help-38

[candid sequoia]

no i know the answer is -1 but the answer my teacher wrote was different

[trim lichen]

no the answer is not -1

(-1)^2 is 1, not -1

[candid sequoia]

well she ticked -1, -1

maybe i need to ask her

thanks

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[fluid jay]

can someone tell me what im doing wrong here???

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[fluid jay]

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[fluid mountain]

Hello,i would like to ask a simple question about Reels Number's

[fluid mountain]

idk how to spell it

damn

basically it says to type these Numbers on the Form in the Question

[fluid mountain]

now in A i am stuck on how to simplify 5 racine 12

now the solution should be 4 racine 3 and 10 racine 3 and 10 racine 3

but i can't simplify 5 racine 12 into 10 racine 3

btw it's french language if anyone wondering

[olive condor]

4 racine 3=2racine 4*3 right?

[fluid mountain]

4*3 will stay like this

but 5*12 should be

10*3

[olive condor]

could u explain what ur supposed to do?

is it not simplification?

[fluid mountain]

idk how to explain in English

💀

all i know i should make these number's look like a*b

[fluid mountain]

damn that's how

thnx very much i was stuck on this one

love ya

close

#close

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[cold aurora]

hi may i know how to know this eqn

[wraith hinge]

Hint: rewrite u = e^x

[cold aurora]

thank you

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[hazy mist]

how do i find the vertex of

[hazy mist]

i know the formula

but ow do i use it for this

[merry bluff]

-1/3x^2 = -1/3x^2 + 0x + 0

but you don't need even need the formula

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[dusk beacon]

can you use the direct comparison test with this series?

The answer key used limit comparison but I don't see why you couldn't just say that thing is less than 1/rad(n)

(when taking the absolute value since it converges conditionally)

nvm

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[molten vessel]

cryptography

[molten vessel]

im having trouble understanding how they got the last line

[wraith hinge]

they just combined the exps?

or u mean like the whole equation from <alph, beta>_~g

[molten vessel]

yea

<@&286206848099549185>

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@molten vessel Has your question been resolved?

[molten vessel]

No

help

<@&286206848099549185>

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@molten vessel Has your question been resolved?

[molten vessel]

help

@tight mural

<@&286206848099549185>

<@&286206848099549185>

how they get the last line

<@&286206848099549185>

<@&286206848099549185>

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[keen plover]

!15m

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[keen plover]

"Ping helpers ONCE"

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[north mirage]

they just combined the exponentials????

$e^ae^b = e^{a+b}$

[solid kiln]

Camilleone

[north mirage]

like that??

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[meager night]

I'd think of it as x and y
x = higher number
y = x - 4

[versed silo]

remember they need to be even

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[astral mortar]

sin^2(x) +2sinxcosx =15cos^2(x) this was a multiple choice where we had 1 minute per question the options are 72 108 79 101 two of these are correct

[astral mortar]

I’m very confused because the angles are not nice

Obviously it’s gonna be 72 and 108 or 79 and 101

Oh wait

Maybe not because of the cos

[upbeat lynx]

add cos^2(x) to both sides

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[bitter crypt]

What do the circles mean in these problems

[night zodiac]

You mean $O$ and $\Theta$?

[solid kiln]

dldh06

[bitter crypt]

Yes

[night zodiac]

That's Big O notation and Big Theta

[bitter crypt]

What do they do

[night zodiac]

https://www.geeksforgeeks.org/difference-between-big-oh-big-omega-and-big-theta/

That should help

[bitter crypt]

Thank you

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[fervent flower]

Hi, What would be the equation for this wave?

[marble wharf]

for example (sin(3x)+1)sin(0.1x)

where you can play around with the 3 and the 0.1 for more/less bumps

[fervent flower]

I get this

[marble wharf]

scroll a bit to the right

[fervent flower]

Oh thanks

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[wraith hinge]

how do i prove that this is an onto function?

[wraith hinge]

such that im(f) = S

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[sudden lagoon]

Hi I've a question in relation to probability theory. Comsideryou have two random variables X1,X2 that both follow an exponential rv with the same rate lambda. Consider then you take the rvs Y1 = X1 and Y2= X2/X1, will this Y1 and Y2 be independent?

