help-23

[thin bridge]

e.g. for the first empty box,

substitute x with -3

use replace all x with (-3)
parentheses included to be safe

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[lean otter]

9.reopen

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[twilit garden]

can someone help me with this please?

[grave flint]

where are you stuck

[twilit garden]

I dont know how to find the line BA

and also finding the scale factor thing

[grave flint]

dont need to

[twilit garden]

you dont?

[velvet nebula]

similar triangles?

[twilit garden]

yep

i dont get this topic much thats why im asking

[grave flint]

similar triangles sides have same ratio

so

$\frac{8}{6} = \frac{x}{9}$

[flat frigate]

dog.

[grave flint]

where CE = x

[twilit garden]

is ratio the scale factor?

[hexed dune]

yh

[twilit garden]

oh ok

[hexed dune]

acc

[twilit garden]

so how do i find the ratio thing

[hexed dune]

the scale factor would be

[grave flint]

yes the ratio of one known side to the other

[hexed dune]

9/6

[grave flint]

one side from each triangle so 9/6 = 3/2 = 1.5

[grave flint]

yeahj

[twilit garden]

so the scale factor is 1.5

ok

8cm x 1.5= CE ?

[grave flint]

yes

[twilit garden]

then CE- 8cm= DE?

[grave flint]

yes

u dont need DE tho

i mean u can solve it that way if u want

[twilit garden]

is it simpler?

[grave flint]

idk

similar

[twilit garden]

or is there a easier way to sole this

[grave flint]

doesnt make a big difference

[grave flint]

idk easier

they are abt the same

ur way is fine

[twilit garden]

then how do i find the area of the trapezium

[grave flint]

ur way is fine

u can also do it as a difference of area of the triangles

[twilit garden]

i need to find the area of the trapezium though

i just dont know how to that

[grave flint]

theres a formula

u can also break it down into a rectangle and triangle even derive the formula

its quite elementary

[twilit garden]

ooh

ok

thanks for the help

.close

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[twilit garden]

.reopen

[safe radish]

✅

[twilit garden]

wait a minute

[twilit garden]

how to find the sides on the triangle though

[grave flint]

u only need two sides

one is given DE

other is AE - BD = 9 - 6 = 3

[twilit garden]

oooh

yea

god im dumb

thanks

.close

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[plush lava]

The question is to calculate the integral. Any help is appreciated. Thank you

[plush lava]

This is my work until now

[stray socket]

Another u sub

But wait

You didn't replace dx correctly either

Fix that first

[plush lava]

ahh

yeah

I see

Does it correct now?

[stray socket]

Yeah

[plush lava]

So, what I should do next?

[stray socket]

Recall that $\sin(2x) = 2\sin(x)\cos(x)$

[flat frigate]

Umbraleviathan

[stray socket]

Therefore you can simplify sin^2(x)cos^2(x)

[plush lava]

ahh

I see

[stray socket]

And then there's the double angle identity

A little bit of torturous handiwork

[plush lava]

Yeah, let me try solving that

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[severe fox]

Hello, I just want to know if what I said for a physics question is correct.

[severe fox]

I said the following: Engineers design roller coaster loops using clothoids instead of circular loops because clothoids require less velocity at the top as the radius is smaller. Because of this, the centripetal force to complete the loop is less, and the G force is less. Once they reach the bottom, there is a higher velocity, which makes the ride more fun, but the low G force at the top also reduces the risk of injury.

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@severe fox Has your question been resolved?

[buoyant shadow]

too hard

too much stuff going on at the same time

if we assume that the entering velocity is the same, there's more G at the top in the second picture

both more curved and faster

or it's slower?

ok i've no clue

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@severe fox Has your question been resolved?

[safe radish]

@severe fox Has your question been resolved?

[safe radish]

@severe fox Has your question been resolved?

[gritty dawn]

Pretty sure you would find help in a physics discord help

[safe radish]

@severe fox Has your question been resolved?

[misty bay]

this is incorrect

at the top of the loop, if you've designed your roller coaster with absolutely no safety margin, the riders should feel like they're in freefall

the radius of curvature is smaller at the top, because it increases the required centripetal acceleration for a given speed, and therefore makes it easier for the roller coaster to complete the loop without dropping right off

[severe fox]

oh okay, but there is less g force at the top which then makes u feel like youre falling down. but how can i implement that clothoids decrease injury (like breaking your spine), is it because the acceleration downwards lessens the stress on your bones?

i mean like

[misty bay]

they don't

for a given speed, you experience less acceleration in a circular thing than in that clothoid thing

[severe fox]

so since the radius is smaller, you dont have to suddenly increase the speed so you can complete it, so since you dont have to increase it and it is easier to clear the loop, then the speed would cause less strain on the muscles and bones

[misty bay]

no

[severe fox]

oh

[misty bay]

I suppose you would have less strain on the bottom and sides, but all you need is a mere 2g at the bottom to successfully go over the loop

and 2g is very low for a roller coaster anyway

[shadow cargo]

How do people get this low

Actually how did it not get timed out

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[silent narwhal]

Do you know your log rules

[lean finch]

This is my work

No my teacher didn’t teach it to us

[silent narwhal]

I think you could simple pull the squared out in front of the log instead of distributing it

Would be a simpler expansion

But your still correct

[lean finch]

Oh okay

Thank you

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[echo yoke]

I need some help on what probably is a simple calculus and vectors assignment. I will send a pic here asap.

