help-39

[vestal tapir]

right

[gilded hollow]

and that's it?

[vestal tapir]

yes

not "sum", you don't add up the remainders, just write them as digits

[gilded hollow]

and it's not euklides algorithm?

It's longdivision?

[vestal tapir]

that's the same thing apparently

it's called divmod in programming

the point is that you care about both the remainder and the quotient

you want to use both so the algorithm has to tell you both

@gilded hollowyou don't look at remainders from each step

for example 101 / 4 is 25 with remainder 1

you would get several remainders along the way, they don't matter, only the final 1

and then you do 25 / 4

[gilded hollow]

Okay that works for smaller values but what happens if i have
(BA0BAB)16 = (1011, 1010, 0000, 1011, 1010, 1011)

I can't take 1011 * 10^5 ...

and then do euclides

As we aren't allowed calculators

[vestal tapir]

i'm confused

if you want to convert it to base 2 you already did

[gilded hollow]

Sorry i want to convert it to base 8

[vestal tapir]

it works when one base is a power of another, like 216 and 6

ok, 8 and 16 aren't powers of each other

[gilded hollow]

ok

[vestal tapir]

i need to think

okay, we can go from 2 to 8 then

1011, 1010, 0000, 1011, 1010, 1011
you just divide this in threes

101, 110, 100, 000, 101, 110, 101, 011

[gilded hollow]

which base is it in now?

2 or 8?

[vestal tapir]

i'm maybe confused, let's see if it works

[gilded hollow]

It should be base 2

i think we need to use the euklides method

and take the remainders

to "translate"

[vestal tapir]

so (BA0BAB)16 = (56405653)8

ok it workes

so because 16 is a power of 2, we can convert between them digitwise, each digit of BA0BAB just defines 4 binary digits

and the same is true for 2 and 8

so the binary 101110100000101110101011 is easily translated to base 8 56405653

@gilded hollowyou got it?
just like you know the 1 to 4 correspondence between binary and hexadecimal there's the same thing for binary and octal

[gilded hollow]

hm

Nah i don't get it

[vestal tapir]

well at least you have a better idea of what's there to get

basically we don't use base 10 here

[gilded hollow]

because base 10 would make the numbers really really large, right?

[vestal tapir]

right

it's a rare case where we can use an easier way

by converting each digit separately

1 001
2 010
3 011
4 100
5 101
6 110
7 111

this is base 8 to base 2 table

by analogy with what you showed

and you could use this trick to go base 216 to base 6 for example

it would be 215 lines long whatever

[gilded hollow]

so something like that?

[vestal tapir]

maybe

and you can't do this to go from base 18 to base 5 for example
if it's not a large number you can do the long way we discussed with base 10

and if it's large you give up

[pearl pond]

@gilded hollow Has your question been resolved?

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[midnight haven]

why is that repeated differentiation even necessary

surely if you got a linear combination of x,... x^n then it will equal the zero vector?

and then u can just compare co efficients?

[pearl pond]

@midnight haven Has your question been resolved?

[midnight haven]

oh yea the qu is to prove that set is linearly independant

nvm i got my answer

.close

[pearl pond]

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[tiny wadi]

any help with Q8, im stuck. all i was able to calculate is that the area of the triangle is $\sqrt{3}$

[jolly parrot]

hoax

[pearl pond]

@tiny wadi Has your question been resolved?

[tiny wadi]

<@&286206848099549185>

[tiny wadi]

<@&286206848099549185>

do y'all just hate me or wut omg

literally, why are my questions the only one that are ignored?

[tiny wadi]

<@&286206848099549185>

[tidal zealot]

hi

[tiny wadi]

hi

[brisk latch]

Same tho fr

[brisk latch]

Sameeee thoooo fr

Can relate

[tiny wadi]

dude i've asked three questions today

all ignored except the second one

the second one was partially ignored

[brisk latch]

Same

I’ve been getting ignored a lot

recently

[tiny wadi]

like

[brisk latch]

I feel like I’m on some list of something

[tiny wadi]

literally every other channel is being well answered

[brisk latch]

“Ignore These People List”

[tiny wadi]

why is this one being ignored in particular

[brisk latch]

Fr

[tiny wadi]

same fr

[brisk latch]

Relate on so many levels

Ok let’s start with the helping lol

[tiny wadi]

my math knowledge hasn't reached to the point of helping ppl yet

[brisk latch]

Wait

Is the circle fully covered

[tiny wadi]

rather it's at the point where i need help

[brisk latch]

.rotate

[brisk latch]

,rotate

[jolly parrot]

Couldn't find an attached image in the last 10 messages.

Jan 7, 2024, 10:54 PM

[brisk latch]

Send it one more time

[tiny wadi]

ok

[brisk latch]

,rotate

[tiny wadi]

i think the circle is fully covered

[brisk latch]

Ok

So

Gimme a sec

Ok I think I got it

So draw a diagram cl

From the center of the circle

To points where the circles touches the side of the triangle

[tiny wadi]

ok

which side tho?

the left side, the right side, or the bottom one?

[brisk latch]

All sides

[tiny wadi]

ok

now what?

[brisk latch]

Ok so

Now the angle between each of the lines

Is 120 degrees

[tiny wadi]

hold on

[brisk latch]

What

[tiny wadi]

how do you know it's 120 degrees

[brisk latch]

Ok so

A circle is total 360 degrees right?