[sudden lagoon]

My thinking is that no they're not independent since y2 can be written as x2/y1 but I'd like some clarification

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[zinc ginkgo]

just use the definition of independence

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[sudden lagoon]

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[solid kiln]

7r0lM$5t3r

[fleet halo]

how do i solve this shit

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[fleet halo]

nevermind

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[wraith hinge]

I need help with verifying my answer

[wraith hinge]

So if you do the question like this 1x(0.25) 2x(0.25) for all of them and the chance you’d end up with 29 over 10 then we simplify that by dividing therefore the answer is 2.9

Am I right in that regard

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@wraith hinge Has your question been resolved?

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@wraith hinge Has your question been resolved?

[wraith hinge]

That’s not the expect value, that’s the total value

Are you looking for the expected value or the total?

Value of a random variable

If you’re doing question a then it looks like they want expect value

I guess

But that is the expected value 2.9

Oh wait yeah you’re right

I made a mistake

Nevermind

And then for the second question you gain 50 cent and lose 10 cent

Would that be correct

Yes

2*0.25=0.50

And the expected value is 0.10 below the price to play

I agree with you

Thanks then the rest are pretty easy questions I’ll send now

It’s just basic stuff the rest I’ve completed all of them

But thank you for your help appreciate it

I just wasn’t sure if the first two

All good :)

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[wraith hinge]

is this correct?

[deft drift]

It's correct

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[restive quiver]

b^2x+by=0, bx+b^2y=0
parameter b belongs to a set of real numbers

[restive quiver]

How to solve this systeM?

Its not very clear to me

[gaunt echo]

b is any real number?

[restive quiver]

It says to solve this homogenous system depending on the parameter b

Its homogenous because all equations within it are equal to 0 right?

[gaunt echo]

no, homogeneous means all the terms have the same degree

[gaunt echo]

here there's no x^2, xy, x^3, etc. It's just a single x or y in each term

[restive quiver]

Well yes

So how would i go about solving this?

[gaunt echo]

hmm

maybe elimination?

do you know how that works/

[restive quiver]

Like gauss theorem?

[gaunt echo]

what?

no I mean like

adding the equations together, and canceling a variable

[restive quiver]

Yeah

thats gauss

Right?

[gaunt echo]

gaussian elimination has to do with matrices

I guess this is the same thing

[restive quiver]

Gaussian elimination

Yes

Thats whats its called

[gaunt echo]

well we don't have to do any matrix stuff

[gaunt echo]

if you multiply the bottom equation by -b

then add the equations together

the x should cancel

[restive quiver]

ok yes

wait uh

-by^3

Plus by?

[gaunt echo]

what?

[restive quiver]

Let me take picture of my paper

[gaunt echo]

yeah nice

[restive quiver]

Do i have to use Bezuots identity now?

or wait no

[gaunt echo]

what? factor stuff out maybe

y(b - b^3) = 0

so either y = 0, or b - b^3 = 0

[restive quiver]

I got that y = 0

[gaunt echo]

wait what about b - b^3

[gaunt echo]

you can't just divide both sides by (-b^3 + b) though. What if it's zero? Then you'd be dividing both sides by zero!

so you have to separate it into two possibilities, two cases

[restive quiver]

ok ok

keep going

[gaunt echo]

either we have y = 0

or we have b - b^3 = 0

(or maybe both, but I'm not sure that case will matter)

let's say y = 0

[gaunt echo]

then this turns into the following equations:

b^2 x = 0
bx = 0

so y = 0, x = 0 is a solution

[restive quiver]

If y = 0

[gaunt echo]

well we just figured out the y = 0 case

now we have to see what happens if b - b^3 = 0

[restive quiver]

ok so for b

b1 = -1 b2 = 0 b3 = 1

[gaunt echo]

yep

so now you gotta check each of those cases

I think we should've started with this situation

but whatever

[restive quiver]

So what where do i put them in

-b^3+b=0

i get that

2 = 0

and -2 = 0

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[grave spruce]

Help

[wraith hinge]

Domain is all the possible x values right

[grave spruce]

So what it will be [,]

[wraith hinge]

Possible not positive my bad

So what x values (as a range) are allowed upon the graph

[grave spruce]

Because the graph has a open circle

[wraith hinge]

Open circle means that point is included or excluded?

[grave spruce]

Not included

[wraith hinge]

Okay

So it starts at the point -11, which isn't included itself

And ends at what point

[grave spruce]

11

[wraith hinge]

Which is in/ex-cluded?

[grave spruce]

Yes

Because it’s a closed circle

[wraith hinge]

Yep

So the domain is from -11 (excluded) to 11 (included)

Does your teacher want interval notation

[grave spruce]

So what it will be like this[ -11, 11]

?

