help-27

[heavy pewter]

thats crazy

bruh my teacher

[shy osprey]

Hold on, where is your 1

[heavy pewter]

didnt teach me none of this bs

wdym

i cancelled

it out

[shy osprey]

x+1/x+1 isnt 0 is it?

[heavy pewter]

oh wait

is this it

[shy osprey]

Thats correct!!

[heavy pewter]

thank u so much

[shy osprey]

And it looks just like what you wanted too!

[heavy pewter]

i have a worded question aswell

[heavy pewter]

yes yes thankk uuuu

what is this method called

[shy osprey]

just remember this

and this

[shy osprey]

oh it may be regional differences but in my country, its called splitting the numerator

[heavy pewter]

what country

[shy osprey]

or partial fractions if we apply this in integration

[heavy pewter]

i havent got up to integration yet

i have another question

[shy osprey]

Sure go ahead

[shy osprey]

I am from india, and i am guessing you are from Aus/NZ?

[heavy pewter]

yess Australia

[shy osprey]

Ah i see

Maybe sydney

[heavy pewter]

nah melbourne

[shy osprey]

banger place

[heavy pewter]

my area is socio economically disadvantaged tho

[shy osprey]

part c?

[heavy pewter]

😭

[heavy pewter]

yep part c

i will send solutions

can u help fix my error

[shy osprey]

erm is this your working?

[heavy pewter]

no no solutions

its just to make ur life easier

[shy osprey]

Huh the solution seems wrong for part b?

Or maybe its contradicting whats stated in th question?

[heavy pewter]

yes i am confused

please help

[shy osprey]

Ah i see the question is wrong in part b

[heavy pewter]

is my working correct

[shy osprey]

here i think you messed up in part a

expand th brackets carefully

[heavy pewter]

what do u mean where

[shy osprey]

-((y-20) 12)

open this

[heavy pewter]

should it be y+20*-12

[shy osprey]

nope

first, simplify everything inside the bracket

[frank musk]

Alr gang 3 question you all talking about ?

[shy osprey]

so it would give you something like -(12y-240)

[heavy pewter]

but dont i * the minus by whats inside the brackets

so everything inside

[heavy pewter]

??

[frank musk]

What u all solving?

[heavy pewter]

these 3

[heavy pewter]

rn im trynna understand where i faulted in a)

solutions r above feel free to read em

[frank musk]

Alr let me see

Emm so u have to make area equation for (a) right ?

[heavy pewter]

yes

its functions

[shy osprey]

yeah

amd so you get -12y **+**240

not -240

[frank musk]

Umm 12 thing is in vertical or horizontal measurement?

[shy osprey]

vertical

[shy osprey]

@heavy pewter

[heavy pewter]

i got +12y now

[shy osprey]

how did you get -(-y+20)12)??

[frank musk]

For (a) why don't u break this big square into two reactangle thing and add there area u will get equations of area easy

[shy osprey]

his method inst wrong

[frank musk]

I am just saying for easy idk what his method ?

[shy osprey]

yeah i am asking where you got -y+20 from

[heavy pewter]

the 2 minus

[shy osprey]

there is only 1?

[shy osprey]

his is total area of rect xy - cut out piece area

[shy osprey]

should i send my working ?

[heavy pewter]

yes please

[frank musk]

Xy+12y+240 what I am getting -12y there ?

[heavy pewter]

@shy osprey gotta teach us both

[frank musk]

Huh?

[heavy pewter]

im so fried bruh my test is tommrow

and i cant do a single worded question

[frank musk]

Just make 2 diff reactangle and solve it diff and add area simple

[frank musk]

Yee this is rightttt

[shy osprey]

the first step is the same for both of us

[heavy pewter]

ohhhhhhhh

[shy osprey]

After that this i what i meant by first calculating inside the bracket

okay now with that out of the way, this is the area term

now do you see that you would have to make some corrections in part b asw?

[heavy pewter]

let me redo part b

[shy osprey]

okay

[heavy pewter]

seems correct to me

[frank musk]

I can't understand (b) question my english is hella weak 😭

[shy osprey]

again its **+**240

[heavy pewter]

where did i get the 240 from

wait

the rule

neverrmind

[shy osprey]

not -240

[shy osprey]

.

You simply substituted the value of y in terms of x in the expression of area right?

[shy osprey]

and previously, you had made an error of -240 in your area

[heavy pewter]

ohh yeah i got it

[shy osprey]

okay

[heavy pewter]

equation is wrong

[shy osprey]

which one

[shy osprey]

pause, where did you get = sign from😭

[heavy pewter]

accidental

i was thinkig about something

[shy osprey]

thats right

[heavy pewter]

but the question is written wrong

[shy osprey]

yeah thats the questions error

your answer is right

[heavy pewter]

and last question

is quadratic

tp x value and y value

-b/2a

[shy osprey]

i uh

[shy osprey]

Okay lets go ahead with this not really sure but okay

Generally to find the max/min we differentiate the Area expression

But i am figuring that you havent studied differentiation yet

[heavy pewter]

we havent

save me

[shy osprey]

Thats correct

I thought you meant introducing new variables a and b

forgot you were talking about coefficients in a QE

[heavy pewter]

no nooo the formula

[shy osprey]

Yeah mb sorry!

[heavy pewter]

all gooood thank u soo much thoo

[shy osprey]

Anytime mate!

[heavy pewter]

appreicate it laddd

.close

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[devout snow]

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[vestal vortex]

How

[frank musk]

Ig u have to use similarity of triangle or smthing I am not good at it but ik u have to use that

[last parrot]

Thàles Theorem

[vestal vortex]

Using vectors

[gloomy aurora]

I'll give you a very simple proof method.

[last parrot]

O using vectors?

No worries you got Annie cover this😂

[vestal vortex]

Wut

[gloomy aurora]

1. Draw a line from C parallel to AB

2. Extend PQ to cut that line at D

3. Can you show some relation(congruence) between QCD and APQ?

it'd be better if you drew it down

[vestal vortex]

but its vectors so im pretty sure have to use magnitude to show its half and idk how to prove its parallel

And i never learnt congurcene never heard of that

[gloomy aurora]

hm fair point.

does the course strictly require you to use vectors?

