help-19

[clever fjord]

Shuba

[gray hearth]

oh big triangles

sick

[odd edge]

@gray hearth Has your question been resolved?

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[gritty fable]

Hi Im in australia and my math methods draft is due tmrw. We have to graph a beach using desmos and showing our working. I can't figure it out for the life of me. Is there anyone who can help me?

[stable sequoia]

If its fine, you can make quadratic and linear approximations

especially on desmos, just as easy as say typing $y_1 ~ a_2x_1^2+a_1x_1^1+a_0$

[clever fjord]

parabolicinsanity

[stable sequoia]

didn't render the ~ 🥀

[gritty fable]

i have to use something called the point method for the working out. I just don't know how it works

[tawny geyser]

math methods? is that vic?

[gritty fable]

QLD

[tawny geyser]

oh alr im nsw

hsc

[tidal matrix]

~ is a space character in latex, try \sim

[tidal matrix]

does that just mean connecting all the points? if so, try parabolic's suggestion

[odd edge]

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[hollow bluff]

how do i find the equation of this graph

[hollow bluff]

how do i get the period

i have v = 350 sin bt

[atomic hornet]

What’s the definition of period?

[hollow bluff]

2pi/b

[atomic hornet]

No

[hollow bluff]

what

[atomic hornet]

“definition”

[hollow bluff]

oh

like the full cycle

[atomic hornet]

Can you make it a bit more precise?

[hollow bluff]

like that

there's 10 of them

in this

[atomic hornet]

Pretty close, but I got what you mean

[hollow bluff]

0.2 seconds

huh

pretty close

is it not that

[atomic hornet]

10 cycles in 0.2 seconds

[hollow bluff]

yeah

[atomic hornet]

Ig you can find the period with this information

[hollow bluff]

yeah but how

it's 0.02 for 1 cycle right

[atomic hornet]

Correct

[hollow bluff]

yeah what do i do after that

[bold geode]

so then all you need to find the amplitude and the trignometric function

since you have the period

[hollow bluff]

amplitude is 350

so 350 sin bx

[bold geode]

ok

[bold geode]

you have to incoporate the period though

[hollow bluff]

yeah idk how to do that

can i do smth with 2pi/b

since i'm trying to find that b

[bold geode]

it was already stated that it was 0.02

that's your period i believe

unless I'm doing something wrong

[hollow bluff]

oh so it's just 0.02x?

[bold geode]

let me double check real quick

[hollow bluff]

would you normally write that as a fraction

or is it ok to leave like that

[bold geode]

it follows like this: asin(b(t+c))+d

a=350

c=0

d=0

and the period is 0.02

[hollow bluff]

do i need that in radians?

[bold geode]

yes

so 2pi/0.02

=100pi

and just plug it in

[hollow bluff]

so it's

350 sin 100pix

[bold geode]

yes

but I think amplitude is 300 but its hard to judge from the photo

[hollow bluff]

has that not gone higher

[hollow bluff]

wait how does this work again

[bold geode]

i think they didn't think too much of it and just put it there so you can see it

[bold geode]

2pi/b

b was the period

[hollow bluff]

no like

2pi/0.02

uh

[bold geode]

0.02 was the period and 2pi is the radians since its a trigonmetric function

[hollow bluff]

yeah but like

with radians

how does it work

i forgot

[bold geode]

wdym

[hollow bluff]

2pi/0.02 normally is

[bold geode]

yes

so its 100pi to simplify it

[hollow bluff]

oh wait

in this

pi is treated like x right

so it's not like 6.28/0.02

[bold geode]

x is a variable has nothing to do with pi

we use pi so we don't have to use decimals and it matches with radians

[bold geode]

it is

its just 2pi is simplified

[hollow bluff]

yeah i'm saying pi is treated as a variable

right

[bold geode]

no

[hollow bluff]

i'm so brainrotted yo

[bold geode]

if you are new to trignometry its hard to explain but overtime you'll get it

pi=3.14

so its just a number

[hollow bluff]

yeah i forgot it

[bold geode]

its like saying 2(3) but instead of 3 you would say 2(k)

But since pi is a universal constant its always going to be refers to as the greek symbol pi instead of a latin letter

Over time you'll get it

[hollow bluff]

yeah nvm

i figured it out

by expanding it

i'm braindead 💔

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[undone grove]

2sinx=secx

[undone grove]

Boundary -pi<_x<-pi

I got to sin2x equals 1

[amber schooner]

and then

[lean yew]

correct thus far

[amber schooner]

there’s some restrictions of course

[undone grove]

2x boundary is -2pi<_2x<_2pi

[amber schooner]

x ≠ npi where n is an integer

[undone grove]

Where do u get that from?