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[echo yoke]

<@&286206848099549185>

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[echo yoke]

.reopen

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✅

[river wharf]

@echo yoke

that drawing is a mess

[echo yoke]

Hi

[river wharf]

you got a picture of the problem without the lines you dreW?

[echo yoke]

Yeah gimme a sec

acctually give me a minute or 2

mb

https://docs.google.com/document/d/1MtAIUiFvrThpwtrsFG-0xRnUwTorQbZYLXSEq9m8mCo/edit?usp=sharing

Does this link work?

Nevermind, here

@river wharf

[echo yoke]

dammit i missed my chance lol!

Reminds me of the one episode of Spongebob where he kept missing the nightbus

[river wharf]

no i was helping someone else

lemme see xd

@echo yoke okay what you need help with ?

your drawings seem appropriate as derivates to the image

[echo yoke]

really?

So do u see the blue function, the original f(x)?

I didn't complete that

The grey one is f'(x)

The red one is f''(x)

[river wharf]

@echo yoke third part to me seems like a third power function

x^3

[echo yoke]

Are you talking abt f'(x)?

How do I complete f(x)?

@river wharf

[river wharf]

yes it looks like x3

[echo yoke]

So what would f(x) look liek lol?

That's what i need help with cuz u said everything else seemed fine

Don't get how to do f(x), the blue one.

@river wharf

[river wharf]

automaticc

[echo yoke]

I'm abt to fall asleep

yea

[river wharf]

problem says that the graph represents a derivative

so you integrate to get fx

[echo yoke]

yeah

[river wharf]

which is x^4

the green on the image i sent

[echo yoke]

Yeah but there's another portion

for f(x)

I did 1/3

yk what im saying

pls help

[river wharf]

Red section f'x seems like x^2, yellow part f'x seems like -x^2, blue part seems like f'x= x^3

derive for f'' integrate for f

[echo yoke]

idk what integrate means dude

we haven't laernt that yet lol

[river wharf]

how do you find the potential original function

you cancel the derived function

[echo yoke]

Bro we haven't learnt that stuff yet

idk what ur talking about

we're literally just graphing derivatives at this point

[echo yoke]

i understand this, obviously

[river wharf]

okay

what function has it's derived function like this:

[echo yoke]

parabola

x^2

[river wharf]

perfect

so here you literally integrated this function

that's what integration means

[echo yoke]

okay..

understood

[river wharf]

anyway here do the same thing in your mind you did for the blue graph i did to find the original functions of 3 sections

[echo yoke]

Yeah I understand what the original functions would be but drawing them makes no sense

for the 2nd f(x), the inflection point occurs at the maximum of the parabola, how am i supposed to draw a cubic function in that space?

u get what im saying?

[river wharf]

oh so you are saying drawing them won't make a good repesentive line

[echo yoke]

ig?idk

[river wharf]

tbh man i never encountered a question like this one where you need to guess derivatives and integrations specially when you don't know what integration is

it's a dumb question to ask

[echo yoke]

lol

[river wharf]

what education you in right now?

[echo yoke]

gr 12 lmao

calc and vectors

got a new teacher so he's doing things "differently "

[river wharf]

so you in calc and you haven't encountered integration yet?

[echo yoke]

than the rest

[echo yoke]

up to now we've just been drawing derivatives and we just got introduced to limits

whil eback

[river wharf]

i mean this problem is a good way for students to create an idea what second order derivative function represents but it's not a proper problem with proper solution

[echo yoke]

yeah idk what's going on

[river wharf]

like i can literally find a tan function that looks similar like that and say it looks like one

either way the sharp turning point that touches graph y

[echo yoke]

yaeh..

[river wharf]

imagine deriving it

it's not a flexion it's a sharp point

[echo yoke]

yeah i already understand it's a x^4 function

[river wharf]

not it's not as a whole this function would be all over the place with powers and other terms 💀

so i don't really get what prof is trying to get

[echo yoke]

man im so tired

[river wharf]

but if you just split it

to 3 sections

it's a bit easy ig

[echo yoke]

h o w t o f i n d f ( x ) ? ?

Maybe im just stupid

been here since 7

cuz i can't do shit mysself

i had 3.5 hrs of sleep

😭

Think im just gonna hand in waht i have

[river wharf]

nah

it's just a pointless problem ngl

[echo yoke]

It's a unit assignment lol

yeah...

idk either mate

Instead of a unit test we're having this....idk what's going on with that teacher

anyways thanks ig, i handed it in, it's over lol

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[thorny dock]

is this a typo?

[thorny dock]

can a vector $v$ really come from a field?

[flat frigate]

normalAtmosphericPa=101,325

[thorny dock]

I don't see how vector multiplication is defined (cross or dot product? and isn't (cross product non commutative abelian)?

