it is in BG

[rough talon]

even for 8th grade comp its still p hard

[ebon glade]

I mean its a cyclic quadrilateral and then inscribed angles or symmetry. certainly doable with enough time

[stiff fossil]

If AMBP was concyclic then your "Q" should be the center, correct?

[ebon glade]

yes

[stiff fossil]

Okay yeah I definetly get it from here

Thanks

[stark wedge]

ok

yall

i talked to a colleague about this

he went

"idk of any 8th grader who would have solved this"

absolute FUCKING cinema

aight

thank yall once again

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[little jungle]

can someone help me with this question

[little jungle]

in particular j)

I don't know how to find the number of trials or which probability to use

I am allowed a non-cas engine calculator (like numbworks) to solve this question

I know I should be probably using this formula but it still doesn't help me much because I don't know what I am looking for or which value to use

[hollow sage]

you're looking for the value of n, using p = 0.877 and k=12

and you know that the probability is 0.95

So you use the binomial distribution for P(X>12) > 0.95

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[still steppe]

where do they get y = x-2 from

[leaden cargo]

if z is such that arg(z) = π/4, what can you say about Re(z) and Im(z)?

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[still steppe]

the real part is equal to the imaginary part?

[leaden cargo]

yes, just apply that to arg(w-2)=π/4 with w = x+iy

[still steppe]

yo basically y=x-2 because when y=0 x=2

and it has a constant gradient?

[leaden cargo]

wdym by gradient and y=0 x=2?
you just have to take the real and imaginary parts of w-2 and they'll be equal by what you said

[still steppe]

i mean if you draw it

it starts at 2,0

and the line has a gradient of 1

[still steppe]

but i think i see what you mean

thanks

[leaden cargo]

I see, you were thinking geometrically that arg(w-2)=π/4 draws the the line y=x+2 on the plane

[still steppe]

ye

thank you

[leaden cargo]

you're welcome

[still steppe]

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[orchid blaze]

Quick question : I'm really confused by the variable u there:

[orchid blaze]

(Topic is continuous random variable)

Is that a typo or it's something I'm missing?

[mild kernel]

its called dummy

[turbid valve]

You integrate the function with an input of u

[ebon glade]

the x appears in the upper bound

the name of the variable in an integral doesnt matter

[turbid valve]

(wow at least 3 people here)

[ebon glade]

could be u, could be t, could be a smiley

the important part here is that its not x, cause x already has another meaning

[autumn lantern]

c’est une variable muette

[turbid valve]

(oh now I get why it's called a "dummy" variable now)

[mild kernel]

oh maybe dummy is an englishism

[turbid valve]

(as in dumb, mute - et donc le mot français "muet(te))

[sleek nebula]

think we wait for OP to say smth then continue

[orchid blaze]

By adding one new variable, doesn't that change the whole meaning of the random variable?

[ebon glade]

this isnt about random variables at all

$\int_0^1 x^2 dx = \int_0^1 u^2 du = \int_0^2 t^2 dt$

[rocky lotus]

Denascite

[ebon glade]

the name of the variable doesnt matter

[mild kernel]

dummy in this sense means 'missing'

[ebon glade]

but the x appears already in the upper bound

[orchid blaze]

Ooooh

[ebon glade]

so you cant use the x

[orchid blaze]

Because it was already used somewhere else, we had to use a different variable name.

[autumn lantern]

yes to avoid confusions

[orchid blaze]

Aaaaah that makes sense.

Just to be clear

[sleek nebula]

Indeed

[ebon glade]

(whoops, typo in the last upper bound)

[orchid blaze]

We use the same "definition of number". Just a different name.

We don't integrate on a different "space".

[ebon glade]

yes

[orchid blaze]

Gotcha, that makes sense I did not notice that variable usage.

[turbid valve]

(You might have seen something similar if you've ever looked at the Fundamental Theorem of Calculus)

[orchid blaze]

Alright, thank you every one 😄

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[pseudo rampart]

5.3 Trigonometric Graphs
5.4 More Trigonometric Graphs
8.1 Polar Coordinates
8.3 Polar Form of Complex
Numbers; De Moivre’s
Theorem
8.4 Plane Curves and
Parametric Equations
9.1 Vectors in Two Dimensions
9.2 The Dot Product
13.1 limits numerically and graphically
13.2 limits algebraically

I’m grade 11 tomorrow I have my pre calculus final and i really need help.
If yall don’t mind can u explain every single lesson in here ( simplify it ) and also simplify the main rules

Especially 8.3 / de moivres rule

[tawny estuary]

can someone help me with 2 problems pls

[pseudo rampart]

Brooo
Ur timing was a few seconds late mb
Ig u have to take another channel

[tawny estuary]

oh ok

thanks

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[pseudo rampart]

<@&286206848099549185>
Sorry this is complex/long/unclear mb

[scarlet siren]

This is a large variety of topics.
I recommed speedrunning whatever you can through YouTube.
Makes most logical sense for you to just ask any doubts / thibgs you didnt understand here

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@pseudo rampart Has your question been resolved?

[pseudo rampart]

Okay okay sorry
How bout just ** DE MOIVRE **

[wraith heart]

there are youtube videos about de moivre as well

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[sharp crag]

ive got to find d-g

and I know a-B=73

and thats what i've got

[glass kelp]

Note that radii of circle are always equal

So u have two isoceles

[sharp crag]

ive tried basically everything

[sharp crag]

i know

<@&286206848099549185> pls guys

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[noble saffron]

The triangles are isosceles, so there are two equal interior angles for each.