[amber schooner]

because of secx

[lean yew]

the secant

[undone grove]

O ye

[amber schooner]

you have to remove the points where cosx = 0 from consideration

[amber schooner]

also the simplest way to think of this is determine the set of values u such that sin(u) = 1

[amber schooner]

can you find a general form for u in this case

[undone grove]

Inverse sin 1

Which is 90degrees

[amber schooner]

ok well what angle is that

yes

pi/2

but we add 2pi k for some integer k since sin repeats

[undone grove]

The period?

[amber schooner]

also if we wanted u to be negative then what would the angle be?

[amber schooner]

yes

[undone grove]

-3pi/2

[amber schooner]

mhm

so our argument is 2x

therefore 2x = pi/2 + 2npi or 2x = -3pi/2 + 2npi

which means x = ?

[undone grove]

Pi/4 and -3pi/4

But there's a negative and positive respectively of each in the answers

[amber schooner]

wdym?

[undone grove]

-pi/4 and 3pi/4 are answers

[amber schooner]

no -pi/4 would give -sqrt(2) = sqrt(2) and 3pi/4 would give sqrt(2) =-sqrt(2) so clearly not solutions

[amber schooner]

just by substituting into the original equation

[undone grove]

I didn't get negative

[amber schooner]

the solution set in general is pi/4 + npi where n is an integer

[amber schooner]

2sin(-pi/4) = 2 * -1/sqrt(2) = -sqrt(2)

cos(-pi/4) = 1/sqrt(2) so sec(-pi/4) = sqrt(2)

[undone grove]

Nvm I did but they were both negative

So still worked

[amber schooner]

they are not both negative

[amber schooner]

-pi/4 is not in this set

{…, -7pi/4, -3pi/4, pi/4, 5pi/4, …}

[amber schooner]

it’s sec

not csc

[undone grove]

O flip

[amber schooner]

sec is the reciprocal of cos

[undone grove]

Mb

[amber schooner]

all good

[undone grove]

So I had the answer and the textbook was wrong like it has been a couple times now

[amber schooner]

get a new textbook

👍🏻

[undone grove]

It's the schools

Can't do anything

[amber schooner]

tough

[undone grove]

Thanks anyways

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[undone grove]

2cos(2x-60)=sqrt3

[undone grove]

0<_x<_180

It's multiple choice

A is 45
B is 145
C is 30,330
D is 15,145

[lean yew]

hint: move the 2 over first. then common angle

[undone grove]

I got A

Is it A answer says d

But I tried and didn't work

[amber schooner]

definitely is 45

[undone grove]

Flipping book again

Actual rage bait

Thanks for confirming

What does that symbol mean

Nvm

But 15 works 2 right

[amber schooner]

but this is a bad question because 15 also works. 2(15) - 60 =-30 and cos(-30) = sqrt(3)/2

[undone grove]

Yr I noticed

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[wooden python]

yeah?

Determine whether A ⊆ B in each of the following cases.

[lone elbow]

2.0000001 in A, then (2,0001)^2 > 3

iii is false

what about iv)?

[summer wave]

What do you think

[lone elbow]

the empty set is a subset of every set

[wooden python]

that is correct as stated, but be careful!

[lone elbow]

the empty set is not a subset of the empty set?

[silk fox]

A is the set containing the empty set, so it has one element
B is the empty set, so it has no elements

so ∅ isn't an element of ∅ because ∅ has no elements

[lone elbow]

crazy

[gentle heron]

any set can be subset of itself....?

[lone elbow]

xD

[lone elbow]

this true?

[silk fox]

yes

[gentle heron]

except empty set ig

[silk fox]

actually, it still holds for ∅

[gentle heron]

?