[vestal vortex]

Yes

its vector proof

[gloomy aurora]

ohk

can you find the vector PQ then?

[vestal vortex]

is it p - q

or q - p

[gloomy aurora]

which one do you think it is?

its PQ

not QP

[vestal vortex]

q - p

[gloomy aurora]

and what is vector BC?

[vestal vortex]

couldnt tell you bro

[gloomy aurora]

oh.

what is vector AB?

[vestal vortex]

actually AC - AB

[vestal vortex]

Im stuck there

[gloomy aurora]

ohk

so observe

1. AB has the same direction as AP

2. |AB| = 2|AP| = 2p right?

$AB = 2\mathbf{p}$

[woven radish]

Annie Maqionde

[vestal vortex]

How do you know its exactly 2

[gloomy aurora]

its the midpoint

P is the midpoint o

of AB

[vestal vortex]

ohh

True

[gloomy aurora]

now what is AC?

and what is BC?

[vestal vortex]

AC = 2q

2q - 2p = BC

[gloomy aurora]

so $BC = 2 PQ$ right/

[woven radish]

Annie Maqionde

[gloomy aurora]

and what does that tell you about BC and PQ?

(BC is PQ multiplied by a scalar multiple)

[vestal vortex]

Its half the length

[gloomy aurora]

if vectors $\mathbf{A}$ and $\mathbf{B}$ are such that $\mathbf{A} = k\mathbf{B}$, is A and B not parallel?

[woven radish]

Annie Maqionde

[gloomy aurora]

here $\mathbf{BC} = 2\mathbf{PQ}$

[woven radish]

Annie Maqionde

[vestal vortex]

I dont get that

[gloomy aurora]

that's a property of vectors. do you mean to say its not in your course?

you can derive it component-wise anyways, if you want to.

[quasi vault]

…qqq

[vestal vortex]

It probably is but i havent revised this since last year

[gloomy aurora]

It'd be a good idea to go over everything before solving problems

the above property is for example, the only straightforward way to solve this

[vestal vortex]

How

[gloomy aurora]

$A = a_1 i + a_2 j$ and $B = ka_1 i + ka_2 j$

[woven radish]

Annie Maqionde

[gloomy aurora]

now $a_1/ka_1 = a_2/ka_2$

[woven radish]

Annie Maqionde

[gloomy aurora]

this is the condition for parallelity.

[vestal vortex]

Is that dividing

[thick lotus]

two vectors can be parallel if you can express one as a scalar product of other/ they are linearly dependent

[vestal vortex]

So you mean just make it cos0

[thick lotus]

um yh you can say angle b/w them is 0

[vestal vortex]

How do u even use this on the question

[devout snow]

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[gilded glade]

What are some transformations I could apply to a data set to make it more normal in its distribution? I am working in R.

More Detail: I am working on the ISLR2 data set Weekly and focusing on the predictors Lag1, Lag2, Lag3, and volume. I then split the data by training my models on pre 2008 observations and then testing it on the remaining using Logistical Regression, LDA, QDA, and KNN. I don't really know what transformations would work to improve the accuracy of the models on the data.

[raw cedar]

I recommend applying a logarithmic transformation or Box-Cox to the Volume variable to reduce its asymmetry.

[gilded glade]

This is the question itself and some notes I wrote based on conversations I had with my professor. I originally had taken a FAR too broad of an approach to it.

[gilded glade]

I am sick in bed and trying to push through this so I dont fall further behind, so I feel really dumb asking this. But how do I apply those?

Transformed.Volume <- log(Weekly$Volume)

would that work?

I am likely butchering syntax

[raw cedar]

Yeah, you are correct.

[gilded glade]

can you explain how Box-Cox works? I am unfamiliar with this.

[raw cedar]

it is a method that automatically searches for the best power ($\lambda$) to transform your data and make it look as similar as possible to a normal distribution.

[woven radish]

Ga³¹Br³⁵I⁵³9000✞

[gilded glade]

Thank you, I will test them both and see if there are any improvements.

[raw cedar]

You are welcome!

Do you have more questions?

[gilded glade]

probably, but that one has been answered. I will be back, just gunna keep working through this.

[raw cedar]

Good.

!done

[devout snow]

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

[gilded glade]

.close

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[jade wraith]

the blue is my drawing, i thought accel would be zero whenever the graph of velocity has a sharp corner or when it just equals zero

[jade wraith]

origincal graph

[lunar harbor]

If the derivative of velocity is undefined, then how can the acceleration (which is the derivative of velocity) be zero?

[jade wraith]

ohhh wait so sharp corners is when accel is undefined ?

[lunar harbor]

That is one case where acceleration is undefined, yes.

[jade wraith]

so at x=1 x=5 and x=8 acceleration is undefined

[lunar harbor]

Correct.

t, not x though

[jade wraith]

okay that makes sense and the slope when t = 3 is zero so that would be the only instance where the acceleration is zero

[lunar harbor]

mhm

[jade wraith]

alright tysm!!

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[lunar harbor]

[devout snow]

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[ionic jasper]

Hello, im looking for an website where i can focus training following tasks. Does somebody knows an good site in which i can do that for analasis. The level i need is around advanced high school or in german Abitur.

[winter patrol]

Khan academy

[gloomy aurora]

1. learn from an outside source like khan academy
2. ask if you have a doubt in a specific problem/concept

[ionic jasper]

thanks

i will look onto it

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[indigo coral]

Yo guys what is that one formula

I am not able to understand it

[thick lotus]

kakashi

[indigo coral]

Yes

[thick lotus]

you can occupy an available help channel see #❓how-to-get-help

[indigo coral]

Idk what to do

[indigo coral]

It's that one formula thats messing me up

[thick lotus]

which one

[gloomy aurora]

just post the formula. we can't quote help you otherwise.

