help-13

[ornate sequoia]

of gelijk aan 5 alsje dat nodig hebt

[crimson sedge]

Dat is logisch man

Ja

dat is logisch

ik snap hem

[ornate sequoia]

het hoeft niet een verschilfunctie zijn, je kan dit ook doen met gewoon f(x) of g(x), dit geldt voor alle functies

bij de verschilfunctie weet je dan gwn extra waar f>g en waar f<g

[crimson sedge]

dus eerst bereken je alle snijpunten

[ornate sequoia]

ja

[crimson sedge]

en dan vul je m in op de format die ze willen

[ornate sequoia]

dan zet je die erbij en zet je een 0 in het stukje van de functie

en dan doe je een testwaarde voor elk stukje na een 0 iykwim

ook voor de eerste 0 en na de laatste

[crimson sedge]

ja

ik snap hem eindelijk

dus eigenlijk is het heel logisch

[ornate sequoia]

dit is idd vrij logische wiskunde

alsje wilt moet je me maar dm'en alsje nog wiskunde vragen hebt

[crimson sedge]

is goed zal ik zeker doen

ik denk dat ik toch echt een docent moet nemen

al is het maar 1 uurtje per week

want dit is niet te doen

ik moet nog 18 hoofdstukken

van circa 50 bladzijdes per

[ornate sequoia]

ik wil gerust helpen maar afhankelijk van hoe zwaar de wiskunde is denk ik dat het misschien wel nodig is

waarom hebje geen docent?

of doe je thuisonderwijs

[crimson sedge]

nee

ik heb mn havo diploma gehaald

maar nu wil ik graag naar de universiteit

daarvoor moet ik wiskunde b doen en een jaartje hbo

ik weet niet of je weet wat die shit is

maar in ieder geval ik moet dus wiskunde b staatsexamen doen

om naar de universiteit te mogen

[ornate sequoia]

ik kan de wiskunde zeker want ik ga in september naar de universiteit maar ik weet niks over die Nederlandse examens

[crimson sedge]

dus dit is eigenlijk gewoon totaal los van school of iets

[crimson sedge]

als je op zoekt wiskunde b eindexamen vwo

dan kom je op examenblad.nl

en dan staan er opgaves.pdf

als je erop klikt

dan kan je zien waar ik naar toe aan het werken bent

[ornate sequoia]

dat is wat je moet kunnen?

[crimson sedge]

Ja

[ornate sequoia]

oke lemme check brb

[crimson sedge]

dat is het eind doel

maar ik doe het hoofdstuk per hoofdstuk zodat ik de basis heb

https://www2.cito.nl/vo/ex2023/VW-1025-a-23-1-o.pdf

ik heb een jaar om dit te kunnen

nja tot mei

3 uurtjes per dag

[ornate sequoia]

hebje verder nog dingen of is het echt enkel dit

[crimson sedge]

dit is het ongeveer

[ornate sequoia]

het is zeker pittig

[crimson sedge]

haalbaar in een jaar?

of is het krap

[ornate sequoia]

ongeveer wat ik moest kennen maar dan wel in 1 toets natuurlijk

hebje nog fysica en wetenschappen en gewoon school? of is dit echt het enige dat je gaat doen

[crimson sedge]

nja hbo

maar daar doe ik niets

het is behoorlijk makkelijk

qua echt heel makkelijk

dus het enige wat het kost is tijd maar geen inzet

daarom 3 uurtjes per dag 's avonds

[ornate sequoia]

ahja zo

[crimson sedge]

Mijn concentratie is nog goed 's avonds enz want ik doe vrijwel niets behalve komen op dagen

en dan 's avonds hier veel tijd in stoppen

[ornate sequoia]

ik denk alsje zeg maar minimaal werk doet voor je hbo dat het wel kan

[crimson sedge]

het is wel beetje dom plan van me

maar ja ik wil echt graag universiteit doen

dus dan maar zo

nja heel erg bedankt man

ik snap m

.close

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[shadow kite]

Hi, I need help in understanding how step 1 has proceeded into step 2

[shadow kite]

Why and how did the variable change from t to x

[long arrow]

it doesn't match with their substitution but variable in general does not matter, it's still same integral

and bounds are still same even after coming back to the x

as you can see

(already changed according to the previous one)

[shadow kite]

Alright! got it

.close

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[viscid pendant]

I'm trying to find the maclaurin series representation of cosh(x)

[tropic oxide]

x^n + (-x)^n is 0 if n is odd and 2x^n if n is even

[viscid pendant]

Idk how they got rid of the negative

[tropic oxide]

reindexing

[viscid pendant]

I'm not sure what that is

[viscid pendant]

right so how do Ik what to put

[tropic oxide]

okay lets put it this way

$e^{x} = \sum_{n=0}^{\infty} \frac{x^n}{n!}$

[wraith dagger]

Ann

[tropic oxide]

$e^{-x} = \sum_{n=0}^{\infty} \frac{(-1)^n x^n}{n!}$

[wraith dagger]

Ann

[tropic oxide]

with me so far?

[viscid pendant]

yes

[tropic oxide]

$\cosh(x) = \sum_{n=0}^{\infty} \frac{1 + (-1)^n}{2} \cdot \frac{x^n}{n!}$

[wraith dagger]

Ann

[tropic oxide]

agree or disagree?

