help-27

[jaunty mantle]

I don’t quite have it yet because I don’t really get it

[wooden veldt]

Ah okay yeah its related to this

[jaunty mantle]

But soon I’ll get it and I won’t need the laplace one anymore (until they tell me to find the determinant of this matrix in an exam)

[wooden veldt]

There's only one multi linear alternating map that is 1 on the identity

And its a diff form turns out (not surprising bc alternating)

[wooden veldt]

Yeah these more fundamental defns aren't suitable for calculations lol

[jaunty mantle]

So for the determinant we’re looking for a multi linear map that’s totally antisymmetric

Ok we need the “1 on the identity” part too probably, but idk how you express that

[wooden veldt]

Yeah the first two requirements don't give you something unique

But they're all multiples of eaxhother

[jaunty mantle]

Well if it’s a multi linear map firstly

It’s clearly a (s,r)-tensor

I’m guessing antisymmetry tells us it’s a differential form in particular?

[jaunty mantle]

Then this gives uniqueness

[wooden veldt]

Roughly Yeah

In a LA setting you're looking for a multi lin ant sym map on the cols of a matrix

That's 1 on the identity matrix

[jaunty mantle]

Yeah but then I need a basis 😦

[wooden veldt]

Yeah a basis free defn is more complicated but seems like it's what youre headed towards

[jaunty mantle]

I’m looking for a multi linear anti symmetric map on k column vectors to give me a number

[wooden veldt]

Since you can view a (0,n) tensor as a function on the cols of a matrix once you pick a basis

[jaunty mantle]

Right

It eats n (column) vectors and spits out a real number

As a (0,n)-tensor should

[wooden veldt]

Long story short something like $dx \wedge dy \wedge dz$ should give you det

[woven radish]

ΣAC

[wooden veldt]

Alteast on R3

[jaunty mantle]

Ok I’ve got more to learn, haven’t really gotten through the wedge dude yet

[wooden veldt]

Yeah wedge is what is capturing the anti symmetricness

[jaunty mantle]

I’ve been told it’s like the cross product but not scuffed

[wooden veldt]

Wedge is secretly just tensor but on a subspace of anti symmetric guys

[jaunty mantle]

Shhh

[wooden veldt]

So that a wedge b = - b wedge a

But anyway I'll keep checking in when I see your channel as itll make me actually learn this stuff properly lol

Again I recommend moritas book and the michael penn playlist

[jaunty mantle]

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[bronze urchin]

Find f1 and f2 size if the ball has a mass of 4.5 kg

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[muted barn]

So im working with integration and i always thought the other function gave you the area of the original one.

but in the case of sine and cosine it doesnt really seem to be like that. I can see that sin(x), the black one indeed is the antiderivative but it doesnt tell the area the same way really

[muted barn]

(The lines are blue = cos(x) & black dotted = sin(x)

And the rule integrating cos(x) = sin(x)+C

[jaunty mantle]

There’s a difference between definite and indefinite integration

[muted barn]

Please explain

[jaunty mantle]

Indefinite integrals aren’t really talking about area

[muted barn]

So when you need +C?

[jaunty mantle]

They are more concerned with a primitive function, whose derivative is the integrand

[muted barn]

Is integrand the function boutta be integrated?

[jaunty mantle]

Yes

[muted barn]

Ok

[jaunty mantle]

When you look at it this way, the + C makes a lot more sense

[muted barn]

Yes

[jaunty mantle]

Because any function F, whose derivative is f, you could add some constant to F and it’ll still differentiate to f

So indefinite integrals actually produce a family of functions that satisfy some constraint

[muted barn]

Yes cuz all the points are realtive?

[jaunty mantle]

The constraint being they differentiate to the integrand

[muted barn]

relative

[jaunty mantle]

I’m not sure what you mean here

[muted barn]

Their relation is the same if you add C to all of them

[jaunty mantle]

What do you mean their relation

[muted barn]

For example distance

to each other is same no matter what +C is

But then definite integration is for area and stuff???

[jaunty mantle]

What distance?

[muted barn]

Between the points

on the line

[jaunty mantle]

Why would there be a distance…?

I don’t get what you’re saying

[muted barn]

Forget it. Ok lets just say the line has the same acceleration at every point no matter what C is

[jaunty mantle]

You mean derivative?

[muted barn]

Yes

[jaunty mantle]

Ok that is true

[muted barn]

Yeah and then definite integration

[jaunty mantle]

Definite integrals are completely different things

Completely different to indefinite integrals

Notice that when you integrate a function (indefinitely), you get a function back

[muted barn]

Yes

[jaunty mantle]

When you do definite integration you get a real number as your output

Those are very very different

[muted barn]

But what would you do when doing definite integrals w sin and cos

[jaunty mantle]

Now it turns out, that there is a very famous and fundamental theorem that does link these 2 concepts together, namely the fundamental theorem of calculus

It states that if $f$ has a primitive $F$ on the closed interval $[a, b]$, then $\int_a^b f(x), dx = F(b) - F(a)$

[muted barn]

ok

Looks familiar, ive worked with it

[woven radish]

Frosst

[muted barn]

And there the C is basically eliminated

[jaunty mantle]

Ok it seems on Wikipedia the precise theorem is a bit more complicated

[jaunty mantle]

But kinda

It’s more like they cancelled out

[muted barn]

Well yes

thx

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[restive river]

need help

[twilit comet]

what's your question?