[final halo]

A vector space is not a field if that's what you're asking

The scalar multiplication is from elements of a field

[thorny dock]

yeah

but above it says:

that's a typo right?

[final halo]

Why would that be a typo?

[thorny dock]

$v$ is just a scalar in this context?

[flat frigate]

normalAtmosphericPa=101,325

[final halo]

There is no mention of vectors at all in the picture you posted

[thorny dock]

I thought it meant like a geometric vector liek: $\rightarrow$

[flat frigate]

normalAtmosphericPa=101,325

[thorny dock]

I don't like this notation

[final halo]

1. it's a $\nu$ not a v technically

[thorny dock]

ok thanks

[flat frigate]

ΣAC

[final halo]

And second even if it was a v, the letter v does not automatically mean its a vector

You can use letters for whatever you want them to mean

[thorny dock]

yeah

oops

thank you!

.close

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[slow ivy]

Can someone please help me with this? I do not understand on what to do, I was out of school when our teacher taught this and I'm lost 😓

[grim plover]

Easy enough

its based on transversals

do you know what a transversal is?

@slow ivy

[slow ivy]

yes

[grim plover]

Can u think of the reason why SR, MD and PQ are all parallel? and the transversals intersecting these 3 lines?

Dont tell me the reason btw, just say yes if yk

[slow ivy]

No

[grim plover]

Okay, u should atleast know why SR | PQ

[slow ivy]

they're in the opposite of eachother ?

[grim plover]

nope

its a trapezoid

/trapezium

A trapezium has one set of sides parallel

[slow ivy]

Ahh okay

[grim plover]

an isosceles one is same, just that the two non-parallel sides are equal

however that doesnt matter to us much in this case I believe

anyways

by corresponding angles

can u list out the equal angles

[slow ivy]

spq, rqp
smd, rdm

[grim plover]

mhm

how would you find the 4th missing angle PMD, in a trapezoid PMDQ?

[slow ivy]

is the same as its equal angle??

[grim plover]

no, what is the sum of all angles of a quadrilateral

[slow ivy]

360

[grim plover]

if ur given sum of 3 angles = x

what would the fourht angle be

[slow ivy]

90

[grim plover]

How?

[grim plover]

that would be the case if x = 270, no?

[slow ivy]

yes

[grim plover]

So in general we write it as 360 - x, yes?

[slow ivy]

yes

[grim plover]

also this aint right

[slow ivy]

🥲

[grim plover]

RDM = DPQ, can u think why ? (yes/no)

[slow ivy]

No

[grim plover]

similarly why SMD = MPQ?

[slow ivy]

they're parallel

??

[grim plover]

corresponding no? ignore the top lines, just MD and PQ, and the transversal, they look corresponding no?

[grim plover]

corresponding angles...

[slow ivy]

ohh🥲

[grim plover]

lastly what would MDQ be? (hint, linear pair)

[slow ivy]

Rdm?

ahh wait noo

[grim plover]

yk how linear pairs work?

[slow ivy]

nooo

[grim plover]

Bruh

two angles which lie on a straight line, and thus sum up to 180 deg, are in a linear pair. so if ur given one, say y, the other one would 180-y, yes?

[slow ivy]

sorry, i did not get to learn these because i transferred to school late 🥲

[grim plover]

its okay

[grim plover]

understand this

[slow ivy]

Yes

[safe radish]

@slow ivy Has your question been resolved?

[safe radish]

@slow ivy Has your question been resolved?

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@slow ivy Has your question been resolved?

[safe radish]

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[lean otter]

This is my question

My logic

But people are saying there should be 9 cases only are they right or am I right pls tell ??

Can someone pls help fast I am in a hurry <@&286206848099549185>

[inner root]

I’m here to help

So I don’t know an equation for it although I’m sure that their is one. I just don’t know it but this should be simple enough.

Since we only need to match up 2 variables

@lean otter

Essentially you just want to go in order and keep it organized to avoid mistakes.

[lean otter]

I have given my logic

Here

Pls tell my answer is correct then I will challenge it to cbse board

I need confirmation on this

[inner root]

Ok

So their is only none

Nine*

[lean otter]

Why not 14

Yeah 9 cases only

[inner root]

Where your logic is wrong is that GB is the same as BG

[lean otter]

But

Take for example a dice

We count the possibility as

See listen

3,5 as seperate

And 5,3 seperate

You are spin the both spinner together

No

You do

We r spinning one at a time

[lean otter]

Read it is written

That means one spinner rotates one time

It's your understanding

But as per the question

The spinners rotate together

Eh?

It's mentioned each spinner rotate one time

[glad whale]

I think the problem is thinking of it as 2 different cases

[lean otter]

Yeah he/she is

[lean otter]

Exactly you spin each spinner one time

So it's valid to rotate 2 together

Or else tell me who will you able to form combinations

I mean

My logic is in one case I spin spinner 2 first and in one case I spin spinner 1 first

So accordingly I have different cases

Mentioned above

[glad whale]

The spins are not dependent on each other

[lean otter]

Yeah listen so one time you rotate both the spinners at a time

Yea

[glad whale]

So they shouldn't be treated as different cases when one is spun before the other

[inner root]

We see your point with their being 14 outcomes but the message I am replying to is the reason why GY and YG are the same thing. Since you are spinning each spinner once it doesn’t matter which one goes first if you get G on one and Y on another. With dice it’s different because you roll more than once.

