help-42

[remote mural]

ok i got it

wait

i got it

[spark relic]

Not log - log = log

[spark relic]

Yep

Now you can cancel out logs

[remote mural]

yeah i suck at logs

[spark relic]

You just have to practice more

[remote mural]

trust me

i practice a lot of problems

daily

but

logs just dont work out sometimes

i practice all kinds of shits

polinoms

integrals

like everything but when i have a log equation like this i just DONT

[spark relic]

.

[remote mural]

its like i am trying to read chinese even tho i know the definition of log

wait what do i do now

i got like

[spark relic]

Solve the exponential equation

[remote mural]

a exponential equation

ok

let me try

isnt this wrong

shouldnt i keep the +

[spark relic]

Keep +

Doesn't make sense if you change it

[remote mural]

true

[spark relic]

Btw this is wrong

In a log equation always make log = log

Expand is take steps backwards

[remote mural]

thank you

[spark relic]

Worse if you do it wrong

[remote mural]

yeah i cant solve this exponential

im blanking

[spark relic]

Because log(a + b) = log(a)log(b)

And

You can't solve it because LHS is wrong too

x(1-lg) = x - xlg

[remote mural]

no?

because then

it means

x - lg^{x}

which is not

[spark relic]

What?

x - xlg(5)

That's it

[spark relic]

Oh yeah sorry

My bad

Yes, it's lg(5^x)

[remote mural]

wait

this is how it looks

right

no mistake

[spark relic]

Yep

[remote mural]

and that is indeed, the answer

correct answer

well thank you @spark relic

[spark relic]

Good job

[remote mural]

have a blessed day:)

.close

[calm coral]

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[brittle siren]

this is more of an optics question

[brittle siren]

check out this scenario where we have lighbulbs A and B (blue and red)

we also have an obstacle in the middle and screen behind it

the lighbulbs are throwing light beams all over the place but i have marked the "interesting one" as they just touch the obstacle forming a shadow

based on the distances given, can anyone help me find the value of all the shadow on the screen?

( the obstacle is parallel to the screen )

[calm coral]

@brittle siren Has your question been resolved?

[brittle siren]

<@&286206848099549185>

[calm coral]

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[brittle siren]

.reopen

[calm coral]

✅

[timber imp]

hi

[brittle siren]

hello

[timber imp]

can you help me with lenear equation

[brittle siren]

what about it?

[timber imp]

the basics of it

[brittle siren]

do you have any specific question?

[calm coral]

@brittle siren Has your question been resolved?

[calm coral]

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[elfin shadow]

Yall i aint got a scooby doo how to do this one

[trim cliff]

oh bro use the uhh

law of sines

okay

[elfin shadow]

It is non calculator btw

[royal basalt]

Yeah the angles are good. You won't need a calci.

[trim cliff]

these are special triangles i think

so that means

[royal basalt]

Nope. It's just that sin30=1/2

[trim cliff]

the length of b to the intersection point is 3 right

[royal basalt]

Yep

But you don't need it. Just find BD.

[trim cliff]

so then just do umm

1/2bh

[royal basalt]

And you already have AC

[elfin shadow]

How could you do that using bd when ac isnt a line in the triangle BAD?

[kind kraken]

Half of it is

[royal basalt]

They've given AC =10.4 cm for a reason.

[kind kraken]

Half of AC = the height of BAD

[trim cliff]

Cd is 10.4

[elfin shadow]

How do u know it is half like without assuming

[royal basalt]

It's a property of kit

[kind kraken]

AC is 10.4.. not CD

[trim cliff]

if adb and cdb are both 30° then bd is def a bisextor

[royal basalt]

This is there for a reason @elfin shadow

[trim cliff]

using 30 60 90 triangles, cd is 10.4

cuz from the center to c is 5.2

[elfin shadow]

How do u know that? Is it a rule?

[trim cliff]

yes

[elfin shadow]

But for what reason

[kind kraken]

Yes

[kind kraken]

Bisector of isoceles triangles

[kind wagon]

bro you got a lot of right triangles and know almost all angles

[trim cliff]

so umm whats 5.2 times root3

crap

[elfin shadow]

But i have to do it without calculator

[mint verge]

(5.2√3×5.2/√3)/2 i think

[trim cliff]

yeah thays a problem

[elfin shadow]

What is kit?

[mint verge]

maybe wrong but

i meant multiply of diagonals divided by 2

[royal basalt]

Kite.

[mint verge]

just find their length using angles

[elfin shadow]

What is the property?

[trim cliff]

the final solution should be (3 * 5.2) + (5.2 * 5.2root3)

[elfin shadow]

Yeah but how'd u do that without using a calculator

[trim cliff]

thats the final answer

[kind kraken]

Dude.. you have two sets of triangles that add up to two rectangles

The area of a rectangle is length * width

[trim cliff]

PH YEAH

[mint verge]

Call O the intersection of lines AC AND BC

[trim cliff]

and then u wont have to worry about root 3

[mint verge]

if AC = 10.4cm

then AO = 5.2cm

[elfin shadow]

I dont see how they add up to two rectangles

[mint verge]

and OC too

[trim cliff]

yepp

[elfin shadow]

Excuse my stupidity

[mint verge]

because it is isosceles

as i see

no problem my bro

you are not

exactly

[trim cliff]

cuz like two right triangles if u put them together they make a rectangle

ykwim

[mint verge]

but i still think the question is incompleted itself

[elfin shadow]

Can you plz go over it again once more i've lost myself

[trim cliff]

area of triangle abc is def 15.6

[elfin shadow]

We are finding the areas of Bad and bcd right

[mint verge]

What do we care if it is a kite or if its two angles are ninety degrees?

