help-23

[safe radish]

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[chilly valve]

how on earth do i answer this?

This is what I tried doing

[safe radish]

@chilly valve Has your question been resolved?

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@chilly valve Has your question been resolved?

[chilly valve]

<@&286206848099549185>

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@chilly valve Has your question been resolved?

[chilly valve]

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[kindred slate]

how do they get beta = 0??

[pliant egret]

if beta=0 here you cant divide by e^-aysin(bx)

beacuse e^-aysin(bx)=0

[kindred slate]

so surely that means beta cant be 0 right?

[pliant egret]

no

for example x^2=x

x=0 is a solution

even tho you cant divide both sides by x when x=0

and you solve this like youd solve x^2=x

put everything on one side and factorise e^-aysin(bx)

you get e^-aysin(bx)(a^2-b^2)=0

[kindred slate]

oh yh

cuz e^y can never equal 0

[pliant egret]

yeah

[kindred slate]

so only sin(Bx) can = 0

and thats when B = 0

or x = 0

[pliant egret]

yes

[kindred slate]

but here we only care about B = 0

[pliant egret]

yes

[kindred slate]

thanks

[pliant egret]

you're welcome

[kindred slate]

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[surreal fable]

here at the and I end up with.
$64(-1+i)$
The solution is -64.
If I multiply everything out I get:
$-64+64i$
I don't know how to get -64 from what I've got.

[flat frigate]

Tomi

[faint seal]

!show

[safe radish]

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

[thin bridge]

how exactly are you "multiplying everything out"

[surreal fable]

$\cos(5\pi) = -1$
$\sin(5\pi) = 0$

[flat frigate]

Tomi

[faint seal]

Precisely

So you are getting -64

How did you get 64(1-i)

[thin bridge]

0 * i isn't i

[surreal fable]

oh

well that was a stupid mistake that I made

.close

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[zinc yoke]

this is supposed to be the area of something and I'm getting a negative result. Any ideas? :/ ( συνχ = cos(x) )

[zinc yoke]

it just hit me f(x) is supposed to be in its absolute value eh

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[burnt wyvern]

<@&268886789983436800>

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[slow fog]

How to find the centre in a matrix transformation and how do I know if the centre is origin or non origin?

[umbral swan]

the center in a matrix transformation?

if the transformation is linear then the center is always the origin

[slow fog]

Could you elaborate

[umbral swan]

firstly I'd like to know which transformations you refer to

linear trans.?

and how do you define center of a transformation

@slow fog

[slow fog]

That's what I dunno, I think I skipped that lecture😭

[umbral swan]

,rotate

[slow fog]

I'm kinda confused on what this whole thing is

[umbral swan]

uh ok

[umbral swan]

which object do they refer to

[slow fog]

Wait

[umbral swan]

it's completely unclear what the context is

I presume your task is that you are given some object in 2D

and the centre, angle and direction of a rotation

in 2D

around a point

[slow fog]

Let me show you a question

And Then there's a graph

Basically the four types of transformations are translation, rotation, reflection and enlargement.

And rn I'm stuck with rotation

[umbral swan]

these are not all but yh

[umbral swan]

in 2D I presume?

there still needs to be more context

we don't have an object and we don't have a center

[slow fog]

I'm at igcse level

[slow fog]

Yes

[umbral swan]

so I can't "Write the distance of th eobject from the centre in matrix form"

as mentioned before, what does the task provide?

[slow fog]

I don't have a specific problem, I'm not sure how to do the rotation part

I'm confused about the whole thing

[umbral swan]

well without more context I can't help :/

[slow fog]

I'm so sorry for not being able to express the problem properly, should've paid attention in class😭

Can you help with this tho?

[umbral swan]

which part

[slow fog]

Part d

[umbral swan]

well you know A

and you know that it gets scaled at center (4, 0)

so do a projection of each corner of the triangle

at (4, 0)

with the factor -2

@slow fog

this would be the first corner

of the image

[slow fog]

How do I know its getting scaled at (4,0)

[umbral swan]

it's written in the task

[slow fog]

OHHH

[slow fog]

But why did you choose the coordinate (-2,-4)? You could've stretched the line more

[umbral swan]

bc it's scale factor -2

🐛

[slow fog]

Then what would the other corner be like

Wait

Is it something like this?

[umbral swan]

yes!

[slow fog]

Oml thanksss😭

[safe radish]

@slow fog Has your question been resolved?