[lean otter]

Ahk

[lean otter]

Does this violate the given condition

[lean otter]

No I told it's 9, ig wrong mention

[inner root]

I was agreeing

[lean otter]

OK I got it

Then I would get I guess at least 1 mark for attempting

[inner root]

I was just replying to that message to show that you had said that. Credit where credit is due.

[lean otter]

My 3 marks gone

Are you contradicting me or supporting @inner root

😭

[inner root]

Supporting, I replied to your message to give you credit for saying the idea first not me.

[lean otter]

No worries brother hopefully you won't repeat such mistakes , try to think of how can you take possible combinations

[lean otter]

Ah OK lol I didn't get that sorry

[inner root]

No problem.

[glad whale]

.close

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[tawdry magnet]

I feel dumb for not knowing this but do you multiply in the parenthesis or outside

[tawdry magnet]

ignore the bad writing skills its on my pc

[vague raptor]

use the distributive law $a\cdot(b+c)=a\cdot b+a\cdot c$

[flat frigate]

ImToxiiq

[tawdry magnet]

so it would be 10/7?

or "6+y=-5/7x+10/7"

[vague raptor]

yup

[tawdry magnet]

then you subtract

so "y=-5/7x+ -32/7

[vague raptor]

correct

[tawdry magnet]

tysm

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[tired jacinth]

Hello everybody. I need help with a Probability problem, namely I'm not sure how to calculate the part b) properly?

[tired jacinth]

my idea was to split up my two ""conditions"", but I don't know if I am allowed to even do that

The solutions are as followed (I'm again not sure how to get to both or if solution for a is wrong?)

[compact ferry]

<5

P(even) = 7/15

and keep in mind a to solve b

[tired jacinth]

ahh I forgot the 10

[tired jacinth]

but my answer differs for "a" to the solution as well, what am I overlooking?

[compact ferry]

they mistyped

[tired jacinth]

ahh okayy

[compact ferry]

its 2/15

[tired jacinth]

thank you so much I understand now

because I need to P(even or less than 5)=P(even)+P(<5)-P(even and <5)

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[fluid condor]

(rational functions) i need help on knowing when some factor is a hole or some factor is a vertical asymptote

[fluid condor]

i know that when top has a lesser exponent it is shared factors but for rational functions with oblique asymptotes i am unsure

here is the particular problem i am working on

f(x) = (x^3 - 3x^2 + 2)/(4x^2 - 24x + 32)

wait nvm im dumb

.close

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[fluid condor]

.reopen

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✅

[fluid condor]

i thought i figured something out but nvm 😭

so i long div to f(x) = x/4 + [(x^3 - 3x^2 + 2)/(4x^2 - 24x + 32)]

and that gets me an oblique asymptote of y = x/4 but the answer key says it is not a oblique asymptote

and factoring (x^3 - 3x^2 + 2)/(4x^2 - 24x + 32) gets me (3x)(x-2)/(4)(x-2)(x-4)

where x = 4 is a vert asymptote

and i think x = 2 should be a hole bc it shares factor with numerator

however the answer key says x = 2 is another vert asymptote

and now i am just very confused

update: so now i get why my oblique asymptote is wrong and its bc i didnt divide enough

and new obl asym is y = (x + 3)/4

[patent vine]

how did you get this?

[fluid condor]

factoring

[patent vine]

,w factor x^3 -3x^2 + 2

dang

bot's still down

[fluid condor]

oof

wait no my factoring is wrong

but i dont think it matters bc i did polynomial long division wrong and so i redid it

and i got

f(x) = [(x + 3)/4] + [(12x + 18)/(4x^2 - 24x +32)]

oh my godd im so dumbb

uhh thank you for the hekp

[patent vine]

I think you're overthinking it

[fluid condor]

no so i just did it wrong

[patent vine]

okay

[fluid condor]

thank you and sorry for wasting your time

[patent vine]

nah you're good

[fluid condor]

.close

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[narrow vault]

how do me make the conclusion that sec^2a=2 or cosa=+/- 1/sqrt(2)?

[narrow vault]

of course i understand that y'=sec^2 a

I would appreciate if someone went through the logic with me!

[bleak pendant]

Basically for y= mx , where m is slope and slope has to be same for parallel.
Since y = 2x.
So the slope of Tanx is also taken as 2

[narrow vault]

because they are paralell

of course..

thank you..

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[narrow vault]

.reopen

but

[safe radish]

✅

[narrow vault]

how do we get a=pi/4 from sec^2a=2?

the code doesnt seem to work

[dusty fiber]

$\sec^2a=2$

must be down

[narrow vault]

can you guys still help me out?

how does this hold? isnt 2^2=4?

oh i get it

that is a pretty tidious derrivation

but thanks

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[narrow vault]

what was done here?