[trim cliff]

ok wait

[trim cliff]

YEAH

so look at abc

[mint verge]

So is this proof that it's a deltoid?

[trim cliff]

if u use the two trianfles there and make a rectangle

its 3 * 5.2

so thats 15.6

[mint verge]

@trim cliff man just say it to me

xD

[mint verge]

.

[mint verge]

.

[trim cliff]

wdym

[elfin shadow]

Im legit not following , im cooked for tomorrow's exam

[trim cliff]

as in… shoulder?

[trim cliff]

ok wait listen

lets start with the area of abc

[elfin shadow]

Yes

[trim cliff]

so look at this

so we have a 30 60 90 triangle

so since the hypotenuse is 6

[mint verge]

So how can you tell me that this shape is a deltoid? The intersection of the diagonals forms 90°

[trim cliff]

we can assume the length from B to the center of the triangle is 3 right

[mint verge]

why we said 90° to them

[trim cliff]

ive never heard of a shape called deltoid

[elfin shadow]

How so

[trim cliff]

look at the rule

for every 30 60 90 triangle

[elfin shadow]

Oh yeah so you do 1 * 3

Innit

[trim cliff]

the side opposite to the 30° is gonna be half of the hypotenuse

[trim cliff]

yuhhh

[kind wagon]

opposite side of the 30° angle is half the hypotenuse

[trim cliff]

what he said^

so THEN

[kind wagon]

its 7th degree geometry

[mint verge]

It is a quadrilateral formed by two isosceles triangles with coincident bases

[trim cliff]

oh yeah a kite

[elfin shadow]

And what the adjacent side to it

[mint verge]

maybe i said it wrong but yk

[elfin shadow]

What is 7th degree

[mint verge]

how do we know this?

[kindred estuary]

This image might be better and clearer to use

[elfin shadow]

Im following

[mint verge]

7th grade he meant i think

[kind kraken]

a^2 + b^2 = c^2

[kind wagon]

i dunno just use apply cos to it and find out

[trim cliff]

well we can assume that BC can be cut in half at the center point

so 10.4 divided by 2 is 5.2

[kind wagon]

bro deltoid is the muscle near the neck

[trim cliff]

it says in the problem

[trim cliff]

THATS WHAT I SAID

[kind kraken]

He doesn't have a calculator, and you'd have to know that cos(30) = .866 or sqrt(3)/2

[trim cliff]

yeah

[elfin shadow]

This is a gcse question so idk what that is in america

[mint verge]

That there are two isosceles triangles with coincident bases?

[elfin shadow]

Yes i follow

[trim cliff]

thats what a kite is yes so we know its that

[mint verge]

really?

xD

[trim cliff]

so we can do area of a triangle 1/2 bh right

[kind wagon]

oh i didnt even notice 😭 european moment

[trim cliff]

yes that is how kites work

[mint verge]

lol i'm asian

what does this have to do with

[kind wagon]

english is not my first language

[mint verge]

well that's not mathematical at all loll

[elfin shadow]

Waittttt wouldn't BO be 3

[mint verge]

me too

[trim cliff]

yes it is

[trim cliff]

bro the definition of a kite is what you said was a deltoid

[elfin shadow]

So which side is 5.2

[kindred estuary]

It also says in the problem that it's a kite too btw

[trim cliff]

CO

[mint verge]

@trim cliff okay, but is there such an element/shape as a kite in mathematics?

It's just the name of an object

non-mathematical

[trim cliff]

in the united states we call these shapes kites, i dont know what the proper term is for them

[mint verge]

pretend I told you the same

[trim cliff]

deltoid kite whatever

[elfin shadow]

So how can bc be cut in half at the centre point , did u mean AC?

[trim cliff]

YEAH mb

[mint verge]

oh really

[elfin shadow]

Same in the UK

[mint verge]

ahahaha

broo

sorry

[trim cliff]

AC is cut in half

[mint verge]

AHAHAHA

[trim cliff]

no problem

[mint verge]

okay okay

it's good then

[trim cliff]

👍

[elfin shadow]

Ok so AO / CO is 5.2

[mint verge]

@kindred estuary sorry i wasnt know that

[trim cliff]

no no AC/2 is 5.2

cuz ac is 10.4

and BD cuts it in half

[elfin shadow]

Yeah so AO and CO are 5.2 innit?

[mint verge]

i didn't know that "kites" are quadrilaterals formed by isosceles @kindred estuary

[trim cliff]

YEAH

[kind kraken]

You have 4 similar triangles.. they are all 30/60/90... the unknown length, B to center is 1/2 of the short hypotenuse

[trim cliff]

yes

so given all of that

[kind kraken]

And the unknown length center to D is a ratio of the 5.2 base of the smaller triangle

[trim cliff]

the area of ABC is 15.6

cuz each triangle is like 7.8

right?

[elfin shadow]

Yea so what then

[kind kraken]

Set up some ratios with your known side lengths and solve for x

[trim cliff]

okay now lets tackle ADC

[trim cliff]

this would work better i bet

[trim cliff]

so according to thr right triangle poster, can you tell that CD is 10.4?

[elfin shadow]

How is that so, dont you do 10.4 * 5.2 all divided by 2?