[restive river]

can you give me a perfect square matrix

[twilit comet]

?

i mean

depends on your bounds

is this just outta your own curiosity, or is it like a question

[restive river]

how do i get the reverse matrix of a coefficient matrix?

[twilit comet]

what...?

that's not the same

[restive river]

i mean in linaer algebra

the reverse matrix (A^-1)

[devout snow]

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[dark sable]

let 25 degrees=x
50 degrees=2x
cosx+sinx=k
use double identity

cos2x=cosx^2-sinx^2=(cosx+sinx)(cosx-sinx)

[devout snow]

@ebon citrus Has your question been resolved?

[ebon citrus]

what is (cosx-sinx)?

[dark sable]

idk either

but if it helps

[ebon citrus]

-_- bruh

[dark sable]

(cosx-sinx)=sqrt((cosx-sinx)^2)=sqrt(cosx^2-2sinxcosx+sinx^2)=sqrt((cosx+sinx)^2+4sinxcosx)

(sinx+cosx)^2=1+2sinxcosx
hope this helps

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@ebon citrus Has your question been resolved?

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[strong forge]

Any chance someone can help me with part b)

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[strong forge]

<@&286206848099549185>

[strong forge]

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[woven radish]

GenK1 Ψ

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@dawn bay Has your question been resolved?

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[ionic swift]

in my book it says that a classical example for a matroid is for a graph G=(V,E) the Matroid M = (E, F) where F is a family containing all the subsets of E that make up a forrest in G.
Then it says : All maximal independent sets of a matroid have the same cardinality.
I get that kind of but I dont get in regards of which relation these sets in F are supposed to be independent.
Can someone explain this to me ?

[ionic swift]

from what I understood, this would be the example of the Matroid of the K3 . But then regarding what relation are these sets of edges independent ?

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[silent sandal]

This is what I got so far

[silent sandal]

I'm not quite sure how to figure out f(-1)

How would I figure it out?

[mortal tinsel]

do you know what the open hole and closed hole mean?

[silent sandal]

yes

but why?

[mortal tinsel]

can you tell me what they mean

[silent sandal]

open means its excluded, closed means included

oh i guess f(-1) would be 4 then

i guess in this scenario this means its approaching 2 and 3 but never reaches it?

[mortal tinsel]

yes it approaches 2 from the left but never gets there, jumps at f(0) = 4, then jumps back to some point infinitely small near 3 then continues

[silent sandal]

ok thank you

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[restive river]

i know that i have to find the area under the curve, but how?

[restive river]

do i set it up as piecewise? do i use some other equation? whats the best way

[visual hazel]

seems like you can just use regular geometric area formulas

[restive river]

would that be something like this

[visual hazel]

yeah thats for a trapezium

which is the area of x = 3 to x = 8

[restive river]

so would b1 be 3, b2 8, and h 9?

[visual hazel]

h 9?

[restive river]

or 8?

does it represent the highest y value in that range?

[visual hazel]

isnt it 5?

[restive river]

i have no clue.

why is it 5

[restive river]

oh. so h represents the x values??

[visual hazel]

its the height of the trapezium

that shape is a trapezium right?

[restive river]

i think so yes

[visual hazel]

then you remember how to find the area of the trapezium?

add the two parallel sides

multiply by its height

divide by 2

[restive river]

i never took geometry

[visual hazel]

wat

[restive river]

nobody told me it mattered

[visual hazel]

well ok..

the h is just the vertical height

or in this case horizontal

[restive river]

oh so it's just flipped on its side

[visual hazel]

how far apart the two parallel sides are

yeah thats a good way to say it

[restive river]

so i'd use the equation (5(3+8))/2

[visual hazel]

yes

[restive river]

which is 27.5 and then i add that to the area of the rectangle in 0-3?

[visual hazel]

yes

[restive river]

36.5 in total?

[visual hazel]

yeah

[restive river]

do i have to use the trapezium equation for the x value of 9 too?

is it just a really small one?

[visual hazel]

x value of 9?

oh for b?

[restive river]

this area

[visual hazel]

yeah same thing

it just has a height of 1

[torpid stump]

surreeeee buddyyyyyyyyy

[restive river]

so (1(8+9))/2?

[visual hazel]

yeah

[restive river]

and so the right side is 71.5

in total

[restive river]

8 to 16 i mean

[visual hazel]

yeah

[restive river]

and for center of balance i just need to find where it all adds up to 54 on both sides? because thats 108/2

which is just x=10?