[tiny wadi]

yea

[brisk latch]

And the circle is in an equilateral triangle

[tiny wadi]

yea

[brisk latch]

So actually

Think of it this way

So draw a line from the midpoint of the line

Of each side

The midpoint of each side

Draw a line so that

It connects to the center of the circle

And extend it forever

[tiny wadi]

uh huh

[timid heron]

couldn't you do it this way and then just calculate the circle's area and just write the probability being Circle's area over Triangles Area?

[tiny wadi]

lemme try

w8

how do you know the angles for the smaller triangles tho?

and what is $\frac{a\sqrt{3}}{2}$?

[jolly parrot]

hoax

[timid heron]

a = 1

because the side of a triangle is 2

[tiny wadi]

how do you know that?

[timid heron]

isn't like under the photo 1? or it's just a random line

[tiny wadi]

so i get this

but idk how we prove a =1

[timid heron]

it's the same triangle but 2x smaller

because it's base is 1

[tiny wadi]

how can you be sure?

[timid heron]

all the sides of the triangle are 2

[tiny wadi]

the larger triangle, yes

[timid heron]

i now realised

I would need to assume that the circle's side is touching triangle's sides

[tiny wadi]

the circle is touching the triangle

[timid heron]

then the circle is touching the triagnle's sides perfectly in the middle

deviding all the sides by 2

[tiny wadi]

is there like a theorem or something for that?

so ur saying in this case the center of the circle is also the center of the triangle?

[timid heron]

yes

I think it's obvious to see

I only remember my teacher saying it one time but didn't show the theorem

something like that

[tiny wadi]

nonetheless how do i calculate the area of the circle?

[timid heron]

I'm drawing it

[tiny wadi]

ok

[timid heron]

for a = 1

[tiny wadi]

$r = \frac{1}{\sqrt{3}}?$

[timid heron]

it's (pi/3)/triagle area

[jolly parrot]

hoax

[timid heron]

1/(sqrt(3)) = sqrt(3)/3

$r = \frac{\sqrt{3}}{3}?$

[jolly parrot]

឵឵឵kaczuch

[timid heron]

(pi/3)/sqrt(3)

[tiny wadi]

bru i used r = $\frac{1}{\sqrt{3}}$ and i got the correct answer somehow?

[jolly parrot]

hoax

[tiny wadi]

if we let r = 1

i mean

1 /(sqrt3)

[timid heron]

then circle's area is pi/3

[tiny wadi]

then the area is A = $\pi\left(\frac{1}{\sqrt{3}}\right)^2$

[jolly parrot]

hoax

[tiny wadi]

yea

[timid heron]

yup

[tiny wadi]

and the area of the triangle is $\sqrt{3}$

[jolly parrot]

hoax

[tiny wadi]

so the probability becomes

Area of circle/ Area of triangle

[timid heron]

yes

[tiny wadi]

$\therefore$

[jolly parrot]

hoax

[timid heron]

checked it with Wolfram (should I delete my messages to make your sentence clear?)

[tiny wadi]

$P =\frac{ \frac{\pi}{3}}{\sqrt{3}}$

[jolly parrot]

hoax

[tiny wadi]

nvm

[timid heron]

which equals to pi/3sqrt(3)

[tiny wadi]

i don't think so

if we multiply the top and bottom by sqrt3

we get

w8

hold on

what am i doing

[timid heron]

already checked it

[tiny wadi]

im tryna evaluate the complex fraction myself

[timid heron]

k

[tiny wadi]

oh yea

we can multiply the top and bottom by 3

wow u actually solved it

ty!

[timid heron]

np

what's funny is that I didn't use trigonometry

[tiny wadi]

there probably is a simpler way to solve this, right?

[timid heron]

except the 30 60 90 triangle

[tiny wadi]

because if a question like this comes in my exam, im screwed

[timid heron]

propably, but it was the first that came to my mind

oooh

wait

I got confused

[tiny wadi]

?

anyways imma go take a shower

[timid heron]

nothing

[tiny wadi]

i have my math exam in like 4 hours

[timid heron]

my calculations are correct but firstly we had to calculate the triangle's sides using the sine that the question gave us

after that everything is correct

now 8. b)

[tiny wadi]

we can just first subtract the area of the circle from the area of the triangle

then we find the probability of a dart landing outside the circle

by $\frac{new area}{area of triangle}$

[jolly parrot]

hoax

[tiny wadi]

then multiply it with 100

[timid heron]

yup

[tiny wadi]

aight, once again

thx for the help

.close

[pearl pond]

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[spring crystal]

How the hell do these numbers in front work

[graceful lava]

,tex $y = - \frac{2}{3} (x - 1)^{2}$

[jolly parrot]

@graceful lava

[graceful lava]

@spring crystal we need to use PEMDAS for this

first, you do the parantheses and the exponents

[spring crystal]

Isn’t it supposed to go down 2 and then right/left 3

[spring crystal]

I mean for graphs

[graceful lava]

thats the wrong kind of transformation

[spring crystal]

Why

[graceful lava]

Have you heard about:

• Vertical shrink/compression
• Vertical stretch/expansion
• Horizontal shrink/compression
• Horizontal stretch/expansion

[spring crystal]

I’ve heard of them

[graceful lava]

ok perfect

[spring crystal]

Fraction is stretch and big numbers r shrink

[graceful lava]

is stretch
vertical or horizontal

[spring crystal]

The number in front

[graceful lava]

i get that pard

*part

but is -2/3 a vertical stretch or horizontal stretch

[spring crystal]

Horizontal

[graceful lava]

ok perfect

but you must remember the -

this flips the entire parabola upside down

[spring crystal]

Yeah so it goes down 2 and then left/right 3?