[trim cliff]

no remember our height was 3

BO

[elfin shadow]

Oh no i know now

Yes

[trim cliff]

yess

[elfin shadow]

3*5.2

/ 2

[trim cliff]

yes

and then multiply that by 2 cuz there’s two of those triangles

[elfin shadow]

Yeah

Thanks i never knew about that 30 60 90

[kind kraken]

or just.. don't divide by 2

[trim cliff]

yeahh they’re pretty useful

[kind kraken]

since those two triangles make one rectangle that's 3 x 5.2

[trim cliff]

thats what i did but i wanted to explain it better to him

[elfin shadow]

Yea i never got taught it like this so its hard to conceptualize even when seeing the pic u sent

So we're on CAD Triangle innit

[trim cliff]

yess

[kindred estuary]

What did you get as a final answer?

[trim cliff]

so according to this, since CO is 5.2, AC should be 10.4 right?

[kind kraken]

So now you have a 3x5.2 rectangle that is similar to a 5.2 x ? rectangle

[elfin shadow]

The same thing just divided it by 2 and * by 2 again coz triangle areas are / 2 then theres 2 triangles

But i understand what apple pi means

[trim cliff]

ooh yeah

3 is to 5.2 as 5.2 is to what?

[kindred estuary]

That's not an answer, I'm asking what value did you get?

[elfin shadow]

15.6

[kindred estuary]

No

[elfin shadow]

Wdym

[trim cliff]

he’s still solving it

[kindred estuary]

You guys went in the worse way possible to solve this

[elfin shadow]

OK WHAT IS THE BEST WAY 😭😭😭 😭😭😭

[trim cliff]

YEAH

[kindred estuary]

You have this right triangle here

[trim cliff]

oops

[shrewd sequoia]

just use kite area formula

[mint verge]

Why does this confuse me?

[kindred estuary]

Using this, you can find the orange side

[mint verge]

it should be very simple

[kindred estuary]

Then the kite formula

[mint verge]

12×10.4/2 then?

[trim cliff]

12

[mint verge]

it must be wrong how

[kindred estuary]

Let the OP do it

[elfin shadow]

Let me draw it out

[mint verge]

6×6√3 is also the area

but 12×10.4/2 is also the area

how

how can i go wrong with this lol

[kind kraken]

(12x10.4)/2 != 12x10.4/2

[mint verge]

they are same

actually

[kind kraken]

Ehh.. in this case, yeah

[elfin shadow]

I got 62.4 cm^2

Is that aight

[elfin shadow]

What is this exactly

[kind kraken]

but not seeing parentheses where there should be gives me the heebs

[mint verge]

oh okay

[elfin shadow]

Never seen this either

[mint verge]

wrong question guys

exactly

i can tell this to you

[kindred estuary]

Special right triangles

[mint verge]

he just wanted it to be rational

with 10.4

actually it is not 10.4

or the 6 is not 6 whatever

one of them is wrong

[elfin shadow]

Special right triangles is the rule for just these specific 2 types with the angles in them exact places?

How do yall even find these rules we dont get taught them here

[mint verge]

36√3 is ≈ 62.3 12×10.4/2 = 62.4

[kindred estuary]

Doesn't have to be in the exact same places, just the same configuration

[elfin shadow]

Wdym by configuration

[kindred estuary]

Same rule, different configuration

The side across 30 degrees will always be a

Hypotenuse is 2a, and side across 60 degrees is a * sqrt(3)

[elfin shadow]

So if 30 degrees was swapped with 60 degrees , the sides' formulas would swap?

[kindred estuary]

Fyi $6 \sqrt{3} \approx 10.4$

[potent lotus]

CaptainNova22

[kindred estuary]

Yes

[elfin shadow]

Thanks bro

Did u get taught that special triangle rule in school or nah

Just curious

[kindred estuary]

Yes

[elfin shadow]

We do not get taught that 👹

How can i find more rules like it

[kindred estuary]

You were probably taught something similar based on that

[kindred estuary]

Google

[elfin shadow]

I go to all my classes and i didnt know what the hell to do when i just stared at this question

Am i allowed to send another question which im struggling with? Or do i open a new tab

Imma take that as a yes lemme just find it

As u can see that is how far i got

[calm coral]

@elfin shadow Has your question been resolved?

[elfin shadow]

<@&286206848099549185>

[mint verge]

okay i got it already

thanks

but it's still wrong question

yk

but not that important maybe

[calm coral]

@elfin shadow Has your question been resolved?

[elfin shadow]

Wait i solved it thank God

.close

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[maiden hull]

Hi- if Bill had 75 peaches, each day he kept a fraction of peaches, and gave the rest away, then ate one. these are the fractions he decided to keep. 1/2 , 1/4, 3/4, 3/5, 5/6, 11,15

[maiden hull]

in which order whould he use the fractions so that

bill would only be left with one

[calm coral]

@maiden hull Has your question been resolved?