No she doesn’t want interval notation

[wraith hinge]

Okay then she wants an inequality?

[grave spruce]

She says do this [,] or >

[wraith hinge]

So interval or inequality will work

[grave spruce]

Yes

[wraith hinge]

So (-11, 11] bc the first is excluded "()" and the 2nd is included "[]"

[grave spruce]

So like this (-11,11]

?

[wraith hinge]

or $-11<x<=11$

[solid kiln]

LigmaStigma

[wraith hinge]

I thought it would do less than or equal to symbol but you get the point

[grave spruce]

Is (-11, 11) acceptable?

[wraith hinge]

No

[grave spruce]

I mean (-11, 11]

[wraith hinge]

The bracket means that the value next to it is included or allowed

The paranrheses means not allowed

Yes that would work

That or the inequality

[grave spruce]

Got it right

Can you do this for me?

My guess is [-9,-4]

Is it right @wraith hinge

Nvm I got it right

How do I do this one? @wraith hinge

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[jade smelt]

Describe the region, $V$, within $x^2+y^2+z^2=4$, above $z=-\sqrt{x^2+y^2}$ and below $z=\sqrt{3x^2+3y^2}$ using spherical coordinates and evaluate $V$

[solid kiln]

RockLEE

[jade smelt]

Im trying to setup the triple integral

so $x^2+y^2+z^2=ρ^2$ so we have that $ρ=2$

[solid kiln]

RockLEE

[jade smelt]

we are going to be integrating the whole 2d circle so theta should be from 0 to 2pi right

so the timepiece kinda region is the one we are trying to find right??>

like up and down cone

theta is angle from z-axis to where?, the x-axis?

oh so its gonna be 0 to pi for theta if we start from z axis

cuz its basically the red vertical outer part

So so far my limits have been 0<ρ<2, 0<phi<2pi, 0<theta<pi

is this looking correct?

i dont know what u mean by that, this is what we use

so its $\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{2}\rho^2\sin(\phi) d\rho d\theta d\phi$

[solid kiln]

RockLEE

[jade smelt]

something feels wrong...

isnt $\int_{0}^{2\pi}\sin(\phi)=0$?

[solid kiln]

RockLEE

[jade smelt]

so isnt the whole thing gonna be 0? cuz im getting 0 and it shouldnt be 0 right???

phi seems to be vertical angle

if i just swap phi and theta it should work out fine right?

u mean 0<=phi<=pi?

so its $\int _0^{\pi }\int _0^{2\pi }\int _0^2\rho ^2\sin \left(\phi \right):d\rho :d\theta :d\phi$ right?

[solid kiln]

RockLEE

[jade smelt]

yeah now im not getting 0 so i think its correct lol

thx a lot!

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[brazen pewter]

can somebody give me an example of (x-h)^2 + (y-k)^2 = r^2? I can't think of a question that can be used using that

[brazen pewter]

as in I mean a test question for example

what would be an example question that I would have to answer using that formula

[digital bison]

What is the equation of a circle centered at (4,2) with radius 3

[brazen pewter]

would that be

(x-4)^2 + (y-2)^2 = 3^2?

[digital bison]

yesa

[brazen pewter]

is there any simplification

or is that just how you would answer it

[stark bison]

Find the equation of the unit circle touching y = 0 and y = x

[brazen pewter]

what

I have no idea how to do that

[stark bison]

Well firstly r = 1, since it's a unit circle

[brazen pewter]

in the origin?

is that what you meant by y = x

[stark bison]

No

[brazen pewter]

oh

[stark bison]

The line y = x

[brazen pewter]

oh

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[swift pawn]

Any tips for how to solve B? I’ve got to factor it

[swift pawn]

Didn’t get the right answer, any takes on what I did wrong? I’m not supposed to have leftovers

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[surreal eagle]

would the answer be y''=0?

this would be equal to 1 at(1,1)

h=x

then this would be equal to 0/1

[zinc ginkgo]

what's h?