[safe radish]

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[ember finch]

I have a question

[ember finch]

I was cooking pretty hard with this integral

and i substituted x^2 + 1 = t

then i continued the integral

You will get this

here comes the cooking part

I realized that I could do x^2 = t - 1

so I did

After working the integral out, i got this

then i removed the t subsitution and returned it to its original state

Makes sense so far right?

[ember finch]

According to wolframalpha however, the answer is actually this

It's almost the same but the big difference is that the x^2 is single

according to my calculations it should be 3/2(x^2 + 1 - log(x^2 + 1))

and not 3/2(x^2 - log(x^2 + 1))

[safe radish]

@ember finch Has your question been resolved?

[ember finch]

<@&286206848099549185>

[mellow iron]

@ember finch

[ember finch]

oh my god yes

pls help me

[mellow iron]

Do u need help with your method or how the solution

Lmao I'll do my best

Or how your solution compared to the actual solution

[ember finch]

everything should be correct but then something weird happens

[mellow iron]

What's the weird thing

[ember finch]

the + 1 is nowhere to be found in the wolframalpha answer

you can look through the photos if you want

everything i did should be correct

[mellow iron]

Oh that's obvious

BTW nice methode

Because of the constant there buddy

Since you did a variable change

The bounds have to change

Thus a different constant

You are trying to find the primitive

Then just put the 3/2 + C = C'

Clear?

[ember finch]

i'm not sure i understand this

[mellow iron]

Ok

I'll try to explain in much simple

I just néed the correct words 1s

[ember finch]

Do we automatically assign all constants to the + C?

is that what you're trying to say

[mellow iron]

Yeah

[ember finch]

so if we had for example 10x + 11 + ln(x) + C it'd be 10x + ln(x) + C

[mellow iron]

Because +1 is a constant is it not?

[ember finch]

yes it is

so the C absorbs it

[mellow iron]

Yeah

[ember finch]

ok thanks alot bro

[mellow iron]

No problem

[ember finch]

been struggling with this for 30 mins lol

.close

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[mellow iron]

Actually

...?

You are very good bro

Keep it up

[safe radish]

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[gloomy prism]

Hi i dont understand how to sketch graphs of subsets of a complex plane when they equal each other

[gloomy prism]

Like this for example

[rough storm]

think of it more like a geometric picture. It may also help to use definition of modulus, and express z as x+iy

| | can be interpretted as distance.

|z-c|=|z-d| can be thought of as the set of points in the complex plane that are the same distance from c and d

[gloomy prism]

If i do the modulus - will x - yi for example become x + yi?

Or is that not how it works

[rough storm]

just combine real parts and imaginary part

so if you have |z-2-3i| this will be |(x-2)+(y-3)i|=sqrt((x-2)^2+(y-3)^2)

[gloomy prism]

Oooh ok like modulus as in ur finding magnitude

Omg ok i got the answer thank you

[rough storm]

np :)

[safe radish]

@gloomy prism Has your question been resolved?

[gloomy prism]

Sorry i just have another question

I have found q and thats correct and i found x in my working which somehow is the answer to P

Im not sure how to find p but did i do it the wrong was or something

[safe radish]

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[pine cape]

How do you solve for $3=2\cos^2(\theta)+3\sin(\theta)$ over 0 <= theta <= $2\pi$ and end up with multiple answers?

[flat frigate]

guy

[pine cape]

im studying for my final rn and have complety forgotten smh

[dense prism]

let sinx=t and solve the quadratic

[runic pine]

Convert cos²x in terms of sin²x

[dense prism]

then use arcsin and arrive at the equivalent angles

[pine cape]

this sounds the most familiar to what i did (i think), do you do that by using $sin^2(x)+cos^2(x)=1$?

[flat frigate]

guy

[runic pine]

Yep, then as the above guy said take sinx as t and solve the quadratic

[pine cape]

why are you converting it to sin^2 though?

[runic pine]

To get every term in terms of sinx...

[pine cape]

oh yeah

and then you factor or smth right?

[runic pine]

It's better to have one similar variable

[pine cape]

ohh yeah

[runic pine]

Right

[pine cape]

ok im starting to remember this

shouldve taken notes tho 😪 lmao

[runic pine]

So you will be able to it now right ?

[pine cape]

lemme write this out and make sure i got it correctly

[runic pine]

Aight

[pine cape]

so since $sin^2(x)+cos^2(x)=1$, $cos^2(x)=1-sin^2(x)$ which means

[flat frigate]

guy

[pine cape]

$3=2-1sin^2(\theta)+3sin(\theta)$

[flat frigate]

guy

[pine cape]

right?