[visual hazel]

is this physics?

[restive river]

no its calc

[visual hazel]

oh huh

i actually dont know sorry

[restive river]

its alright

thank u for ur help!!

[visual hazel]

but wait hmm ill think about this for a bit

[restive river]

i feel like x=10 kind of makes sense

because 36.5+8.5=45+9=54?

[visual hazel]

yeah its somewhere around there

[restive river]

im just estimating so i think that works

[visual hazel]

i got like 9.9

[restive river]

thank u again for ur help! i appreciate it

[restive river]

yeah im just rounding to whole number

thank you!!

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[cold hornet]

i need help on a question

[cold hornet]

<@&286206848099549185>

also i have to evaluate

[near sand]

1/5^-4

1/ ( 1/5^4)

= 5^4

[cold hornet]

huh that makes no sense

the negative four disapears?

[near sand]

let me break it down then:

suppose you have two numbers, a and b:

then:

$$a^{-b} = \frac{1}{a^{b}}$$

[woven radish]

guizin

[near sand]

example:

$5^{-4} = \frac{1}{5^{4}}

oop

$5^{-4} = \frac{1}{5^{4}}$

[woven radish]

guizin

[near sand]

@cold hornet

[cold hornet]

ye

thanks

[near sand]

now the questions asks you how much is 1 divided by that

so just do

$\frac{1}{\frac{1}{5^{4}}

$\frac{1}{\frac{1}{5^{4}}$

[woven radish]

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

[near sand]

ugh

[cold hornet]

its good

[near sand]

1 divided by 1/5^4

ok ok

u get it then

[near sand]

which in other words is just inversing it

[cold hornet]

yeah kinda

[near sand]

lol

[cold hornet]

so i get 1/625 then i just divide it by 1

right?

@near sand

[near sand]

yeah, which is 625 anyway

when you divide 1 by a fraction, you can reverse it

for example

1 / 2/3

it's 3/2

[cold hornet]

i have another question

these are the instructions

[brittle burrow]

$n^{-1}+n^{-2}$ where $n=2$

[cold hornet]

"subsititute the given value into each expression, then evaluate"

yup

[brittle burrow]

or is it n^-2 for the second one

[cold hornet]

thats what it says

oh wait

no yeah ur right

[woven radish]

chlamydia

[cold hornet]

yup

[cold hornet]

thats what it says

[cold hornet]

<@&286206848099549185>

[little tendon]

aight

so

we're trying to find the value of that expression i'm assuming

[little tendon]

right?

hello?

leaving this for when you get back, ping me if you have any questions:

n^3=n times n times n, n^2 is n times n, so n^3 ÷ n = n^2

n^2 ÷ n is n^1

and n^1/n = n^0

but n^1=n, and n/n=1, so n^0=1

n^0÷n=n^-1, so n^-1 = 1/n, n^-2 = (n^-1)/n= 1/(n^2)

[cold hornet]

sorry i was doing something else

i ddint think someone would come

[little tendon]

using this knowledge, we can turn n^-1 into 1/n, and n^-2 as 1/n^2

plug in 2 for n

[little tendon]

i'll help anyone (as long as its not anything to do with calculus)

[little tendon]

what do you get?

[cold hornet]

oh ok thanks

[cold hornet]

pretty understandable

is there any more steps?

or is that it

[little tendon]

remember, we're finding the value of this expression when n=2

[cold hornet]

oh

ok give me one sec

[little tendon]

so thats 1/n + 1/n^2

hope that helps

[cold hornet]

its 1/n +1/

im confused

[little tendon]

1/n + 1/n^2

plug in 2 for n

[cold hornet]

oh

[little tendon]

1/(2) + 1/(2)^2

calculate that

[cold hornet]

1/2 + 1/4

?

[little tendon]

nice, great

.....i hope you know how to add fractions haha

[cold hornet]

yeah of course lol

[little tendon]

so that gives you?

[cold hornet]

1/6

wait

shit mb

2/6

[little tendon]

....

[cold hornet]

uhh wait gimme a sec

[little tendon]

what time is it your side

[cold hornet]

12:48 am

[little tendon]

midnight or

[cold hornet]

midnight

yup

almost 1

[little tendon]

fair

1/2 = 2/4, correct?

so 2/4 + 1/4 =?

[cold hornet]

3/8

[cold hornet]

yeah

oh shit i forgot about lcm

and all that stuff

my fault i think i got caught up in doing it with the same denominators

[little tendon]

its almost 1 AM on your side

understandable lol

[cold hornet]

yeah lol

what time is it for you?

[little tendon]

10

pm

[cold hornet]

oh

[little tendon]

"thats 3 hours past my bedtime"

[cold hornet]

lol nah

[cold hornet]

i usually sleep 11;30ish

but i guess not today

[cold hornet]

oh

3/4

im in grade 9

[little tendon]

8th, but i know a tiny bit of calc already lol

welp, hope that helps, gotta help some other peepsies

[cold hornet]

youre in grade 8??