[surreal eagle]

h is meant to be x, i just messed up with the calculator

[sharp arrow]

i’m p sure it shouldn’t be 0

cause its similar to hypebola no?

or i’m completely mistaken

in which case

[surreal eagle]

confusion

[sharp arrow]

,w second derivative of x^(1/4)+ y^(1/4) =2

[solid kiln]

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[sharp arrow]

[surreal eagle]

???

what

[sharp arrow]

,w second derivative of x^(1/4)+ y^(1/4) =2 at x=1,y=1

[solid kiln]

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[surreal eagle]

naw jits tripping

[sharp arrow]

[surreal eagle]

i think my second derivative is right

[sharp arrow]

https://www.desmos.com/calculator/ig7xlbqxyd

[sharp arrow]

[sharp arrow]

desmos says 1.5

[surreal eagle]

well shiiiiiit

[sharp arrow]

also this

-1 no?

[surreal eagle]

yea that makes sense

[sharp arrow]

[surreal eagle]

but it would still be 0/1 when subbed into the second deriv that i got

but that doesnt make sense

if the graph says 1.5

[sharp arrow]

,w -3/4 -3/4

[solid kiln]

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[sharp arrow]

check again perhaps

its 1.5

[surreal eagle]

wait how can it y'=plus/minus 1 since if it were -1 then you would get y^(3/4)=(-1^3)^(1/4)=-1^1/4 which isnt possible

[sharp arrow]

what?

[surreal eagle]

you said that y' can be equal to plus/minus 1 at (1,1)

[sharp arrow]

do u get y’ at (1,1) will be -1?

no not plus/minus

only minus

[surreal eagle]

oh but how?

[surreal eagle]

oh shit

yea you right, i was looking at my notes and forgot to carry over the negative

[sharp arrow]

[surreal eagle]

😅

thank you master

[sharp arrow]

so now do u get it?

[surreal eagle]

yea everything makes sense

[sharp arrow]

np

[surreal eagle]

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[silk mauve]

Given that we define $\lim \sup$ of a sequence $(a_n){n=m}^\infty$ as $\inf(\sup(a_n){n=N}^\infty){N=m}^\infty$, and define $\lim \inf$ of $(a_n){n=m}^\infty$ as $\sup(\inf(a_n){n=N}^\infty){N=m}^\infty$, how would I show that any limit point of $(a_n)_{n=m}^\infty$, denoted $L_0$, would exist between $\lim \inf$ and $\lim \sup$??

[solid kiln]

dulg

[silk mauve]

Id est, $\lim \inf \le L_0 \le \lim \sup$

[solid kiln]

dulg

[silk mauve]

The book I am using, Analysis 1 by Terence Tao, defines limit point (or accumulation point) of a sequence $(a_n)_{n=m}^\infty$ as $L_0$ such that
\\
$\forall \varepsilon >0, \forall N \ge m, \exists n > N, |a_n - L_0| < \varepsilon$

[solid kiln]

dulg

[silk mauve]

But the problem is, I just cannot find a way to connect and compare this definition of a limit point to limsup and liminf

.
Ping me when you've responded, and thanks in advance!

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[cerulean hearth]

a quick question : $(uv)'' = u''v + 2u'v' + uv''$

[solid kiln]

Max.cc

[cerulean hearth]

is it true?

[reef plaza]

put in a random numbers

and see if it is

[cerulean hearth]

i don't find the same

that's why i'm asking

[crisp lantern]

yes its true

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[cerulean hearth]

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[wise dove]

Can someone help I’m stuck on this

[wise dove]

Bottom one is what I’ve done

[light coral]

who can help me

[long basin]

Stop

#❓how-to-get-help

Don't post it in here

[long basin]

Move 3s to the left side and then find the discriminant

[light coral]

im totally confused on that

[long basin]

It legit gives you visual directions

[light coral]

ok where is the “available section”

[long basin]

Legit above this category

[light coral]

oh sorry

im cockeyed

[wise dove]

3 s's

[long basin]

What

No that's not the discriminant

If you have a quadratic, $f(s) = as^2 + bs + c = 0$, then the discriminant is $b^2 - 4ac$. The quadratic will have two solutions when $b^2 - 4ac > 0$

[solid kiln]

Umbraleviathan

[wise dove]

yea

so what should i write n>0

i expanded the brackets too

and for the disciminant what parts would I have to use

@long basin

[long basin]

Not n > 0

The discriminant > 0

And then solve for n

[narrow talon]

hey

[wise dove]

im confuzzeled still

[narrow talon]

how old are you

guys

[long basin]

First, move everything to the left side

So it equals to 0

Arrange the terms accordingly

[long basin]

And if you can read, I spoonfed you the inequality

[wise dove]

yea i got it

i didnt realise that i could stick in n-2

thanks

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[jolly forge]

how do i start solving this? D:

[jolly forge]

no idea rn, i was hoping drawing in the map this would work, but this one it's pretty thought

this one it's more hd

[jolly forge]

thx for the cheat sheet, but still no idea 😦

[wraith hinge]

prolly just graph it ngl

[jolly forge]

i mean, i've already decomposed the equation to sen and cos terms, but no idea still

[wraith hinge]

i wouldnt even know what to do with the cheat sheet

[jolly forge]

i'm already trying it :c

[rugged latch]

Send what u have so far

[jolly forge]

i have no charge on my phone, so i gotta write it, wait

this is all i've got

i've just wrote it

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[jolly forge]

it seems that i can do it like this

tan(x)-cot(x)=0

so then the answer will be when tan=cot

when x = 45, x = 135, x = 225, x = 215

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[gaunt axle]

Why are the files weird now

But anyways, number 47. How in the world do I set that up?

Also if anyone wants to tell me why I can't just send normal photos anymore that would also be epic (@ pls)

<@&286206848099549185>

[fringe minnow]

i dont think anyone wants to open a weird file

[gaunt axle]

cry idk how else to send it

I promise it's not like corrupt or anything, my phone isn't letting me send the actual photo which sucks

[trail ingot]

i can't even open it with anything on my computer ☠️

[gaunt axle]

ok so it basically says

If g(x) = xf(x), where f(3) = 4 and f'(3) = -2, find an equation of the tangent line to the graph of g at the point where x = 3

there we go

as in like how do I set that up

[fringe minnow]

lol

[gaunt axle]

Like idk my phone has just decided to like not send photos normally anymore

dang just one of those days 😭

tbh mood

yk what it's fine

King classmate coming in clutch

.close

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[gentle sleet]

just wondering

[gentle sleet]

is it possible to do dis with chain rule instead of product rule

,rotate

[gentle sleet]

or do I just stick with product rule for this one

the derivative

[digital bison]

you need both

[gentle sleet]

chain rule and product for this?

[digital bison]

yes

[gentle sleet]

I seen someone do it with only product rule

[digital bison]

oh yeah. I was thinking weird.

[digital bison]

product rule. Not chain rule.

[gentle sleet]

but can I also use chain rule doe

oh

[digital bison]

no

[gentle sleet]

why is that? could I just not do f(x) = 4^x and g(x) = (5^x-x)

[digital bison]

sure.
But that gives you f(x)g(x) not f(g(x))

[gentle sleet]

oh yeh

im slow asf

tysm

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[gentle sleet]

just wondering is aln(x) == ln(ax)

[sonic coyote]

Aln(x) = ln(x^a)

[gentle sleet]

yeye thats what i meant sorry

bet ty

for clarifying

[sonic coyote]

When we work with logs, we’re in the land of powers

So yeah u r right

[gentle sleet]

ty

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[subtle geode]

Hi can someone tell me how to get angle EHY of this problem

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[reef plaza]

https://gyazo.com/923bbaa0663350b8ce85b92bcd45e554

[reef plaza]

i need to find

$(a\cdot b)c$

[solid kiln]

Mortta

[reef plaza]

how to find

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[topaz thistle]

I’m confused on how to explain this properly

[tidal crane]

I'm not entirely sure, but I think the assumption here is that interval (a,b) is continuous on [a,b] and differentiable on (a,b)

Since x_1 and x_2 are contained in (a,b), that makes also makes (x_1, x_2) differentiable and [x_1,x_2] continuous

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[topaz thistle]

ok thanks

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[wraith hinge]

Solved first part and the smallest integer is a = 3, but I am not sure on what the question wants when it says " Use the function to encrypt the message "NEED HELP""

[marble wharf]

well it encodes H as C

what does it encode N as

E?

D?

etc

[wraith hinge]

So basically plug in the number for the letter and see what the output is?

[marble wharf]

yes

[wraith hinge]

well N is 13 so i would plug in f(13)?

[marble wharf]

yes

[wraith hinge]

I get this then

[marble wharf]

and a=3

[wraith hinge]

oh right

so just simplify this and the answer should be the encrypted letter?

[marble wharf]

yes

and then repeat for the other ones

[wraith hinge]

alright thanks ill try to do them quick

[marble wharf]

but it could help to first calculate f(A), f(B), f(C)

to see the pattern

[wraith hinge]

dont understand

[marble wharf]

smaller numbers -> easier to calculate

and you will notice a pattern which might help you

[wraith hinge]

oh but how would I go about calculating those?