The interior angles sum up to 180:

2*(Alpha-90) +(360-gamma) =180

2×(90-beta) + (360-gamma-delta) = 180

So:

2*Alpha=gamma (1)

2beta +gamma+delta = 360 (2)

Let gamma-delta be x and alpha-beta =73, lets rewrite (1), (2), (3) wrt x and alpha-beta

(1) 2alpha-2beta = gamma -2beta => 146 = gamma -2beta =>
146-delta = gamma-delta -2beta=>146 = x +delta-2beta

(2) 2beta +gamma+delta = 360 =>
2beta + gamma-delta = 360-2delta => 2beta+ x = 360-2 delta

Im also stuck on this ... one equation is missing

<@&286206848099549185>

[static birch]

360*2 doesnt essentially mean its equal to 360.

its true that sin(360*n)=sin(360)=sin(0)=0

[golden gate]

!occupied

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Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[static birch]

but for example

360*3=3 alpha

you can think that its true, alpha=360/3=120

but angles arent really like that

if you have, say

alpha=360+5

you can see that alpha=5

or, alpha=2*90, you may say alpha=180

but dont jump to conclusions

now, i see that you said here ...=720

and then the same is equal to 360

thats a classic case of that exact thint.

thing.

[golden gate]

@static birch this is someone else's thread, I love your spirit to help but please remember that the reason this server has threads is so that people can get personalised help for their questions

[static birch]

its for his problem. why?

[golden gate]

If you can't help the person who opened the thread

[static birch]

oh i do. bc that solves the problem here

please, i have some experience doing triangles and trigo and all the geometry in the world. i fell on these mistakes too.

[noble saffron]

Im not sure that what you are saying is true @static birch . Even though sin(360+5) =sin(5), that, of course, does not mean that 365 = 5 .

[golden gate]

I thought @noble saffron asked their own question and you were solving that lmao

[static birch]

thats my point.

[golden gate]

I'm sorry for the confusion

[static birch]

theres several mistakes here.

it begins at that part

720°=360°

[noble saffron]

Also trying to solve it @golden gate ahah. But got stuck I guess

The author might be long gone lmao

[golden gate]

happens to me more often than I'd like to admit, np

[static birch]

alpha-beta=73
then:
alpha+beta=73+2 beta

[noble saffron]

Keep going

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[brittle spire]

(question redirected from other channel that got closed)

g(x) is a function that outputs a PDF created using x, such as a normal distribution with variance x.

given two arbitrarily close numbers n and m, g(n) and g(m) will also arbitrarily close.

f(x) randomly samples from g(x)

given that, would $\sum_{n=0}^{ax}\frac{f\left(\frac{n}{a}\right)}{a}$ approach a single number as a approaches infinity?

[brittle spire]

originally i thought it would be another pdf but after thinking about it more im fairly certain it would just be a number

[rocky lotus]

sodium fluoride

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[brittle spire]

The old channel #help-46 message

@tribal helm new channel if you’re still interested

[tribal helm]

one thing you can do to easily figure out if it converges is to look at the variance & mean of the sum

[brittle spire]

I feel like it would converge for any g(x) though

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[brittle spire]

<@&286206848099549185>

[remote frost]

Yep, it converges.

It's basically just definition mashing with an integral definition like reimann sums or darboux integrals

It's relatively easy to prove using general reimann sum

And since we're working with stats everything is integrable pretty much lol

[brittle spire]

Cool

[brittle spire]

Is it?

[remote frost]

It's just definition mashing

[brittle spire]

I just figured that repeated convolutions of a single pdf would continuously decrease the variance, and since we’re doing it an infinite alosint of times it would become a single value

[brittle spire]

What’s that?

[remote frost]

Wdym? This isn't a convolution?

[brittle spire]

Isn’t adding random samples of PDFs convolution?

[remote frost]

Wait so is f(x) a random variable?

[brittle spire]

Bruh

[remote frost]

Sorry, I didn't find your original question very clear

[brittle spire]

Sorry

Is there anything specific I can explain

[remote frost]

Sure, so is f(x) a random variable? Since your sampling it from g(x)?

[brittle spire]

Yes

How were you interpreting it?

[remote frost]

I though f(x) was a pdf.

That's generally the notation we use

Okay, but now how is the random variable dependent on x?

[brittle spire]

Because the pdf outputted by g(x) changes based on x

f(1) could equal f(2) but g(1) is a different pdf from g(2)

The example I gave was a g(x) that outputs a pdf with variance x

[remote frost]

Yeah, okay. It's just a weird notation I would have done it differently. It implies that the random variable is a function

[remote frost]

Okay, if the varience is x, then what if the actual variable in the pdf?

[brittle spire]

Wdym

[brittle spire]

Yeah sorry part of the motivation for me asking these questions is so I can learn the proper notation

[brittle spire]

Now that I think about it I could have just removed f(x) from the notation

[remote frost]

Okay, generally we write a pdf as $f(x; \theta) $ where $\theta$ is the parameter vector and x is the value in the support that we want the probability of.

[rocky lotus]

theaveragejoe6029

[brittle spire]

Good to know

[remote frost]

But anyways, your thing still holds since and it approaches some constant if the support of f is finite

The infinite case is a little trickier.

[brittle spire]

Also is this true?