[viscid pendant]

idk what you did there

[tropic oxide]

i added e^x and e^-x termwise and multiplied the 1/2 in

[viscid pendant]

where did the 1+ come from

did you factor out x^n

actually nvm I see what you did

[viscid pendant]

agree

[tropic oxide]

ok

$\frac{1 + (-1)^n}{2} = \begin{cases} 1 & n \text{ is even} \ 0 & n \text{ is odd} \end{cases}$

[wraith dagger]

Ann

[tropic oxide]

agree or disagree?

[viscid pendant]

agree

@tropic oxide

[tropic oxide]

yeah so

what happens from here is that all the terms with odd n are killed

and the ones with even n stay

so you get $\sum_{n \text{ even}} \frac{x^n}{n!}$

[wraith dagger]

Ann

[viscid pendant]

oh I see they're killed becuase it results in 0 right

is that why the negative disappears?

[viscid pendant]

why did they get x^2n

[tropic oxide]

yeah sure

[tropic oxide]

you understand this?

[viscid pendant]

yes

[tropic oxide]

you can rewrite this as a sum for k from 0 to infinity

with n =2 k

n = 2k

[viscid pendant]

how come the sum still starts at 0

for the answer

[tropic oxide]

well the first even number is 0 eh

[cedar kiln]

@viscid pendant Has your question been resolved?

[cedar kiln]

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[crimson sedge]

,w variance(4, 1, 5, 2, 8, 6)

[wraith dagger]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by toby____#0

[crimson sedge]

Do you guys have some tips to speed up the proccess?
var(X) = E[X^2] - E[X]^2 is probably faster to compute than the other definition

are you able to calculate the mean of (4, 1, 5, 2, 8, 6)?

yeah

now calculate the mean of the squares

[crimson sedge]

of the data

[crimson sedge]

hmm wolfie calculated sample variance

did you want a sample of population variance?

mean of (4^2, 1^2, 5^2, ...)

ah ok, thats population variance then

,w population variance(4, 1, 5, 2, 8, 6)

[wraith dagger]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by toby____#0

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Upgrade to WolframAlpha Pro!

Requested by brotmithonig#0

[crimson sedge]

put commas between the numbers and use a decimal point instead of comma

yes

,w population variance(.4, .2, .5, .15, .75, .5)

[wraith dagger]

Results provided by WolframAlpha

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

Requested by toby____#0

[crimson sedge]

yes, now put them into E[X^2] - E[X]^2

thats for population variance

[crimson sedge]

the sample variance version is not as nice

E[X^2] is the mean of the squares. E[X] is the mean of the original data

yes

,calc 24.333 - 18.777

[wraith dagger]

Result:

5.556

[crimson sedge]

,calc 4.333^2

[wraith dagger]

Result:

18.774889

[crimson sedge]

should be 18.7777 (remember to use enough decimal places before rounding)

its the same as wolfie

[crimson sedge]

^

yes, population variance is 5.555 for the first one

thats for the second set you had

yw :)

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[tropic glacier]

how to integrate 7/5^x

[gray juniper]

Ignore the constants for a second

[tropic glacier]

what's the constant

7?

[gray juniper]

how do you integrate 1/5^x at least

[tropic glacier]

ln?

is it this formula

[gray juniper]

That's for 1/x

[tropic glacier]

ok

so no idk how

oh wait

let me try something

[gray juniper]

$\int a^x dx = \frac{a^x}{ln(a)} + C$

[wraith dagger]

despairful_deltoid

[tropic glacier]

when it says ln a

do I ln a

[gray juniper]

Yeah you take the natural log of the base

[tropic glacier]

like ln 2 is .69

[gray juniper]

In our case it is 5

Just leave it as ln(5)

[tropic glacier]

ok wait

is 1/5x

5x^-1

[gray juniper]

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

[wraith dagger]

despairful_deltoid

[gray juniper]

so you need to change -x to x

then apply what I sent earlier

[tropic glacier]

how do I do that

[gray juniper]

sub u = -x

[tropic glacier]

so 5x^u/ln 5

[gray juniper]

not quite

that's true if we had 5^x

Do you know u substitution?

[tropic glacier]

idts

[gray juniper]

?

[severe void]

i dont think so

[gray juniper]

Oh

[tropic glacier]

mb

[dire geode]

,tex .int rules

[wraith dagger]

rie.mann

[dire geode]

the bottom row isn't familiar?

[tropic glacier]

what

nahhh

[gray juniper]

Ok so

When we sub u = -x

we need to differentiate both sides

so u becomes du and -x becomes -dx (because d/dx(-x) = -1)

du = -dx

[tropic glacier]

damn

I think I'll try it on my own first

[gray juniper]

Let me know if that was confusing

Ok sounds great

[tropic glacier]

I'll reopen a channel if I do

ty

.close

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[viscid pendant]

Is my answer correct for representation of taylor series?

of ln(1+x^2)

[cedar kiln]

@viscid pendant Has your question been resolved?

[viscid pendant]

<@&286206848099549185>

[south tundra]

It is correct, yeah, though you can cancel that 2 out in the end

[viscid pendant]

How do I change the indices

[south tundra]

You want to start from n = 1?