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[molten sundial]

helloo could anyone help point out the mistake i made doing this integral? please dont solve it your own way, just tell me where i went wrong 🙏
$$\int{ \frac{x^3} {\sqrt{x^2 + 100}}dx}$$
$$x = 10\tan\theta \text{, } dx = 10\sec^2\theta\ d\theta$$
$$\int{ \frac{1000\tan^3\theta} {\sqrt{100\tan^2\theta + 100}}10\sec^2\theta\ d\theta}$$
$$\int{ \frac{10000\tan^3\theta\sec^2\theta} {10\sec\theta}\ d\theta}$$
$$1000\int{ \tan^3\theta\sec\theta\ d\theta}$$
$$1000\int{ (\sec^2\theta - 1)\tan\theta\sec\theta\ d\theta}$$
$$u = \sec\theta \text{, } d\theta = \frac{du}{\sec\theta\tan\theta}$$
$$1000\int{ (u^2 - 1)\tan\theta\sec\theta \cdot \frac{du}{\sec\theta\tan\theta}}$$
$$1000\int{ (u^2 - 1)du}$$
$$1000(\frac{u^3}{3} - u)$$
$$1000(\frac{\sec^3\theta}{3} - \sec\theta)$$
$$\sec\theta = \frac{1}{\cos\theta} = \frac{\sqrt{x^2 + 100}}{10}$$
$$1000(\frac{(x^2 + 100)^\frac{3}{2}}{30} - \frac{\sqrt{x^2 + 100}}{10})$$

[celest sinew]

yall help i dont understand this

[potent lotus]

declspecl

[celest sinew]

nvm

[calm coral]

@molten sundial Has your question been resolved?

[molten sundial]

<@&286206848099549185>

[rancid crater]

I really cant spot anything wrong with it, do you know what the correct answer is suppose be?

[molten sundial]

integral calculator says $$\frac{ (x^2 - 200)\sqrt{x^2 + 100}}{3}$$ and when compared against it numerically, my answer is not equal 😦

[potent lotus]

declspecl

[molten sundial]

and they are visibly not equal when graphed as well

i also found a youtube video for this question, but their answer was also wildly different

https://youtu.be/H6y7_6uciqg?si=iY2OE1lTyLczYfD2

[calm coral]

@molten sundial Has your question been resolved?

[rancid crater]

the more I look at this question the more confused I get, wolfram alpha indicates that maybe it because sqrt(sec^2(x)) isn't sec(x) but is |sec(x)|

[molten sundial]

hm thats interesting but i dont think its true. ive never had to use || when cancelling a sqrt for trig sub

not to mention its principal square root

[molten sundial]

i found the example my instructor modeled the problem off of, only changing a 4 to 100

looks like my answer is correct, maybe integral calculator is wrong 🤷‍♂️

thanks for the help @rancid crater , i thought i was going insane 😅

.close

[calm coral]

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[maiden hull]

.reopen

idgaf-

[calm coral]

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[dry isle]

guys i have a doubt in the solution of this question

[dry isle]

should i choose x^2-1 or 1-x^2

[royal basalt]

Find BC. That'll answer your question.

[calm coral]

@dry isle Has your question been resolved?

[calm coral]

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[blazing coyote]

kind of lost here

[blazing coyote]

so the point of discontnuity can be a

but doesn't it depend on the values of a and b?

!status

[calm coral]

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

[blazing coyote]

4

[graceful dust]

why would it be discontinuous at a?

[blazing coyote]

oh yeah, not differentiable at a

[graceful dust]

then yes sure

does that change your question or did you mean to type differentiable originally

[blazing coyote]

meant differentiable

read the option wrong

so it's not differentiable at exactly one point, right?

[graceful dust]

it's def not differentiable at a

what about 2?

[blazing coyote]

yeah, when a=2 it isn't differentiable either

the LHD would be -1

and the RHD would be zero

[graceful dust]

ok where are you at in the problem now then?

[blazing coyote]

If D is right , I'm done with it

[graceful dust]

no we just established D is not right lol

'exactly one'

[blazing coyote]

ah

so what do I do now?

try finding the condition.for continuity at x=2?

[graceful dust]

yea that looks relevant to all the other options

[blazing coyote]

the LHL at x=2 would be a-3

the RHL would be 0

so a has to be 3

for the function to be continuous

[graceful dust]

sounds like you have your answer then

[crystal forge]

but it also depends on b

[blazing coyote]

how, the function approachs 0 from the right

b(0)=0

[crystal forge]

oh nm its 0 right : )

[blazing coyote]

Thanks!

.close

[calm coral]

Channel closed

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Use .reopen if this was a mistake.

[graceful dust]

(it is A just to be clear lol)

[blazing coyote]

yeah, thanks!

[crystal forge]

don't both a = 1 and a = 3 work?

b/c you have absolute value, there's 2 values of a that work, or what am i missing?

[blazing coyote]

the definition is for x<2

so the function is a-x-1

when x<2

[crystal forge]

what do you mean

.reopen

[calm coral]

✅

[blazing coyote]

when a=2

|x-2|-1 is the function

but the functionf(x) is defined to be this only for x<2

[crystal forge]

oh I see, both a = 1 and 3 work but the problem gives you the constraint that a is in (0, 2) so only a = 1 works

[blazing coyote]

a=3

not a=1

[blazing coyote]

oh yeah

so it isn't continuous

[crystal forge]

lim |x - 1|-1 as x-> 2- = 0
and lim |x-3|-1 as x-> 2- = 0

these a values both work, but because of that constraint only a = 1 works

[blazing coyote]

ah

got it

thanks!

can I close the channel now?

[crystal forge]

.close

[calm coral]

Channel closed

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[calm coral]

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[fierce cradle]

Could someone help me solve this problem

[solar spruce]

there's 4 problems

[leaden thunder]

!show

[calm coral]

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

[fierce cradle]

Sorry rieman, I am going to expound on this. I have to use the bathroom quickly bb.

brb*

[leaden thunder]

yea you don't need to ping me to tell me that

[graceful dust]

hi riemann i have to use the bathroom bb

this is going to be a garbage channel

[leaden thunder]

there's usually a garbage can in the bathroom

[fathom shuttle]

the fuck is wrong with you

[leaden thunder]

but not maximo

[graceful dust]

i blame maximo

[fathom shuttle]

!redir @exotic falcon

[calm coral]

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

[solar spruce]

ight guys lets purge @exotic falcon @fathom shuttle @graceful dust @leaden thunder

[graceful dust]

i'm not a helpful i don't have to follow the rules

[coral osprey]

wtf man

[marsh agate]

It's woman btw

[calm coral]

@fierce cradle Has your question been resolved?