Isn’t it the slope

@graceful lava

[graceful lava]

um no

thats a translation

[graceful lava]

ya, it is the slope

but the slope is not what makes a function move left/right/up/down

[spring crystal]

Isn’t functions like 3/4x+2 means the graph starts at 2 and then move 3 up and 4 left/right? And it’s the slope

[spring crystal]

So what do I do

[cloud zephyr]

here's a site about your question.

https://www.mathsisfun.com/sets/function-transformations.html

feel free to read it and understand how those +2 or times 3/4 works first

[spring crystal]

So the graph starts at (1,0) and then goes down 2 and left/right 3?

Is that what the graphs saying ?

[cloud zephyr]

firstly, the graph can start at (1,0) but not necessarily.
secondly, it doesn't "goes down 2 and left/right 3", but it compess the graph along y-axis by 2/3

and of course, don't miss the negative at the front

[spring crystal]

I don’t get it

Then how do I know the points in the graph

[cloud zephyr]

hmmm, do you mind telling which grade are you in?
so that i can provide suitable method for you

like if you are around grade 7 to 8, I'd suggest to use a table to draft it first.
like if you are around grade 9 to 10, I'd suggest you can use basic graphs and do transformations on it

[pearl pond]

@spring crystal Has your question been resolved?

[pearl pond]

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[dreamy wind]

How do you find the derivative of e^-1

[near echo]

with respect to what

[dreamy wind]

to x

[near echo]

,w e^-1

[jolly parrot]

Results provided by WolframAlpha

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

Requested by maximofs

[near echo]

does this change your perspective on the problem

[dreamy wind]

oh wait

Mb i meant

e^-x

[near echo]

i see

what derivative rules do you know

or do you need to use the definition

[dreamy wind]

What rule would you use here?

[near echo]

what derivative rules do you know

[dreamy wind]

I think i know all of them

[near echo]

can you name some

[dreamy wind]

product, constant, quotient, multiplying, chain rule,

[near echo]

can you think of any that apply here

do you have a product of functions?
a constant function?
a quotient?
a composition of functions?

[dreamy wind]

i feel like composite

[near echo]

yes

[dreamy wind]

where f(x) = e^x and g(x) = -x?

[near echo]

yes

[dreamy wind]

okok that makes sense

Thanks

.clsoe

.close

[pearl pond]

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[violet pulsar]

Can anyone proof-check this work? I have most possibly made a mistake here, since it should've given an area of the first quadrant of an ellipse.

[violet pulsar]

[jolly shuttle]

As I know, the radius $r \ne \sqrt{(acos\theta)^2+(bsin\theta)^2} at the angle \theta$

[jolly parrot]

Dri111

[violet pulsar]

This is an ellipse

[jolly shuttle]

you thought about making ellipse with $(acos\theta, bsin\theta)$ right?

[jolly parrot]

Dri111

[violet pulsar]

Yes

[jolly shuttle]

is that point on the line with slope of theta?

[violet pulsar]

Trying to find the area of the first quadrant.

so, r goes to the border

like on the ellipse (I guess)

[violet pulsar]

What do you think I should do

[jolly shuttle]

I think Integrate as xy coordinate would be easier

not with polar form

[violet pulsar]

I somehow need to figure out the polar form way, should I do with x and y then swap the values with acos\theta and bsin\theta

[violet pulsar]

but my integration has no mistakes right?

[subtle ginkgo]

on this second page, you missed jacobian

[violet pulsar]

Wait... how does it work

still at highschool didn't really see that ever

[subtle ginkgo]

$\text{If}\text{ }\text{ }x=ar\cos\theta,\text{ }y=br\sin\theta\text{ }\text{ then jacobian is: }\\J=abr$

[jolly parrot]

Joanna Angel

[subtle ginkgo]

hence:

$\left| D \right|=\int_{0}^{\frac{\pi}{2}}d\theta\int_{0}^{1}abr\text{ }dr$

[jolly parrot]

Joanna Angel

[subtle ginkgo]

it can be also written in second form, but more complicated, if you liek i can show

[violet pulsar]

i have never seen jacobian

[subtle ginkgo]

also, let me explain you , this jaocbian thing

as you knwo elipse is not a circle

yo agre eon it )

but

[violet pulsar]

yes

[subtle ginkgo]

evaluting area of region limited by circle is a way easier

you confimr it

so th eidea is

how to transform

ellipse

into easy circle

[violet pulsar]

hm

[subtle ginkgo]

this gola we get with:

x = acostheta and

y = bsintheta

but

jacobian iis such a facgtor of correctness

in lienar algebra it is determinanto of the matrix

so if you always use soem transformaiton

to other coordinates

you need to use this factor jacobian

[violet pulsar]

Is it possible to determine with the highschool knowledge? This is even beyond the IB Math AA Higher Level curriculum...