[devout snow]

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[indigo coral]

Can't find it rn

[indigo coral]

It's used in triangle and trapezium

[untold ravine]

area?

[indigo coral]

And also kite

[winter patrol]

Are you able to post specific questions where that formula is supposed to be used?

[indigo coral]

Let me see

[devout snow]

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[polar chasm]

. @agile badger you can use any of these channels, including this one, to ask your question

. @agile badger just type your question in here and send and bot will do the rest

[agile badger]

ok

[polar chasm]

nice, now you've got a channel and you can ask your question

next time, your question should ideally be the first message you write in the channel

[agile badger]

i got part a

and uhh for part b do i equate the two

wait lemme try this

[polar chasm]

if you mean the right sides of the equation for C and the equation of l, then yep

[agile badger]

yup i got part b nice

ok now part c lets see

would (x-6) be a factor?

[frank musk]

What problem gang ?

[supple knot]

,calc 4 * 6^3 - 26 * 6^2 - 3 * 6 + 90

[woven radish]

Result:

0

[polar chasm]

yep, its a solution so it indeed is a factor

[supple knot]

yes 6 is a factor of the polynomial in b)

[polar chasm]

see pins

[frank musk]

Where I can learn this command?

[frank musk]

This one and all

[agile badger]

i thought so ok gotta have dinner brb

[devout snow]

@agile badger Has your question been resolved?

[supple knot]

just copy paste what i typed

but simpler math hopefully

[agile badger]

is this right for part c?

[supple knot]

,w factor 4x^3 - 26x^2 - 3x + 90

[woven radish]

Results provided by WolframAlpha

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

Requested by rie.mann

[supple knot]

,w roots 4x^2 - 2x - 15

[woven radish]

Results provided by WolframAlpha

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

Requested by rie.mann

[supple knot]

looks right

[agile badger]

nice

thx

btw my handwriting sucks yall have any advice to improve it?

[agile badger]

on a scale of 1 to 10 how legible is my handwriting pls be honest

[supple knot]

print and practice https://i.etsystatic.com/19351163/r/il/d4417c/3325050263/il_794xN.3325050263_koub.jpg

[agile badger]

also btw

[devout snow]

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[agile badger]

.reopen

[devout snow]

✅ Original question: #help-27 message

[agile badger]

also

AS Pure Mathematics:
Algebra & Functions: Indices, surds, quadratics, simultaneous equations, inequalities, and graphs.
Coordinate Geometry: Straight lines and circles.
Sequences & Series: Binomial expansion.
Trigonometry: Sine/cosine rules, trigonometric identities, and equations.
Calculus: Basic differentiation and integration.
Exponentials & Logarithms: Laws of logs and exponential growth/decay.
Vectors: 2D vectors.
A2 Pure Mathematics:
Advanced Algebra: Modulus functions, partial fractions, and composite/inverse functions.
Advanced Trigonometry: Radians, reciprocal functions (
), and double angle formulae.
Advanced Calculus: Chain, product, and quotient rules; integration by parts and substitution; differential equations.
Numerical Methods: Locating roots (Newton-Raphson) and Trapezium rule.
Advanced Vectors: 3D vectors.
Statistics (AS & A2):
Data Representation: Histograms, box plots, and measures of central tendency/dispersion.
Probability: Venn diagrams, tree diagrams, and independent events.
Distributions: Binomial and Normal distributions.
Hypothesis Testing: Correlation and testing for mean/proportion.
Mechanics (AS & A2):
Kinematics: Constant and variable acceleration (
).
Forces & Newton’s Laws: Gravity, friction, and normal reaction.
Moments: Equilibrium and turning effects (A2 only)

do yall cover all these topics? asking for future references

[supple knot]

read #❓how-to-get-help

[devout snow]

@agile badger Has your question been resolved?

[devout snow]

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[agile badger]

yes

[agile badger]

ok thx

[devout snow]

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[hexed star]

Right, hello.

I asked this question a few days ago, but in retrospect I think I may need a somewhat better explanation.

Using Lanchesters laws, is there a way to get an isolated value of A or B at a certain time (t?)

[hexed star]

Assuming the starting values of A and B, as well as the values of alpha and beta are known

[lusty sapphire]

Not familiar with the law, but my first instinct is to solve this as a system of ODEs

[hexed star]

like this?

[lusty sapphire]

No. My way is very different. Gimme few minutes

[devout snow]

@hexed star Has your question been resolved?

[lusty sapphire]

I'm still here btw

@hexed star you still here?

[devout snow]

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[lusty sapphire]

🥀

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[distant helm]

$\frac{1}{i}=\frac{1}{\sqrt{-1}}=\frac{\sqrt{1}}{\sqrt{-1}}=\sqrt{\frac{1}{-1}}=\sqrt{\frac{-1}{1}}=i$
Am I.., stupid?

[woven radish]

Player_X_YT

[supple knot]

sqrt(ab) = sqrt(a) sqrt(b) in general only when a > 0 and b > 0

So your third equal sign is incorrect

[distant helm]

how come?

[faint gorge]

it's inconsistent else

[distant helm]

where did a,b>0 come from?

[distant helm]

well yes, but why

[distant helm]

ok...

thanks

.close

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[cursive island]

hello in ap stats, i was wondering how to input the T in PHANTOMS for both the linear regression test for slope and the match pair t interval?

[cursive island]

into the calculator

[devout snow]

@cursive island Has your question been resolved?

[cursive island]

no but i figured itout

.close

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[last parrot]

you can look up on youtube btw

[devout snow]

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[keen sundial]

\documentclass[12pt]{article}

\begin{document}
How is \begin{equation}
\vec{c} =\vec{b} - \vec{a}
\end{equation}

\end{document}

[woven radish]

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

[keen sundial]

\begin{document}
How is \begin{equation}
\vec{c} =\vec{b} - \vec{a}
\end{equation}

\end{document}

[woven radish]

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

[small geyser]

c is explicitly stated to be obtained from the difference of the coordinates of A and B.
not to mention, c is defined as AB, which is OB - OA.

so it's answered in the very statement explaining how that came about.