[runic pine]

2sin²x

2(1-sin²x)= 2-2sin²x

[pine cape]

wait what did you just do

[runic pine]

$3=2-2sin^2(\theta)+3sin(\theta)$

[flat frigate]

ADITI

[pine cape]

oh wait nvm i see

yes i see

okay thank you i get it now

[runic pine]

Aight

[pine cape]

.close

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[slow fog]

N3ed help with part d, e and f

[safe radish]

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@slow fog Has your question been resolved?

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[languid stirrup]

Hi can someone help

[burnt wyvern]

,rotate

[languid stirrup]

Thx

Can sm1 help

[lean otter]

ye

[languid stirrup]

And then?

[lean otter]

get the angle

[languid stirrup]

How

[lean otter]

you have the length ratio

[languid stirrup]

Uhm

I don't know..

Ya

The answer is 47/100

I have sm1s working but I don't understand

,rotate

[lean otter]

whered you get 2/3pi

[languid stirrup]

As i said

Not my working

[safe radish]

@languid stirrup Has your question been resolved?

[languid stirrup]

<@&286206848099549185>

[umbral swan]

start by describing it as formula

ShadedArea = (2*CircleArea + RhombusArea - 2*IntersectionArea)/LargeCircleArea

@languid stirrup

you know that CircleArea = pi * 1²

you know that LargeCircleArea = pi * 2²

you know that RhombusArea = 2 * 4 * 1/2

you know that IntersectionArea = pi * 1² * (IntersectionAngle/2pi)

you know that half of the intersection angle is: 1/2 * IntersectionAngle = arctan(2/1)

so you have all values

[safe radish]

@languid stirrup Has your question been resolved?

[umbral swan]

._.

[safe radish]

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[lime atlas]

This is what I did

[lime atlas]

This is the derivative

Found the same answer as WolframAlpha, but the questionnaire doesn't accept any answer with cos(x)

[obtuse plover]

ngl bro ur writing looks like u wrote with a pen on the other side and the ink bled through

like i can barely see it

[upper rivet]

consider changing pens

or is that a pencil

[lime atlas]

Mechanical pencil

Is this any better?

[upper rivet]

any answer with cos(x)?

[obtuse plover]

can u make it a lil easier to read

[upper rivet]

wdym

[lime atlas]

I can completely see it personally.

[obtuse plover]

aight bro

[upper rivet]

well either way thats not important

[obtuse plover]

realize that ur not the one whos tryna check the work xd

[upper rivet]

since you say its the same answer

as wolfram

[lime atlas]

Cox(x) is not an accepted term on the questionnaire.

[upper rivet]

what is the exact question

partial derivative wrt x of that function?

is this the whole question

[lime atlas]

Yes

That's the whole question

[upper rivet]

this is definitely right

so im not sure what it could be

[obtuse plover]

maybe they want u to take out a 2sinxcosx and make it a sin(2x)

[lime atlas]

This seems more readable.

[upper rivet]

could be that yeah

but would be a lil weird

[obtuse plover]

yea fr

[lime atlas]

The questionnaire also doesn't have any sin(2x) seemingly.

[upper rivet]

can we see the questionnaire

is that possible?

[obtuse plover]

is it multiple choice?

[lime atlas]

It has e^cos(x) but I can't ln cos(x)

[lime atlas]

False and true

[obtuse plover]

Hmmmmmmm

[upper rivet]

did you maybe switch x and y in the question

could it be 11sin^8 (y) e^(5cos(x))

[lime atlas]

Then I reconstruct each function.

But no cos(x)

[lime atlas]

I definitely did not.

That's what's bugging me.

The professor seemingly forgot to get cos(x).

[upper rivet]

wdym

what does f1 and f2 refer to here

[lime atlas]

The function is separated into two functions, f1 and f2.

The f2 is the e^5(cosy) and f1 should be sin⁷(x)cos(x)

However that seems to not be true.

But no option seems to give me a proper answer.

[upper rivet]

Oh so their product is supposed to be the final answer

why n1 and n2 then

[lime atlas]

Yeah, the constant is separate.

[lime atlas]

He wants me to power the functions there it seems.

So it probably would be (sin(x)¹)⁵

With 5 being N1.

Not 5 but 7

Anyways

Cossine is disappeared?

[upper rivet]

yeah regardless of power, cosine disappearing doesnt seem right

[lime atlas]

That's what's bothering me.

Unless I have to use some trig identity, but none seem to fit onto the question.

[safe radish]

@lime atlas Has your question been resolved?

[safe radish]

@lime atlas Has your question been resolved?