[fierce cradle]

Hi guys, I'm available again

[fierce cradle]

Okay, I am stuck on 1. Like, I know how to find the a periodic point of f with prime period k, but how do I prove this?

[solar spruce]

is f^n(x) composition or exponentiation

[fierce cradle]

composition

[solar spruce]

start with the definition
what does it mean for f^n(x) to be periodic

or for a point p to be a periodic point of f^n

[fierce cradle]

Sorry, just looking back. So when we say a function f^(n)x is periodic or a point p is a periodic point of f^(n), this is just a reference to the behavior of the system after
n iterations of the function...

[solar spruce]

that's not what i had in mind

say p is a periodic point of f with period 2pi

what does that entail

f(p) = f(...)

[fierce cradle]

this would mean that applying f to p repeatedly, after every 2pi applications, returns you to the point p....?

[solar spruce]

maybe i am confused here

are you saying that p is k-periodic if f(p) = f^k(p)

if so we can run with that
i thought it was f(p) = f(p + k) but yours would seem to make sense in the way the question was phrased

[fierce cradle]

yes

[solar spruce]

alright

so if we want f^n(p) to be a periodic point with period k, we'd need
f(f^n(p)) = f^k(f^n(p))

[fierce cradle]

yes

[solar spruce]

that was my hint

[fierce cradle]

So far I have this

[solar spruce]

that is just the definition

[solar spruce]

that looks good

i dont see why f^i(p) should be different from p

is that part of the definition you were given?

[fierce cradle]

yes, from notes. maybe i extrapolated that somewhere elsewhere, because it's not actually in the book...

[solar spruce]

we could assume it to be part of the definition and just move forward

[solar spruce]

try and prove the second point by contradiction. assume f^i(f^n(p)) = f^n(p), and see what results you can get

[solar spruce]

why cant n + i be a multiple of k

n < k, i < k

[fierce cradle]

it possible that n and i could be chosen such that their sum is a multiple of k? Would we instead argue based on the definition of a prime period and the fact that
p returns to its original position after every
k applications of the function
f?

[solar spruce]

sort of yes

consider that if

(f^i(f^n(p)) = f^n(p))

[potent lotus]

maximo

[solar spruce]

and if (k = i + \tilde{k})

[potent lotus]

maximo

[solar spruce]

then (f^{\tilde{k}}(f^i(f^n(p))) = f^k(f^n(p)))

[potent lotus]

maximo

[solar spruce]

and since we just showed f^n(p) is a periodic point with period k

f^k(f^n(p)) = f^n(p)

oh shoot i think i wrote the wrong indeces. give me a moment to rethink it

[fierce cradle]

ok

[solar spruce]

yes i wrote it backwards

i meant to say

(k = n + \tilde{k})

[potent lotus]

maximo

[solar spruce]

so then (f^{\tilde{k}}(f^i(f^n(p))) = f^{n+\tilde{k}}(f^i(p)) = f^k(f^i(p)) = f^i(p))

[potent lotus]

maximo

[solar spruce]

since we showed f^i(p) to be a periodic point

[solar spruce]

and now coming back to the original assumption

if you apply f^(tildek) to both sides

you get

f^i(p) = p

which is a contradiction since p is a fixed point with period k =/= i

if that was overwhelming let me know and i can explain again

[fierce cradle]

Am I misunderstanding? I get this:

[solar spruce]

why is f^k(f^n(p)) = p?

hint: it is not

[fierce cradle]

And now?

[solar spruce]

looks fine at a glance yeah

[fierce cradle]

Are you free for one problem?

more*

if you have to go i understand

[solar spruce]

yeah sorry im going to bed soon

[fierce cradle]

Okay, no problem!

I;ll see you

bai

[calm coral]

@fierce cradle Has your question been resolved?

[calm coral]

Channel closed

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[flat inlet]

Need help with the whole unit of Large Sample test in statistics

[flat inlet]

I have notes, but it's really hard to understand the raw math notes smh

It would be great if someone wants to connect and teach me

[flat inlet]

These are my syllabus btw

[tulip briar]

You're looking for a tutor?

[flat inlet]

Kind of, yeah

I suck at math

[tulip briar]

My friend tutors, but he charges

[flat inlet]

price?

[tulip briar]

I shot him a message, I'll let you know when he replies

What class is that for, by the way?

[flat inlet]

College level statistics

Business degree

[flat inlet]

Sure thing, but i'll most probably stick to social help. I'll try considering it on any given circumstances.

[tulip briar]

Looking at your syllabus, it looked a bit similar to CS type statistics

What year are you in?

[flat inlet]

Yeah i know, we study engineer statistics in an business degree smh

[flat inlet]

2nd year last sem, Got my exams coming this friday

[tulip briar]

And this syllabus is for the exam?

[flat inlet]

Yeah actually there's alot of units and most of them covers up the theory part which i don't really struggle in. It's the sums and problems

that i'm stuck in

[tulip briar]

Do you guys have question models?