[remote frost]

Why would decreasing the varience imply that the sum of random variables approaches a single number? It may mean that the probability of getting a number further away from the mean decreases which means that the distribution of each f approaches a uniform distribution. But if you sum infinitely many uniform distributions you get a normal distribution. And if you sum normal distributions you get a normal distribution.

Note that last sentence is fairly helpful.

At least for the case of normal distributions

[brittle spire]

Would a variance of zero not mean the probability density is all concentrated in one spot?

[remote frost]

Oh yeah, no that makes sense.

Let me have a scribble real quick

I think it may result from the slln. I could be wrong tho

So basically what you're trying to prove is that is converges in probability to some number.

I don't think it does

Take for example even if you have X,Y iid ~ U[a,b] ~ N(0, u). Then, X+Y ~ Bates(2, {a,b})

Sorry, not bates. I think it's an Irwin hall distribution

But (X+Y)/2 is a bayes distribution, but whatever.

So I think it converges in distribution to a normal distribution with known parameters.

Unless there is an easier way to prove this, it's getting pretty techy

So, long story short, I'm pretty sure it doesn't hold by a bounding argument.

[brittle spire]

Well not some number

Amount of samples times some number

[brittle spire]

Slln?

[remote frost]

Which is some number when fixing n

[remote frost]

Strong law of large numbers

[brittle spire]

my original question about the infinite sum, right?

[remote frost]

Yup

[brittle spire]

hm

[brittle spire]

wouldnt the slln make it converge?

if you pick a value theres an arbitarily large amount of samples of arbitarily similar pdfs being summed in its neighborhood

[brittle spire]

seems like it would converge to the mean of the pdfs

[remote frost]

We can't use ssln since we don't have constant varience unfortunately.

At least not in that way

[brittle spire]

whats that?

[remote frost]

What specifically are you referring to?

Constant varience?

[brittle spire]

yes

[remote frost]

We need each random variable to iid, which we don't have since the varience changes.

[brittle spire]

iid?

[remote frost]

independent and indentically distributed

its a very common assumption

[brittle spire]

but by the definition of g, the difference in the variance between numbers in the neighborhood of each other is negligible

[remote frost]

only if m and n are close tho

its not good enough

[brittle spire]

and any value has an infinite amount of close values being summed with it

[remote frost]

yeah, thats not really how it work

I think you're using infinity a bit too loosely. you can't really throw it around like that

I mean sometimes you can

but not in this sense

cough cough elliptic curves

[brittle spire]

ok not infinite, but an arbitrarily large amount of arbitrarly close values

[remote frost]

yeah, but arbitrary still isn't good enough. take x = 1000 and n = 1 how many summands are there around that yk?

not that many

and certainly not enough to say infinite

[brittle spire]

i find it hard to believe that summing up all the pdfs with a rational variance between 2.(90 trillion zeroes)001 and 2.(90 trillion zeroes)002 variance are too different to approach a mean when summed

[remote frost]

okay, I see what you're saying. The tail converges, that's probbaly true. But you're forgetting the head of the summand.

Consider an example

take f ~ U[0, 12x10^6]

then factoring out 1/a, the first term in the sum is f ~ U[0, 12x10^6], the second term is f ~ U[0, 144]

we have that two are not even close to each other

[brittle spire]

but since every value is in a neighborhood of values with similar pdfs that when summed all converge to a single number, you arent adding random variables anymore. you're just adding numbers

[remote frost]

again, consider the example I just gave the closest pdf to the first term in the sum is like 12X10^6 away, it's not even remotely close.

Honestly tho I can't keep discussing this one. I have to get back to my assignment

[brittle spire]

well yeah if you only sum two pdfs then its not going to converge

[brittle spire]

godspeed

[remote frost]

My advice is maybe ask it in the stats channel and maybe someone has some more insight that I dont have. But if you do I suggest reformulating you'r original question so it's a bit more clear, notation wise anyways.

[brittle spire]

alr

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[mortal cypress]

I mean... if you have an apple and you share it between 3 people each person get a third of an apple.
But if you have an apple and you share it with 0 people... ?
Wouldn't it make sense to 1 / 0 = 1 ?
So why is division with zero undefined in mathematics?

[sleek nebula]

nop

if there's 0 people than how many apples does each person get

[mortal cypress]

There's still 1 apple even if there's no one to hear the tree fall.

[modern rune]

Even in the case where you divide it in 3, there's still 1 apple

But you don't say that 1/3 is 1

[mortal cypress]

It's a different object, though, when divided.

[modern rune]

If it was instead a sponge cake or something

Where cutting it doesn't really change its nature

Or okay think of it this way

when you say a/b = x what do we mean? It means b*x = a

now 1/0 = x, so 0x = 1

solve for x

[mortal cypress]

So, wait. Do you think math actually exists as a real thing in the world, or do you think that it's utilitarian and invented? If it's invented, isn't it just lazy not defining division by zero?

[tender sphinx]

u could define 1/0 in the system if u wanted

but that would actually be lazy

since it has no meaning or logic

[modern rune]

It's invented imo, it's a series of abstractions on things you observe in order to make broader conclusions about the universe

It's a set of rules to a game in a sense

You can change the rules, but then the game breaks

And stops being consistent

[mortal cypress]

In that case - if I think about the problem computationally, it would be a beneficial convention in my opinion to return the original value when dividing by zero. Because in computing you don't really want to have undefined states.

Would it be harmful in some way to define division by zero?

I guess that's what I'm trying to comprehend.

[tender sphinx]

a system cannot be consistent and complete

[tender sphinx]

how much maths have u done?