[lime atlas]

What

.close

[safe radish]

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[dusk nova]

if i have 0 < x^y < 1 then can i ln both sides

[dusk nova]

ik ln(0) isn't defined but can i work around it using this?

,, \lim_{x \to 0^+} \ln (x) = - \infty

[flat frigate]

nyxie9151

[peak estuary]

well I mean you would get -infty < ln(x^y) < 0 basically

so sure you could do that but you dont gain anything

[dusk nova]

i have a trivial task which is to show that given y is an integer, y can either be greater than 0 or less than 0

[peak estuary]

-infty < ln(stuff) anyway

[dusk nova]

for my task it's okay right?

[peak estuary]

yeah then using ln etc is probably not a good idea

what even is your definition of integer

[dusk nova]

no it's a multiple choice

not a proof

[dusk nova]

whole numbers + negative whole numbers

is there supposed to be another definition?

It's just a multple choice like this:

a) Y > 0

b) Y < 0

c) Y can be >0 or < 0

given that y is an integer and x has no constraints

[peak estuary]

!original

[safe radish]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[dusk nova]

if i have 0<x^y< 1 and y is an integer, then choose the option that is true:

A) y > 0

B) y < 0

C) y = 0

D) Neither

The answer is "neither" i guess lol

[peak estuary]

yes if you know nothing about x then you also dont know anything about y

[dusk nova]

this is why i was curious if i could do:

0 < x^y < 1

ln both sides: -inf < y ln(x) < 0

y > 0 then ln(x) < 0 which means 0 < x < 1

y < 0 then ln(x) > 0 which means x > 1

[dusk nova]

okay yes but can i do the ln(0) as a limit for other questions or whatever

i'm sorry this is kinda stupid but just wanted to make sure lol

0 < x^y < 1

ln both sides: -inf < y ln(x) < 0

y > 0 then ln(x) < 0 which means 0 < x < 1

y < 0 then ln(x) > 0 which means x > 1

y = 0 then ln(x) can be anything to satisfy the inequality

so the answer is D.

[peak estuary]

well if y=0 then x^0=1

so cant be that

but yes fine enough for the rest

[dusk nova]

okay, i just wanted to know the ln both sides

because this is unconventional

but the inequality is preserved because i'm applying an increasing function right?

ln(x) is increasing throughout it's domain

the only problem i had was 0 technically isn't in its domain

lol

[peak estuary]

0<x^y just tells you that you can apply ln to x^y in the first place

the "information" -infty < ln(x^y) is useless

[dusk nova]

so i could've just went

x^y < 1

and then ln both sides

[peak estuary]

yes

with the knowledge that you can apply ln on both sides because 0 < x^y

[dusk nova]

would i always ignore the lower bound btw?

if i ln both sides

if the lower bound is 0

[dusk nova]

yeah fair enough

[peak estuary]

I dont like answering questions with "always" in them

but sure if its the same situation

[dusk nova]

Lol thanks

.close

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[timber cave]

The diagram shows a half cylinder of diameter, 20 cm and length 40 cm calculate the volume of the half cylinder

[timber cave]

Guys is diameter radius here

[somber parcel]

Diagram?

[marsh walrus]

diameter isnt radius

diameter is double radius

radius is from the center to an edge

diameter is from an edge, to the center, and back to an edge

[timber cave]

Is it correct

[sleek hornet]

yes

[timber cave]

But shouldn’t we divide by 2

Since it’s half cylinder

[sleek hornet]

no

general formula for volume of cylinder is pi r^2 height

[somber parcel]

Yes

[timber cave]

One says yes and one says no.

what.

[sleek hornet]

eh nvm

i read wrongly

lmao

yea divide

diagram cut half vertically instead of horiontally

[surreal socket]

Why is there a sharmoota wrap 😭😭

[timber cave]

Zits shamoota wrap..

Not with r

[surreal socket]

Alright sorry🙏

[timber cave]

Guys I have problem with knowing if it’s cm squared

Or just cm

I got 7065

[sleek hornet]

volume

[timber cave]

Huh

[sleek hornet]

if length is cm
area is cm^2
volume is ?

[timber cave]

Volume is 7065

[sleek hornet]

its cm^3

[timber cave]

Why

[sleek hornet]

cm is a unit

cm multiply by cm u get cm^2 right?