[lone elbow]

in this case we arent checking if empty set is a subset of the empty set

[silk fox]

true

[lone elbow]

we are checking if empty set is an element of the empty set

[silk fox]

as i stated earlier, no

[lone elbow]

exactly

this exercise was cool

[silk fox]

agreed

[lone elbow]

.solved

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[pallid tinsel]

I think it's the mean value theorem or rolles theorem idr exactly what it's called

[solar peak]

yeah mean value theorem was my guess as well

f'(c) = f(1)-f(0)/1-0

how do we get one of the options

[pallid tinsel]

Well u don't need to look for an exact equality ig..

[solar peak]

yea

[summer wave]

Does < mean strictly less or less/equal'

[solar peak]

less

[summer wave]

Sooo which one

[solar peak]

strictly less

[summer wave]

Then f=cst means that 1, 2, 3 are false

[solar peak]

do i assume functions and check which options get eliminated?

[summer wave]

It's a good strategy for mcq

[solar peak]

ohk i thought there was some concept behind it

where we can use maths to get an inequality

but no ig

okay ty

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[summer wave]

However, if this is part of a course, perhaps it is just a typo

[solar peak]

no its an exam

previous year question

not a typo

[azure kite]

but why it introduce S

[solar peak]

no idea lol

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[mental kiln]

To find dy/dx in
sin²x + cos²y = 1

[mental kiln]

Apparently the correct solution is (sin2x)/(sin2y)

But isn't this only possible when cos²x = cos²y

Shouldn't dy/dx = 1?

[wooden python]

not quite 1 always

[mental kiln]

sin²x + cos²y = 1 can be written as
sin²x + cos²y = sin²x + cos²x
cos²y = cos²x

[mental kiln]

Oh wow that's a pretty graph

[mental kiln]

But like what's wrong with what i said?

[wooden python]

how exactly do you get from cos^2(x) = cos^2(y) to dy/dx=1

[mental kiln]

i mean x must be equal to y for that?

[wooden python]

nope

[mental kiln]

why not? What am I missing/doing wrong here

Is it because cosine function oscillate

[wooden python]

the cosine function oscillates a lot yes

for example cos^2(0)=cos^2(pi)

[manic sleet]

not necessarily from cos^2(x)=cos^2(y) we have a lot of solutions notably:

well lemme type them out lol

[mental kiln]

Oh thanks I get it I was skeptical of that at first

I tend to forget the general solution for the trig functions a lot

[manic sleet]

(y=\pm x+2k\pi, y=\pm x+\pi+2k\pi)

[clever fjord]

PajamaMamaLlama

[mental kiln]

I thought the cosine general form solution was y = ±x + nπ ?

[wooden python]

same thing except llama's is clunkier

[manic sleet]

(\cos^2(y)=\cos^2(x)\implies\cos(y)=\pm\cos(x)) from the negative branch we have (\cos(y)=\cos(\pi-x)) which gives (\pm x+\pi+2k\pi)

[clever fjord]

PajamaMamaLlama

[manic sleet]

bcz -cos(x)=cos(pi-x)

of course I think it'd be easier to just do implicit differentiation

[mental kiln]

Wait i need to ask something else as well

[manic sleet]

[mental kiln]

if you mod something = mod something else

Can you ignore the mod?

[wooden python]

mod like modulus?

[mental kiln]

Yes

[wooden python]

then no

[mental kiln]

|x| = |y|

[mental kiln]

Id need to consider multiple cases then?

[wooden python]

idk what exactly you're doing but in general yes

[mental kiln]

Okay thanks got it

Just worked it by hand to check and u do indeed get different cases but 2 of them are the exactly so it's just y = ±x

Thanks

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[digital mirage]

Vf is 2.98x10^4 not 2.98x10^5

[odd edge]

@digital mirage Has your question been resolved?

[digital mirage]

<@&286206848099549185>

[glass vault]

Could u show ur calculation

Im getting 10^5

2.98

[hidden token]

,calc (2pi1.496*(10^11))/(3.15*(10^7))

[clever fjord]

Result:

29840.143554097

[lean yew]

,w (2pi * 1.496 * 10^11) / (3.15 * 10^7) in scientific notation

[clever fjord]

Results provided by WolframAlpha

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

Requested by stseia

[hidden token]

Hmm

Ye it should be 10^4, maybe they made a mistake

[glass vault]

Oh yeah i was counting digits my bad

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[shadow lotus]

can someone help and explain this question ?

[tulip mica]

Do you know the Sum and product of roots?