[bleak pendant]

Both numerator and denominator were divided by 2 .
And the powers of both terms was taken common

[narrow vault]

"And the powers of both terms was taken common" how is this done?

@bleak pendant

nervermind

but the 1/2 in the beginning...

[bleak pendant]

@narrow vault both terms were divided by 4 not 2

I was incorrect

[bleak pendant]

Try dividing by 4, you will then then see

[narrow vault]

Still I don’t get h/2 in the denominator

[bleak pendant]

Ok wait

Divide by 4
You will get
[2sin^2(h/2]/(4h^2/4)

It will reduce to
sin^2(h/2)/(2h^2/4 )

(This is where that 1/2 came from)

Now convert 4 = 2^2

sin^2(h/2)/2(h^2/2^2 )

Take all the powers of 2 into one bracket

1/2×(sin(h/2)/(h/2)^2

[narrow vault]

I dont understand why you have a second /4

shouldnt it be : Divide by 4
You will get
[2sin^2(h/2]/(4h^2)

how did you get (4h^2/4) by dividing by 4

[bleak pendant]

I'm dividing both numerator and denominator by 4

the value has to stay the same right

that is why, both numerator and denominator are divided by same number

[narrow vault]

Isn’t this just dividing by 4?

[bleak pendant]

we are dividing both numerator and denominator by 4.

The value has to stay constant right

[narrow vault]

okay makes sense now

thanks for the patience

!

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[lean otter]

I just did this question, following the steps in the picture below this message.

The answer I got is different from the one on the answer key.

So, could someone look over my work and help me understand where I went wrong?

I have shown my work in blue.

[junior smelt]

,calc 68/4

[flat frigate]

Result:

17

[junior smelt]

https://tenor.com/view/theyre-the-same-picture-the-office-pam-the-office-us-gif-20757621

[junior smelt]

(In particular use surd rules to manipulate your answer into theirs)

[lean otter]

So my answer just isn’t simplified ?

@junior smelt

[junior smelt]

Correct

sqrt{68} is 2sqrt{17}

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[wind abyss]

Hi am i on the right track or im doing wrong

[wind abyss]

thanks!

for question #11

[drifting elk]

you can't just replace 20 with sqrt(4) * sqrt(5)

that's sqrt(20), not 20

[wind abyss]

O

ill fix that

is it best to just use a quadratic formula

[drifting elk]

you don't need it here

you can just divide both sides by 15

so y^2 = 20/15 = 4/3

and now you use the provided property

[wind abyss]

wow im gonna practice more because that took too long

i got :

so there is rationalizing denominator in this as well

?

[drifting elk]

is that a question?

your answer seems fine

apart from the fact you are missing =

[wind abyss]

ah

Alright i have practice problem similar to this so thank you and if i get stuck ill do a chat

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[hasty olive]

.help

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Commands:
clopen: .close, .reopen, .solved, .unsolved
consensus: .poll
factoids: .tag
help: .help

Type .help <command name> for more info on a command.

[hasty olive]

):

[hasty olive]

can someone help me with this

I dont hab any idea tbh

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@hasty olive Has your question been resolved?

[hasty olive]

sigh

[mortal sandal]

Do you have a calculator or a z-score table somewhere

calculator with normalcdf

otherwise if 1.54 is a special value I don't know it

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[granite cobalt]

is this correct? Is there an alternative way to do it

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@granite cobalt Has your question been resolved?

[granite cobalt]

<@&286206848099549185>

[spiral bane]

,rotate

,w 1+1

bot still down 😔

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[drifting prism]

(2, 4) + (2, 8) = (4, 12)
(4, 12)/2 = (2, 6)
((2+5)/2, (6+6)/2) = (3.5, 6)
A = 1/2 * base * height = 1/2 * (5-2) * (8-4) = 6
V = A * d = 6 * 6 = 36 cubic units

[drifting prism]

idk how its wrong

i tried regular 36 too

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@drifting prism Has your question been resolved?

[tall bough]

its 6 x 6 x 2pi

u forgot 2pi

[drifting prism]

so 72pi?

[tall bough]

yers

wait no

[drifting prism]

uh

alr

it didnt work

[tall bough]

y is from center of triangle to x-axis

nvm y will be still 6

[drifting prism]

yeah

ive been lost

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@drifting prism Has your question been resolved?

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@drifting prism Has your question been resolved?

[tall bough]

@stable estuary

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[fickle dew]

Would i just solve normally

Im stuck bc idk what to do with the top two values

[icy lance]

you very literally just do a-b and see what you get, a=3x+4 and b=2x-1

[fickle dew]

So wouls i do 3x+4-2x+1

I got x + 5 but is that correct

[icy lance]

yes

[fickle dew]

So would th awnser be a?