[little tendon]

13 haha

[silver crane]

wut happened?

[cold hornet]

oh lol

[little tendon]

so uh hope that helps, bye

[cold hornet]

yeah thanks

[silver crane]

mkt, what help you need?

[cold hornet]

oh i needed help sovietpotato helped already

[silver crane]

k

[cold hornet]

thanks though

.close

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[keen loom]

Hey, need some help deriving this function. f(x)= e^x/1+e^-3x

[midnight grove]

what have you tried

[keen loom]

I used quotient rule to get this, but now im stuck on how to continue

[lavish nimbus]

Expand.

[brittle burrow]

(to simplify, not expanding everything)

[keen loom]

is this correct when expanding?

then distribute the negative?

[brittle burrow]

yup

[keen loom]

Okay

Idk why i always doubt myself lol

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[winter brook]

I'm trying to show that Z is a ring with 1 but I'm not sure how to go about proving that addition is abelian... any tips?

[graceful cosmos]

You have to assume a lot of properties

[winter brook]

I just got an idea

[graceful cosmos]

Addition on Z is a cyclic group, I guess you could say

[winter brook]

even though we are talking about rings, what if say that the group Z with addition is isomorphic to some other abelian group?

therefore that set is abelian under addition?

[winter brook]

Oh true

that makes sense

[graceful cosmos]

It depends on how the ring Z was defined for you

[winter brook]

the issue is it hasn't been formally defined yet

we are asked to define it ourselves first ig

[graceful cosmos]

Yeah it never really is lol

"You all know what the ring Z is let's move on"

[winter brook]

gg

if it helps, we are given 8.1 and 8.2 and must prove 8.3

then given 8.4 and 8.5 but must do 8.6

[winter brook]

actually, I should have not used "prove"

bc I suppose I am not proving it, rather being convincing enough

idk

For 8.6, I first wanted to establish Z is a ring and then show that it a ring with 1

[graceful cosmos]

Indeed, it doesn't really make sense to prove it. I think you want to define Z, and then argue why your definition makes sense. It's pretty open ended though

[winter brook]

Alright

also I just realized btu

wait woah

1a=a1=a for all a in R

that part is fine

but Z doesn't have multiplicativ inverses for everything

rather, for almost all of it

only 0 has a multiplicative inverse, I believe, which would be 0... so would that not be an issue?

[devout snow]

@winter brook Has your question been resolved?

[winter brook]

um

this is what I wrote

[graceful cosmos]

Let's back up a bit.

What's the multiplicative identity in Z?

[winter brook]

1

[graceful cosmos]

Okay. So if 0 has a multiplicative inverse, then there must exist some element g, such that 0g = 1

[winter brook]

0 times 0 is 1

wait

omg

no

😭

listen...

it's 2am... my brain is FRIED

LOL

I see

[graceful cosmos]

All good we all have a long day sometimes

[winter brook]

u have no idea

so 1 is the only element in Z that has a multiplicative inverse then

[graceful cosmos]

But you're right, not many elements have multiplicative inverses

1 has a multiplicative inverse. Any other?

[winter brook]

since the other elements would not be elements in Z, rather Q

well the inverse operation of multiplication would be division, or rather, multiplication by the recipricol of an element of Z to get to 1

for 0 it's not possible as we cannot divide by 0, and for the rest, we cannot do it either since the inverse need be in the set, not outside

i.e. 2g=1 is true for g=.5 but .5 is not an integer

therefore 1 is the only element with a multiplicative inverse

[graceful cosmos]

I like your way of thinking! But there is one more

[winter brook]

NEGATIVE ONE

TRUEEEE

[graceful cosmos]

Yep lul

[winter brook]

nice nice nice ncie

[graceful cosmos]

Those are the only two "units" in Z

[winter brook]

I see

thanks!

.close

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[midnight grove]

Use the vertical line test to determine if it is a function or not

[bitter halo]

?

[regal moat]

The goal of this assignment is to test ur intuition

basically, it asks u to find an example of function k that match the graph, if such function k exist, then answer yes else no

[bitter halo]

first is a linear function

second a quadratic

[regal moat]

Good

[bitter halo]

what about it?

[regal moat]

Then just answer it

They both exist

[bitter halo]

what about this tho 💀

[regal moat]

What do u think?

[bitter halo]

I am not sure, I am new to math

[regal moat]

Do u know how Desmos work?

Its a website that can help with graph

https://www.desmos.com/calculator

[bitter halo]

I cannot use desmos in the test tho

[regal moat]

Doesnt matter

[bitter halo]

what is it needed for

tf?