[mortal cypress]

Can it be complete it something is not defined, though?

[mortal cypress]

I don't even know the multiplication tables by heart.

[tender sphinx]

that doesnt mean u havent done maths

i have a friend who is incredibly good at maths and sucks at arithmetic

[mortal cypress]

Ok, well I'm a programmer, but my knowledge of mathematics is at a very basic level. I imagine many of the common man know more about math than me.

[tender sphinx]

are you in highschool? uni? graduated? somewhere in between?

[modern rune]

If 1/0 = 1, then 1 = 1*0, so 1 = 0

So you either redefine multiplication or reject this definition

[modern rune]

Not really no

[modern rune]

Imagine you have a cat that runs 1 meter in 1 second. This is a normal cat. If you have a cat that runs 1 meter in 0 seconds, this is a teleporting cat that doesn't exist.

But you're saying that they both have speed 1 (m/s)

Idk if my messages are loading, I can't see them

[modern rune]

I'm in 12th and even I don't know them, they're rather useless later on

And I'm pursuing math

[tender sphinx]

im not sure i agree with that

[mortal cypress]

Actually that makes a lot of sense to me intuitively.

[tender sphinx]

but there are undefined states (error handling)

you just treat any undefined state as an error

right?

[modern rune]

You don't need to know the tables if you can multiply in your head quickly imo

[tender sphinx]

what happens if you pass a string into a function for integers

[modern rune]

That 1 is 0? Or the argument makes sense

[tender sphinx]

but that is just inherently knowing ur tables

[modern rune]

Fair enough

[mortal cypress]

That 1 is 0.

It kind of feels like they're opposite sides of the same thing.

[modern rune]

Ah. So if I have an apple and I want to eat it, I have 1 apple. Now timmy the thief comes in. He has 0 apples. He takes your apple. Now you have 1 and he has 0

He says that 1=0 so you both have the same number of apples, so it's fair

[tender sphinx]

they most definitely are not i think ahaha

[modern rune]

Well, you're right that they're intrinsically related, but to say they're identical is a stretch

[modern rune]

Another example that finally got sent (damn internet)

[mortal cypress]

All right. That was certainly interesting and you gave me a lot of new perspectives. Thank you for your time.

I'm not sure if I have anything productive to add. But I'll think about what you all said.

[modern rune]

Yeah it was interesting to try and explain something you take for granted

[modern rune]

Yep, just mull over it and see what new insights emerge

Also there are some systems where division by 0 is defined, but it's usually defined as ±infinity in those systems, not 1

[tender sphinx]

wait wait

before you leave

[mortal cypress]

Yes?

[tender sphinx]

do you know turing machines?

/ the halting problem

[mortal cypress]

I know the concept on a general level.

[tender sphinx]

look into them a bit

then

look at godels incompleteness theorem

they are essentially the same thing

one is maths, one is CS

but it kinda says smth about ur question regarding whether every thing needs to be defined

[mortal cypress]

All right. Thank you shavet.

[crude wolf]

ill give one more argument for why 1/0 doesnt make sense
if we have x apples and y people then we have x/y apples per person
you're interested in when there are 0 people
well in this case any statement we make about these people is vacuously true (because there is no one to make it untrue)
so i could say all of the people have 1 apple or 100 apples or 100000000 apples, i could also say they are all aliens or anything else

in more mathsy language say i have a set A = {} (the empty set) i can make any statement about the elements of A and it will be true

personally i think this is the most intuitive way to think about it

[mortal cypress]

Oh, so are you saying that if division by zero was defined it would break equations with unknowns?

I didn't think of that.

[crude wolf]

ty

did you see what i said before it got deleted lmao

[mortal cypress]

The reason I started thinking about this in the first place was because I was watching 3Blue1Brown on Youtube.

Yes.

[crude wolf]

which video

[mortal cypress]

Oh, well actually he was visiting on StarTalk:
https://www.youtube.com/watch?v=7DEWW1yUN74

[crude wolf]

is there some particular point that caught your interest or ... ?

[mortal cypress]

Just in general. I'm not very good at math, but I've always found it interesting and entertaining.

[crude wolf]

anyhow my point in this was that through this way of looking at it we could assign any value to 1/0 we wanted and it would be equally valid, and that makes none of them valid

[mortal cypress]

Right. So, bottom line: aside from practical niche cases there's no value in defining division by zero in mathematics?

[crude wolf]

thats not necessarily true

but in standard arithmetic, like talking about sharing apples between people, division by 0 is meaningless

[mortal cypress]

The way I feel about zero - and I don't have any math to back me up - is that it's undivided infinity where all numbers originate from.

[crude wolf]

that sounds very philosophical

not a bad thing but its not math

[mortal cypress]

You might think of zero this way as a starting point on the number line, the absence from which positive and negative numbers extend. In that sense all numbers could be thought of as originating from it through addition or substraction.

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[rare shadow]

i know the turning point : (1,7)

i always get confused with the part b

[paper verge]

do you know what root means

[rare shadow]

nope

[paper verge]

it is the point where ur graph intersects with the x line

y=0 line

[rare shadow]

oh

ohh

[paper verge]

which are also the points where your equation is y=0

[rare shadow]

so -1.6 and 3.6?

[paper verge]

yess

[rare shadow]

ah ok

one more if u dont mind

[paper verge]

sure

[paper verge]

are u stuck on a?