[viscid pendant]

Yes

Idk how my textbook answer got this

[south tundra]

Basically replace every n with n-1

[viscid pendant]

right

[viscid pendant]

oh got it

Does this look good

[south tundra]

Yup, and (-1)^(n-1) is the same as (-1)^(n+1)

[viscid pendant]

how come

[south tundra]

Oh and it should be n=1 in the upper summation since you did the replacement there

[viscid pendant]

right

[south tundra]

(-1)^(n+1) = (-1)^(n-1) * (-1)^2 = (-1)^(n-1) * 1 = (-1)^(n-1)

[viscid pendant]

Is it because it alternates regardless?

if its n+1 or n-1

[south tundra]

Yeah and starts with the same sign as well

[viscid pendant]

that makes sense

thanks!

[cedar kiln]

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[hard yoke]

how do i do ii?

[turbid mesa]

What did you try?

[hard yoke]

well i put it into combination notation and thats just messy

[turbid mesa]

Okay, what's the probability that Linda hits a bulls eye in 1 shot?

[hard yoke]

1/5

[turbid mesa]

Yeah, and the probability of missing?

[hard yoke]

4/5

[turbid mesa]

You need 2 hits and 8 misses

How many such combinations are possible?

[hard yoke]

oh im talking about part 2

[turbid mesa]

Oh

Yeah the logic still holds. Say it it takes n hits

What's the probability of taking exactly 1 shot in n hits?

[hard yoke]

(1/5)^n

[turbid mesa]

Why?

That's the probability that she takes all the shots bulls eye

She needs to take EXACTLY one and miss the rest

[hard yoke]

oh yes

[turbid mesa]

Or better yet, what's the probability that she misses all?

[hard yoke]

(4/5)^n?

[turbid mesa]

Yes

[hard yoke]

ohh

[turbid mesa]

So what's the probability that she takes atleast 1? Are the events of taking atleast 1 and none mutually exhaustive and exclusive?

[hard yoke]

so it would be 0.9=1-(4/5)^n then

oh wait

[turbid mesa]

No, 0.9=1-(4/5)^n

Yes

So (4/5)^n=0.1

You can work it out now ig

[hard yoke]

thank u so much

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[severe flume]

can anyone tell me the Basic linear graph skills?

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[severe flume]

.close

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[steady elk]

how did they cloncluded first function and second function

[dreamy sleet]

what do you mean?

[steady elk]

imean why is f(x) is first function there

[royal loom]

first function where

[steady elk]

the notation man

[royal loom]

you sent a picture of an entire page of math

when you say "there"

it is hard to tell which of the 20 instances of f(x) you are referring to

[steady elk]

in the question where roman numerals are written

those roman numerals are first and second function

as in pic mentioned as 1 and 2

and i want to know how they concluded which to be the first and which to be second function

[royal loom]

they chose the first function to be the function that isn't being differentiated, and the second function to be the function whose second derivative is being taken

[steady elk]

ru sure?

tomorow is my test

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[shy pebble]

unable to solve it.

[livid hound]

first step is already wrong

where's $\frac{b^2}{4x^2 +9}$ coming from

[wraith dagger]

ℝamonov

[shy pebble]

oh yeah. I'm sorry.

[livid hound]

or in your other attempt, where's
$$\frac{-b^2}{4x^2 -9}$$
coming from

[wraith dagger]

ℝamonov

[shy pebble]

oh yeah. It's wrong.

am I right?

Is this correct?

[livid hound]

yeh

[shy pebble]

thank you very much

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[shy pebble]

I don't know what to do after finding the scale factor.

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[rapid harbor]

Hello, I have these formulae regarding the transformation of the standardnormal distribution, the transformation from normal distribution

and the calculation of the distribution funktion for the standard normal distribution

I tried inserting 1 for x, and the result is not 0,841

What am I doing wrong?

[vagrant elbow]

You shouldn't replace the variable of integration with 1

[rapid harbor]

How can I make it so it's not an absolute number, but a relative amount (I suppose that 0.8 means it's an 80% of the total)

[rapid harbor]

Oh, what would I insert for x then?

[vagrant elbow]

Now you have nothing more than a glorified constant function being integrated

Just let it be x

I think you're only supposed to change z

[rapid harbor]

here is the full formula I got from the internet

[rapid harbor]

Here on the 2nd picture, you can clearly see he is using the same symbol, and as argument inserting a 1

[rapid harbor]

Which would be the z in this formula

[dawn junco]

yes it's the bound that changes

not the function being integrated

[rapid harbor]

So excuse me, what should I change?

I don't have much experience with integrals

Ahh I'm dumb

I get you now

the x and z are different variables

I treated them as the same

mb

[dawn junco]

yeah

[rapid harbor]

Damn I just corrected it and it's still wrong lmao

oh it's the half?

[rapid harbor]

damn why is my result exactly twice as much as the result I'm searching for?

[dawn junco]

yeah there's a typo on your slides

it shouldn't be 2/sqrt(2pi)

[rapid harbor]

Yea that's what I thought

it would be 1/?

[dawn junco]

yes

[rapid harbor]

jesus haha

thanks

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[vapid sand]

I am watching a video that helps me go thorugh my question book but I am confused why is he drawing the coloured in brackets when the question doesn't use a greater or equal to sign

[jaunty mural]

where exactly

[vapid sand]

the circle

[jaunty mural]

x =< -1/2

[vapid sand]

this is to find the domain

[jaunty mural]

is written at the top

[vapid sand]

yea I know but why

the original question doesn't have a greater equal sign

[jaunty mural]

because thats their choice of cases

they chose to first consider
x =< -1/2

then -1/2 < x =< 3

then presumably 3 < x

[vapid sand]

why is it their choice of cases

[jaunty mural]

look at the original inequality

[vapid sand]

the original just has <

[jaunty mural]

how else can you solve if not split into case?