[icy lance]

yeah

[fickle dew]

YAYYY

Thank you sm

❤️

[steel spruce]

Solve the derivative of y = arc csc x^1/2

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[fickle dew]

.close

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[strong warren]

I am not sure how to continue for this question because I have that I need to choose z such that $e^t((2z_3-z_2)\begin{bmatrix} \frac{\cos(t) - \sin(t)}{2}\cos(t) \cos(t) \end{bmatrix} + (2z_1-5z_2+4z_3)\begin{bmatrix} \frac{\sin(t)+\cos(t)}{2}\ \sin(t)\ \sin(t)\end{bmatrix})$ goes to 0 as t goes to infinity. Below is what I have tried so far for this question:\\
\newcommand{\C}{\ensuremath{\mathbb{C}} }
The characteristic polynomial is $$-\lambda(\lambda^2-4)+7(2\lambda+4)+8(-4-2\lambda)=-\lambda^3+2\lambda-4.$$
Equating it to 0 gives
\begin{align*}
\lambda^3-2\lambda+4 &= 0\
\lambda = -2 \lor \lambda = 1 \mp i
\end{align*}
The eigenvectors are:
\begin{itemize}
\item $\lambda = -2 \implies \begin{bmatrix} 3\2\1\end{bmatrix}$.
\item $\lambda = 1 + i \implies \begin{bmatrix} \frac{1+i}{2}\1\1\end{bmatrix}$.
\item $\lambda = 1 - i \implies \begin{bmatrix} \frac{1-i}{2}\1\1\end{bmatrix}$.
\end{itemize}
The general solution is
\begin{equation}
u(t) = C_1e^{-2t}\begin{bmatrix} 3\2\1\end{bmatrix} + C_2e^{(1 + i)t}\begin{bmatrix} \frac{1+i}{2}\1\1\end{bmatrix} + C_3e^{(1 - i)t}\begin{bmatrix} \frac{1-i}{2}\1\1\end{bmatrix}
\end{equation}
with $C_1, C_2, C_3 \in \C$. Note that
\begin{align*}
e^{(1 + i)t}\begin{bmatrix} \frac{1+i}{2}\1\1\end{bmatrix} &= e^t(\cos(t)+i\sin(t))(\begin{bmatrix} \frac{1}{2}\1\1\end{bmatrix} + \begin{bmatrix} \frac{i}{2}\0\0\end{bmatrix})\
&= e^t(\begin{bmatrix} \frac{\cos(t) - \sin{t}}{2}\\cos(t) \\cos(t) \end{bmatrix} + i\begin{bmatrix} \frac{\sin(t)+\cos(t)}{2}\ \sin(t)\ \sin(t)\end{bmatrix})\
\end{align*}
The real representation of the general solution is
\begin{equation}
u(t) = C_1e^{-2t}\begin{bmatrix} 3\2\1\end{bmatrix} + e^t(D_1\begin{bmatrix} \frac{\cos(t) - \sin(t)}{2}\\cos(t) \\cos(t) \end{bmatrix} + D_2\begin{bmatrix} \frac{\sin(t)+\cos(t)}{2}\ \sin(t)\ \sin(t)\end{bmatrix})
\end{equation}
with $C_1, D_1, D_2 \in \C$.

[flat frigate]

FrankF

[strong warren]

\newcommand{\R}{\ensuremath{\mathbb{R}} }
Filling in $t = 0$ gives
\begin{equation}
\begin{cases}
3C_1+\frac{1}{2}D_1+\frac{1}{2}D_2&=z_1\
2C_1+D_1&=z_2\
C_1+D_1&=z_3\
\end{cases}
\end{equation}
Solving it gives
\begin{align*}
D_1 &= 2z_3-z_2\
C_1 &= z_2-z_3\
D_2 &= 2z_1-5z_2+4z_3
\end{align*}
Since, $z \in \R^3$, we know that with $C_1, D_1, D_2 \in \R$. Plugging it in the general solution gives
\begin{equation}
u(t) = (z_2-z_3)e^{-2t}\begin{bmatrix} 3\2\1\end{bmatrix} + e^t((2z_3-z_2)\begin{bmatrix} \frac{\cos(t) - \sin(t)}{2}\\cos(t) \\cos(t) \end{bmatrix} + (2z_1-5z_2+4z_3)\begin{bmatrix} \frac{\sin(t)+\cos(t)}{2}\ \sin(t)\ \sin(t)\end{bmatrix})
\end{equation}
When $t \rightarrow \infty$, $e^{-2t} \rightarrow 0$, so the first term of $u(t)$ vanishes.

[flat frigate]

FrankF

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@strong warren Has your question been resolved?

[strong warren]

<@&286206848099549185>

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@strong warren Has your question been resolved?

[strong warren]

tldr: How does stability of ode depend on the initial conditions?

[flat frigate]

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

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@strong warren Has your question been resolved?

[safe radish]

@strong warren Has your question been resolved?

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[dusty talon]

Ah

[dusty talon]

what can be combined on the bottom?

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@dusty talon Has your question been resolved?

[dusty talon]

no

[safe radish]

@dusty talon Has your question been resolved?