[keen sundial]

I mean I need a proof because I forgot how thats the case

[small geyser]

AB = AO + OB = -OA + OB = OB - OA.

[keen sundial]

how is it equal to AO + OB

[small geyser]

have you not encountered vector addition at all up to this point?

[young spade]

For TeXiT you prob just wanna write it as

$$\vec c = \vec b - \vec a$$

[small geyser]

but either way, AB is represented by moving back to the origin from A, and then moving to B from the origin.

[keen sundial]

I have

oh do I just add them by components

which compponents do I add?

[young spade]

,tex If you have the vectors $\vec v = (a,b)$ and $\vec u = (c,d)$, then the sum of the two is as follows:
$$\vec v + \vec u = (a+c , b+d)$$

[young spade]

Now, consider, $-\vec v$ is just $(-1)\cdot (a,b) = (-a,-b)$

[keen sundial]

I agree

so how do I use this info to apply to my OP?

[devout snow]

@keen sundial Has your question been resolved?

[keen sundial]

<@&286206848099549185>

nvm all good

.close

[devout snow]

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[indigo coral]

Yo guys what is the surface area of the hemisphere of a trapezium

[devout snow]

@indigo coral Has your question been resolved?

[granite arch]

do you mind elaborating

[twilit comet]

the what

[devout snow]

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[indigo coral]

Surface area of hemisphere

[indigo coral]

For a trapezium

[winter patrol]

your question makes no sense

[winged tapir]

https://tenor.com/view/-gif-7240636

[winter patrol]

wdym for a trapezium

trapezium and hemisphere are completely separate things

!occupied

[devout snow]

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).

[winter patrol]

can you please clarify your question

[devout snow]

@indigo coral Has your question been resolved?

[twilit comet]

troll?

[devout snow]

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[mild sorrel]

Not now

[indigo coral]

Not a troll there is a question

Tangential quadrilateral

[devout snow]

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[summer harbor]

Hello

[summer harbor]

What do I do with the 10 ft?

the equations i immediately figured it out because Its asking for largest/smallest combined areas

[alpine wave]

can you post how you approached it?

[uncut crow]

good username

[summer harbor]

i havent even started the solution part

[alpine wave]

oh ok i see let me try once

[summer harbor]

my intuition is to divide 10 by 2, but that evenly divides it out

you can have a 2.5 inch wire which shapes a circle and a 7,5 inch square shaped

[alpine wave]

ok i got a) x=10pi/4+pi

[summer harbor]

????

[alpine wave]

is it correct?

[summer harbor]

wait why are u giving me the answer 😭

[alpine wave]

wait so you didnt want an ans? 😢

[summer harbor]

its a help channel, your supposed to guide people on the way to the answer

not give the answer

its fine

[alpine wave]

sorry i didnt see

[last parrot]

!nosols

[devout snow]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[alpine wave]

circle
Circumference
𝑥=2𝜋𝑟⇒𝑟=𝑥2𝜋x=2πr⇒r=2πx
so area of circle is pir^2 which equals x^2/4pi​

do the same with square

[summer harbor]

i dont get this

[alpine wave]

the questions asked for combined area so you have to find area of them individually and add them

[summer harbor]

OH crap

[alpine wave]

atlast minimise them

[summer harbor]

the derivative of the area of a circle is the circumference

[alpine wave]

yes

[summer harbor]

i derive both the areas

A'(r) = 2piR and A'(x) = 2x

OH SHIT

okay i get it

[alpine wave]

first one is right second wrong

[summer harbor]

why do we not derive the area of the square>

oh wait

[alpine wave]

you have to but you did it wrong it should come A(s)= (10-x)^2/16

dont differentiate individually first find the area and then add then differentiate

[summer harbor]

C(x) = 2.pi.r; P(x) = 4x; 2.pi.r + 4x = 10

r = (-4x + 10) / 2pi

help please :((

[alpine wave]

sure

are you there?

[summer harbor]

idk if im doing it right

[alpine wave]

nope wrong

[summer harbor]

is this correct tho?

im adding circumference of the circle and the perimeter of the square = 10 inches

[alpine wave]

yes it is right

x = length used for the square
10 − x = length used for the circle

Square:
Perimeter = x = 4s ⇒ s = x/4
Area = s² = x² / 16

Circle:
Circumference = 10 − x = 2πr ⇒ r = (10 − x)/(2π)
Area = πr² = (10 − x)² / (4π)

Total Area:
A(x) = x²/16 + (10 − x)²/(4π)

read the above

Minimize Area:
A'(x) = x/8 − (10 − x)/(2π)
Set A'(x) = 0:

x/8 = (10 − x)/(2π)
2πx = 8(10 − x)
2πx = 80 − 8x
x(2π + 8) = 80
x = 40 / (π + 4)

Minimum occurs when:
Square = 40 / (π + 4)
Circle = 10π / (π + 4)

Maximize Area:
Check endpoints:
x = 0 → all wire is a circle
x = 10 → all wire is a square

Since a circle gives more area per unit length,
maximum area occurs when all 10 ft is used for the circle.

[summer harbor]

wait dont give the answer 😭

[alpine wave]

i have to go tution so i provided it you dont peek 🙏

[summer harbor]

oki

[devout snow]

@summer harbor Has your question been resolved?

[devout snow]

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[drifting agate]

i need channel

[mild comet]

Hi

[drifting agate]

okey

g=hi

hi

im of 4th grade

is it okey?

[drifting agate]

dude

is this channel still for me if im 4th grade ?

[frank musk]

Ye?

[drifting agate]

ok

[tribal patio]

Discord is not for anyone under 13

It's in the TOS

[drifting agate]

;-;;;;;;;;

[rain summit]

You made a really big mistake claiming you're a 4th grader

[drifting agate]

okey im 8th man

i just came here for fun to see yalls reaction

[tribal patio]

what a jump

[frank musk]

🫡

[drifting agate]

exactly

[rain summit]

Fr

[small inlet]

<@&268886789983436800> trolling

[drifting agate]

mn.....