[regal moat]

The point is to build up ur knowledge for ur test, thats why homework exists

So u can play around with Desmos, then when ur test comes, simply asnwer with what u know

For this one

[bitter halo]

,w plot 2pir^2

[regal moat]

https://www.desmos.com/calculator/sscvhf26td

[woven radish]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by xbz24

[regal moat]

Play around with those sliders

[bitter halo]

how does circle work

[regal moat]

Then tell if this can exist

[regal moat]

Good question

Do u know why y = x^2 create this shape?

[bitter halo]

no

I am new

to math

[regal moat]

Simply cause the ^2 term result in "2 possible" x terms to satisfy y

Now

Look at the circle

[bitter halo]

what about it

[regal moat]

By the fact that the ^2 of x give it 2 possible term that mirror the curve through OY axis, make it kinda match the circle right

Now for the curve to finally match the cirle

it also need to be mirrored by the OX axis

which mean we need to...?

[bitter halo]

no idea

[regal moat]

read this

slowly

[bitter halo]

fucking math sucks broo

[regal moat]

Bro lives in a world where math keeps bro alive yet call it suck

The answer is the ^2 the y term too

So it becomes a circle

[bitter halo]

makes sense, thx.

[regal moat]

Yeah

thats the function of a circle

So function that makes circle exist

[bitter halo]

what about this fucker

[regal moat]

watch ur language bruh

[regal moat]

There's something called slope of a function

[regal moat]

basically, this wont exist

[bitter halo]

why doe

[regal moat]

....

[regal moat]

this will not

its a bit complicate

But think of a function like this

for every x term, there's always at most n-term of f(x) for n is the biggest power of x in f(x)

So if f(x) = 3x

For each value of x, f(x) should only return at most 1 value cause x^1

if f(x) = x^2, then for each value of x, f(x) should only return at most 2 value cause 2 is the biggest power of x, in this case (x^2)

Understand?

[regal moat]

These are line function, which mean the biggest power of x is 1

Which mean

[bitter halo]

explain better

otherwise I am not going to get it

[regal moat]

We somehow have 2 values

thus must be impossible

[regal moat]

bruh

Answer this simple question

f(x) = 20x + 2
find x such that f(x) = 0

[bitter halo]

,w 20x + 2 = 0

[woven radish]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

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[regal moat]

Do it urself bruh

[regal moat]

And whats the biggest power of x in this f(x)

[bitter halo]

?

1

I guess

[regal moat]

good

boi

Now

f(x) = 2x^2 - 2

find x such that f(x) = 0

also whats the biggest power x in this f(x)

[bitter halo]

,w 2x^2 -2 = 0

[woven radish]

Results provided by WolframAlpha

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

Requested by xbz24

[regal moat]

?

Whats the biggest power of x bruh

[bitter halo]

idk

[regal moat]

1 is the wrong answer

bruh

What do u call this?

[bitter halo]

power of 2, ofc.

[regal moat]

x power of 2

So 2 is the correct answer

[regal moat]

I saw that

[regal moat]

How many solution do we have for this case

[regal moat]

See the patterns?

[bitter halo]

the solution set is {-1, 1}

[bitter halo]

no.

[regal moat]

How 😭

If a function have the biggest power of x is n; we can only have n solutions for that f(x)?? {n could be 0, 1 , 2, 3 etc}

Does that look like a pattern now...

[bitter halo]

I see.

[regal moat]

Do u see those 2 lines?

Please tell me u spark some sort of "A-HA momment" ....

[bitter halo]

its ilegal

[regal moat]

Because?

[bitter halo]

three solution for a two degree polynomial

[regal moat]

knock them down by 1

2 output values from 1 degree function

[regal moat]

Additional Question: Do u think this could exist?

And why?

[bitter halo]

its strange

[regal moat]

Doesnt matter

Just apply what u've just learnt

[bitter halo]

there is a whole area of parameters for this function where it doesnt exisst

. . .

I am not sure.

there is no AHA moment .

[regal moat]

Just give me an answer

Doesnt matter its right or wrong

If its right then u're a fast learner, else you learn more from ur mistake

Literally no drawback from giving out an answer

[bitter halo]

I wanna be big brain doe

[regal moat]

Is this illegal? Yes or no

[bitter halo]

it is very largely yeah

[regal moat]

Yeah no

Its totally fine

[bitter halo]

. . .

[regal moat]

Its called piecewise function

basically, its a function that contain sub-function that takes action if a condition is met

[regal moat]

so something like this can totally happen, due to the fact that for 1 value of x, we only have 1 value of y AKA f(x) and the biggest power of x is 1, so no rule is being violated here

Well

U've learnt something

I hope what u've learnt here could help u with ur test

[bitter halo]

im stupid

.close

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[median nymph]

We have to cross a distance of 1100km with a jeep that can carry 100litres of petrol. The jeep uses 1 litre per 10km. We need to place a petrol depot that has to be filled by our own tank to cross the 1100km. Make sure the least amount of petrol is used. (harder questions coming soon)

[weak cove]

you're not maximo... :(

[smoky gyro]

he’s an opp

[regal moat]

130l

[median nymph]

place the depot at 100km? with 10l?