[marble wharf]

ah sorry I should have written f(0), f(1), f(2)

which are the letters A, B and C are encoded as

[wraith hinge]

but isnt that what my equation is already doing?

[marble wharf]

just forget that I said anything I guess

[wraith hinge]

alright

this is what I got as the encryption

how can I prove the second part of the question

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[jolly jay]

I calculated q to be 50. But the answer sais 45. Am I wrong or is there something I've missed?

[lethal anvil]

sorry what is the full question?

[jolly jay]

The figures are uniform. Calculate the side marked with q and y

[mighty gazelle]

Are you sure the answer is q = 45?

[jolly jay]

The book sais is. It could be missprinted...

[lethal anvil]

yeah the key takeaway from this is should be that 20 is to 50, so 20 is to q means that q is 50

since the adjacent walls are equal in size

I think it might be a misprint

[jolly jay]

yeah, thanks for the help though!

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[marble wharf]

do you know an exponent law for $\frac{a^b}{a^c}$ ?

[solid kiln]

Denascite

[grim oriole]

No

[grim oriole]

🥲

[marble wharf]

if I tell you $\frac{a^b}{a^c} = a^{b-c}$, does that help you?

[solid kiln]

Denascite

[grim oriole]

Uhh

Where did the a fraction thing come from?

[grim oriole]

Cause I don’t know 🥲

<@&286206848099549185>

[dusty saffron]

a,b, and c are variables

[grim oriole]

Oh

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[brave jewel]

I was working this question out and then looked at the markscheme which had worked it out a way I havent seen before as i usually do it on my calculator. I got 0.034 so you would reject the null hypothesis and there would be sufficient evidence to suggest that the passenger losing his luggage would be higher however they got 0.1271.

[naive crest]

the markscheme doesn't make mistakes here, P(X = 0, 1, 2) are correct

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[naive crest]

and:

,w 1-0.95^25 - 25(0.05)(0.95)^24 - 12*25(0.05)^2 * (0.95)^23

[solid kiln]

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[naive crest]

if you had a doubt

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[hollow niche]

Its asking me find the mean, median and standard deviation of the data sets, i keep getting the wrong when im using the formula std dev = (Σx^2)/n - (x̄)^2

[hollow niche]

any help is aprriciated

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[delicate bobcat]

which part

[verbal cloud]

stuck on a rn and will probably be stuckon the rest too

just want to lnow if im doing the right thing

[delicate bobcat]

$\DoubleCos$

[verbal cloud]

for part a) would the first step be changing cos4x to cos(2x+2x)

[solid kiln]

Modus

[delicate bobcat]

use the last one

but rewrite theta as 2theta

[verbal cloud]

ah

i forgot u can do that

[delicate bobcat]

not really, but I guess it would work as well

[verbal cloud]

so with 1-2sin^2 (2x)

im guessing i do another douvle angle

and expand sin2x

ah i get k to be 4

1-4sin^2xcos^2x

[delicate bobcat]

should be k = 8

[verbal cloud]

sin2x = 2sinxcosx

[delicate bobcat]

yep, then squared and multiplied by 2

[verbal cloud]

is sin^2 2x = 2sin squared x cos squared x

oh so the 2 also get squared

[delicate bobcat]

yes

like

[solid kiln]

Modus

[delicate bobcat]

and * 2

then

[verbal cloud]

got it

[delicate bobcat]

on b) just use that fact

[verbal cloud]

would i rearrange it to

[delicate bobcat]

u have

[solid kiln]

Modus

[verbal cloud]

= sin^2 x cos^2 x

[delicate bobcat]

ah

yep that's it basically

now plug theta = pi/24

[verbal cloud]

so

[delicate bobcat]

ye

[verbal cloud]

am i expected to do that on a calc

cause im pretty sure that gives a nasty decimal

and not a kind number to do

[delicate bobcat]

just stay with the surd

[verbal cloud]

or just leave it i. sur

yea okay

[delicate bobcat]

only simplify it a bit

[verbal cloud]

(2-root3)/ 16

[delicate bobcat]

ye

[verbal cloud]

question c has left me puzzled

[delicate bobcat]

u know what is the 2nd part

sin^2(theta) * cos^2(theta) in terms of cos(4 theta)

from the previous one

for 2cos^2(2 theta) use double cosine again

but other one

$\DoubleCos$

[solid kiln]

Modus

[delicate bobcat]

2cos^2(theta) - 1 = cos(2 theta)
hence
2cos^2(2 theta) - 1 = cos(4 theta)
2cos^2(2 theta) = cos(4 theta) + 1

[delicate bobcat]

you cannot multiply by 3

since it's not an equation

[delicate bobcat]

yes

[verbal cloud]

sorry but how do i get the left side of the eq to cos4x

[delicate bobcat]

and replace 2cos^2( 2 theta) by cos(4 theta) + 1

(look above)

[verbal cloud]

wait sorry still not seeing how that equals cos4x
2cos^2(x) - 1 = cos2x
cos4x =

to get from cos2x to cos4x you would need to..?