[subtle ginkgo]

i do not know, i am from Europe )

jacoian we show to students, not to the pupils

[violet pulsar]

oh

[subtle ginkgo]

you must be +18 when you study at university

then you learn jacobian

anyway

jacobian is necessary

to make yoru transformaiton be correct

and now this area given by integral i wrote for you

is very easy

you have to admit

[jolly shuttle]

I think it is possible to get polar form

[subtle ginkgo]

that is also polar, but we use trig identity and then cosien and sine disappear

[jolly shuttle]

I mean without having to think about jacobians

getting the formula of r(theta)

[subtle ginkgo]

it wil be a way more comlexed in my opinion

complicated*

[jolly shuttle]

I admit it, but from what I see the questioner can't use jacobians while need to use polar integrals

[violet pulsar]

Awful but yes

At this point if it's way too complicated I don't want to ask too much from you, I thank you for the help. But I'd really appreciate the further support.

[subtle ginkgo]

look:

$\text{assume we use classical polar coordinates:}\\x=rcos\theta,y=rsin\theta\text{ }\text{ then}\\\left| D \right|=\int_{0}^{\frac{\pi}{2}}d\theta\int_{0}^{\frac{ab}{\sqrt{\left( bcos\theta \right)^{2}+\left( asin\theta \right)^{2}}}}r\text{ }dr$

that is time consuming too )

and is lookign scary )) smiles

[subtle ginkgo]

so jacobian gives us a lot of comfort, and to be honest, this variable r in double integral is also a jacobian of polar classical coordinates, but almost no one remembers that

[violet pulsar]

I can understand, that the Jacobian really evens things out and helps when working with polar coordinates.

Even tho that is the case, I have not yet learned about matricies (I think thats where Jacobian is)

[subtle ginkgo]

i can bet that in 95% or even 99% of your tasks, your Jacobian wil be J = r, and such tricks like ellipse occurs very rarey, i mean how exams are designed

i can give you a link with solution on DM

[violet pulsar]

Is $r = \frac{ab}{sqrt((acos\theta]^2 + (bsin\theta)^2)}$ here

[jolly parrot]

C

[subtle ginkgo]

yes it is

[violet pulsar]

so i took it as

before

is this false?

[jolly parrot]

Joanna Angel

[subtle ginkgo]

but ab, must be in numerator

[jolly shuttle]

as I know that is not an ellipse

for polar integral r should mean the radius when angle is theta

but problem is $(acos(\theta),bsin(\theta))$ is not usually the radius for $\theta$, but for other angle

[jolly parrot]

Dri111

[violet pulsar]

hm

when i use r instead of ab

the problem is lifted

right

i mean $x^2 + y^2 = r^2$ will be existent

[jolly parrot]

C

[violet pulsar]

therefore i can work with this in a different matter

[jolly shuttle]

use it on where?

[violet pulsar]

instead of the initial definition that i gave

[jolly shuttle]

you can use the method you've originally thought if the shape was circle yes

I hope this is what you mean

[violet pulsar]

This is so confusing

[subtle ginkgo]

$\text{if }\text{ }\text{ }x=rcos\theta, y=rsin\theta\\\left( \frac{x}{a} \right)^{2}+\left( \frac{y}{b} \right)^{2}=1\Leftrightarrow \left( \frac{rcos\theta}{a} \right)^{2}+\left( \frac{rsin\theta}{b} \right)^{2}=1\Leftrightarrow \\r^{2}\left( b^{2}cos^{2}\theta+a^{2}sin^{2}\theta \right)=a^{2}b^{2}\Leftrightarrow \\r=\frac{ab}{\sqrt{\left( bcos\theta \right)^{2}+\left( asin\theta \right)^{2}}}$

[jolly parrot]

Joanna Angel

[subtle ginkgo]

$x=rcos\theta,y=rsin\theta\text{ }\text{ then}\\\left| D \right|=\int_{0}^{\frac{\pi}{2}}d\theta\int_{0}^{\frac{ab}{\sqrt{\left( bcos\theta \right)^{2}+\left( asin\theta \right)^{2}}}}r\text{ }dr$

[jolly parrot]

Joanna Angel

[subtle ginkgo]

$\text{but if }\text{ }\text{ }x=arcos\theta, y=brsin\theta\\\left( \frac{x}{a} \right)^{2}+\left( \frac{y}{b} \right)^{2}=1\Leftrightarrow \left( \frac{arcos\theta}{a} \right)^{2}+\left( \frac{brsin\theta}{b} \right)^{2}=1\Leftrightarrow \\r^{2}\left(cos^{2}\theta+sin^{2}\theta \right)=1\Leftrightarrow \\r=1$

[jolly parrot]

Joanna Angel

[subtle ginkgo]

then

[jolly shuttle]

So this is approach with transforming ellipse into circle, right?

[subtle ginkgo]

yes it is

hence presence of the jacobian

[violet pulsar]

sorry for exhausting you guys

im just in no position to use the jacobian and hence dying in a loop

[jolly shuttle]

I mean using this as r

[violet pulsar]

this is what you mean right

[jolly shuttle]

I mean the one with polar integral

$\int{\frac{1}{2}r^2d\theta}$

this thing

[jolly parrot]

Dri111

[jolly shuttle]

srry same thing

[violet pulsar]

Let me start off from a more certain explained point. My goal is to use two ellipses and compare the areas or r lenghts in order to calculate the seat prices.

Therefore I need to create a general formula for this.

[jolly shuttle]

area between two ellipse?