[dusty talon]

no it has not

thanks for nothin chump

.close

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[green nebula]

Hallo

[plucky elk]

,rotate

[green nebula]

Ahhhhh

the formula is. 1/3 x h x pi, r^2

so

165 = 1/3 x h x pi x 4^2

idk wat to do after that

i know i’m supposed to divide but i don’t know what to divide by

[plucky elk]

start by multiplying both sides by 3, then divide by pi.

then there will be one more step to isolate h

[green nebula]

so 165 x 3

495 = h x pi x 16

so what now

do i multiply pi and 16?

495 = h x 16pi

?

Ohh ok

do i just divide 495 by 16

Okay

it is

30.9375

so 30.9375 = h x pi

[junior smelt]

,calc 495/16

[flat frigate]

Result:

30.9375

[green nebula]

Okay

but what happens to pi

[junior smelt]

Yea, what could we do to get rid of it

[green nebula]

Subtract it

on both side

?

Hmmmmm

divide 30.9375 by pi?!

can i do it by 3.14

9.8527070064 is the answer using 3.14

9.84771210381

answer using pi

Okay

so now i have to round

[junior smelt]

I would try to use pi and not round too early tbh

[green nebula]

so it will be 9.85

Yayyayayayayayayyaya

can you help me with a few more if you don’t mind?

A

[junior smelt]

Any thoughts on how to do it?

[green nebula]

Ammm

first we have to calculate the volume of the cone

[junior smelt]

And that would be?

[green nebula]

let me see

the volume is 3?

3pi

[junior smelt]

How did you get the very very last step?

[green nebula]

Ammmmm

I removed pi it was dumb

[junior smelt]

Happens sometimes

[green nebula]

Unfortunately

Okay

So now we calculate the sphere volume

I think i may be wrong on this one

[junior smelt]

Well the first line is fine

[green nebula]

Hm

[junior smelt]

How did you get from first to second?

[green nebula]

i cubed 1.5

[junior smelt]

Guess what, I’m

[green nebula]

[junior smelt]

,calc 1.5^3

[flat frigate]

Result:

3.375

[junior smelt]

[green nebula]

Okay

[junior smelt]

Second line

[green nebula]

So the second line is okay

so where did i go wrong on the 3rd line

Oh wait i think i see

do i divide 3.375 by 4/3

Oh

So do i multiply pi and 3.375

So i do it right

[junior smelt]

Look at how the reacts progressed

[junior smelt]

In here

[green nebula]

Hm

so

i feel like it might be wrong because the volume would end up being less than pi

3.375pi/4/3

=0.88357293382pi

[junior smelt]

Don’t divide it

[green nebula]

Multiply

?!?!?!??!?!?!?

[junior smelt]

Yeeeaaaa

[green nebula]

14.1371669412

[junior smelt]

,calc 3.375*(4/3)

[flat frigate]

Result:

4.5

[junior smelt]

,calc 4.5*pi

[flat frigate]

Result:

14.137166941154

[junior smelt]

But note how they want the answer in terms of pi

[green nebula]

Oh

So it would be 4.5pi?

[junior smelt]

Yep

[green nebula]

Okay

I need help with one more

it say it wrong

[junior smelt]

Why the sullies

[green nebula]

Oh wait

we have to find the difference

of extra soap

it takes 1.5pi more soap

Okay

wee have completed

it

[junior smelt]

[junior smelt]

And now, for the final one

[green nebula]

A

okay

the volume is expressed in cubic feet i would assume right

It wail take 63 minutes

YESSS

[junior smelt]

Was away and yet you managed to get it

[green nebula]

THANK YOU SO MUCH

[junior smelt]

[junior smelt]

I mean you did most of the work by yourself, I was just sneakily watching! Well done

[green nebula]

I hope you have the best rest of the week ever

.close

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[junior smelt]

You too ❤️

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[fair oasis]

how would u solve this

[dull sequoia]

Which part

[fair oasis]

b and c

[dull sequoia]

With a point and gradient you can define a line

[fair oasis]

how ?

[dull sequoia]

What is the general equation of a straight line?

[fair oasis]

y=mx+c

[dull sequoia]

How many unknowns are in that

[fair oasis]

2

[dull sequoia]

And you have 2 pieces of information

[fair oasis]

oh

do you just substitue

it in

[dull sequoia]

Yes

[fair oasis]

wiat

im kinda confsued

how would u do c then

[safe radish]

@fair oasis Has your question been resolved?

[safe radish]

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[lean otter]

How would I go about solving this question?

First thing I did was find the inverse of the function:

[worthy hemlock]

You should recall functions and their inverses

[lean otter]

what is an inverse

[worthy hemlock]

f(f^-1(x)) results in a very simple solution

[lean otter]

opposite

or reverse

Would the answer be b(3)?

I inputted 3 into the inverse function and I got 0

then i substituted 0 into the original function and got 3

[worthy hemlock]

That's right

But there was a much easier way

[lean otter]

like?

[worthy hemlock]

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

[lean otter]

by defenition, f^-1 is something that f(f^-1(x)) = x and f^-1(f(x)) = x

.close

[safe radish]

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[vale hollow]

Need help with calculus multiple choice questions.