[regal moat]

Yep

Let A is where we start, B is the destination, C is away from A 100km;

Bring 30l from A to C, which used up 10l; left us with 20l; drop 10l at C; go back to A which used up the 10l;

Fuel up to 100l; goes to C; which use 10l; we have 90l; pick up that 10l; now we have 100l which can get us to B

in total 130l

[median nymph]

This is only the beginning...
We're gonna be here for a while boi

[regal moat]

Lmao

Well good luck

I have something to do

[median nymph]

o7

[devout snow]

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[scenic magnet]

can anyone explain why I need to use nPr for part B

instead of just another 5!

[scenic magnet]

ah maybe I shouldve cropped it

but anyways, to solve for A I just had to do 7! - 5!, as there are 5! arrangements of the letters in KITCHEN that start with KI

however, when trying to solve for part b, aren't there also 5! arrangements of the letters in KITCHEN that start with IK?

meaning that the answer to be would be 7! - 5! - 5! ?

why is the answer 7! - 5p5 - 5!

nvm I'm stupid, I mustve input something wrong in the calculator

as 5! = 5p5

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[smoky gyro]

!status

[devout snow]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[smoky gyro]

well you want the equation in terms of one variable which is usually y

so a good start is solving for y

you want y to be the subject of both equations so you can equate them

yes.

looks good

now set those equal to eachother and solve for x

its some algebra and application of log rules

[devout snow]

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[crimson cape]

Hi I don’t get how to do this question

[midnight grove]

so the formula for the cylinder is

V = πr²h right

you can get the raidus from there

and from there, maybe try to use a formula for the area of a segment of a circle

[twilit comet]

*radius

[devout snow]

@crimson cape Has your question been resolved?

[crimson cape]

<@&286206848099549185>

[knotty crow]

Is there a diagram?

[crimson cape]

Yes

[knotty crow]

Pls send

[knotty crow]

I'll try

Is answer 144?

@crimson cape

[crimson cape]

Ok thank you

But could you also help on one more problem

@knotty crow

[knotty crow]

Sure

Send it

[crimson cape]

This what I have

@knotty crow

[knotty crow]

I'll try

Done

Area is 50

@crimson cape

[crimson cape]

Could you show me the work

[knotty crow]

Wait let me do it neatly

[crimson cape]

Ok

[knotty crow]

Any doubt?

@crimson cape

[crimson cape]

Nope, thanks

[knotty crow]

welcome

[crimson cape]

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[solar cedar]

For 2)b) do we use P70 to find m?

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@solar cedar Has your question been resolved?

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[smoky gyro]

I don’t get it what am I doing wrong??.?

[smoky gyro]

the answer is supposed to be $\frac{1}{2}e^{4}+e^{-2}-\frac{3}{2}$

[woven radish]

water beam

[red turret]

e^0 is not equal to 1/2

if that is what you wrote in the last line

[smoky gyro]

i wrote 1/e

[smoky gyro]

this part is 1/e

[red turret]

so you mean e^0 = 1/e?

[smoky gyro]

oh wait

its just 1

[red turret]

[smoky gyro]

wait my brain is in a funk its like 12:30am and my algebra isnt algebra-ing

$\frac{e^{4}-1}{2}+e^{-2}-e^{0}$

[woven radish]

water beam

[smoky gyro]

is this right?

[red turret]

yeah

[smoky gyro]

okahy

good

thanks

[red turret]

[smoky gyro]

.close

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[smoky gyro]

.reopen

[devout snow]

✅

[smoky gyro]

why is C true?

is this because the graph f'(x) = 0

[midnight grove]

nah

f'(x) being 0 at a value only tells us that f(x) has a turning point at that value

think about f''(x)

what do you think f''(x) is in there

[smoky gyro]

if f"(x) = 0 there is a possible point of inflection

[midnight grove]

correct

[midnight grove]

but what do you think f''(x) is according to the graph

[smoky gyro]

hm

[midnight grove]

on x=-4

[smoky gyro]

like what it looks like? or

[midnight grove]

what do you think is the value of the derivative of f'(x) at x=-4

since you can see on the graph of f'(x) at x=-4, there is a turning point

[smoky gyro]

yeah

[smoky gyro]

im not quite sure

[midnight grove]

alright so

look at the point x=-4

the graph turns at x=-4 right?