[solid kiln]

Modus

[verbal cloud]

oh

oh

ohhhhhh

oh

makes perfect sense

thank uou

how would we get the greatest and smallest possible values

goti t

.close

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[alpine fern]

How to prove derivative of arc csc?

[keen plover]

If y = arc csc x

Then csc y = x

Then use implicit differentiation

[alpine fern]

Arc csc and csc -1 are the same too right?

[keen plover]

Yes

[alpine fern]

What does it mean to prove it?

If it’s already there?

Didn’t someone already prove it

Why do we have to do it again

[keen plover]

So that you can see where it comes from

[split chasm]

to confirm whether it's correct

[alpine fern]

Is there a worked example of proving this stuff?

I’m quite confused teacher only skimmed through this quick

[keen plover]

Why do you need to answer any questions when your teacher already knows the answers?

[alpine fern]

Try

Try

True*

I actually don’t know how to start this lol

[keen plover]

.

[alpine fern]

Ok if we dy dx both sides we get (derivative of csc) (dy dx) = 1 right?

[alpine fern]

From this*

[keen plover]

Yes

[alpine fern]

So that’s dy/dx = 1/–csc(x)cot(x)

Yes?

[keen plover]

No

The derivative of csc y is not -cscx cotx dy/dx

Why did y change to x?

[alpine fern]

All I did was isolate the dy dx

[keen plover]

Yeah

[alpine fern]

And I thought that is correct for the derivative of csc y since it’s not the target variable

So I multiple dy dx

[keen plover]

You can't just change the variable cuz you think it's better

d/dx (cscy) = d/dy (cscy) * dy/dx

[alpine fern]

Oof

LOL

so I should change the x es to the y

In the -csc cot?

So dy/dx = 1/-csc(y)cot(y)

[keen plover]

Yes

[alpine fern]

Ah aight

[keen plover]

Then replace y with arccsc x

[alpine fern]

????

[keen plover]

y = arccsc x remember?

[alpine fern]

Oh yeah

Oof this looks hard

[keen plover]

It's not that bad

[alpine fern]

Nested trig functions everywhere

[keen plover]

Only one nested trig function

The other cancels out

[alpine fern]

Csc (-arc is) is that one

-arc csc

-csc(arc csc)

[keen plover]

Yes

That's just x

So you just need to simplify cot(arccsc x)

[alpine fern]

So right now I got dy/dx = 1/-x(cot(arc cscx )

I think

[keen plover]

Yes

[alpine fern]

Idk how to simplify this tbh

[keen plover]

Let's focus on cot(arc csc x)

[alpine fern]

Ok

[keen plover]

arccsc x is just an angle

Let's call it θ

Thus arccsc x = θ--> x = csc(θ)

Can you draw a right triangle with an angle θ such that csc θ = x?

[alpine fern]

Idk what that means

[keen plover]

Draw a right triangle

One of the angles is θ

And csc θ = x

Remembering csc θ is hypotenuse/opposite

[alpine fern]

This?

[keen plover]

Yes

But now you need to label the sides such that the csc of that angle is x

[alpine fern]

This?

That’s csc z

X*

[keen plover]

You want csc θ = x

So h/o = x

What could the hypotenuse and opposite side be so that it's equal to x?

[alpine fern]

Cant it be anything?

[keen plover]

Anything such that h/o = x, yes

But there's one very simple choice

The lengths can be in terms of x btw

[alpine fern]

U mean just rewrite csc as 1/sinx

[keen plover]

No

Let me give an example

Let's say we knew csc θ was 1/2

[alpine fern]

Yeah

[keen plover]

Then I could say the hypotenuse is 1 and the opposite side is 2

[alpine fern]

Yeah

[keen plover]

I could say a whole range of other things as well, but that's a simple one

Now, we know csc θ = x

So what could the sides be?