[rain summit]

And underage too

[rain summit]

Blud this is a help channel

[drifting agate]

okey guys

[split stream]

This is no place to pester the learners

Take an hour off

[rain summit]

Do you mind reading the messages above please?

[small inlet]

most offensively

[split stream]

He's in 8th grade

[rain summit]

.

[split stream]

I saw but he said this afterwards

[small inlet]

Sounds like he was just trying to get out of the ban

[rain summit]

I agree, I think you just fell for the old tricks

[willow helm]

great function

[split stream]

I'll relegate up to the smods

[rain summit]

Alright

[split stream]

Thankyu

[rain summit]

Can you close the channel please?

[split stream]

Yea, sure

.close

[devout snow]

Channel closed

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[last parrot]

Darq saved the day

[rain summit]

We got senior mods involved now fr

[devout snow]

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[jolly narwhal]

no idea how to do part c. did the rest correct

[dull parrot]

The perimeter consists of the line segement AD, Segment on the x-axis from D to the circle boundary and then the circular arc

I'm assuming you found the point D to answer the previous questions?

[jolly narwhal]

D is (1,0), origin to the circle is sqrt(2) units

A is (1,1)

that makes the two lines add to sqrt(2)

but idk how to find the arc length

[dull parrot]

It's in radians so you can use L = Rθ formula

where R is the radius

actually

it's in degrees, i didn't see the 360 degrees

[jolly narwhal]

i don't think that matters since that's for part b not c

[jolly narwhal]

how to find theta in that case?

[dull parrot]

it's 90 degrees

[dull parrot]

hmm true

[jolly narwhal]

how

im assuming theta would be AOD

since O is the centre

[dull parrot]

oh my bad, it's 45 degrees since you have the point A(1,1) making an angle with the x axis

so tan(θ) = 1/1 = 1

θ = 45 degrees

[jolly narwhal]

that seems right but the markscheme is using radians for some reason

[dull parrot]

that's strange

does the mark scheme account for degrees or just radians?

[dull parrot]

Yeah I guess you have to put in radians then

It's also easier that way

so 45 degrees is equivalent to pi/4 radians

and the radius is sqrt(2)

[jolly narwhal]

alright i get it. thank you

.cl9se

.close

[devout snow]

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[devout snow]

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[rough nova]

I'm on GeoGebra and have point C on segment AB such that CA/CB=k. How do I construct point D so that DA/DB=k? Basically i'm trying to draw the apollonius circle

[rough nova]

Segment lengths on GeoGebra are kinda messy

.close

[devout snow]

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[devout snow]

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[devout snow]

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[willow helm]

[dapper mica]

@willow helm my bad i accidentally deleted it

[willow helm]

"as this action is irreversible"

[faint gorge]

You need to start a new channrl this will auto close

[willow helm]

You will need to open a new channel

[dapper mica]

Oh ok

.close

[devout snow]

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[long trout]

Does there exist a natural number n such that there exists a natural number x that satisfies the equation: x^2-x-1 = F_n where F_n is the nth Fibonacci number given that n>11

[rain summit]

Oh god

[dull parrot]

Isn't this essentially looking for Fibonacci numbers that are one less than a product of two consecutive integers? Since x^2 - x = x(x-1)

[rain summit]

Wait what's the 12th fibo number?

[dull parrot]

,w 12th fibo number

[woven radish]

Results provided by WolframAlpha

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Requested by mxrgd.

[rain summit]

Oh

[long trout]

Yep, I did try to interpret it like that but couldn’t get anywhere

[rain summit]

,w 13 th fibo number

[woven radish]

Results provided by WolframAlpha

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

Requested by iaminfinityiq

[dull parrot]

pretty sure the equation can be rewritten by multiplying by 4 and completing the square, to get something like $(2x - 1)^2 - 5 = 4F_n$. so $4F_n + 5$ is a perfect square

[woven radish]

MxRgD

[dull parrot]

don't know if that helps

[long trout]

Yeah I also got that but couldn’t get much further

[dull parrot]

hmm

[thick lotus]

,w 11th fibo

[woven radish]

Results provided by WolframAlpha

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Requested by mis.ter.x

[thick lotus]

11 th seems to satisfy dk if you can prove if there exists n>11

[long trout]

Yeah, that’s why I restricted n to be >11

[rain summit]

89 is prime if im not wrong

So really you can't do anything with F_11

[thick lotus]

?

[long trout]

x^2-x-1 = 89 means that x=10 which works

[rain summit]

Oh mb

[last parrot]

I don't think any number n>11 would satisfies this

[dull parrot]

Let $k = 2x - 1$. You could try checking $k^2 \equiv 4F_n + 5 \pmod{m}$ for various moduli with the pisano period

[woven radish]

MxRgD

[dull parrot]

if you did $\pi(11)$ as the pisano period then the sequence repeats every 10 terms

[woven radish]

MxRgD

[devout snow]

@long trout Has your question been resolved?

[devout snow]

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[devout snow]

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[twilit comet]

DAMN that was fast

[devout snow]

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[viral kernel]

$\int_{-\infty}^{+\infty} \delta(\cos(x)) dx$

[woven radish]

Mr. Penguin

[viral kernel]

is this +infinity?

[supple knot]

do you know what delta is ?

[viral kernel]

yes

[supple knot]

can you write its definition

[viral kernel]

delta(x)=infinity at x=0 , 0 in x different from 0

if the sum is infinite periodically it always equals infinity

[versed juniper]

er what

[viral kernel]

it gives me nothing

i mean wolfram

[viral kernel]

cos(x)=t

[sand quarry]

Yes

[versed juniper]

i think it infinity

[sand quarry]

It just becomes $\ds\sum_{n\in\vb N} 1$

[viral kernel]

thanks

.close

[devout snow]

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[devout snow]

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[frigid rain]

Can someone help me with this? I tried taking the natural log but it just became a whole mess

[viral kernel]

Hi

Try to subtract and add 10 to the numerator

[frigid rain]

can you elaborate

[viral kernel]

$9x+10-10$

[woven radish]

Mr. Penguin

[frigid rain]

what about the outer exponent?