[smoky gyro]

yes

it goes from negative to positive gradient i think

[midnight grove]

thats right

the change of gradient tells us that the derivative of the graph at x=-4 is 0

since that graph is f'(x), and the value of its derivative is 0, we can say that f''(-4) = 0

what does f''(a) being zero have to do with f(a) again?

you already mentioned it earlier

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[coral wraith]

i was reading up on how to do this, but im still confused

[stable storm]

for an alternating series that converges, the error is less than the next term

so if i computed only 5 terms

the error would be less than the 6th term

[coral wraith]

right, but how many terms do i compute to

[fossil locust]

You don't need to know this

You just need to solve an inequality given that you know the nth term

[coral wraith]

yeah i did that

i think i got 12

oh nice thats right

[fossil locust]

Also you should use absolute values, so you will have |...| <= 0.006

Yeah nice, you did it already!

[coral wraith]

thanks for the help

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[hollow furnace]

My question is not really about a specific question. I was recently learning about matrices and was wondering if it is possible to do something like 2^A, where A is a matrix. I was also wondering is it is possible to take a square root of a matrix

[fossil locust]

Yes, it is indeed possible

Write 2^A as e^((ln 2) * A)

And then use the Maclaurin series for e^(A): it works out where you just raise A to various powers and divide by some numbers

[hollow furnace]

ooo alr

and the sqrt?

[fossil locust]

https://en.wikipedia.org/wiki/Square_root_of_a_matrix#By_diagonalization

You can find the square root by diagonalising the matrix

[hollow furnace]

alright thanks

[fossil locust]

The square root of a diagonal matrix is fairly obvious

No worries

[hollow furnace]

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[unkempt quiver]

I'm bored so I made up a math/physics/kinematics type problem

So I want to calculate the difference in the distance over time. As the red dot moves upward, it's not getting much closer but the rate at which the difference is changing increases the most at y = 0 I believe, but im not sure.

How would I derive an equation for this?

[unkempt quiver]

Reposting a question since nobody responded

So f(x) = difference between object 2 and 1

[devout snow]

@unkempt quiver Has your question been resolved?

[devout snow]

@unkempt quiver Has your question been resolved?

[spare crypt]

you'd use the pythagorean for the distance, and say the height difference changes with time as a variable t (you can multiply speed to this) to get distance=sqrt(x^2+y^2)=sqrt(4^2+x^2)

[unkempt quiver]

yeah that's how i calculated all the distances above at different times

ohhhh

wait

y = |sqrt(4^2 + x^2)| hold up

[spare crypt]

when you're far away you're basically just traveling straight away so it's a straight line, but nearby the function bends a bit

[unkempt quiver]

ok lemme try understanding how u got this

so

[unkempt quiver]

yep

[spare crypt]

here's a graph https://www.desmos.com/calculator/ngsobeqxa3

[unkempt quiver]

oh bruh

i've tried making derivatives of graphs in desmos b4

by just typing f'(x)

[spare crypt]

lol

[unkempt quiver]

that's how u do it

ohhh

u can do this too wow i didnt know that

[unkempt quiver]

so we know that for example at t = 1 we are sqrt(4^2 + 3^2)

[unkempt quiver]

we also need to account for the fact that not just the x value but the y value is 4 away as well initially

so

at y = 0 they are 4 away in both x and y value

so

t = 0: distance = sqrt(4^2 + 4^2)

t = 1: distance = sqrt(4^2 + 3^2)

so

the actual equation would be

distance = sqrt(4^2 + (4 - x)^2)

[spare crypt]

yea

[unkempt quiver]

BRUH

I'VE BEEN USING DESMOS FOR YEARS

AND I NEVR KNEW U COULD JUST

[spare crypt]

pfff

[unkempt quiver]

TYPE f(3)

[spare crypt]

yea it's a whole universe

[unkempt quiver]

and get the value

💀

no bruh

i've tried it

so many many many times

but for some reason now it works

maybe i did it incorrectly

TY that's the real help u helped me w/ LMAO

[spare crypt]

np haha

[unkempt quiver]

epic it works

[unkempt quiver]

ty bro ur cool

[devout snow]

@unkempt quiver Has your question been resolved?

[devout snow]

@unkempt quiver Has your question been resolved?

[somber zodiac]

I do not think so

It is more like this green line

Basically my reasoning is that you plot the data

Calculate the best-fit line. Maybe it would be a bad approach, but this approximation is practically perfect. (R^2 is much more precise than for 4 sg)

You get a function like this:

$f\left(x\right)\ =\ 0.0025x^{3}-0.0423x^{2}-0.0177x+0.714$

[woven radish]

mlysikowski

[somber zodiac]

While not a good approximation for any x outside (1, 5) (possibly, as the extrapolated values may be ok), it is basically perfect for this domain

The derivative is clearly not linear

Unless I did something wrong

But this is only given a few assumptions ofc. Is the change continuous or discrete?

If discrete, my way of thinking is wrong

Even though the final answer should be identical, I think

Any extra information besides those data points would be great, eg. where did you take them from? Are those just random values?

Hmmm, I am not sure

It would depend on the nature of this change

Line of best fit may be too much of an assumption in terms of how those values change

Ok, nevermind

I got the question wrong

Basically I get this, where y-axis values is the rate of change of separation between those two points, and x-axis is position of the other point. On the paper below they are denoted as derivative of z and y

Also, ignore the two negative signs in line 2.