[subtle ginkgo]

Are you designing a stadium? 🙂

[violet pulsar]

ratio of inner and outer ellipse

[violet pulsar]

pricing model yes

then compare it to the current price

[subtle ginkgo]

You know, there are many factors that influence this type of construction, and the fact that you increase the number of seats does not necessarily mean that the business will be profitable.

[jolly shuttle]

ratio of area for outer:inner ellipse you mean?

[violet pulsar]

It's not for the profitability, it's just so the prices increase/decrease on a certain rate.

I'm doing this for an assessment (a must when graduating the diploma programme)

totally unrelated*to the existent salloon

venue*

[subtle ginkgo]

ok)

[violet pulsar]

yes

this is why im trying to use complex stuff

[jolly shuttle]

can you just calc inner&outer separately then get the ratio of them?

[violet pulsar]

Yeah this has to turn into some sort of formula afterwards

[pearl pond]

@violet pulsar Has your question been resolved?

[violet pulsar]

@jolly shuttle may I ask a final question before closing this help channel?

[jolly shuttle]

Sure

[violet pulsar]

You think this'll work out?

[jolly shuttle]

I think so

[violet pulsar]

Alright.

Thank you for your time and help.

t!rep @jolly shuttle

[unreal elbow]

No reputation points!

You've already given a reputation point today!
You can send another rep after <t:1704672000:t>.

cemmisali

[violet pulsar]

ffs.

[jolly shuttle]

thx

[violet pulsar]

I'll send it later on.

[jolly shuttle]

idk rep was a thing

[violet pulsar]

Yeah apparently it is, I try to give as much as I can (to show gratitude (its never enough))

have a great day/night.

[jolly shuttle]

have a good timenwith math

[violet pulsar]

much love

.close

[pearl pond]

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[pearl pond]

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[grand bough]

i have to show this using the definition

[foggy fog]

Just insert the value?

[grand bough]

using the definition

[foggy fog]

xd

[opal moss]

plug into the epsilon delta definition, when the limit exists this should be extremely straightforward

[grand bough]

but how can i plug a value inside delta

[prime bramble]

you aren't

you can find the value of delta later

[grand bough]

why not ??? they plugged 0.5 into it to then find that delta = min (0.5, other )

[foggy fog]

Yeah thats common to choose delta as some minimim but you will see what delta needs to be later.

But you start finding some inequalitys

$| x - 2 | < \delta \implies | \frac{2x+1}{x-3} - (-5) | < \epsilon$

[jolly parrot]

Tobi

[foggy fog]

And there are some steps in between where you find out what delta is

[grand bough]

yeah it gives $| x - 2 | < \delta \implies 7 | \frac{x-2}{x-3} | <\epsilon$

[jolly parrot]

RulzerFly

[grand bough]

and what is those steps

[foggy fog]

Algebraic manipulations. Try some you know and see where it leads

[grand bough]

the thing that i don't understand the most is why they chose in the correction $| x - 2 | < \frac{1}{2} why not | x - 2 | < \frac{1}{4}$

[jolly parrot]

RulzerFly

[foggy fog]

often in such exercise they choose $\delta = \frac{1}{\epsilon}$

[jolly parrot]

Tobi

[foggy fog]

Uhm

but there is no clear way to choose

most things work

so 1/2 over 1/4 or the other way is arbitrary. It just needs to be close enought for convergence

[grand bough]

so it should work with $| x - 2 | < \frac{1}{4}$

[jolly parrot]

RulzerFly

[grand bough]

okay but look

@foggy fog

[foggy fog]

I think, my practice is some years old xd

Why they choose 1/2 as delta

and if you could choose smth different, I guess

$|x-2|<\delta = 1/2$ in the first line

[jolly parrot]

Tobi

[foggy fog]

implicitly define I guess, @grand bough is there a clear choosing?

[grand bough]

why ??

[pearl pond]

@grand bough Has your question been resolved?

[grand bough]

okay okay i understand thanks for help !!

okay

[pearl pond]

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[pearl pond]

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[pearl pond]

@midnight haven Has your question been resolved?

[pearl pond]

@midnight haven Has your question been resolved?

[pearl pond]

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[marsh lava]

Can you please hlep with this please

[marsh lava]

just the last part please

[pearl pond]

@marsh lava Has your question been resolved?

[marsh lava]

no

[pearl pond]

@marsh lava Has your question been resolved?

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[midnight haven]

I don't understand how I would convey d)

c) would be f composed with itself 4 times

making it x-12

[nocturne plover]

well what happens everytime you add another one

[midnight haven]

u should establish a pattern

from a - c

[nocturne plover]

you know 2, 3, and 4

and 1

that being x - 3 itself

[midnight haven]

-3 gets added every time

[nocturne plover]

so if f is composed with itself n times

[midnight haven]

4*3 is 12

so I guess maybe

n(-3)?

this would hould true for c)

4*-3

-12

b)

3*(-3)

-9

n(-3) where n is the amount of f's there are

[nocturne plover]

yes

so f composed with itself n times is

in terms of x

?

[midnight haven]

I'm not sure

[nocturne plover]

what's f composed with itself 2 times?

[midnight haven]

x-6

[nocturne plover]

quick thing, i don't know if they mean composed with itself 2 times being 2 of those composition symbols or 2 of the functions

[midnight haven]

idk if it helps but this is the answer

[nocturne plover]

ok so it's the number of circles

so at a f is composed with itself once

and b is twice

and c is three times

so n times is n of those circles

[midnight haven]

when did circles come into this lol

[nocturne plover]

that's the symbol for composition of functions

looking at c

what's the answer, how many compositions are there and how many f?