[rare shadow]

no

wait let me do part a

[paper verge]

okk

[paper verge]

so what are u stuck on

[rare shadow]

the third one

[paper verge]

it is asking the same thing as this questions b

[rare shadow]

ohj ust different wording?

[paper verge]

yeah if u think about it when ur graph intersect the x line what is its y value?

[rare shadow]

0

[paper verge]

yeah so if a question is asking for solutions on y=0 its basically the same meaning as asking for roots

[rare shadow]

oh

wouldnt there be two solutions?

[paper verge]

yess

[paper verge]

its asking for multiple anyway

[rare shadow]

doesnt the line go through the x line once though

[paper verge]

well it will intersect with x line again but thats after x=5

[rare shadow]

yea

[paper verge]

it doesnt say you are bounded to -1, 5 or anything so

its still a root, just one that u havent drawn lol

[rare shadow]

so how do i find it 💀

[solar arch]

do you need help for this?

[rare shadow]

yeah

i only have one solution

wait no

[solar arch]

yeah?

[rare shadow]

not that one

[solar arch]

which one

[rare shadow]

part c on this

i only have one solution

and i dont know how to get the second one

[solar arch]

ohh

I see

wait a second

have you

drawnn the graph>

[rare shadow]

yea

[paper verge]

do you know how to solve 0=7x-x^2?

[rare shadow]

here

as a quadratic equation?

[solar arch]

for c

yes

[paper verge]

start with simplfing the equation so the roots are clearer

[solar arch]

you convert it to a qudadrattic

equation

and factorize it

[solar arch]

yes

waitt

make x square

positive

[paper verge]

?

[solar arch]

keep the squared variable in a quadratic equation to be positive always

like write x^-7x

=0

it would be x(x-7)

[rare shadow]

that

[solar arch]

yess

noww

what are your values

of x

[rare shadow]

x = 0

and im guessing u have to solve the bracket?

idk what thats called

[solar arch]

yes

itss the values of x

that you solvedd using factorization

[rare shadow]

x = 7 and x = 0

[solar arch]

yes

those

[rare shadow]

oh

[solar arch]

are your roots

[rare shadow]

oh ok

thanks a lot

[solar arch]

no worries =DD

[rare shadow]

@paper verge thank you too

ill close the ticket now

bye

.close

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[polar oxide]

changed a function expression and now im getting different values for the derivatives

[polar oxide]

,, f(x)=\frac{e^{2x}-1}{6e^x} = \frac{e^x-e^{-x}}{6}

[rocky lotus]

<rajel />

[polar oxide]

is this normal ?

[abstract island]

Show your work

[polar oxide]

i.e if im looking for the derivative should i derivate the initial expression ?

[polar oxide]

its verified with wolfram

[worn sparrow]

theyre equal functions , so they have the same derivative

[abstract island]

,w (derivative of (e^(2x) - 1)/(6e^x)) - (derivative of (e^x - e^(-x))/6)

[rocky lotus]

Results provided by WolframAlpha

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

Requested by alonerbean

[fickle rose]

(\frac{e^{2x}-1}{6e^x}=\frac{\frac{e^{2x}}{e^x}-\frac{1}{e^x}}{\frac{6e^x}{e^x}}=\frac{e^x-e^{-x}}{6})

[rocky lotus]

PajamaMamaLlama

[inland nymph]

You might have entered it wrong in wolfram

[abstract island]

Can you show what wolfram outputted to you?

And your input

[polar oxide]

nvm i get it now , thx

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[obtuse swallow]

ORE GAAAAAAAAAA

[obtuse swallow]

MONKEY D LUFFY

[gloomy harness]

silence

got a math question or not?

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[mild spoke]

If the radius of the circle is 1

What is the length of the side of the square

The small triangle is equilateral

And the big one is right-angled

Im struggling to find any rigorous way of solving this simple puzzle, any ideas?

[glass kelp]

i found all the angles

i just need all the sizes

maybe u could work with that

so the big right triangle is a 30-60-90 triangle

[mild spoke]

How can you find the angles?

[glass kelp]

well note that it is an equilateral

so all angles are 60

[mild spoke]

Yep

[glass kelp]

wait lemme pull up a diagram

[mild spoke]

I got what you mean

I think

[glass kelp]

yes

[mild spoke]

That helps

So we're kinda stuck here now

[glass kelp]

u have 3 similar right triangles

but i have been able to establish ratios

oh wait

i can make 60-30-90 using that circle

👀

[mild spoke]

what is 60-30-90

You lost me

[glass kelp]

angles of triangle

[mild spoke]

[glass kelp]

a lot of trig

but doable

[mild spoke]

well the problem is that the 2 you're showing

doesn't go to the bottom-left corner right

[glass kelp]

oh ye

mb

[mild spoke]

So I don't see how we're getting any lengths in the game

[glass kelp]

is this illegal

[mild spoke]

I feel like it is lol

but

it looks right

what would be the rigorous argument?

[glass kelp]

let me try to see

if i can do it with power point

[mild spoke]

Yeah that's true

They are "congruent"

[glass kelp]

https://k12.libretexts.org/Bookshelves/Mathematics/Geometry/06%3A_Circles/6.18%3A_Tangent_Lines

[mild spoke]

But does that help?

[glass kelp]

sas congruency?

we have

1. DC = BC (side, two tangents theorem)
2. ADC = ABC (90 deg = 90 deg)
3. AD = BA (side, radius of circle)

therefore ADC congruent to ABC?

[mild spoke]

damn i started with other letters

[glass kelp]

there we can see that they both must be 30-60-90?