[viral kernel]

Then you use the distributive property

Do you know the limit of e?

[faint gorge]

clever

[frigid rain]

im giving it a try its just taking a bit

[viral kernel]

Okay!

[frigid rain]

okay hmmm im totally doing smthn wrong give me a second and ill post my work

[viral kernel]

Okay

[frigid rain]

technically its ln(y) = all that but i got lazy

[viral kernel]

Mmm

Knowing that $\epsilon_x$ is a function that tends to 0 you can use

[woven radish]

Mr. Penguin

[viral kernel]

$(1+\epsilon_x)^{1/ \epsilon_x}=e$

[faint gorge]

remove the second $

also ur missing the lim

[woven radish]

Mr. Penguin

[viral kernel]

If x tends to infinity

Do you know this ?

[frigid rain]

no i didnt, i dont even recall hearing abt it in class

i have no idea how to use that

[viral kernel]

Okayokay

Do you know that

$\ln(1+\epsilon_x)$ Asymptotic to $\epsilon_x$

[woven radish]

Mr. Penguin

[frigid rain]

we havent really talked about epsilon at all in my class very much, only the very basic proof of a limit type situation. is this something that is like, supposed to be in calc 1?

[faint gorge]

maybe you could show how this applies here

[viral kernel]

Have you made any notable limitations?

[supple knot]

just expand here and find the highest powers of each numerator and denominator

[frigid rain]

ill try that

[frigid rain]

At that point is it okay to have the +10 and -10 cancel? Because im ngl i dont really see their use at this point

[viral kernel]

Yes

[frigid rain]

Having those cancel out has it turn out to this, which is another infinity / infinity indeterminate form, so lhopitals rule would be repeated iirc

[viral kernel]

What Is this

He doesn't have to do that

[frigid rain]

why not? isnt it an infinity / infinity indeterminate form?

[viral kernel]

This indeterminate form can also be solved without hopital, but I don't understand what it has to do with your question

[viral kernel]

.

[frigid rain]

can you elaborate on what u mean by that question? like are we talking in terms of this current question or in the overall class

[viral kernel]

Yes

[viral kernel]

Where did this come from?

[viral kernel]

Current question

[frigid rain]

I havent really noticed anything for the current question, no

[frigid rain]

from expanding what pi_day said

[viral kernel]

How did you remove the logarithms?

[frigid rain]

its still in the form ln(y) = that

im jsut too lazy to write it out

[supple knot]

you really shouldn't need l'hopital to find the limit of something like ax^2 / (bx^2 + x) as x goes to inf

[viral kernel]

How did you go from the last line of the first page to the first line of the second page?

[frigid rain]

taking derivative of top and bottom like how my prof explained l'hopitals rule

eh ill just try it till i get it right

.close

[devout snow]

Channel closed

Closed by @frigid rain

Use .reopen if this was a mistake.

[viral kernel]

I don't think you can use hopital in that case

@frigid rain

[frigid rain]

i mean thats plausible, since we are just only barely starting on it, i mightve got it wrong

but like idk what else can be done really so

[viral kernel]

Did you only do De Hopital's theorem?

@frigid rain

[frigid rain]

I mean I took the natural log of both sides to get the exponent down in front, but from there all changes were just to try and get it in the indeterminate form for Lhopitals rule

[viral kernel]

Wait

I think there is an error in the calculation of the derivative

[devout snow]

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[subtle storm]

How does bowers extended array function work for values like {n,n,n,n,n} n{n}n=n(↑^n)n n{{1}}3=n{n{n}n}n} n{{2}}3=n{{1}}n{{1}}n n{{{n}}}n can be written as n({n}^3)n n({n}^n)n can be written as {n,n,n,n} so what is {n,n,n,n,n} defined as in bower's extended array function

[devout snow]

Please don't occupy multiple help channels.

@subtle storm Has your question been resolved?

[subtle storm]

<@&286206848099549185>

note: a({c}^d)b={a,b,c,d}

[subtle storm]

.close

[devout snow]

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[devout snow]

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[visual stag]

How to do?

[tender cobalt]

you here?

[visual stag]

Yes

[visual stag]

I'm here

[tender cobalt]

take a look at these angles

call angle BOC x

can you find angle COD and angle BOE

[visual stag]

No?

I don't have angle DOE

[tender cobalt]

consider the following

we can consider, for any line, how long of it is inside the circle

we can also consider how much of an angle the two points through the circle make

now lets say the line is always vertical

[tender cobalt]

the moment I showed that the line has this much going through the circle,

[tender cobalt]

does its corresponding angle seem locked in place?

are there any other possibilities for that angle?

[visual stag]

No if it is centre

[tender cobalt]

very interesting

since you can always rotate any line to look vertical this way,

does that mean that, for any line through the circle, its angle is locked in place by how much of it is inside the circle?

[visual stag]

Yes

[tender cobalt]

now youve got three lines here where BC = CD = BE

and BC had its angle BOC = x

[visual stag]

So we know that angle BOC = angle COD = angle BOE due to equal chords, equal angles

[tender cobalt]

yep

this makes things much easier I imagine

[visual stag]

I have no point on how 44 degree can help

[tender cobalt]

you need a way to connect this x to that 44

there are a few ways you can go about this, but the general idea is:

find a way to express angle A in terms of x

that way, you knowing that angle A should be 44 would give you an equation

(in terms of x) = 44

for you to solve

[visual stag]

It expressed me and I will try

.close

[devout snow]

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Use .reopen if this was a mistake.

[tender cobalt]

np

[devout snow]

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[devout snow]

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[grave idol]

Is my solution correct? and if it is correct is there any solution that is faster than this?