[midnight haven]

this is the answer

[nocturne plover]

that's for d

not c

[midnight haven]

no that's just

the general answer

[midnight haven]

this is

question d

and that is the answer

for d

[nocturne plover]

this is incorrect btw

c is f composed with itself 3 times

according to the answer for d that you showed

there are 3 compositions

[midnight haven]

okay

[nocturne plover]

and 4 f

so n compositions is n+1 f

meaning it's x - 3 (n+1)

[midnight haven]

btw are you meant to foil this?

[nocturne plover]

foil?

you mean multiply?

[midnight haven]

FOIL

its short for

expanding

[nocturne plover]

i know

i wouldn't

[midnight haven]

then why is the answer written like that

[nocturne plover]

it's a lot more clear if you have a single -3 term

it's clearer what you mean if you write x - 3 (n + 1) than writing x - 3 - 3n

[midnight haven]

ok so lets say

n was 3

x-3(3+1)

x-3(4)

4x-12?

[nocturne plover]

no

[midnight haven]

or is it just the 3 term

[nocturne plover]

the x is not inside brackets

[midnight haven]

being multiplied

okay

so the 3

is the only thing

being multiplied

[nocturne plover]

it's not (x - 3)(n + 1)

it's x - 3(n + 1)

[midnight haven]

got it

so x-3(3+1

x-3(4)

x-12

what purpose does the 1

serve

[nocturne plover]

it's if you have f alone

if you do not compose f with anything

you have 0 compositions

aka just having f(x)

and f(x) = x - 3

[midnight haven]

so f(x) is the lowest you can have it be

which is at minimum 1 composition?

[nocturne plover]

0 compositions

[midnight haven]

I think we mean different things

but is the lowest f thing

1

because f(x)

hence why you add 1

[nocturne plover]

in the exercise you're asked to compose f with itself

f is given

being x - 3

[midnight haven]

okay

[nocturne plover]

and so you compose f with itself

[midnight haven]

so x-3(1) is just itself

[nocturne plover]

which means doing (x - 3) - 3

[midnight haven]

when n=0

[nocturne plover]

yes

[midnight haven]

which is the base

okay

[nocturne plover]

yes

that is just f

[midnight haven]

I get it now, thanks

[nocturne plover]

ok 👍

[midnight haven]

.close

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[lament radish]

can anyone tell me why some functions are expressed as an integral?

[lament radish]

for example

$W_{0}\left(x\right)=\frac{x}{\pi}\int_{0}^{\pi}\frac{\left(1-t\cot\left(t\right)\right)^{2}+t^{2}}{x+t\csc\left(t\right)e^{-t\cot\left(t\right)}}dt$

[jolly parrot]

DaanHun

[plush bramble]

because they can?

[lament radish]

why do i have to integrate it

[plush bramble]

!original

[pearl pond]

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.

[lament radish]

wouldnt it be easier if it wasn't?

there isnt really a problem i was just wondering why some functions are expressed as integrals it seems to me that would make it more difficult to calculate?

[plush bramble]

yes. go find the expression without the integral if you can

[prime bramble]

sometimes there aren't closed forms for functions

[elfin stirrup]

most functions cant be expressed as e, ln, sin, cos functions and so on

[prime bramble]

and they are best represented by an integral

[lament radish]

ohh so you cant really easily express it as an integral?

so i really just have to use a calculator for it?

[prime bramble]

you cant easily express a lot of functions without an integral

[lament radish]

alright thanks

[elfin stirrup]

numerical integration if you want most of the values

or complex analysis if that works

[lament radish]

!close

.close

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[midnight haven]

[0.8\textwidth]\let\z\Card
\textbf{Question:} If each coded item in a catalog begins with 2 distinct letters followed by 3 distinct nonzero digits, find the probability of randomly selecting one of these coded items with the first letter a vowel and the last digit even.

\vs{3 mm}
\textbf{My attempt:} so we have $\z S = (26)(25)(9)(8)(7)$ and $\z(A\cap B) = (5)(25)(4)(8)(7)$ so we get [
\m\P{A\cap B} = \f{\z(A\cap B)}{\z S} = \f{(26)(25)(9)(8)(7)}{(5)(25)(4)(8)(7)}
]

[midnight haven]

is this legit

[plush bramble]

why is it 9,8,7 instead of 10,9,8

oh nonzero i can't read

your fraction is flipped

[midnight haven]

uwa

aighty thanks

.close

[pearl pond]

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[plush bramble]

.reopen

[pearl pond]

✅

[plush bramble]

i'm not sure about the distinct and even condition.

[midnight haven]

why is that

[plush bramble]

i was doing it this way: P(at least 1 even) = 1 - P(all odd numbers)

[midnight haven]

oh but like

[acoustic path]

only the last digit is required even though

[midnight haven]

its not at least 1 even

its the last digit being even

[plush bramble]

you can just divide by 3! since order doesn't matter

[midnight haven]

a bit confused

why is at least 1 being even something we are considering

[plush bramble]

me too. i'm not entirely sure.

because of distinct

Maybe I'm wrong

[midnight haven]

hmmm

i think my answer is correct checking the manual so its all good thanks

.clsoe

.clsoe

.close

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[ebon matrix]

What would the restrictions for this trig identity be, would it include sin and cos or just sin

[pearl pond]

Please don't occupy multiple help channels.