So "does the rate they come closer change"? Yes, it is a nonlinear function of the current position of particle y

[somber zodiac]

You can also notice it based on the graph

The blue function is the distance between points

It clearly does not change linearly

Or by considering how does the distance change for the case when y = 0. They do not come any closer, so the instantaneous change is 0

[somber zodiac]

I am not sure what you are asking for tbh. Each sentence seems to ask about something else, including the photo. I believe that I did provide the solution you asked for. The rate at which the difference is changing increasing is another derivative of the above, graphed below. So you are correct that it increases the most at y = 0

[unkempt quiver]

LOL sorry yeah you did

I clicked the wrong button oops it kept pinging me

Btw tysm for that new explanation

I didn't realize the derivative answered my other question

[unkempt quiver]

Wait hold up

[somber zodiac]

I hope that it is correct, at least 🤣 You may check the details

[unkempt quiver]

Uh

Someone above solved it very simply and it was correct

[somber zodiac]

Everything above the line "-----------------------------------------------------" is not fully correct

[unkempt quiver]

distance = sqrt(4^2 + (4 - x)^2)

This was the equation

I'm not exactly sure what u did

But we know we just use the distance formula

The distance 4 stays constant cause we're traveling in a linear path

X value does not change

Y value does

[somber zodiac]

Yeah, basically it is the same

[unkempt quiver]

Ohhh ok

[somber zodiac]

But I thought that it was the starting point

[unkempt quiver]

Oh right u can rxpandc

16 + 16 - 8x + x^2

X^2 - 8x + 32

Huh

[unkempt quiver]

Oh

What if it wasn't traveling in a linear path

The red dot

I think that rhe easiest thing to do would just be to rotate the x and y axis

Since we only care about distance not position

[somber zodiac]

It is travelling in a linear path. I calculated the rates of change, etc. you mentioned in the question

[unkempt quiver]

O ok

Tysm

[somber zodiac]

It was some very useful work 😂. Well, it is late here.

[unkempt quiver]

Alr! Gn!!!!

I'll read it over

I just thought of another cool problem but u can sleep now if you want

Anyways I wanna figure out the distance of the red dot from the blue dot in this equation now where x is time

[somber zodiac]

No, bring it on!

[unkempt quiver]

Red dot travels in a parabolic shape

I should've done an actual kinematic equation

[somber zodiac]

Like this, I think:

[unkempt quiver]

How'd u get the second equation?

With x - 2 and 3

[somber zodiac]

Substituted the y-value of the red point

in the equation for distance between 2 points

You provided the y-value for this point:

[unkempt quiver]

O ok

[unkempt quiver]

Wait hold up

U could just subtract the two equations to find the distance but then again

0, 0 isn't an equation

Damm math is so weird

[somber zodiac]

I don't get it. Subtract what exactly?

[somber zodiac]

This formula:

[unkempt quiver]

Nvm lol

[unkempt quiver]

Oh yeah I'm dum

Nvm

[restive river]

close the room

[unkempt quiver]

k

.close

[devout snow]

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[shrewd plume]

4/6.88x = 10/43

I got this

[coarse pivot]

Look at this,

You know that it's a 2d shape, so it's a squared relationship, but this time

r^2 (16) times an unknown factor gives 10^2 (100)

[shrewd plume]

Oh wait what

Hoe do you know it's squared

Is a side squared the area?

[coarse pivot]

You're right, sorry that's on me

The ratio it grows should be relatively close if squared, but not accurate. that's on me

100/16 is 6.25, which is very close to your 2nd option

[shrewd plume]

A side squared would be the area of a square

[coarse pivot]

It would

[shrewd plume]

Oh wait I see

Cuz penta is 5 sides

Prettu xlose

[coarse pivot]

You could solve for the area of a pentagon with 1 side, but the formulas long and weird

It's easier to use just a square and approximate it

I hope this helps

[shrewd plume]

Wait it could be 16/100

@coarse pivot

So how do I know

[coarse pivot]

Because you move the 16 to the other side, you have to divide by 16

[shrewd plume]

.close

[devout snow]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[glass jetty]

Can somebody help me with part d?

Of question 3

Not too sure how to get all of them but I do know that there are two elements of the domain that must share a value in the codomain

[eternal tapir]

that's injective

surjection is when the range and codomain coincide

[glass jetty]

What do you mean?

[eternal tapir]

consider the function f: A -> B given by f(x) = 5 for all x in A

what is the range of f? what is the codomain of f?

[livid carbon]

Stirling number of second kind is what you need to know to count all surjective functions

But I guess you can do this problem without knowing that

As these are tiny sets

Hint : count all the possible functions and then subtract the number of functions which are not surjective

To start with, what is the total number of functions here? @glass jetty