[shadow lotus]

yup

[vivid wyvern]

Well since alpha and beta r the roots of the 1st equation
U can say that (x-alpha) and (x-beta) r it's factor

[shadow lotus]

thats rigth

same for the other equation with

delta and the other symbol

[vivid wyvern]

So multiplying the two factors u should be able to find alpha ,beta,gamma and delta with respect to p,q,P and Q

[shadow lotus]

umm which ones?

[vivid wyvern]

Multiply (x-alpha)(x-beta) first

What do u get

[vivid wyvern]

Take x common as well

[shadow lotus]

something like this

[vivid wyvern]

From the 2nd and third term

To get the coefficient

[shadow lotus]

Yea

Thats -(alpha+beta)

[vivid wyvern]

Yea

[vivid wyvern]

Now compare with the 1st equation

[shadow lotus]

We get p and q

[vivid wyvern]

Yes

[shadow lotus]

I get it till now

[vivid wyvern]

Find the value of alpha and beta in term of p and q

[shadow lotus]

Wait

Sorry my wifi went off

[vivid wyvern]

I tried to get values of aloha and beta in only terms of p,q
Struggling a lil

[shadow lotus]

Yea

Same

[vivid wyvern]

Well we will see what happens let's figure our gamma and delta too

[shadow lotus]

I dont think its possible

Okay

[vivid wyvern]

Yeah i suppose

[shadow lotus]

Well gamma and delta would be the same roo

We cannot get them in only terms of p and q

[vivid wyvern]

It might be okay
Let's try finding the values of (alpha -gamma) and the rest separately first

We can try,otherwise we have other helpers

[shadow lotus]

That's fine too

[vivid wyvern]

I got an idea

What if we applied quadratic equation here

It gets complex but we can find individual values that way

[shadow lotus]

How do u suggest to apply quadratic here ?

[hidden token]

There is a way to do it without finding the values of alpha, beta gamma and delta

[vivid wyvern]

If we subbed alpha in the value of p with beta/q

(x-gamma)(x-delta)

2e could use that form

Too

[hidden token]

Using purely the sum of roots, product of roots and some clever usage of the properties of roots

[vivid wyvern]

Well help him then,thanks

[hidden token]

Fine

Firstly, we know that $\alpha + \beta = -p \ \alpha \beta = q \ \gamma + \delta = -P \ \gamma \delta = Q$

[clever fjord]

Sarin

[shadow lotus]

i hv a senior who helped me with this
but i didnt get his solution either

[hidden token]

Secondly, if alpha and beta are the roots of the quadratic, then $\alpha^2 +p\alpha + q =0$ Do you agree?

[clever fjord]

Sarin

[shadow lotus]

yes

[hidden token]

Ok so lets expand the factors $(\alpha-\gamma)(\alpha-\delta)$ first

[clever fjord]

Sarin

[hidden token]

When you do this, you get $\alpha^2 - (\gamma + \delta) \alpha + \gamma \delta$ and similarly, $\beta^2 - (\gamma + \delta) \beta + \gamma \delta$

[shadow lotus]

thats correct

[clever fjord]

Sarin

[hidden token]

Now we know what the sum of gamma and delta and the product is in terms of the coefficients

[shadow lotus]

yes P and Q

[hidden token]

So substitute those and show me what you get

[hidden token]

-P and Q

[shadow lotus]

right its negative p

wait no

[hidden token]

Substitute gamma + delta = -P and gamma delta = Q in this

[hidden token]

,rccw

[hidden token]

Uhh, can you explain what this is?

[shadow lotus]

The sum and product of roots

Uhh

Idk what im supposed to do

[hidden token]

ok let me explain the relation between roots and cooefficients of a quadratic

[shadow lotus]

Okay

[hidden token]

if you have a quadratic $x^2 + bx + c$ with roots $\alpha$ and $\beta$, using the factor theorem, $x^2 + bx + c = (x - \alpha)(x - \beta)$

[clever fjord]

Sarin

[hidden token]

Right?