[mild spoke]

I think so actually

[glass kelp]

we can find AC

using trig

[mild spoke]

OB is 1

so the rest next to A is 1 too

[glass kelp]

yes

[mild spoke]

so if the two triangles are 30-60-90 it's a win

I'm trying to convince myself of it though

[glass kelp]

it should be

[mild spoke]

So, if I can prove this "theorem":

• A circle with center O'.
• Points A and B on the circle.
• Tangents CA and CB meet outside the circle at point C.
• Let’s prove that CO' bisects ∠ACB

I'm good

[glass kelp]

it should bisect

cuz they are congruent

its actually an incircle

krazy

https://www.math-only-math.com/incircle-of-a-triangle.html

[mild spoke]

ok i got it

it's true because the two triangles share a hypotenuse

they also share a size (radius of the circle)

[glass kelp]

ohhh

[mild spoke]

and they share an angle (90° angle)

[glass kelp]

how did i miss the hypotenuse

[mild spoke]

and I think three degrees of liberty is enough

[glass kelp]

ye

sas

[mild spoke]

what does sas mean

[glass kelp]

side-angle-side

[mild spoke]

ah okok

what's the rigorous argument for AD = 1?

(it's not AB = AC typo sorry)

[glass kelp]

Make another point

On the equilateral

[glass kelp]

Aoc = 60 not 30

[glass kelp]

Then u have a right angle

In fact u will have 3 right angles

And that will force AO that point to be 90 as well

[mild spoke]

ah it's a square

[glass kelp]

Hence a rectangle

Since A to that point is parallel to AD

They must have the same length

[glass kelp]

Yes

[mild spoke]

and do you need that extra point down there for the rest of the side of the square?

surely I can just use AO'C with trig right?

tan(60) = AC/AO'

tan(60) = AC

sin(60)/cos(60) = (sqrt(3)/2)/(1/2)

AC = sqrt(3)

[glass kelp]

Yes

[glass kelp]

I would

[mild spoke]

ngl it's weird cos it feels like AC is bigger than sqrt(3) on the drawing lol

[glass kelp]

But not sure if it’s the most efficient way or not

But certainly one of a way

Sqrt(3) is around 1.7

So around twice the radius

[mild spoke]

yeah actually makes sense

I just stuck my ruler on my screen

the ratio between AD and AC is around 1.7

[glass kelp]

Sigma male method

[mild spoke]

can i say thank you anyhow on this server lol

!thanks

again thanks for the help

.close

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[high sequoia]

Hey guys I was wondering how these two are forming a right angle triangle I appreciate the help

I can’t rlly visualise it

[glass kelp]

Try drawing

An altitude

[high sequoia]

Ohhhhh I see it now that makes much more sense thank you!

The second one would also have a similar deal right?

[glass kelp]

Yup

Make ur own altitude stuff

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[topaz tree]

-"How many ways is there to such values for k1.k2.... k6 such that their sum is equal to 28 and all k values are positive integers"

[topaz tree]

I am unsure whether my thinking is correct and would appreciate some guidance

I first noted that the first element can be any value from 0 to 28

[flint sparrow]

are you familiar with the stars and bars method

[topaz tree]

The second one has any value from 0 to 27

and so forth, all the way to 28-6 = 22

so 28!/22!

[topaz tree]

No, I am not

[flint sparrow]

So basically, the idea is you want to divide 28 stars into four piles, using 3 "bars"

You can think about it as cutting a log of length 28 into 4 pieces using 3 cuts

[topaz tree]

Hmm, okay

[flint sparrow]

Since each k_i has to be a positive integer, you can only cut inbetween the stars

[topaz tree]

So, no half stars

[flint sparrow]

right

so you have 28 stars, meaning there are 27 places where you can place your 3 bars

so it's 27C3

[topaz tree]

Why does it have to be 3 bars?

[flint sparrow]

oh it should be 5 bars

Idk why I got 3 lol

because you want to have 6 positive "pieces"

[topaz tree]

Is it because the last bar has to add up to 28?

so you can really only choose the first five bars

[flint sparrow]

The idea is you're trying to make 6 pieces, so you make 5 cuts

Like if you're trying to cut a cake into 6 slices, you make 5 cuts

[topaz tree]

Okay, that I can understand

But why are we interested in the combinations (since you wrote 27C3 last time) and not the permutations?

I am wondering since the question asks about the possible values, i.e possible ways of creating these 5 bars

[flint sparrow]

Because the bars are basically ordered already

so the number of stars to the left of bar 1 is k1
the number of stars between bars 1 and 2 is k2
and so on

Or ig another way to put it is the order of the bars doesn't matter

Because if you were to say, switch bar 1 and bar 4, the values of k1 thru k6 don't change

[topaz tree]

Right, the values do not change but the way you ccut the bar did, no?

[flint sparrow]

Sure, but we only care about the values of k1 ... k6

[topaz tree]

Am I making the mistake of thinking about the total ways of adding up to 28, rather than the possible values?

I think I am

[flint sparrow]

Not really

the total ways of adding to 28 is actually a lot less

You just need to understand why the order in which the bars are placed doesn't mater

because all that matters is their placement in the end

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@topaz tree Has your question been resolved?