[timber cave]

Ya

Then we divide it

By 2

[sleek hornet]

2 doesnt have a unit -

[timber cave]

Ok then wha

[sleek hornet]

cm x cm u get cm^2

[timber cave]

Ya

[sleek hornet]

pi r^2 height its cm^2 x cm

which gives cm^3

dividing by 2 doesnt change the unit

cuz ur juz halving the volume

[timber cave]

Oh ok

[safe radish]

@timber cave Has your question been resolved?

[safe radish]

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[wooden oyster]

where exactly is p

is p just a smaller circle within r

[safe radish]

@wooden oyster Has your question been resolved?

[wooden oyster]

<@&286206848099549185>

[steel stag]

p is within the circle of radius r, close enough to a so that you can pick the epsilon you want

[wooden oyster]

but why

it only says (0,r)

it isnt even specified it is within omega

i understand is is smaller than r and bigger than 0, but i dont see why that circle couldnt be outside of omega

[steel stag]

well you can always make the circle small enough to be within omega right? (from the definition of open set)

and they say that here, the circle is in omega

[wooden oyster]

i am aware the circle is in omega

and that p is smaller than the circle that is is within omega

i just dont see why that the smaller circle ough to be placed within r

why would this not also h old

circle is smaller than "r" circle in omega

[steel stag]

does it make sense in the 1 dimensional case? picking a small enough interval delta away from a

[wooden oyster]

yes, but now it is away from a

[steel stag]

there's a circle of radius r such that f(x) is fully defined (not outside omega), and then within that they get to make a smaller circle of radius p to apply continuity

[wooden oyster]

yes

that can only make sense in that way

but how does the notation in any way tell me that

[steel stag]

if there's any part of the p circle not in omega things break

you can't define f

so they go through all this trouble to find a spot that lets them do normal analysis stuff

[wooden oyster]

so because they are continiuous you can find such a p

i just find the notation very weird

(0,r)

why doesnt this just mean a circle with smaller radius than r and bigger than 0 radius

[steel stag]

the picture is ||y-a||<p which is a circle

sure they just don't describe a circle with radius p at all explicitly

they could've said circle of radius p centered at a to make it clearer beforehand, but that comes up at the end with ||y-a||<p

idk it seems like a style issue, it's implied that K(a,p) in omega as well

[wooden oyster]

they give you that at least

also, what is the difference between || and |

absolute value which sometimes means distance and the norm which also sometimes means distance?

[steel stag]

they're the same idea of a distance function yea

| is usually 1 dimension absolute value and || is higher dimensional

sometimes people use | for higher dimensions anyways but whatever

[wooden oyster]

so p is basically delta in this proof

[steel stag]

right

[wooden oyster]

alright

where y is just a point close to a

that is epsilon/2 close

[wooden oyster]

but doesnt this just tell you that the distance between y and a are smaller than p

y-a < p < r is what we have

this only talks about the size of p not its location

[steel stag]

right, p is just a number

[wooden oyster]

how does given the size of a number ensure me that it is in r

[steel stag]

||y-a|| < p < r means you're looking at y's within the circle K(a,r)

[wooden oyster]

wh

y

[steel stag]

||y-a||=r is the geometric definition for a circle of radius r at point a

[wooden oyster]

yes

but we are still only talking about radius

where in what is given would it be wrong for the radius to be placed elsewhere

[steel stag]

like just looking at this

the only thing telling you that is the a in y-a

it means you're looking at the center a

[wooden oyster]

yes

[steel stag]

maybe it'd be better to rephrase it as p is chosen such that the circle of radius p at a is in a valid domain

[wooden oyster]

this is how i imagine it

[steel stag]

right that's the correct picture

[wooden oyster]

i cant not see how that could possibly be extracted purely from the text

since the text only takes a bout the size of p and nothing else, but i suppose its alright

[steel stag]

I see how it can feel weird because the only equation of a circle with radius p is the very last part of the last row of the proof, but that's where it comes up

sometimes proofs are written in a sort of backwards way where you choose delta and epsilon so that 10 steps later it works out

[wooden oyster]

well the proof isnt finishe

d

[steel stag]

oh

[wooden oyster]

i can show the rest

let me write it into the document

[steel stag]

I don't expect it to matter tbh

[wooden oyster]

it is about a page tho

[steel stag]

sure go for it if a part there doesn't make sense

[steel stag]

looks like the rest is going from the 2d circle interval to 1-dimensional limits in both directions

[wooden oyster]

i think the idea is that

we've onyl defined the mvt for 1 dimension

so you just hold one take on variable as a constant

im just not sure about the rest after that

what happens with the t

[steel stag]

t is basically wherever g() is but they quickly move on to only using the xi they get out of it

like it's a dummy variable for the function constant in the y-direction

[wooden oyster]

yeah

how do we get the last line

f(a1+deltax,a2)-f(a1,a2) = D1f(xi,a2)deltax1

[steel stag]

looks like it's both converting from g back to f: g(t)=f(t,a2) and plugging in the D1