[shadow lotus]

yes

[hidden token]

If you expand the RHS, $x^2 + bx + c = x^2 -\alpha x - \beta x + \alpha \beta \ \implies x^2 + bx + c = x^2 -(\alpha + \beta) x + \alpha \beta$

[clever fjord]

Sarin

[shadow lotus]

yeah

[hidden token]

the coefficients of x must be equal, so $-(\alpha + \beta) = b \ \implies \alpha + \beta = -b$.
Similarly, the constants have to be equal so $\alpha \beta = c$

[clever fjord]

Sarin

[hidden token]

This is the general relation between the coefficients of a quadratic and its roots

If your quadratic is instead of the form ax^2 + bx +c, you can divide all the terms by a to get x^2 + b/a x + c/a

[shadow lotus]

i understand that

[hidden token]

Ok now that you understand that we can go back to the question

[shadow lotus]

i guess so

[hidden token]

Using what we just learnt, we know that $\alpha + \beta = -p \ \alpha \beta = q \ \gamma + \delta = -P \ \gamma \delta = Q$

[clever fjord]

Sarin

[shadow lotus]

yup

thats true

[hidden token]

So they are asking us to write $(\alpha - \gamma)(\beta - \gamma)(\alpha - \delta)(\beta - \delta)$ in terms of $p,q,Q,P$

[clever fjord]

Sarin

[hidden token]

If we expand $(\alpha - \gamma)(\alpha - \delta)$ and $(\beta - \gamma)(\beta - \delta)$, we get
$$(\alpha^2 - (\gamma + \delta) \alpha + \gamma \delta)(\beta^2 - (\gamma + \delta) \beta + \gamma \delta)$$

[clever fjord]

Sarin

[hidden token]

do you understand till this part?

[shadow lotus]

Yes

Some of those term can be substituted

[hidden token]

Yes

We know what gamma + delta and gamma delta is

[shadow lotus]

Same for alpha.beta and alpha+beta

[hidden token]

Ye we do know what those are but we can't substitute for that just yet

Substitute for gamma + delta and gamma delta and tell/ show what you get

[shadow lotus]

Lemme do it wait

Something like this

[hidden token]

Almost correct, (gamma + delta) is -P so it should be +Palpha and +Pbeta not -

[shadow lotus]

Okay ill fix that

Done

[hidden token]

Now think about what it means for alpha and beta to be the roots of a quadratic

[shadow lotus]

p and q ?

[hidden token]

What is the first property that comes to mind when you think about the roots of any polynomial, not just a quadratic?

What is the definition of a root?

[shadow lotus]

i havent done polynomials yet
but i think they should be equal to 0

[hidden token]

Exactly

[shadow lotus]

mhmmm

oh

[hidden token]

They are values of x such that a polynomial(or a quadratic) is 0

[shadow lotus]

yes yes

roots are the values of x in a polynomials

right \

[hidden token]

So you agree that $\alpha^2 + p\alpha + q = 0$ and $\beta^2 + p\beta + q = 0$?

[clever fjord]

Sarin

[shadow lotus]

yes

[hidden token]

Now think about how you can use that with this result

If you don't know, tell me, I will help you

[shadow lotus]

i think its like (alpha -delta)(alpha-gamma)

[hidden token]

uhh, no?

[shadow lotus]

shi

[hidden token]

We can say $\alpha^2 = -p\alpha - q$ right?

[clever fjord]

Sarin

[shadow lotus]

oh u mean this

[hidden token]

Try substituting for alpha^2 (and beta^2) in this

[shadow lotus]

wait

[hidden token]

?

[shadow lotus]

did

[hidden token]

Show me

[hidden token]

,rccw

[hidden token]

Good

Now you can take alpha and beta common to get this $$(\alpha(P-p) + (Q-q))(\beta(P-p) + (Q-q))$$

Now, expand this, but treat (P-q) and (Q-q) as a single term

[shadow lotus]

i think its P-p

[hidden token]

Oh right my bad

[clever fjord]

Sarin

[shadow lotus]

right

[hidden token]

Now expand this but treat P-p and Q-q as a single term

[shadow lotus]

can i asumme them as some variable ?

[hidden token]

Ye I was just about to say that

[shadow lotus]

okay wait

x and y

[hidden token]

I would go with something like i and j to avoid confusion but sure

[shadow lotus]

(alpha x +y)(beta x+y)

[hidden token]

Yep, now expand it

Something good will happen

[shadow lotus]

done

[hidden token]

show?

[shadow lotus]

Oh wait

[hidden token]

Did you figure it out?