[shy osprey]

Holy handwriting btw

[grave idol]

thanks

the cons is that the writing speed is slow af lol

[shy osprey]

worth it if youw ask me

[shy osprey]

Anyhow this seems correct

[grave idol]

mmm alr thx

[shy osprey]

Idts as for this specifc case you have one of the components of v as 0 so its quick

[grave idol]

i see

[grave idol]

also another question, r1 and r2 are considered to be different unit vectors right? (Their direction is not the same)

[shy osprey]

Yup!

[grave idol]

alr

anyways, thanks a lot, appreciate it

[shy osprey]

Just a quick fun qsn if you want, how many such orthogonals vectors can be there?

[grave idol]

the ones that are orthogonal to u and v right?

[shy osprey]

Yeah

[grave idol]

mmm infinite?

[shy osprey]

Why?

(ans is right btw)

[grave idol]

because there is inf amount of vectors that can be derived from the unit vector that is orthogonal to both u and v.

[shy osprey]

Good!!

[grave idol]

alright thx again

cya

.close

[devout snow]

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[devout snow]

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[outer isle]

I am sorry everyone I have asked this question before. 😭

Someone called Civil Service Pigeon helped me already and I already got the answer to this problem. However, the more I think about it, the more I got confused on how these seats/points actually move.

Is it like this?

A --> F (goes down by 0,5 m, now is at the lowest height) --> E (goes up by 0,5 m and got back to the starting height) --> D (goes up by 0,5 m, now is at the highest height) --> C (goes down by 0,5 m and got back to the starting height) --> B (goes down by 0,5 m, now is at the lowest height).

However, when I try to plot (using a graphing calculator) the answers given in the options, why does it feel like the graph looks different from the motion above?

Is it the problem being inconsistent? Or am I just completely wrong?

Also, does anyone have any idea what this part mean?
--> "the vertical motions of the three pairs of opposite points on the roof move in opposite directions."

Thank you very much 🙏

[devout snow]

@outer isle Has your question been resolved?

[devout snow]

@outer isle Has your question been resolved?

[outer isle]

<@&286206848099549185>

[runic grove]

The line you're confused about means that if F goes up then C goes down

And vice versa

They're a pair since they lie opposite to each other

Same for AD and BE

[outer isle]

Right, I see that now. Thanks

How about the motion of the seats? Let's say point A for example, can you help me with that?

[devout snow]

@outer isle Has your question been resolved?

[devout snow]

@outer isle Has your question been resolved?

[pallid robin]

hey you still there ?

[devout snow]

@outer isle Has your question been resolved?

[outer isle]

Yes, I am still very confused about the motion of the seats.

[devout snow]

@outer isle Has your question been resolved?

[pallid robin]

okay so i'll try to explain the motion of one point, let's say A

let the height of all the points at t=0 be the mean position

now when the wheel rotates by 60° i.e. point A is now where point F was initially, it has moved 0.5m downward from its mean position

after another 60° rotation, point A now reaches the position of E. during this time A was going up, which means it is now at the height it started with initially i.e. the mean position.

then point A continues moving up for another 60° till it reaches the position D, at this moment A is 0.5m above its mean position.

as you can now guess, A now moves down until it reaches point C, and which is now at its mean height.
it continues moving down till it reaches point B, meaning it is currently 0.5m below its mean height.
after that, it moves up to complete the 360° rotation and is finally at the mean height it started with.

ping me when you are back!

[outer isle]

@pallid robin Alright, thanks a lot. I agree with the motion you presented.

However, I am very confused. I used Desmos to plot the graph for each function in the answer options (A - E). Yet, none actually represented the motion as you've described.

For example, for option C, every 60°, the seat would always be back at the starting height (mean position) and have completed the motion of going up and going down by 0,5 metres once.

Am I missing something? Is the graph actually supposed to be that way?

[pallid robin]

could you share what the graph looks like ?

so from the data given, i think the mean height is 2m from the ground

[pallid robin]

nvm about that

i plotted them

[outer isle]

Oh, thanks

[pallid robin]

is that what you calculated as well ?

[outer isle]

No, the mean height I got was somehow 1.5 metres

[pallid robin]

how?

[outer isle]

I know that it should be 2 metres, since point A was at B (1.5 m) based on the problem

[pallid robin]

yes

[outer isle]

But the graph said the mean height is 1.5, is it not? I am so confused 🥲

[pallid robin]

which graph exactly?

[outer isle]

Oh yeah, option C

[pallid robin]

i mean none of the options match if we consider 2m at t=0

[pallid robin]

is c the correct option acc. to the answer key ?

[outer isle]

I got no answer key unfortunately. Sorry. I was just biased on option C and D since they are the only option which give h = 1.5 meters at 300°

Is the problem just bad? If it is, then it's okay.

[pallid robin]

no but 1.5 at 300 is valid for point A only ?

no wait, you're right in some sense

but the question is just inconsistent imo

the values don't agree with each other

i might be wrong tho, you could cross check with a teacher or anybody else

[outer isle]

Yes, thank you very much. Do you have any other insights?

Alright, I'll close this channel. Thank you very much. Have a good rest of your day

.close

[devout snow]

Channel closed

Closed by @outer isle

Use .reopen if this was a mistake.

[pallid robin]

do let me know if the problem actually turns out to be right and there was just a flaw in our understanding, that's an interesting problem : )

[outer isle]

:). Alright.

To be honest, the problem will most likely be wrong. This is a problem I found on the internet as an example for my country's university entrance exam math test. And who made the problem itself is unclear.

Thanks a lot

[pallid robin]

no problem! have a nice day!

[devout snow]

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[dark sable]

what topic/concept is behind this in statistics ._.

[sand quarry]

skewness?

i mean more generally it is probability theory

i dont know what you are looking for exactly

[dark sable]

I'm a bit confused
how do I do the skewness from the PMF

in that form too

[sand quarry]

\lm You mean, how to evaluate $\ds \vb E\9{\8{\4{X-\mu}\sigma}^3}$ generally?