[jaunty mural]

nothing to do with this.

[vapid sand]

but the original inequality shows that

[jaunty mural]

the original question could be any of <, =<, =, >, >= signs

and you still need to split into cases

[vapid sand]

I don't get it...

[jaunty mural]

how do u solve |x+1| > 1

as an example

[vapid sand]

x+1 = 1 and -x-1 = 1

[jaunty mural]

> not =

But thats not all.

You are splitting into 2 cases based on the sign of x+1

in particular you should recall definition of absolute value

[vapid sand]

yes

[jaunty mural]

sec

,, \abs{x} = \begin{cases}x, & x \geq 0\-x, & x < 0\end{cases}

[jaunty mural]

@vapid sand sorry so this ^ you familiar with it

[vapid sand]

yep

[jaunty mural]

For |x+1| > 1, we apply the definition on |x+1|

,, \abs{x+1} = \begin{cases}x+1, & x+1 \geq 0\-x-1, & x+1 < 0\end{cases}

[jaunty mural]

you see how this splits into 2 cases based on the sign of x+1

[vapid sand]

yes I see now

but can you show in graph way or something why it has the >=

[jaunty mural]

This is from the definition

it actually doesnt have to be <=

[jaunty mural]

if you look at this

[vapid sand]

oh so it can just be >

[jaunty mural]

we include the equality case

inside x >= 0

but we could instead include ot on x < 0

x =< 0

and the definition is valid

[vapid sand]

oh I see

[jaunty mural]

you need to include the equality case somewhere

[vapid sand]

does the >= exist to say "this number is also included don't forget it"

[jaunty mural]

Your original question does these 3 cases

x =< -1/2
-1/2 < x =< 3
3 < x

[vapid sand]

because if it was x>0 -x<0 it would have like a gap in the middle

[jaunty mural]

you need to include all real numbers however you chop your cases up

notice the union of all 3 intervals is the real line

[vapid sand]

mhm

[jaunty mural]

This is also valid
x < -1/2
-1/2 =< x < 3
3 =< x

[vapid sand]

ok I see

so let me summarise this

you have the >= to show that that number is included in the whole total segment of the number line correct?

because it if were all >

there would be gaps in the graph if we were to plot it

@jaunty mural

[jaunty mural]

You have an equality you want to solve over all real values of x

however you split your cases, the cases must still encompass all possible real values

like you say, no gaps.

[vapid sand]

perfect then

[jaunty mural]

one sec

|2x + 1| < |x-3|

this was your original

that is equivalent to

[vapid sand]

I get it now

[jaunty mural]

|2x + 1| - |x-3| < 0

So we're looking at this

[vapid sand]

uh huh

weird funky bend going on

[jaunty mural]

and thats how they split the cases

Green, Black, Purple

It must encompass all of the x-axis

[vapid sand]

I see

so its just showing that all of that is included

I see

thanks!

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[cedar kiln]

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[hearty flint]

Please help

[gilded elm]

have you found the inverse yet

btw that's not how it works

you need to find the inverse of f first, in this case g

THEN you can differentiate it

[dusty hazel]

I mean, you can't always find the inverse just like that. And the result g'(b) = 1/f'(g(b)) also is correct for as long as g = f^-1

So you don't really want to find the inverse...

[hearty flint]

So im looking for a really smart way to rewrite that function ?

[gilded elm]

you find the inverse first

right?

eh

letse see if you made a mistake while differentiating

yup

you did

spot the mistake

wait

[jaunty mural]

[gilded elm]

no

ur right

lol

we kinda do need to know what g is here tho apparently

seems to ask for it in terms of g

[dusty hazel]

Eh, there's a way.

[hearty flint]

I’ve had a similar question but it was way easier there it’s unfortunately in German but the idea was the same

[dusty hazel]

sub x -> g(x) in the original equation.

And you'll get the result.

Without having to find g.

Though of course, if you'd like to find something like g(x) you're free to do so. But it's unnecessary.

[hearty flint]

what do you mean by sub x -> g(x) I don’t have g(x)

[dusty hazel]

f(x) = e^(x-1) + 2 + 2x

f(g(x)) = e^(g(x)-1)...

That's what it means to sub.

If you get g(x) anywhere, write it as g(x) only.

You don't need to know what it is.

[hearty flint]

ok what would be the next step then ? I just don’t see any way to eliminate the e since i cannot pull a ln() out of the air ?

[still barn]

The solution they asked you to show doesn't make sense to begin with. The function f' is a function of x, then you are evaluating it at g. There is simply no way to have an x floating around in the denominator like that. You're never going to get around the fact that at some point you are plugging in g into all the x's.

[dusty hazel]

The whole idea is to get e^(g(x)-1) because that's in your working. And you need to remove it.
From here you can get its value in terms of other terms.

[dusty hazel]

What are you talking about?

[hearty flint]

I mean ok I subed the g(x) in f(x)

Now I have

f(g(x)) = e^(g(x)-1))+2+2g(x) how does me having the e^help tho

Do I maybe have to integrate the derivation of f‘(g(x) and see if I get anything ??