[topaz tree]

Ok so after some thinking with this method

28 stars, 6 cuts

the cuts represents k1 to k 6

or my bad, 5 cuts results in 6 bars

each bar contains 0 stars or more

so CSSSCS represents two bars made by 2 cuts

first bar has 3 stars, the second bar has 1 star

We have 5 cuts and 28 stars

So, we can make these cuts in 28C5 places

A cut can be made on one position and a star rcan be put on one position. There are 28+5 = 32 positions. I can put the cuts in 32C5 possible positions

aghh im lsot

[flint sparrow]

This is only if you allow the k values to be 0

also 28+5=33 not 32

But for your problem, everything has to be positive, so every bar has to be between 2 stars

and you can't have 2 bars in the same "slot"

So you have 27 slots for the bars (between each pair of the 28)

so 27C5

[topaz tree]

I am just a tad confused as the values are allowed to be 0

An example from my textbook'

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[opal spruce]

What do you think is the null hypothesis in this question

In the question, they are trying to prove that the probability is >= 60%, so the null hypothesis should the probability is < 60%

why?

[covert acorn]

hmm

thats a hard one

[opal spruce]

i felt like in the first case type 1 error will never happen because it happens only when the null hypothesis is true

the probability can be 0

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[orchid jewel]

Help help help

[wraith heart]

!da2a

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No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

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[fading hill]

can someone walk me through these notes? i missed that day and im confused about the notes because there is no video only this

[verbal badger]

Is there anything specific you're confused about? Try to go through the notes and ask us when you get stuck.

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[fading hill]

is there a difference between approacing -infinity and infinity?

im doing homework rn and the thing has me find both for the same equation

and i go tthe same answer based on the notes

but im not sure if thats correct

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[fading hill]

this is the key- can someone explain to me why 16/x and 1/x are 0?

[storm perch]

What is 16/99999999999999999999?

Or, what is $\lim_{x\to\infty}\frac1x$?

[rocky lotus]

;(

[fading hill]

um

i am not sure

hold on

0

yeah

okay okay okay

so

i got to lim(x->infinity)x

but wouldnt the answer just be one?

why is it infinity?

[wild linden]

The limit of x as x -> infinity?

[fading hill]

yes

[wild linden]

I mean...

[fading hill]

hold on a seconf

[wild linden]

What's the limit of x as x goes to 10000?

[fading hill]

i think im simply stupid

[wild linden]

Not stupid, just learning

[fading hill]

10000...

so

its infinity

that makes sense

okay wait though for the x->-infinity one whats the difference in the answer

ohhhhhhh wait nope

its the same thing again

[wild linden]

Hahah yeah 😄

Sometimes math is so simple it trips you up

[fading hill]

yup

im.. going to fail this test lol

[wild linden]

Nah, you're doing great

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[magic sentinel]

Yo chat I got the inverse equation, but how do I know if its the positive or negative result

[magic sentinel]

These were the two results I got, but how would I know which answer would be the correct one given the restricted domain

[native plinth]

It is stated here that x≤-1

[magic sentinel]

nvm my domain is 3 not 2

[magic sentinel]

for the original function

im talking bout the inverse

[noble saffron]

Notice that Df = (-infinity,-1]

[magic sentinel]

yeh

[noble saffron]

So the image of the inverse should be Df

[native plinth]

Well, the domain of the original function is(when it exists) the image of the inverse function, and the image of f should be the domain of the inverse

So that's something you should consider, the function's had its domain restricted

[magic sentinel]

these are the two equations I got, how do I know which one is correct

[native plinth]

Whichever satisfies the conditions we gave you

[magic sentinel]

whats the condition, the restricted domain?

for the restricted domain for the inverse, i just found the range of the the original function

[noble saffron]

Yeh, exactly. Its Df = (-infinity,-1], so the range of the inverse is also this

These are all negative numbers

[noble saffron]

So which of these expressions would he correct in that case?

[magic sentinel]

idk negative?

[magic sentinel]

is the restructed domain of the inverse this

[noble saffron]

Sorry, I was checking if you got the domain of the inverse right. But lets focus on your question first.

The domain of the original function is the range of the inverse

If the range is only negative values, only your expression on the right is valid

[magic sentinel]

oh ok

[noble saffron]

The one on the left will sometimes evaluate to poaitive balues

[magic sentinel]

negative range = negative answer?

[noble saffron]

In this case yes. But its a bit more complicated than that if the range wasnt only negative.

But give me a sec, i think u got ur domain of the inverse wrong

[magic sentinel]

i did

i forgot to put the second x in after the 2

[noble saffron]

How did you get to the answer $x\geq 3$

[rocky lotus]

Rui Martins

[magic sentinel]

this is my new domain

[noble saffron]

Right!

[magic sentinel]

i forgot to put the second x

[noble saffron]

And for sanity check, the domain of the inverse should coincide with the range of the original function. Is that the case?

[magic sentinel]

yes

[noble saffron]

All good then

[magic sentinel]

ok thanks

[noble saffron]

You want me to explain this better? Or is this enough for you?

[noble saffron]

Np

[magic sentinel]

if u could that woould be great

but if u can't could u send a yt video

[noble saffron]

Sure. Lemme think of an example where the range is neither negative or positive.

[magic sentinel]

also, could u give a way where I wouldn't need to graph it

[noble saffron]

Yes, ill do my best.

Give me like a couple of minutes to think of the whole thing through

[magic sentinel]

ight

[noble saffron]

Consider this example:

F(x) = x^2 -2x, Df=(-infty,1)

First, do the same as you did in the first exercice, and reach two possible expressions for the inverse

[noble saffron]

Exactly! Now lets try to understand that only one of these expressions makes sense

[magic sentinel]

would be the one where the total is really low

[noble saffron]

What is the domain of the inverse first?