[dark sable]

probably

I'm confused abt the topic in general since my stat is so to speak, extremely basic

[sand quarry]

maybe you need to dial it back a notch before tackling something like this

do you know how to evaluate E[x], generally?

[dark sable]

iz why I'm asking about the topic so I can build on it

[dark sable]

expected val yeah I think

hopefully I know 🥀

[sand quarry]

\lm it is $\ds\vb E[X] = \sum_ix_i\2f(x_i)$

[sand quarry]

so for your problem, try to identify the missing variables

you need to know f(x_i), mu, and sigma

try to first retrieve f(x_i) from the set of three equations given to you regarding it

[dark sable]

I can only imagine that since PMF
f(0)+f(1)+f(2)+f(3)+f(4)=1,
2(f(0)+f(2)+f(4))=1 => it's equal to 1/2, so f(1)+f(3) must be equal to 1/2,
so f(1)=3/10, f(3)=1/5, then I'm kinda clueless for f(0) f(2) and f(4)
and I know variance is 1.3, so

[sand quarry]

yeah f(1) and f(3) are right

you know f(0) = f(4)

you also know f(0) + f(2) + f(4) = 1/2

[dark sable]

yes but that doesn't help with f(2)? I think

[dark sable]

yes

[sand quarry]

so like

if you write that in terms of f(0), you would have f(2) = 1/2 - 2*f(0)

[dark sable]

yes

[sand quarry]

you can retrieve what f(0) should be from the mean and variance now

what is the mean, first and foremost?

in terms of f(0)

[dark sable]

1/5?

[sand quarry]

no

can you show me how you calculated that

[dark sable]

f(0)+f(1)+f(2)+f(3)+f(4) divided by 5 😔

[sand quarry]

no, not statistical average

\lm like, $\ds\sum x\2f(x)$

[dark sable]

so is it like f(0) * chance+f(1) * chance /5

[sand quarry]

no /5

but yeah

[dark sable]

0

[sand quarry]

no

just write it out

call f(0) = a for simplicity

[dark sable]

I forgot this but this has been taught to us 😔 I'm sorry

[sand quarry]

you have f(0) = a, f(1) = 0.3, f(2) = 1/2 - 2a, f(3) = 0.2, f(4) = a

[dark sable]

so the average is 1.9

[sand quarry]

yaeh!

great

so, you have to calculate the variance now

this is gonna be mega annoying tho

,, \lm \ds \Var[X] = \sum(x-\mu)^2\2f(x)

[dark sable]

variance is 1.69, stdev is 1.3
so I have to like

[dark sable]

🥀

[sand quarry]

yes

you have to solve for a

solving for a gets you all the f(x_i)

[dark sable]

in a 90 second question is crazy

[sand quarry]

lol

[dark sable]

2.1^2 x a+1.1^2 x 0.2+0.1^2 x (1/2 - 2a)+0.9^2 x 0.3+1.9^2 x a=1.69
(3.61+4.41-0.02)a=8a
8a+0.242+0.005+0.243=1.69
=> 8a=1.20 a=3/20=0.15

so the other one becomes 0.2

0.15 0.3 0.2 0.2 0.15

[dark sable]

isn't this E(x)

[dark sable]

so then how do I do this

(-1.9/1.3)^3+(-0.9/1.3)^3+(0.1/1.3)^3+(1.1/1.3)^3+(2.1/1.3)^3
3005/2197
trust we have a more than 100% chance to do that

[sand quarry]

thats correct

[sand quarry]

yes

[sand quarry]

you are forgetting to multiply with the probaibltiy

liek

[dark sable]

how do we get the probability for said things

[sand quarry]

(-1.9/1.3)^3*0.15

you alreayd did

[sand quarry]

.

[dark sable]

wait so it's ((X-u)/o)^3 * f(X)

[sand quarry]

yes

[dark sable]

(-1.9/1.3)^3 x 3/20+(-0.9/1.3)^3 x 3/10+(0.1/1.3)^3 x 1/5+(1.1/1.3)^3 x 1/5+(2.1/1.3)^3 x 3/20

first I'm gonna take out 1/5(1.3)^3 and multiply 1000 just to make it ((-19)^3x3/4+(-9)^3 x 3/2 + 1^3 + 11^3 +21^3 x 3/4

[dark sable]

what's a good book on this

[sand quarry]

probability?

i can suggest the book i used back in my probability course a few years ago

i liked it

this is it

[dark sable]

3x2402/4+ 2662/2-2187/2+1=7203/4+475/2+1=7203/4+950/4+4/4=> (8157/4)5*13^3
or 8157/43940
which isn't the answer

[dark sable]

thanks, I will be exploding over the next 3 years

[sand quarry]

99% sure you miscalculated something but let me try myself

[dark sable]

I hate calculators but they are damn convenient 🥀

unfortunate I lacked the brain capacity to think that maybe I should do this with a calculator first

here is the answer

wait I'm missing 3 units somewhere

like just 3

OHHHHH

2402 NOT 2401

8160 which is 408 IS THIS THE INTENDED SOLUTION????

suffering 🥀

[sand quarry]

this is what i get

[sand quarry]

this is right

[dark sable]

hey code is cheating grrr >:(

[dark sable]

is there like any easier way this is literally impossible for 90 seconds

there might be tbh

[sand quarry]

yeah idk

[sand quarry]

you are inevitably gonna have to compute this

your teacher might be trolling you

[dark sable]

I spent like 5 minutes of my life calculating the values you should also suffer /j

[dark sable]

oh it's a competition

https://klipy.com/gifs/ishowspeed-darren

truly this was my face when I saw that question 2 days ago

[sand quarry]

i have a trick in case it is multiple choice

is it multiple choice

[dark sable]

I'm gonna be noting this, idk about the f(x) thing but I have no knowledge in pmf and pdf

[dark sable]

you saw the question format 😭

[sand quarry]

you should deffo read up on that