[dusty hazel]

No.

What's f(g(x))

[hearty flint]

e^(g(x)-1) +2+2g(x) ?

[dusty hazel]

no, other than that.

[hearty flint]

If that’s not it I have to get from f‘(g(x)) to f(g(x)) ?

[dusty hazel]

Do you remember the proof for g'(x) = 1/f'(g(x))

[hearty flint]

Yes

[dusty hazel]

How does it begin?

[hearty flint]

By finding the derivate of f ?

[dusty hazel]

f(g(x)) = ??
f'(g(x)) * g'(x) = d/dx(??)

That's how it goes.

You should check your notes...

[hearty flint]

So I am looking for the derivation of f(g(x))

[dusty hazel]

Just check your notes, please.

[hearty flint]

Is this what I’m looking at coach ?

This should be the same I somehow should now be able to just transform them they must be equal 100%

[dusty hazel]

Are we just going to ignore everything I said so far?

g(x) = f^-1(x)
f(g(x)) = f(f^-1(x)) = x

[hearty flint]

Do you want me to transform the f(g(x)) so I will result at the to g(x)=..:?

[dusty hazel]

I want you to transform the f(g(x)).

But because then you'd get the result e^(g(x) - 1)

=

g(x) is fine because you need it in your final answer.

Think for yourself. Why do you want to get rid of g(x). We're working only because we NEED g(x).

[hearty flint]

How do I put out g(x) from e with kb but the rest is in ln then

Log and ln

[dusty hazel]

...

[dusty hazel]

Let's back up for a moment.

[hearty flint]

This is it when I put it in

[dusty hazel]

Precisely, let's start from there.

What does f(g(x)) equal

[hearty flint]

X

[dusty hazel]

Good!

Then write that.

[hearty flint]

👍

[dusty hazel]

What's up with discord recommending these random stickers ...

[hearty flint]

Now I need g(x) on one site right?

[dusty hazel]

No.

You need e^(g(x)-1) on one side.

[dusty hazel]

You'll put that in the last line here.

[hearty flint]

-2g(x)

[dusty hazel]

...

[dusty hazel]

And you're done.

[hearty flint]

Dude thanks a lot for taking that time and being a tremendous help

[dusty hazel]

Honestly I was going to leave much sooner. But because it was I who said you don't need g to prove this. I kind of stayed. Lol.

[hearty flint]

.close

[cedar kiln]

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[cedar kiln]

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[steep sinew]

Let A, B and C be sets such that A is a subset of C and B is a subset C prove that A union B = C if only if A =C or B=C. How should I approach proving this problem

[gilded elm]

you have an iff statement

prove => and <=

[prisma gull]

Usually equality of sets is proved by showing an arbitrary element taken belong to both sets

[gilded elm]

suppose $A \cup B = C$. Show $A = C$ or $B = C$

[prisma gull]

to prove A= B, we have show both cases, ie A is a subset of B and B is a subset of A

[steep sinew]

yeah my issue is that when I try to prove C is a subset of A or C is a subset of A I do not know if I am doing anything wrong because I would say that let x be an element of c meaning x is an element of B or x is an element of A which would show the subsets?

[prisma gull]

If x element of A then x element of B implies A subset of B

you can think, every arbitrary element you choose from A belongs to B.

[steep sinew]

yeah I know that but I guess I am more so worried that my proof is wrong

[prisma gull]

Do you see the iff statement here?

[steep sinew]

yea I do

I know if implies that you have to prove both directions

I am just mroe so saying that

when I am proving --> way

because we are already given one way

[prisma gull]

First we have to suppose A U B = C and then show A= C and B= C

right?

[steep sinew]

yes I agree with that

[prisma gull]

Let x element of A U B, then x element of A or x element of B

[steep sinew]

yep I agree with that

[steep sinew]

no its show A=C or B=c

[prisma gull]

If x element of A, x element of C. ( given A subset of C)

[steep sinew]

yep I agree with taht

[prisma gull]

since A U B = C, if x element of C then x element of A or x element of B right?

since they are equal

[steep sinew]

yep

I agree

[prisma gull]

so we need if x element of A, x element of C aswell if x element of C x element of A

which is A= C right?

since it’s OR, there is one more similar case

[steep sinew]

yep

I guess what if x is an element of a aand b

would we need to prove that

[prisma gull]

not here

Given A U B = C, so both sets are subsets to each other

using this we can show A= C or B= C

notice OR

[steep sinew]

okay

[prisma gull]

do you understand the procedure ?

Can you do it now?

or still confused

[steep sinew]

yep I think so

[prisma gull]

maybe write the proof and post here

[steep sinew]

I will type out what I think is correct and tell me if I am right

[prisma gull]

pls write in paper

or tex

@steep sinew

I have a better way

[steep sinew]

Latex?

[prisma gull]

We need to show A= C first

right?

[steep sinew]

yep I am typing ti out right now

[prisma gull]

let x element of A, then x element of A U B right?

[steep sinew]

let x be an element of C. B/C we know that C=A union B then we know that x is an element of A or B which shows that C is a subset of A or C is a subset B.

[prisma gull]

maybe try this way

first to show A= C

[steep sinew]

did I do it wrong?