[wooden oyster]

mhh okay

but how does t become xi

[steel stag]

xi is a certain t value

I'm a little confused yea I think it should read g'(t)=D1f(t,a2)

[wooden oyster]

idk

thye keep using it throughout the proof tho

[steel stag]

yea D1 should always be applied to a function here

[wooden oyster]

first we have

$$ g(a_{1} + \Delta x_{1}) - g(a_{1}) = D(g)(\xi_{1})\cdot \Delta x_{1} $$

[flat frigate]

// mav

[wooden oyster]

which is just the mean value theorem

because it is differentiable at a1 and a1+delta x1

So f(a) = a1 and f(b) = a1+deltax1

and our b-a = delta x1

$$ f(a_{1} + \Delta x_{1}, a_{2}) - f(a_{1}, a_{2}) = D_{1}f(\xi_{1}, a_{2}) \cdot \Delta x_{1} $$

[flat frigate]

// mav

[wooden oyster]

I am just sure how it turns into this

[steel stag]

does the left side make sense? it's just that going from g to f is adding in that ,a2 parameter

[wooden oyster]

i can see since g(t)=f(t,a2)

then g(a1) = f(a1,a2)

oh

i think i see it now

i have more questions @steel stag but i have to go

can i tag you later or dm you privately when i have returned?

[steel stag]

uhh you can ping me tonight but 50/50 on if I'll be able to come

[wooden oyster]

its fine. There is always tomorrow

[safe radish]

@wooden oyster Has your question been resolved?

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[lean otter]

https://www.scribd.com/document/709550643/8300-1H-QP-Mathematics-G-8Nov23-AM-pdf

[lean otter]

can someone help with solving this exma paper

from question 15 till end

[lean otter]

can you share a picture?

not many people would like to click on links

sure

oh sorry

i need workings aswell

We can't give you answers or do the work for you. Also, you should tell us what you tried/know and what is confusing you at least

oh

wait whats the point then?

i dont even know what a box plot is

to help you get to the answer yourself, and not give it to you on a silverplate

lol i just know i have an exam and i need to cram this stuff

[lean otter]

ill seem like an idiot if i say give it to me on a gold plate but lemme cook rq

is the fastest time29?

[lean otter]

i mean doing it yourself will make you remember it much much easier. Reading whatever answer ill give u will like, not help that much ig

[lean otter]

well thats the person that finished the last

so like, the slowest lmao

NAWH

so i am dumb

dont do this to me

17 MINUTES

yes

next question we move aheaddd

okay so on this one

ill cook

in 2019 the l.q was 20 minutes, the median 25 minutes and U.q the same as 2019 and 2020 but in 2020 the lq was 21 minutes the median 24 minutes and u.q the same showing that on avg. times were faster in 2019 that 2020

DID I COOK>

?!

@qlyo

@lean otter

how is the median in 2019 25 mins, isnt the vertical line meant to denote where the median is?

[lean otter]

oh shoot

22.5

mb mb

except that im correct right?>

next oneeee

this one

@lean otter i really dont understand this one

can u explain some

oh uh are we done with the previous one already?

yepp i asked m mum she said its 2 marks leave it

lol

i said bett

anyways help

:<

aight

wai

i got somehting

ill ping u when i cook abit

[lean otter]

you should probably figure out what x is first

i defo cooked

x= 19

ok sounds legit

waht next

anyways just divide the amount of people that bought a hoodie with the amount of people that were there to begin with, so like 80

19+8/80

?

27/80

thats gonna bee the first answer?

sounds fine yeah

second one 23/80

okay WHAT THE FLIPP

is this

[lean otter]

dont think so

uh humm

its conditional probability so u need to do like [
\6\P{A\where B} = \4{\6\P{A\cap B}}{\6\P B}
]

[lean otter]

ohh

so whats the answer gonna be?

p a and b

hmm

and is + right?

wot is +

plus

wot

huh

can u gimme the answer

ill cook

no

bru atleast help abit

u just gave me an equation i saw for the first time

second maybe

can someone who acctually helps come

.close

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[lean otter]

so a quick q: if im trying to evalute when a t interval method is an option for calculating the confidence interval, and I see that the sample has some outliers, does that mean that I cant use the t interval because the sample is not normal?