[ebon matrix]

.close

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[midnight haven]

is there a general formula for the inclusion exclusion principle

[midnight haven]

like to avoid overcounting

[pearl pond]

@midnight haven Has your question been resolved?

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[blissful orbit]

Is it just me or the answer isn't in the choices

[pearl pond]

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[midnight haven]

hmmm soooo

i made 1 into c/c

and multiplied the c cos x - v by c

[pearl pond]

@midnight haven Has your question been resolved?

[midnight haven]

<@&286206848099549185>

[grizzled swallow]

Yea?

[pearl pond]

@midnight haven Has your question been resolved?

[midnight haven]

helpp

[pearl pond]

@midnight haven Has your question been resolved?

[midnight haven]

<@&286206848099549185>

yes

?

whats is it

can u help

maybe

with the question

wahst ur question

im not that advanced yet unfortunately

im not good at alot of stuff lol

[midnight haven]

why did u reply then

i tohught i could help

thats kind of the point of helpers

dam right

<@&286206848099549185> can someone help me?

[subtle swan]

!noping

[pearl pond]

Please do not ping individual helpers unprompted.

[astral cairn]

my bad

[subtle swan]

!redir

[pearl pond]

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

[subtle swan]

!redir

[pearl pond]

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

[subtle swan]

!noping

[pearl pond]

Please do not ping individual helpers unprompted.

[subtle swan]

!redir

[pearl pond]

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

[subtle swan]

!occupied

[pearl pond]

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).

[subtle swan]

<@&268886789983436800> please help we are under attack

i repeat we are under attack

[unborn abyss]

i don't understand

[buoyant plinth]

I think the amogus posts

[opal lantern]

dealt with

[subtle swan]

thanks

also im getting attacked in dms

[unborn abyss]

,prune 100 --from 996869619417624618

[jolly parrot]

Couldn't find a member matching 996869619417624618!

Jan 15, 2024, 10:46 PM

Couldn't find the requested user, cancelling purge.

Jan 15, 2024, 10:46 PM

[pearl pond]

@midnight haven Has your question been resolved?

[pearl pond]

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[signal lynx]

I'm confused on what theyre asking in c. for a, I figured no because the pair fails to include 0. for b, i said they were equal because sets cant really have duplicates. however, for c im just confused as to what theyre stating. wouldnt the two be identical simply because of a lack of parameters surrounding the variables?

[fleet sky]

hi there

[signal lynx]

hi

[fleet sky]

yeah your intuition is correct for c

there are no days with 40 hours

there are no weeks with 6 days

so they're both the empty set

and thus they're equivalent!

[signal lynx]

thats awesome thank you

[fleet sky]

for (b) yeah you're right there are no duplicates so allowed so you only count them once

so both sets end up equating to {1, 2, 3}

[signal lynx]

tyvm for the help

i was just confused bc i hadnt seen a question like c before and was confused on what they like wanted from me

[fleet sky]

yeah for sure, kind of tricky. It's a good lesson on how things will be excluded out of sets 🙂

as a practical example

say if we had this: {p | p in Z, p is prime}

so this is the set of prime numbers, they're required to be integers

so things like 4, 6, 8, etc, won't belong to a set

[signal lynx]

that makes sense thank you

[fleet sky]

yep no problem

any other questions?

[signal lynx]

nope! thanks

.close

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[languid pier]

hello, this is a question of the principle of finite induction, but i'm having some problems to understand what happened from the line i draw the arrow to the next one. i noticed they grouped 2k + 3 (to fit the finite induction equation), but what happened to the other (k+2), since there was two? i'm feeling really dumb rn but i just don't know lol

[feral leaf]

Factor out the k + 2

2k(k + 2) + 3(k + 2)
If you do something like a = k + 2, you can rewrite that as 2ka + 3a
Then if you notice that a is common in both terms so you can factor it out therefore: 2ka + 3a -> a(2k + 3)
Plug back in k + 2, since you substituted so a(2k + 3) -> (k + 2)(2k + 3)

[languid pier]

ohhhh

thanks

the "a" really helped lol

when there are too much numbers i get a bit confused

i think thats all

ty again

.close

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[midnight haven]

how can i read this in a way ik what is the corresponding sides

[midnight haven]

like how is 24 corresponding with X?

[simple jasper]

bowtie theorem, the angles where the triangles meet are equal, which means since it is a right triangle all the angles are the same, and since you can find congruency of a triangle through the AAA theorem.

[midnight haven]

ohhh

i need to look into bowtie theorem more

[simple jasper]

bowtie theorem works if the triangles have a side that is straight and connected

[midnight haven]

oh i see

so much like the name

it looks like a bowtie

when connected

[simple jasper]

yeah, although it doesn't work if it looks more like a bowtie, it majorly works through the fact that a straight line is 180 degrees

and a bisector at any angle will have the same angles on opposite sides on the other side, if that makes sense?

[midnight haven]

yes total sense!

you’re lots of help thanks

[simple jasper]

no problem, do you have any further questions?

[midnight haven]

as of right now, no. i’m gonna research bow tie theorem now and solve this problem. i’ll lyk if i come into any troubles

have a great day!

[simple jasper]

you too

[midnight haven]

.close

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[pearl pond]

@vale turtle Has your question been resolved?

[earnest stratus]

A+B is a matrix lol.