[prisma gull]

i mean, im telling a better way

short aswell

[steep sinew]

okay

[prisma gull]

First to show A= C

[steep sinew]

we need to show that C is a subset of A

[prisma gull]

let x element of A, then x element of A U B, since A U B= C, x element of C

[steep sinew]

because we know that A is already a subset of C by the given

ok

[prisma gull]

similarly the other way for A= B, if x element of C, x element of A U B then x element of A or x element of B, if x element of A then A = B

if not other case

that’s why we see an OR

x element of C can either mean x in A or x in B

[steep sinew]

so is that the simple way of doing it

the one comment above

[prisma gull]

yes

Do you see that if element of C mean x is either in A or C right?

[steep sinew]

yep

[prisma gull]

If x is in A , A = C, if x is in B, B= C

do you see?

[steep sinew]

yea

[prisma gull]

if x in A or x in B both cases it’s in union always

[steep sinew]

ok

[prisma gull]

so the only cases is when x in C

i think you get a sense of

write it, prolly i can help from that more

[steep sinew]

okay so you were saying that there are two cases

[prisma gull]

Can you attempt a proof?

[steep sinew]

yea

[prisma gull]

I can help you make it rigour on the go dw

[steep sinew]

i dod do it

let x be an element of C. B/C we know that C=A union B then we know that x is an element of A or B which shows that C is a subset of A or C is a subset B.

for the -->

[prisma gull]

What’s B/C?

b minus C?

[steep sinew]

because not st sifference

[prisma gull]

where did we get B/C here?

[steep sinew]

b/c standign for because

[prisma gull]

Suppose $A \cup B = C$. Show $A = C$ or $B = C$

This is the statement we have

[wraith dagger]

dotdoc.

[prisma gull]

You can only suppose A U B= C

[steep sinew]

I mnetion a is asubset of C and b is a subset of c by the givens

[prisma gull]

you cannot assume that now

[steep sinew]

why cant i

[prisma gull]

Well what you have is a two way implication

[steep sinew]

what I am sayign is that we already know that A is a subset of C always and b is a subset of C always so we only need to prove that C is a subset of A or C is a subset B

[prisma gull]

You have suppose A U B = C and then prove either A= C or B = C, similarly suppose A= C or B = C show A U B= C

[steep sinew]

yea I was more so meaning --> for that direction only

[prisma gull]

When proving the —> direction, only think we have is A U B = C

we don’t know if A subset of B or any of that matter

we need to show it

[steep sinew]

no A subset C

is given before hand

[prisma gull]

no, that’s what we need to prove

supposing A U B = C

that’s our —>

$A \cup B = C$ iff $A = C$ or $B = C$

[wraith dagger]

dotdoc.

[steep sinew]

Let A,b and C be sets such that A is a subset of C and b is a subset of C prove that A Union B =C iff A=C or B=c

I might of typed the question wrong

[prisma gull]

This is your statement right?

[steep sinew]

no

[cedar cypress]

Why do I think this problem is wrong? Because its not necessary for A or B to be equal to C for this to hold.
For example, A={1,2}
B={3,4}
C={1,2,3,4}

A and B are subsets of C here and their union is C. And neither of them are equal to C

[steep sinew]

so the problem is false?

[cedar cypress]

Yes

I just proved it above

[steep sinew]

ok

thank you

[cedar cypress]

Where did you get this problem from

[steep sinew]

I was basing it off of another problem

about power sets

and I want wondering if I could provei t using the knowledge that

[cedar cypress]

Well...

[steep sinew]

if x is a subset of y, then power set of x must be a subset power set y.

the orginal problem was

I have the solution for it

I was more so wodnering if there was another way using that idea but i guess not

how do I end the help section

/end

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[torpid urchin]

can i solve this by taking the cross product of the two vectors supposedly spanning, and see if it is the normal?

[untold torrent]

The set of vectors is obviously linearly independent since they are not scalar multiples of eachother

Just check to see if they lie on the plane

[torpid urchin]

i can do that by just subbing them in correct

[untold torrent]

Yes

[torpid urchin]

ok thank you

.close

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[celest heath]

Help

[celest heath]

Here is my work

[cedar kiln]

@celest heath Has your question been resolved?

[celest heath]

<@&286206848099549185>

[cedar kiln]

@celest heath Has your question been resolved?

[cedar kiln]

@celest heath Has your question been resolved?

[lunar lynx]

Umm... You have made a few errors in calculation.

Firstly while calculating the length of major axis, you have written 11 -5, while it must be 11 - (-5) which will compute to 16. Thus, major axis will come to be square root of 256 which is 16.

Also, you aren't properly computing the expression. You have to square the terms inside and then take root. While calculating minor axis, (10-6)^2 is 16, whose square root will be 4. Not square root of 4.

If you rectify those, i believe you should be reaching to the answer correctly.

[cedar kiln]

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[limber stag]

I have this equation and I want to animate it unrolling. Any ideas how I can do this?

[limber stag]

it makes this shape

it is also modeled by this equation on the complex plane

I have calculated its perimeter to be 4

[cedar kiln]

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@limber stag Has your question been resolved?

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[crimson sedge]

Why do they have it as -log(cos x)

does the function change if you move the -1 inside the log?

Desmos shows the same graph

Does it have something to do with domain/range?