[lean otter]

i figured it out, the t interval cant be used the sample is not normal

.close

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[urban eagle]

Can any1 help with Q14 b

I need the normal right?

But how do I find the normal

Centripetal force = vertical weight - Normal

But how do I get vertical weight

[crude coyote]

i came here looking for the answer to this exact bwsi question lol😭

[safe radish]

@urban eagle Has your question been resolved?

[fiery ibex]

💀

[fiery ibex]

i have it i can dm it to u

[crude coyote]

yes PLEASE lol im searching far n wide

[fiery ibex]

ya lol ill send it rq

have u done the pizza shop exercise for the python prereq? im stuck on how to differntiate between two pizzas in the function params 😭

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[silver cape]

Hello, am working on my calc 2 question here for uni, not sure if this is correct? Apologies for my incompetence.

[safe radish]

@silver cape Has your question been resolved?

[silver cape]

<@&286206848099549185>

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@silver cape Has your question been resolved?

[silver cape]

<@&286206848099549185>

[silver cape]

<@&268886789983436800>

[white umbra]

Please only ping moderators for server-related concerns, not math help

[silver cape]

Apologies, seen someone else do that. Thanks

[safe radish]

@silver cape Has your question been resolved?

[eternal carbon]

is close sort of but no

u need the "K" term in ur error bound formula (the maximum value of the second derivative of the integrand on the interval of integration)

also not incompetence... everyone agrees error bound is irritating

[silver cape]

I see, so I’m missing the K term, so I need to find the second derivative of e^-x^2, and then how do I get the maximum value and where do I place the K value in the formula?

[eternal carbon]

u put it right here

[silver cape]

Oh shit missed that

Thanks

[eternal carbon]

and for an error bound problem, the maximum value normally occurs at the endpoints of the interval or at like some simple point like x = 0 or something

if there's a situation where u aren't 100% sure u can check by analyzing the 3rd derivative of smth and seeing where it is 0

[eternal carbon]

welc

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[crude grove]

how is the curve on the right and the line on the left..

[crude grove]

i can’t tell

[proud belfry]

Can you give some context

What is this about

[crude grove]

it’s abt area between curves

[proud belfry]

Well in that picture, the curve is to the right of the y axis

[crude grove]

it looks like it’s going in the same direction

i can’t rlly tell how to see that

oh wait

nvm

if u do it

wait let me show

[proud belfry]

Whet

What

No i mean like

[proud belfry]

In the graph

[crude grove]

r u supposed to look at it like this

the curve on the right

[proud belfry]

????

[crude grove]

and then the

uhm

[proud belfry]

Thats the left

[crude grove]

the curve

is on the right

the line is in the left

[proud belfry]

In the rotated picture

The curve is on the left

[crude grove]

this is the line i’m talking abt

it’s on the left

of the curve

[proud belfry]

Thats not the y axis

What

[crude grove]

Oh ok wait

i’m confusing myself

ok back to here

[proud belfry]

Ok, they are trying to find l

[crude grove]

jow can u see left and right on this

it just shows top and bottom same direction

😭

[proud belfry]

Ok do you know what the y axis is

[crude grove]

yes

[proud belfry]

Draw an arrow to it

Or just describe it

[crude grove]

oops

x axis

my bad

WAIT

THATS X AXIS

y axis is the vertical

one

oh

wait but

ik whatur talking. abt

[proud belfry]

Ok

We are trying to find L

[crude grove]

y axis

[proud belfry]

Brain farts be like

[crude grove]

the curve is on the right of the y axis

[proud belfry]

Yea

[crude grove]

earlier

i showed it here

like this

to see the y axis

it was correct 😭

[proud belfry]

What

Ok now the curve is on top of the y axks

Axis

[crude grove]

like that?

[proud belfry]

Yes

[crude grove]

ok

got it now

thank u

.close

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[plucky juniper]

Sorry for the messy work but can anyone help me with this

[safe radish]

@plucky juniper Has your question been resolved?

[plucky juniper]

<@&286206848099549185>

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@plucky juniper Has your question been resolved?

[left gyro]

not sure about this but try rephreasing the question to be:
Use Lagrange multipliers to find the point that maximizes z with the constraint (x^2 + y^2 + z^2 - 34)^2 + (2x + y - z - 2)^2 = 0

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[plucky juniper]

.close

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[covert acorn]

[covert acorn]

23/10 how

23/10 where

nvm

im dumb

supposed to be minus

mbbbb

.close

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[dry stone]

Could you show me in steps how to get Y:

[bleak verge]

I'll give you a small hint

You can take 1/(1+y) to be equal to some variable (say a)

After which you will get a quadractic equation

Upon solving it you will get the value of a from which you can find the value of y (dont forget this step)

[dry stone]

ill try rn

[bleak verge]

Ok

lmk once your done

[dry stone]

i keep getting stuck

[bleak verge]

can you tell me where?