[flat inlet]

Yeah we do

[tulip briar]

Solved examples have saved my life on multiple occasions

I suck at Physics, a week before the exam, I downloaded a PDF and practiced all the types of questions that they could ask

[flat inlet]

You're right, but the fact that studying it from a raw textbook is pretty difficult as everything is pasted super dry and difficult to understand even the formulas and stuff, so i feel like the example models resembles the same so if someone could look onto and explain me in human language would be easier and i can pick up from there.

[flat inlet]

I can relate to you man

[flat inlet]

<@&286206848099549185>

[calm coral]

@flat inlet Has your question been resolved?

[flat inlet]

.close

[calm coral]

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[sharp badge]

A pair of dice is rolled, what is the probability of a number less than the second number

[sharp badge]

(1,2)

1/36?

Or 2/36

[winter elbow]

Do you roll them simultaneously?

[sharp badge]

Yea

Numbers less than the second number is 1 and 2 I guess

[winter elbow]

Just calculate the chances of them being equal

And deduct them with 1

[sharp badge]

wdym can you elaborate

[winter elbow]

There are only two scenarios, two are equals and they’re different

[sharp badge]

Oh so

1-1

[winter elbow]

Do you agree?

[sharp badge]

2-1

3-1

[winter elbow]

Too complicated

Just deduct it from 1

[sharp badge]

(1,2,3,4,5,6)

I deducted them already

[winter elbow]

yeah, that’s all

The chances of rolling two equal numbers is 1/6

So the chances of getting different is 5/6

[sharp badge]

I prompted chat gpt and it input this

To find the probability of rolling a number less than the second number on a pair of dice, we need to consider all the possible outcomes.

There are 36 possible outcomes when rolling two dice (6 outcomes for the first die and 6 outcomes for the second die).

Let's list the outcomes where the first number is less than the second number:

• (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
• (2, 3), (2, 4), (2, 5), (2, 6)
• (3, 4), (3, 5), (3, 6)
• (4, 5), (4, 6)
• (5, 6)

There are 15 such outcomes. Therefore, the probability of rolling a number less than the second number is 15/36 or 5/12.

[winter elbow]

yeah it’s right

I haven’t done my solution

[sharp badge]

Oh okay

[winter elbow]

Different numbers can be (1,2) or (2,1)

The latter is what we want

So divide it into 2

You’ll get the answer 5/12

[sharp badge]

Thank you

[winter elbow]

Np

!done

[calm coral]

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

[sharp badge]

.close

[calm coral]

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[rough arrow]

sorry bro

[ivory steeple]

oh

[uneven shell]

Lol, was this sent at the same time??

[rough arrow]

yes

[raw dirge]

take the derivative

of the function

[celest stream]

y = mx + a/m

general form of tangent to parabola

[raw dirge]

n1n1

[rough arrow]

this topic im learning at school is just quadratics not derivatives just yet

[rough arrow]

what do i do from that point

[celest stream]

substitute the given point in that general eqn

also u can find 'a' using the general form of parabola x^2 = 4ay

[rough arrow]

oh alr

may i ask how the general form of a tangent to a parabola is derived

y=mx +a/m

how is that derived

[celest stream]

assume that the tangent line is y = mx + c where m is a known constant,
now iif this line is tangent, it touches the curve at one and only one point
so u substitute mx + c for y in the parabola eqn to get a quadratic in y
since it touches only at one point this quadratic shud give only one sol.(equal roots)
so u find discriminant and equate it to 0 and find c in terms of m

[rough arrow]

sorry for my messing working out but this is what i get when solving for c in terms of m

[celest stream]

yes yes this is correct

[rough arrow]

alr

[celest stream]

the general eqn i gave was for parabola of the form y^2 = 4ax

my bad

[rough arrow]

oh

so how would it work instead?

[celest stream]

exact same way

its the same for most curves
assume the linee y = mx + c to be tangent
substitute x as (y-c)/m or y as mx + c
equate discrimnant to 0

[rough arrow]

oh alr, so i solved for the discriminant above, what do i do after that?

m^2+4c=0 is the discriminant

i understand everything up to there

[celest stream]

now u have the line as y = mx - m^2/4

substitue the given point in this eqn

solve for m

[rough arrow]

can you verify what ive done is right?

i get irrational solutions, not so sure if ive done it correctly

[celest stream]

looks correct only

[rough arrow]

so you reckon if i plug the two m values into y=mx+c and solving for c will give me the equations of the two lines?

[celest stream]

yes

[rough arrow]

alright thanks alot

.close

[calm coral]

Channel closed

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[celest stream]

the graph for that question

[rough arrow]

oh wow thats cool

ill say it again, thanks a lot for helping me w/ this question lol

[celest stream]

you are welcome 😀

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[remote mural]

Hey, how do I find the parametric equations for the tangent line to the curve with the given parametric equations at the specified point? x = e^-tcos(t), y = e^-tsin(t), z = e^-t; (1, 0, 1), started by doing the following

[calm coral]

@remote mural Has your question been resolved?