[short blade]

they are the same thing

[crimson sedge]

so they just didn't do it because

okay

thank you

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[real crypt]

Hello,
I am trying to find a way to meaningfully display a set of given data. It's about plane crashes and their affects on the pilot. I have the amount of G's experienced and the time they were experienced for. I assume I would want to refer to this data based on the force the person experiences, but how would I used the data I have to do that? I could cancel the seconds in acceleration, but that leaves me with the change in m/s which doesn't really mean much in this case as far as I am aware

[gentle flower]

you need to figure out when the emergency took place by referring to the g-force experienced by the pilots/passengers?

[real crypt]

Not necessarily, sorry I worded it kind of poorly. I have the g-force experienced by the pilots and for how long at the point of impact, I am trying to find some way to manipulate that to show the force (in newtons) they felt so I can compare that to other data points

For example, one crash they may have felt 40 G's for 0.05 seconds and in another they felt 75 G's for 0.032 seconds
Which pilot experienced a greater force (And therefore most likely sustained greater injuries)?

I don't know if newtons is really the units I'm looking for here, that's just what makes the most sense to me

[gentle flower]

do you know how much the pilot weighs?

[real crypt]

It's arbitrary, so I'd just go with 175 pounds

[gentle flower]

for example: if a pilot weighs 200 lbs and pulls 10g’s, approximately 2000 lbs of pressure would be exerted on his body

[real crypt]

That's calculated with F=ma I assume?
Is time ever a factor when considering the force?
Because from what I understand, you can experience high amounts of G's but for only a short period of time before blacking out or further injuries
So there must be some connection between the amount of G's and the time experience that affects you

[gentle flower]

the longer you experience the amount of g’s for, the more likely you’ll be severely injured

you can spend an entire life time experiencing 1g, but you can only last a short period of time alive at, for example, 10g’s or 20g’s, the body can’t withstand that much force

but a person did survive 214g’s, iirc

[real crypt]

Yeah, that was during an F1 crash I believe

[gentle flower]

yeah

indycar i think

[real crypt]

So what would be the best way to write that given the data I have?

[gentle flower]

hmm, you can multiply the amount of g’s by the pilots weight and then include the total time spent by the pilots body exerting that force

[real crypt]

If I multiply G by 9.81 to get m/s^2 and then divide by the time I get m/s^3
That's a value I can compare, but I can't really decide what that means physically

[real crypt]

Not a bad idea, though that would be difficult to graph I'd imagine

Since it's not just one value

[gentle flower]

it’s two values, so you could have the force on the y-axis and then time on the x-axis

m/s^3 is jerk

i think

[real crypt]

Oh true, I think you're right

Using the two examples from before I get this:

So clearly the second pilot experience a lot more jerk

[gentle flower]

oooh

yes

but how would you graph this

jerk and time?

[real crypt]

Hmm, possibly

[gentle flower]

that should be good enough

[real crypt]

Yeah, I think I can work with that

Thanks!

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[crimson sedge]

I need help please

[cedar kiln]

@crimson sedge Has your question been resolved?

[flint plinth]

!status

[cedar kiln]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

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[floral vigil]

I need some help
Jean puts 12 books on a shelf in random order. Three are by the same author.

What is the probability that those 3 books will be next to each other?

[floral vigil]

this is a permutations and combinations question but i dont know how to get it into probability

i might be good

I am not good

dear god its evolved

Chloe's bookshelf has 6 books by Smith and 8 books by Jones.

If she removes 4 books at random, what is the probability of removing 2 by each author?

i still have no clue where to start if someone could give me an equation that'd be great

alr nvm this class is awful

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[crimson sedge]

Can somebody please verify that I did this correctly?

[cedar kiln]

@crimson sedge Has your question been resolved?

[stark quest]

you evaluated du/dx incorrectly

d/dx(1+x^3)=3x^2, so the coefficient in front of the integral should be 12pi

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

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[hybrid wren]

still confused by this

[dire geode]

what did you learn from the last time you asked this question

[hybrid wren]

first of all, why am i being bullied.

im just trying to understand it.

[dire geode]

No one's bullying you

Just answer the question so people don't tell you what you already know

[hybrid wren]

i learned that S_0 is in units of meters

the si unit of length is meters

so s= meters

t= seconds

if the _0 means nothing

then a) should = meters

now P_0 is being multiplied by t^2

so that means its seconds are being multiplied by seconds ^2

thats about all i understand about it

[dire geode]

To finish p_0, you know
[meters] = [units of p_0] [seconds^2]

Do you know how to solve for [units p_0] using division

[hybrid wren]

yeah would i multiply both sides by seconds^2

[dire geode]

That would get you seconds^4 on the right

[dire geode]

Read the last word here

[hybrid wren]

division

divide both sides by seconds^2?

[hybrid wren]

so for k_o

[dire geode]

Yes?

Yup

[hybrid wren]

so what else was this question asking?

just for that?

like its asking for the si units

i understand that the primary one is meters

the second one is seconds

and then the dimension is m/s^3

[dire geode]

What

[hybrid wren]

is this all that this question was asking for?

there is hardly any reference to it in my book

[dire geode]

P_0 is not seconds

[hybrid wren]

its centiseconds?

or its hectoseconds?

[hybrid wren]

this because its in second^2?

[dire geode]

.

[hybrid wren]

so its because the meters are divided by the seconds?

so its less than a meter?

or because its seconds being divided?

so its double the seconds?