Please show the entire problem, whatever options they've given you and stuff.

But something that will probably help you is this, recall the definition of an orthogonal Matrix. What does that say about its determinant?

[pearl pond]

@vale turtle Has your question been resolved?

[vale turtle]

$|A||A^t|=1$

[jolly parrot]

krish15662

[pearl pond]

@vale turtle Has your question been resolved?

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[vale turtle]

.reopen

[pearl pond]

✅

[pearl pond]

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[vale turtle]

.close

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[worthy grail]

can someone help me with this one?

[mighty oriole]

hmmm

lets ee

see

[west sapphire]

can the absolute value of something be negative?

[minor fog]

No.

[west sapphire]

was asking the OP but sure 😁

[minor fog]

Sorry.. just rolled in, didn't realize your strategy

[west sapphire]

ha, nw

[worthy grail]

lol

no though

ryan is right

i would think no solution

but i think there is a solution

just not here

[west sapphire]

|x-3| is an absolute value

so it can't be negative

so in particular it can't be <= -4, right?

[worthy grail]

mhm

[west sapphire]

therefore it must be > -4, right?

[worthy grail]

yees

[west sapphire]

and everything i wrote above is true for all values of x

[worthy grail]

yeeeees?

[west sapphire]

therefore every x is a solution to the inequality

[worthy grail]

what about the negatives though

[west sapphire]

even negatives

because you're taking the absolute value of x-3

that's always gonna be a positive number (or zero)

e.g. if x is -100 then x-3 is -103 and |x-3| is +103

[worthy grail]

oh so the one i randomly clicked on was righ

oops

[west sapphire]

yep

that is the right answer

[worthy grail]

that makes sense though thank you!!

.close

[pearl pond]

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[glass remnant]

17 is the questiong I get the first part but can somebody explain how you go from the first part to the red

[blissful cloak]

which step exactly are you confused on, the red part itself or how they got there?

[glass remnant]

How do you get there

[glass remnant]

My apologies for not being clear in the original question

[blissful cloak]

you're good

so first off do you know the Pythagorean identity?

[glass remnant]

I know them but I dont see any

[blissful cloak]

sec^2-1? 👀

[glass remnant]

Oh yeah I got that part

But im confused on how you go from the tan/sin to the red

[blissful cloak]

$\frac{\tan^2\theta}{\sin^2\theta}=\tan^2\theta\cdot\frac{1}{\sin^2\theta}=\frac{\sin^2\theta}{\cos^2\theta}\cdot\frac{1}{\sin^2\theta}$

[jolly parrot]

PajamaMamaLlama

[blissful cloak]

this is how :)

recall that $\frac{a}{b}=a\cdot\frac1b$

[jolly parrot]

PajamaMamaLlama

[glass remnant]

Ah I see

[glass remnant]

I forgot about this

Thank you for the help I appreciate it a lot

.close

[pearl pond]

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[pearl pond]

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[pearl pond]

@pine compass Has your question been resolved?

[pearl pond]

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[sweet bone]

it's the same thing

augment them into a matrix and row reduce

it's the exact same row operations

nothing about row reduction depends on the matrix being square

you're not getting it to the identity matrix, you're getting it to (reduced) row echelon form; all zero rows below non-zero rows and pivots for each row are strictly to the right of the pivots of any rows above (for rref, pivots additionally have to be 1 and elements above pivots are zero)

no, you have 3 vectors sitting in 4d space

they are unlikely to be dependent

geometrically, you have more vectors than dimensions

so any more than the dimension must certainly sit in the span of the others

no?

a single vector [1,2,3,4]^T has linearly dependent rows as a matrix

but consists of a single column and isn't linearly dependent

just row reduce the matrix

you also don't need to write out the zero vector on the right or multiply the matrix by an arbitrary vector

you can already tell if the vectors are linearly independent by reading the number of pivot columns

you only need to do that if you're finding the kernel explicitly

for a square matrix

also that sheet is for invertibility

non-square matrices squash down dimensions or embed into a subspace so they can't be invertible (same reason they don't have a determinant)

yeah, that's not really a helpful way of thinking about it

invertibility is a sufficient condition

but it's not necessary

yes

what did you get exactly

the rref is always going to have at least one zero row because you have 4 dimensions and 3 vectors

whether its LI or not depends on how many zero rows you get

[jolly parrot]

Mycobacterium

[sweet bone]

yeah, so you have 3 pivots

so 3 independent vectors

[jolly parrot]

Mycobacterium

[sweet bone]

yes

the pivot columns of a rref matrix are linearly independent

furthermore, they form a basis of the image of the transformation the matrix represents

the point is that row operations preserve linear independence

and those vectors are clearly independent in the rref form

so the original vectors are also independent

do you know that matrix represent linear transformations?

I would recommend getting a more geometric grounding in linear algebra then

most of the objects you encounter will make a lot more sense in a geometric context

[pearl pond]

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[sweet bone]

https://www.youtube.com/watch?v=fNk_zzaMoSs here's a good video series on it

[pearl pond]

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[vale ivy]

can someone help me with this question below?

"Chora has a special power. She can find out all the divisors of a number instantly. Golpo wants to
defeat her. So, she came up with a large number 𝑃𝑃 = 7! × 8! × 9! × 10! × 11! × 12! and asked
her to find all its divisors that are perfect squares. How many divisors does Chora need to find?"

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[pearl pond]

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