[random dragon]

feel free to ping helpers

[remote mural]

<@&286206848099549185>

[remote mural]

Usually doesn't work haha

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[calm coral]

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[brazen elbow]

$f(x)$ is differentiable on $\mathbb{R}$

[potent lotus]

FungusDesu

[brazen elbow]

$x^6[f'(x)]^3 + 27[f(x)-1]^4=0$, $\forall x \in \mathbb{R}$

[potent lotus]

FungusDesu

[brazen elbow]

$f(1) = 0$, evaluate $f(2)$

[potent lotus]

FungusDesu

[brazen elbow]

i dont know how to approach this ode

as usual, i converted the ode into leibniz notation

[mortal orbit]

been a while

[brazen elbow]

$x^6\left(\frac{dy}{dx}\right)^3+27(y-1)^4 = 0$

[potent lotus]

FungusDesu

[brazen elbow]

but from here im stumped

[blazing coyote]

you'd probably want to take the (y-1) to the right

and then take the cube root of both sides

[brazen elbow]

hm

$x^2\left(\frac{dy}{dx}\right) = -3(y-1)^{\frac43}$

[potent lotus]

FungusDesu

[brazen elbow]

...oh

now its separable

[winter loom]

you can separate the variables and directly integrate

yeah

[brazen elbow]

oh lol ive been trying to solve this as if its a cubic equation

alright brb imma solve this rq

$-\frac3{(y-1)^{\frac13}} = \frac3x + C$

[potent lotus]

FungusDesu

[brazen elbow]

subbing f(1) = 0 in, this yields me C = 0

subbing f(2) in, this yields me f(2) = -7

and it seems to be correct, nice

alright, thanks yall!

.close

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[surreal pollen]

I would like some help understanding this step

[bronze adder]

definition of a derivative

[potent lotus]

NEON

[surreal pollen]

oh ok got it

yeah and in the exercise it tells us g(-2)=1

so makes sense

wow huh

thank you

.close

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[frosty mica]

I think I was going somewhere with this, but I lost my way. Anyone able to nudge me back on track?

[frosty mica]

I got that there is some integers a and b such that they multiply to x^2 to get yz.

Am I able to just use that as fact, and say that x divides yz, or do I need to get rid of the square

From what I'm seeing, since x^2 divides yz, then x would divide yz since it's just a factor

[tender coral]

The contrapositive is: x|y or x|z. If either one is true then you can immediately conclude x|yz

[frosty mica]

So we're assuming that it divides either, then we say that it divides yz because it's still a factor somewhere in there?

Can I say then, that ax=y, or bx=z, then multiply them together to say that yz=abx

[tender coral]

*I am a bit confused why the opposite of x does not devide y and x does not devide z should be x devides both of them

[frosty mica]

I changed it to an or

[tender coral]

Anyways, suppose x|y then y = ax => yz = axz = (az)x => x | yz

[frosty mica]

Then I'd have to do the case where x|z, but it's the same thing more or less

[tender coral]

Exactly

Btw. I think your pfp is the most beautiful piece of art I am familiar with. I love Van Gogh's style

[frosty mica]

Thanks! It used to be a cow lol

I like the art more though, since I took an administrative role in a discord server.

Looks a bit more professional.

[frosty mica]

Did I state this in words correctly?

[tender coral]

Yes

[frosty mica]

And then the therefore, etc, and then the other case

[tender coral]

If you are having two similar cases which are basically the same you can use wlog

[frosty mica]

This is a foundational course, so my teacher wants to see everything, as annoying as it is.

[tender coral]

Yes, I remember. That was annoying in the beginning

[frosty mica]

Just to be thorough:

[tender coral]

Looks great

[frosty mica]

Cool, then just copy paste, move the letters around. Badabing badaboom. Appreciate it.

.close

[calm coral]

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[hearty glade]

i have a hard time understanding SSS theorem and vectors

[calm coral]

@hearty glade Has your question been resolved?

[calm coral]

@hearty glade Has your question been resolved?

[hearty glade]

<@&286206848099549185> sorry but the assignemnt is allready late and dont want to get in more trouble

[hasty narwhal]

are you read department

[hearty glade]

HEH

sorry caps

am i reading?

[hearty glade]

sorry i gtg thanks for trying ig

.close

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[calm coral]

@remote mural Has your question been resolved?

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@remote mural Has your question been resolved?

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[calm coral]

Please don't occupy multiple help channels.

[summer tangle]

.close

[calm coral]

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[weak osprey]

So from my information the sign of a only determines the convexity of the curve

And the other part is the term under the root in the quadratic equation formula so it can't EVER be -ve

Yet the answer is c?

And BTW is this curve shaped as a parabola like the x^2 or no?

[mellow crater]

rather as -x^2 if a < 0

but might be

[weak osprey]

I got that from the answer guide idk why

[weak osprey]

Ah k

[mellow crater]

Don't you understand what you've just written down?

I mean your explanation

[weak osprey]

I do and its the opposite of the answer?

The answer says the term u der the root is < 0

How's that?

[mellow crater]

Aah I misread

Okay, so

Determinant tells you how many roots does the quadratic equation have

but first of all, f(x) will be decreasing if f'(x) =< 0

So we should think about the parabola which is whole under x-axis (or touches it at exactly one point)

In other words its values are less than 0 (or 0 at tangency point)

When is this possible?

When a < 0 (the parabola opens downwards) and D = b^2 - 4ac <= 0 (so no roots or exactly one root)

You can visualize taking -x^2 as an example as it was already mentioned

[weak osprey]

But it can open downward and still be above the x axis, right?

[mellow crater]

Yeah, a < 0 only guarantees that it opends downward

And if we had only that condition it would be possible

But we have D = b^2 - 4ac <= 0 which tells us it has no roots

(or exactly one root obv)

so it cannot be above x-axis, right

Because if it was then it would have two distinct roots

[weak osprey]

The roots of the curve are its intersection with the x axis right?