[hidden token]

Ok so we can take xy common to get $\alpha \beta x^2 + (\alpha + \beta)xy + y^2$

[clever fjord]

Sarin

[hidden token]

Now do you know what to do?

[shadow lotus]

Factorisation?

[hidden token]

No

[hidden token]

Use this

[shadow lotus]

qx2-pxy+y2

[hidden token]

Upon seeing alpha beta and alpha + beta, your mind should instantly think of the sum and product of roots

[hidden token]

Yes, now if you substitute for x and y, you will get the required expression

[shadow lotus]

Wait

This looks fine

11

[hidden token]

interesting

one sec

,w expand -p(P-p)^2 + q(P-p)(Q-q) + (Q-q)^2

[clever fjord]

Results provided by WolframAlpha

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

Requested by sarin.5063

[shadow lotus]

the answer might just be wrong

[hidden token]

The p^p is definitely wrong

[shadow lotus]

yuh thats a typo

[hidden token]

Im trying to match the other stuff

Hmm, there are some things that definitely don't match. If you don't think the answer given is wrong, you can ask another helper to check whatever we have done although I doubt there was a mistake

[shadow lotus]

i believe so too

but wow that was long

thanks a lot too

[hidden token]

Ye it seemed long because I was explaining everything as we went but in reality it should only take like 5-10 steps

[hidden token]

Np

!done

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[shadow lotus]

.close

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[odd edge]

@livid juniper Has your question been resolved?

[livid juniper]

.close

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[gaunt flicker]

i dont really understand what you're asking

[quasi sparrow]

#study-discussion

Yes

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[sage helm]

Is my interpretation correct? This matrix has the basis of the input space as columns, but expressed in terms of the output basis. Thus, the matrix expects to be given vectors expressed in the input basis [v]_Bv, but the resulting vector will be expressed in terms of the output basis [v']_Bw.

[odd edge]

@sage helm Has your question been resolved?

[trail horizon]

!15m

[odd edge]

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.

[trail horizon]

@sage helm ^^

(i dont know how to do it)

[sage helm]

Thanks. I'm not sure how to 'ping' helpers. Maybe @ them? will try.

[trail horizon]

yeah just do that

[sage helm]

<@&286206848099549185> : Need help on linear algebra question, on room help-19.

[tidal matrix]

since T is linear, you do get a set of n L.I. vectors on the other end, and this matrix representation will give you a vector in terms of this set

however, there is a nitpick: this set will only be a basis if T is an isomorphism

but the understanding seems good, and a similar idea to your understanding is used to construct the change-of-basis matrix later on

[sage helm]

"T is an isomorphism", as in, if T is bijective?

also, what is "L.I."?

[low locust]

lin independent

that part isnt relevant to your question...

your interpretation is basically correct, yes

[sage helm]

thanks, both.

.close

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[copper pollen]

help woth this please

[copper pollen]

help pls?

[modern thicket]

you can see the clock as a unit circle

you can use trigonometry to find the coordinate

[copper pollen]

yea but I dont understand how to get the cordinate

[modern thicket]

you have 60 minutes for a whole circle

and in trigonometry a circle is 360 degree (2 pi)
therefore for each minute you have 6 degree

[copper pollen]

yea but so dose the thing start at 12 o clock?

[copper pollen]

oh okay so for every mins past i multiply 6 degs?

[modern thicket]

for instance for the 30 minutes

you have a angle of 180 (or 1 pi)

[copper pollen]

okay

[modern thicket]

sin(pi) = 0, cos(pi) = 1

and because it start at 12 and not a 3, you have to rotate the axis

[copper pollen]

so its (0,4)?

[modern thicket]

my bad

cos(pi) = -1

[copper pollen]

so how do I figure out like the cordinate cuz I am reallly confused

[modern thicket]

let's try for 50

[copper pollen]

okay

[modern thicket]

you have an angle of?

[copper pollen]

idk man I am like fresh into trig so idk

[modern thicket]

50 * 6 = 300 degree

then you need to convert this in radians ,
$rad = \theta \cdot \frac{\pi}{180}$

[clever fjord]

Amiso_

[modern thicket]

here we have $sin(\frac{5\pi}{3}) = - \sqrt{3}/2$
(this is like a common value)

[clever fjord]

Amiso_

[modern thicket]

and for $\cos(\frac{5\pi}{3}) = \frac{1}{2}$

[clever fjord]

Amiso_

[modern thicket]

therefore the coordinate is (1/2, -\sqrt{3}/2)

[copper pollen]

so for 5 mins is it pi/60?