[bleak verge]

Give me a second

[dry stone]

Okay

[bleak verge]

Take a look at this

Beyond this you may need a calculator i suppose

You can use the quadractic formula to solve

[dry stone]

Oh I see what u did

[bleak verge]

👍🏼

[dry stone]

Thank you!

[bleak verge]

Np!

Good luck on solving!

[dry stone]

Ty

Y should be 7.9

%

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[dry stone]

.reopen

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✅

[dry stone]

<@&286206848099549185>

How did they get the yield

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@dry stone Has your question been resolved?

[dry stone]

<@&286206848099549185>

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@dry stone Has your question been resolved?

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[vagrant yew]

is it fine if i have a question from physics?

[umbral swan]

despicable

yes is fine

[desert pasture]

physics and math go hand in hand, sop yes

[vagrant yew]

this is ez but um i want to check my answer

[fleet orbit]

its A if you want the answer

[vagrant yew]

ah tysm

.close

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[umbral swan]

yes, but rarely I feel like this with some physics problems

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[timid scroll]

Is this wrong or right

[frigid locust]

should be -2 in the exponent place

then the denominator would be -2 and combined with the negative would be 2

integral power rule is
$$\int f^n(x) \cdot f'(x) = \frac{f^{(n+1)}(x)}{n+1}$$

[timid scroll]

Ok

But look what I get then

[flat frigate]

JustToPro

[timid scroll]

Or am I doing something wrong

[frigid locust]

that isnt the answer u did something wrong

[runic pine]

You mixed up diff and int

[timid scroll]

On purpose ^

[frigid locust]

thats u sub , i think

[runic pine]

No

Like

[frigid locust]

oh diff power rule u mean

[runic pine]

Yeah

[frigid locust]

im confused

[timid scroll]

Oh but im mixing it with

[frigid locust]

which question are we doing?

[runic pine]

@timid scroll u did smwhat right but

[runic pine]

The powers ain't right

[timid scroll]

This question

I think im mixing two formulas which im not uspposed to

[runic pine]

There's a easier way tho

[frigid locust]

yeah this is right , but applying the integral power rule was wrong

[runic pine]

Just take cosx as u

[frigid locust]

but ur second attempt was something else

[timid scroll]

Can we do it this method

Integration by substitution in next chapter

[runic pine]

This is fine too

[timid scroll]

aka in 15 questions

[runic pine]

That u did correct

But u messed up powers

[timid scroll]

ok?

ill try again

[runic pine]

1/t³ isn't 4/t⁴

It's -2/t²

[frigid locust]

$$\frac{1}{t^3} = t^{-3} = \frac{t^{-3+1}}{-3+1} = -\frac{1}{2t^2}$$btw

[flat frigate]

JustToPro

[frigid locust]

the 2 doesnt go up

it cant go up

[runic pine]

Yeah mb

[timid scroll]

,rotate

[timid scroll]

But that’s not ^-3

But ^-1

So huh

I can get rid of the 2 ez tho

[runic pine]

The question?

[timid scroll]

1/2

[runic pine]

Has 3?

[runic pine]

See

U took u as 2+cotx right

[timid scroll]

yea

[runic pine]

and u got du as -cosec²x dx right

[timid scroll]

yes

[frigid locust]

why do u not write the rest of the steps?

im pretty sure if u write them , u would get the answer correctly

[timid scroll]

du x dv

[runic pine]

So that becomes -1/u³ right

[timid scroll]

???

Ill write all the steps again

Very nicely < 3

[runic pine]

int of -1du/u³

Just replaced all the terms wrt u

[timid scroll]

I dont understand

lemme send pic

,rotate

[timid scroll]

Ok hit me

@runic pine @frigid locust

[timid scroll]

It’s perfect wym

[frigid locust]

i dont understand what u mean by this

is this the "u-sub" part?

[timid scroll]

Yea

[frigid locust]

should i solve it and send a pic and then if u dont understand i can explain

cuz u are somehow going further away from the right answer

[timid scroll]

[frigid locust]

ok wait

[timid scroll]

OH I REALISED

dy/dx of u^-2

[runic pine]

Integral

of

u^-2*

it's like int of dy=-u^-2 du

[timid scroll]

.