[magic sentinel]

should be the right one?

[magic sentinel]

negatrive infiniti to +1

[noble saffron]

Im actually not sure yet, let me follow this through with you and we should both arrive at the correct conclusion

[noble saffron]

Thats not correct

Remember. The range of f is the domain of f^-1

What is the range of f?

[magic sentinel]

wait i thought u was asking original

range of the original will be

x > -1

[noble saffron]

Exactly. You mean the domain tho, right? The domain of the inverse

[magic sentinel]

yeh soz

[noble saffron]

Np

[magic sentinel]

Domain of inverse is x > -1

[noble saffron]

Now lets evaluate your left and right expressions on this domain, and remember that the range of the inverse should be the domain of the original function x<=1

[magic sentinel]

do i sub in any values

[noble saffron]

If you want. But what I am really asking is what is the range of the left expression and what is the range of the right expression, given the domain x>= -1

[magic sentinel]

range on left would be x > 1

[noble saffron]

Only one of these will give you the correct range, that should be the domain of the original function

[magic sentinel]

range on right would be 1 < x < -infiniti

[noble saffron]

Thats not right. Sub x=-1 on the right, and you get 1 -sqrt(0) = 1

Then, as x grows, 1 - sqrt(x+1) goes to -infty

[magic sentinel]

im tweaking

[magic sentinel]

ight yeh

[noble saffron]

The range will be (-infty,1)

This coincides with the original function domain, if you notice.

[magic sentinel]

for the one on the left?

the left on is 1 + sqrt(x+1)

[noble saffron]

If you do the same with the left expression, it wont coincide. So it will not be the correct expression for the inverse.

[magic sentinel]

wouldnt that be y > 1 for the range on the left

[noble saffron]

For the right one, for 1-sqrt(x+1)

[magic sentinel]

is the left's range this

[noble saffron]

Yeh, so [1,+infty)

[magic sentinel]

and the rights range 1>y>-infiniti

[noble saffron]

This does not coincide with the domain of the original function

[magic sentinel]

so that means only option is the second one

we just match domains with ranges and ranges with domains

i think i understand it now

[noble saffron]

Yeh, because the domain of f should be range of f^-1 and viceversa. When u have two options, only one will ever be correct, if the function is invertible

[noble saffron]

Exactly

!

[magic sentinel]

ok thanks

[noble saffron]

Nice doing buiseness with you 🙂

Eheh

[magic sentinel]

one more question

[noble saffron]

Good luck

Yeh sure

[magic sentinel]

part ii

can i just use auxiliary form

i got part ii and i solved if thats the case, i just wanted to know if auxiliary angle form is adequet

[noble saffron]

I think ure supposed to use the result in the first question to solve the second one.

You can only really solve sin(ax) = b or cos(ax)=b, so you need to convert the sines to cossines or vice versa

What is the auxiliary form you mention?

[magic sentinel]

U may not know my syllabus so if u don't know if it will be adequate proof dw

coz sometimes in exams, i dont wanna like provide insuffiicient reasoning

@noble saffron dw about that question

thank you for helping me i gtg

[noble saffron]

Np, gl

[magic sentinel]

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[random thistle]

How do I calculate the volume?

[random thistle]

The answer is supposed to be 4.5pi but I don't understand how they got to that point

[noble saffron]

I suppose you have to construct the volume integral of that shape right?

[random thistle]

Yea that is correct

[noble saffron]

Okay, think of the slices of this paraboloid.

Take one of these slices at height y. What would be its volume dV, in terms of dy and y?

[random thistle]

You mean how I would write the integral?

[noble saffron]

Well, yeh, if you can give me the expression of the full integral go ahead

[random thistle]

This is kinda how I wrote it, not sure if this is the correct way to go about it

[noble saffron]

Not really I think.

Try to follow my reasoningz we can discuss why this is wrong afterwarda if you want

[noble saffron]

Can you answer this question?

Bear in mind, that each one of these slices has height dy, and its a cylinder. How would you compute its volume?

[random thistle]

Honestly I have no clue what I'm doing really so I'm not sure how to answer your question

[noble saffron]

Dont worry. Lemme try to explain this a bit better

[random thistle]

Alright

[noble saffron]

The idea is to slice the paraboloid into several cylinders. And then sum their volumes, to obtain the volume of the full shape.

Do you understand this idea?

[random thistle]

Yes I do understand that part

[noble saffron]

Okay. But you will make the cylinders have an infinitesimal height, so you can make infinite slices, and then sum them all up

Their height will be dy

Like in the figure. Do you understand?

[random thistle]

So in this case dy should be 3 since that's the total height, am I understanding you correctly?

[noble saffron]

Not really. The limits of your integral will indeed be 0 to 3, but dy is a symbol

It represents the height of an infinitesimal cylinder

Dont think of it as a number

[random thistle]

Oh okay I think I get what you mean

[noble saffron]

Okay, so now we get to my first questio

To compute the volume of a cylinder, you need the height and the area of the basis right?

[random thistle]

Yes that's correct

[noble saffron]

We have the height dy, its the same for all cylindera

The basis area will be pi*R^2, where R is the radius yeah?

[random thistle]

Yea

[noble saffron]

Does the radius of the sliced cylinder depend on the height at which you cut it?

[random thistle]

Yes, in the picture the radius is bigger at the top and then it gets smaller toward the bottom right?

[noble saffron]

Yes, exactly right