[dire geode]

meters / seconds^2 is in terms of SI already

[hybrid wren]

so i dont gotta touch it anymore?

[hybrid wren]

?

[dire geode]

Yes

[hybrid wren]

so aslong as i got those variables by themselves they good?

[hybrid wren]

?

[dire geode]

I already said yes?

[dire geode]

.

[hybrid wren]

Okay ty

.close

[cedar kiln]

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[inland blade]

(f(b)-f(a))/b-a

I'm not sure where to start

[livid hound]

(f(b)-f(a))/(b-a)

[inland blade]

oh

[livid hound]

note that this is pretty asking you for the slope between two points

[inland blade]

so can I just do

y2-y1/x2-x1x

x1

[muted bear]

yes, that is the same thing

just different notation

[inland blade]

ohhh

what is average rate of change?

(11+4)/(3+2)

15/5 = 3

wait- why did they say it as 'average rate of change'?

[tropic oxide]

as opposed to what

[inland blade]

.close

[cedar kiln]

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[inland blade]

ty I got it

[inland blade]

as oppose to the slope

[tropic oxide]

if your function was anything but linear you could not talk about its "slope"

[inland blade]

?

so it only works on linear functions?

[cedar kiln]

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[gusty mason]

Hello can anyone help me find the length denoted by ‘?’ ?

[gusty mason]

I don't know how to start

[ember mesa]

may be blind but i don't see a question mark

[ember mesa]

lmao

[gusty mason]

<@&286206848099549185>

Can someone confirm my answer
$? = \frac{(l^2-b^2)\sqrt{l^2+b^2}}{3}$

[wraith dagger]

lmaowhat

[dire geode]

!show

[cedar kiln]

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

[gusty mason]

I didn't find any geometrical way of doing it so I used polar from and rotated the point by an angle theta where theta is the angle of inclination of the plank

I needed to find rsin(theta + alpha), but you get the point

? comes out to be $\frac{2}{3}lb\sqrt{l^2+b^2}$

[wraith dagger]

lmaowhat

[gusty mason]

Made a stupid mistake here, it's supposed to be $\frac{1}{3}\frac{l^2-b^2}{\sqrt{l^2+b^2}}$

[wraith dagger]

lmaowhat

[cedar kiln]

@gusty mason Has your question been resolved?

[gusty mason]

Alright, it's correct

.close

[cedar kiln]

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[lethal summit]

I don't exactly need help I just need an assurance that this doesn't make sense right?

[lethal summit]

The very bottom number is 20.5

[cedar kiln]

@lethal summit Has your question been resolved?

[cedar kiln]

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[primal lodge]

How would i "solve" this?

[lunar lynx]

I think you can make use of the fact that since sum of two odd numbers is even, then at least one of a or b has to be an even prime that is 2.

Basically, if a and b both are odd - then a^x and b^y will be odd as well for any x and y. Then two odd numbers can't sum to an odd number i.e. 129 in this case.

[primal lodge]

okay makes sense

we want odd + even to get odd

only way to get even is exponentiating even so 2 is the only even prime number hence b or a has to be 2

[lunar lynx]

Yes.

Now, you can just make different cases and solve.

There are only 7 cases to be checked.

[primal lodge]

isn't that a bit ineffective?

wdym 7 cases?

[lunar lynx]

Wait. Not even 7 cases. there are only 4 cases.

[lunar lynx]

There are only four primes for x i.e. 2, 3, 5, and 7 for which 2^x is less than 129.

[primal lodge]

oh i forgot the exponents have to be prime too lol

[lunar lynx]

Perhaps. However, i think that these problems are often solved this way.

[primal lodge]

okay so technically

3 cases

2,3, and 5

[lunar lynx]

Maybe there are other methods too. Alas, i can't think of any.

yeah.

[primal lodge]

cuz a^x can never = 1 unless x = 0

[lunar lynx]

Yeah

[primal lodge]

x = 0 not prime

hmm okay so

now what?

[lunar lynx]

Check for these three cases one by one.

[primal lodge]

b^x = 125
b^x = 121
b^x = 97

[lunar lynx]

Yes.

[primal lodge]

and identify which are perfect squares, cubes, or whatever?

hmm okay

[lunar lynx]

Exactly.

[primal lodge]

i did get 2^2 and 5^3 before but wasn't sure how to systematically approach these type of questions

tbh

but okay this makes some sense i guess

[lunar lynx]

Okay. Well, i think that in these questions - you rule out the possibilities and constrict the solution space until it's small enough to check manually.

[primal lodge]

hmm okay makes sense

[lunar lynx]

Here, only 121 = 11^2 and 125 = 5^3 are possible.

[primal lodge]

you have another question in mind with similar theme?

[lunar lynx]

Using that, you can get two possibilities:
(a, b, x, y) = (2, 11, 3, 2) and (2, 5, 2, 3)

[primal lodge]

mhm makes sense

didn't notice b^x = 121 lol i'm blind

[lunar lynx]

Well, not exactly on primes.

[primal lodge]

anything works

that's somewhat related

obviously lol

[lunar lynx]

It's okay. Actually, you can go systematically about this too.

Biggest number is 121, so if y has to be 2, you have to check for primes less than or equsl to sqrt(121).

[lunar lynx]

So 4 primes to check there.

For cube, range will be even smaller. Primes less than or equal to cube root of 121.