[vale hollow]

<@&286206848099549185>

[spiral bane]

many things

!status

[safe radish]

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

[spiral bane]

!show

[safe radish]

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

[spiral bane]

!15m

[safe radish]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[vale hollow]

I just want confirmation if those r the only correct options

I have chose them based off prior definition memory

[safe radish]

@vale hollow Has your question been resolved?

Channel closed

Closed by @vale hollow

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[safe radish]

Available help channel!

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Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[mint bluff]

When I integrate/differentiate a power series, its the same series but what happens to the bounds like n +-1?

[plucky elk]

Show an example

[mint bluff]

I dont really have one its just like

dont the bounds change of the series when u integrate/differentiate

[plucky elk]

Make one up

[mint bluff]

uhhhhhhhhh

ok lulw

series of 2(x-3)^n from 0 to infinity

if i take the derivative

is it now from 1 to infinity

[plucky elk]

Where's the series index

You just gave one term

[mint bluff]

oops

i changed ^2 to ^n

n = 0 to infinity

[plucky elk]

Differentiate term by term

The first term is a constant

What's the derivative of a constant?

[mint bluff]

https://math.stackexchange.com/questions/1737697/when-taking-derivatives-of-power-series-why-do-we-shift-the-index-up

i found the answer

.close

[safe radish]

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• Show your work, and if possible, explain where you are stuck.
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[wide jetty]

If anyone is willing to check my answer that'd be much appreciated! 🙂 also does c need to be /x^5 or can I leave it as + c

[safe radish]

@wide jetty Has your question been resolved?

[wide jetty]

<@&286206848099549185>

[last wren]

In your fourth line, it should be 5x^4, not 5x^6

Since x is in the denominator

[wide jetty]

this part?

[last wren]

Yep!

[wide jetty]

Sorry that's my hand writing it is 5x^4 * y

xD

[last wren]

Oh LOL okay lemme check again

[wide jetty]

Haha i know it's awful, broke both wrists multiple times so i've never been neat XD

[last wren]

Okay so are you just checking your work? This looks correct to me

[wide jetty]

Yeah 😄

Thank you very much

also

C/x^5

should i just write it as C?

or does it need that division on it

[last wren]

Oh well c is a constant, so just keep it how it is and you're good. The x^5 in the bottom there is important if you have an initial condition

[wide jetty]

So it would be incorrect to leave it as +c on it's own?

[last wren]

It would yes

[wide jetty]

I vaguely remember learning something about c being arbitrary or something so it stays as just C

That's the only reason I ask

[last wren]

You can only rewrite stuff as another constant if EVERYTHING it is multiplied by is a constant

If you had like 3C at the end

[wide jetty]

Ohhhhhhhhhhhhhhhh

[last wren]

You could just rename it as C

[wide jetty]

So for division its important

[last wren]

Or multiplication yes, if you had like Cx^2 you can't just Absorb the x^2 into the C

Since x^2 isn't a constant, it can change

C represents a constant, a value that never changes

[wide jetty]

So it's to do with the x that causes it to have to stay

[last wren]

Yes, if you have no x attached to it you can do all that junk you said lol

[wide jetty]

so say for instance I had 5y=15+c --> y=3 + c because there is no x

wow

you taught me something very valuable right there!

[last wren]

Lol yes that's correct

[wide jetty]

Much appreciated!

[last wren]

Happy to help :D

[wide jetty]

What does G+ mean

out of interest in your roles?

[last wren]

Oh

I just got here like 10 mins ago lol, I think it means graduate pending

Im a grad student but I need to confirm it or something weird

[wide jetty]

Ohhh! What level of study did you achieve

Yeah I seen that when I joined haha

[last wren]

Oh I'm a master's student in math rn haha

Wayyyy easier and faster than PhD, not doing that lol

[wide jetty]

Definitely trust your word then XD

I'm undergrad in comp sci currrently

[last wren]

Ooooo that's exciting, what year are you

[wide jetty]

1st! XD

[last wren]

I considered comp Sci but the teachers make or break it lol, it's so risky

[wide jetty]

Though i am a mature student cause of covid

so I'm 21 in 1st year XD

[last wren]

Oh wow well if you're in Differential Equations first year that's super solid

[wide jetty]

I think they're preparing us for the worst

[last wren]

OHHH yeah covid ruined like everything lmao, that sucks

[wide jetty]

XD

legitttt

But i got work experience n stuff before uni so it was good in some way

[last wren]

Oh yeah that's amazing on resumes and stuff

And hopefully you got paid for it lol

[wide jetty]

Oh yeah healthy amounts thankfully

XD

Well Imma do the next question, I'm sure you'll see it in the help section in like 10 minutes but thank you for your help! 😄

.close

[safe radish]

Channel closed

Closed by @wide jetty

Use .reopen if this was a mistake.

[last wren]

Lol well happy to help, I'll probably be sleeping by then so good luck :)

[safe radish]

Available help channel!

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Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• 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.

[atomic notch]

Can someone please walk me through this

[atomic notch]

I don’t understand how to create a normal distribution graph using just mean and sd

[dull sequoia]

how many SD is 265 and 279 away from the mean

use z score to figure out the proportion of the %