[alpine fern]

2 and 2 let’s say

[keen plover]

That wouldn't work, because then csc θ = 2/2 = 1 ≠ x

[alpine fern]

X/1?

[keen plover]

x = x/1, yes

[alpine fern]

So hypotenuse x opposite 1

[keen plover]

Yep

Then you can find the adjacent side

Now, arccsc x = θ

So cot(arccsc x) = cot(θ)

So using the right triangle, what's cot θ

[alpine fern]

Cot is adjacent over opposite

[keen plover]

Yep

[alpine fern]

Something over 1

[keen plover]

What's that something?

[alpine fern]

Pyth theorem gives me y = square root of x minus 1

I think I’m not on my tablet right now

Oh wait it give me square root of x squared minus 1

[keen plover]

You sure?

a² + b² = c²

Nvm you're right

[alpine fern]

Ok

[keen plover]

So what's cot θ?

[alpine fern]

Square root of x squared minus 1

[keen plover]

Yes

So cot(arccscx) = √(x² - 1)

[alpine fern]

I forgot what we were doing

Tbh

[keen plover]

Simplifying this

You can now replace cot(arc csc x) with √(x² - 1)

[alpine fern]

That’s all good and all, the question is HOW THE HELL DO I REMEMBER THIS ON AN EXAM

[keen plover]

Just follow the basic ideas

[alpine fern]

☠️☠️☠️

[keen plover]

Typically, if you're taking the derivative of an inverse function y = f^-1(x), you can do implicit differentiation on f(y) = x

Then you just need to remember how to simplify nested trig functions

Anything like trig(arctrig(x)) can be simplified using the exact method we just used

[trim jolt]

@alpine fern Has your question been resolved?

[trim jolt]

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[slender shard]

Opposite angles are equal.

Therefore, PS = QR and PQ = SR. Which implies 2PS + 2 PQ = C. Because PS is twice the length of PQ, 2(2PQ) + 2PQ = C.

[cerulean ember]

Notice that 144pi=2(PS+PQ)

[trim jolt]

@sand bough Has your question been resolved?

[sand bough]

thanks :)

[trim jolt]

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[oak lion]

i was wondering how i’d go about finding the domain and range of this?

[fringe minnow]

,rotate

[fringe minnow]

@oak lion are u meant to use derivatives or any specific method

do you know what domain and range mean?

[oak lion]

all im familiar with is domain is x axis and range is the y axis

[fringe minnow]

ok

what class is this for

[oak lion]

algebra 1

[fringe minnow]

ok

so

for domain

its the set all of x values we can input

[oak lion]

and y values are outputs

[fringe minnow]

so because this is just a standard polynomial its going to be the set of all real numbers

so (-inf, inf)

thats your domain

[oak lion]

would the y axis be less than/greater than or equal to -1 then?

[fringe minnow]

so whats the range of y=x^2

range is set of y values

[oak lion]

so would it be all real numbers because x is (-inf, inf)?

[fringe minnow]

lets look at a graph of y=x^2

the domain as we said before is all real numbers so (-inf, inf)

thats because we can input any value for x and get an output y

and any polynomial will have the same domain of (-inf, inf)

so for range

its the set of y values

looking at the graph what would that be

[oak lion]

y is greater than or equal to 0?

[fringe minnow]

yeah

[oak lion]

so the output can be any number greater than or zero

[fringe minnow]

so we write that in interval notation as

[0, inf)

because its inclusive of 0

hence [

instead of (

[oak lion]

would i be able to find the domain of x^2 without a graph?

[fringe minnow]

the domain of any polynomial is just the set of real numbers (-inf, inf)

do you mean range ?

[oak lion]

ah, no, that makes sense

[fringe minnow]

ok

[frozen plover]

Just add an asterik to that

[fringe minnow]

ok now for y=x^2-1

have u done transformations of graphs?

like do you know what the -1 does?

[oak lion]

y intercept?

[fringe minnow]

urm yes but more generally

it shifts the graph down by 1 unit

[frozen plover]

A constant term shifts the function correspondingly

Keep that in mind it's very useful

[fringe minnow]

so what would the range of this function be

[fringe minnow]

huh?

[oak lion]

y is greater than or equal to -1

[fringe minnow]

yes so what would that be in interval notation

its inclusive so [

[oak lion]

[-1, inf)

[fringe minnow]

yes!

[oak lion]

[fringe minnow]

so the domain is (-inf, inf) and the range is [-1, inf)