[mellow crater]

ah typo

x-axis obviously

[weak osprey]

OK, can't it still be above the x axis and open downward?

If it has no roots or 2 roots

[mellow crater]

parabola which is above x-axis and opens downward would have 2 roots

(always)

[weak osprey]

I see

[mellow crater]

here we have D <= 0, so it's impossible

[weak osprey]

And because the roots would be imaginary so we say there are no roots

[mellow crater]

yeah

These two are sufficient to say that the whole thing is under x-axis

[weak osprey]

OK I got it

[weak osprey]

But can I understand why is this a requirement?

It can't have any roots and be above the axis

And open downwards

[mellow crater]

Yes? Give me an example (I mean sketch)

[weak osprey]

You want me to draw it?

It's just a curve

Oh I get it I was thinking as if it was in an interval

[mellow crater]

an example

no matter what, we have 2 roots

[weak osprey]

Yup I got it

[weak osprey]

Thanks dude

[sudden ivy]

I cannot understand what to do can someone help

[mellow crater]

#❓how-to-get-help

[weak osprey]

Uhm do I close it @mellow crater

[mellow crater]

.close

[calm coral]

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[sudden ivy]

Oh mb

[calm coral]

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[sudden ivy]

I don’t understand, can someone help

Don’t know where to start

Someone

<@&286206848099549185>

Hello

.close

[calm coral]

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[chilly oak]

how do i do part g?

[potent igloo]

I've not seen a comma operator before in matrix algebra. If I had to guess, you concatenate the matrices by joining them into one matrix. I could be wrong though

Maybe your notes say something

[calm coral]

@chilly oak Has your question been resolved?

[barren dawn]

just multiply the matrices as you do normally the resulting matrices willbe of 2x1 and 2x2

[calm coral]

@chilly oak Has your question been resolved?

[calm coral]

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[hollow sphinx]

can someone explain why what I wrote above is incorrect

[remote mural]

the angle ABC is 60°, can you see why?

[hollow sphinx]

yeah

thats what I used in the equation above

[remote mural]

Then what you wrote is a relation in any triangle: $\frac{sin (a)}{A} = \frac{sin (b)}{B} = \frac{sin (c)}{C}$

[potent lotus]

A human being

[hollow sphinx]

its incorrect tho

[remote mural]

?

my formula?

[hollow sphinx]

no mine

using this method AC is 9.86

but in the solution its 7.76

(for context AC is the tension in the cable)

[remote mural]

in your formula AC represents the length of the cable

[hollow sphinx]

11 is also tension

in cable BC

[remote mural]

the sine relation works for length

how do you mesure tension?

Is it proportional to length?

[hollow sphinx]

no

[calm coral]

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[calm coral]

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[fiery quail]

anyone can help me with these?

[fiery quail]

im totally lost

[strange lichen]

draw a triangle

dibuja un triangulo

[upbeat venture]

first hint: sin(-t + pi) = sin(t)

[strange lichen]

idk if i remember that correctly

[upbeat venture]

that's correct lol

Unless you wanna be picky about accents :p

[strange lichen]

yea that doesnt matter

accent on the a

QondaWEYITO

[fiery quail]

can i do it without drawing the triangle ?

[strange lichen]

quizas

pero no se

haz el triangulo

[fiery quail]

i have to do it with identities ☠️

[strange lichen]

u will end up using identities

why dont u wanna do the triangle

[fiery quail]

professor told me we have to use identities on the test

like

its a must

[strange lichen]

ok, you will use identities. u just need to do a triangle first

[fiery quail]

triangle is done

[strange lichen]

muestramelo

[strange lichen]

iirc

have u never used a triangle to solve this type of stuff?

like have u drawn it on the coordinate plane then marked the angles with the side lengths?

[fiery quail]

yes yes

wait a sec

im actually doing a lot of stuff 💀

[calm coral]

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[mental merlin]

Im having a huge problem understanding LU decomposition with pivot search in particular how we get the components of L

[kindred estuary]

Do you have an example?

[calm coral]

@mental merlin Has your question been resolved?

[mental merlin]

I do in particular it is this matrix

[kindred estuary]

And the goal is LU decomposition?

[mental merlin]

this is L

[kindred estuary]

And you are confused on getting the L matrix?

[mental merlin]

well yes cause we were taught a method of calculating L in class

but with that the signs in the first colmun dont add up

do explain we calculate, after swapping since |-2| > |+-1|

II = II - 0.5 * I

[kindred estuary]

What's the U matrix?

[mental merlin]

ths

[kindred estuary]

I meant U

Sorry

[mental merlin]

and calculating U or P is no issue at all

but the main issue remains why there is a sign problem, I was battleling with chatgpt the whole day, I forgot this existed

he gets the solution but cannot explain it and corrects himself once he reliazes he cant argue for it, or changes the calculation entirely

[kindred estuary]

So that LU decomposition does result in the A matrix but it's a permuation of A not the original

[mental merlin]

yeah its LU with pivot search

[kindred estuary]

Oh missed that my bad

[mental merlin]

so the result is LU = PA

A bigger problem is also that my Matrix L would be correct if I dont use the pivot search then it is not swapped but in the calculations that shouldnt matter

[kindred estuary]

I've never dealt with LU with pivot search sorry, I thought it was just general LU

[mental merlin]

@sly geyser anyone else did?

well thats fine its super frustrating I just need a rule to make it

I see no system in the calcs, the other I found, he also has the right solution, is super time intense and would not help for an exam

[calm coral]

@mental merlin Has your question been resolved?

[calm coral]

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