[copper pollen]

how did you get cos (5pi/3)?

[modern thicket]

the angle I found in radian is $\frac{5\pi}{3}$

[clever fjord]

Amiso_

[copper pollen]

okay

so I just sub that vaue into cos and sin?

[modern thicket]

yes

because it is a unit circle

[copper pollen]

okay thank you soo much man I apreciate the help

[modern thicket]

I think this is \pi/6

[copper pollen]

yes your right

[modern thicket]

nice

[copper pollen]

tysm

[copper pollen]

hey for this I got 0.009125.....

how did you get it into a sqrt answer

[modern thicket]

this is like common values

[copper pollen]

yea but how do I get that beccause wehn I put sin (5pi/3) it gave me this

[modern thicket]

I have this
-0.86602540378

[copper pollen]

wait how ? I put this in the cal

5pi/3

[modern thicket]

you can use wolfram alpha maybe?

haaaa

I know

you can try sin(300)

[copper pollen]

oh okay

but still how do I get the sqrt answer

[copper pollen]

this

[modern thicket]

I can send you a table

[copper pollen]

yea that would be nice

[modern thicket]

this one have all the values

[copper pollen]

ok

[modern thicket]

but you cal is in degree that why it didn't work (you have to put 300 degree directly instead of 5pi/3 radian)

[copper pollen]

hmmmm

[modern thicket]

your cal is currenlty speaking in degree

not in gradiant

[odd edge]

@copper pollen Has your question been resolved?

[odd edge]

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[opaque hearth]

need help with calculus 2

[opaque hearth]

i believe the question wants us to use FTC part 1, but i always thought the lower bound had to be a constant

so how woould i approach this

[quasi sparrow]

Use difference of integrals

Say 8/x to 1 and 1 to x^2

[opaque hearth]

so all i need to do is plug in the upper bound for t then substract after plugging in the other bound

[junior ledge]

Take the derivative, remove the integrand, then multiply your function with the derivative of the upper limit minus the function with the derivative of the lower limit.

[opaque hearth]

ok

[junior ledge]

Then plug in two

[opaque hearth]

alright thanks i got it. -24

that's right im p sure.

im doing a whole calc 2 review btw so i may need more help

there's no answer key to this question so i'll post my answers please check it

false
true
false
true
true
true
false
false
false
true
false

[junior ledge]

Second one is sometimes false

[opaque hearth]

oh ur right. harmonic series

[junior ledge]

Yup

[opaque hearth]

wait but that's a sequence

same thing?

[junior ledge]

Not 100% sure

[opaque hearth]

thanks for being honest

im surrounded with cocky people who can't say "i don't know"

[quasi sparrow]

b) is practically the definition of convergence

[opaque hearth]

alright preciate it

so any other mistakes or no

[junior ledge]

So by looking at some of my notes, a series is the sum of terms of the sequence. And A_n = 1/n is considered a sequence

[opaque hearth]

ye that's right

[junior ledge]

I always thought that if the limit as n goes to infinity = 0 for an infinite series is inconclusive. So d is sometimes false?

And on the other hand e is always true since if the limit is anything else than 0 then the series is divergent

[opaque hearth]

you may be right as i'm losing my brain power

[junior ledge]

😂

[opaque hearth]

but about this

im pretty sure it's the other way around

for example, the harmonic series has a limit that goes to 0 as n goes to infinity, but that doesn't mean the series convereges, hence the nth term test says its inconclusive

[junior ledge]

Yeah

[opaque hearth]

but for every series that does converge, it's limit as n goes to infinity must go to 0

[junior ledge]

I see what you’re saying now

[opaque hearth]

that makes sense

[junior ledge]

I thought that was for alternating series though

[opaque hearth]

ye same with that. the AST includes the nth term test and it states it should be decreasing

[junior ledge]

Oh wait I see how the question is worded now, it states that it’s convergent, so the convergent series must have a limit that goes to 0

[opaque hearth]

ye exactly

but vice versa isn't always true. if the series has a limit that goes to 0, that doesn't mean it converges (remember harmonic series)

[junior ledge]

Yeaaa

I just made a rookie mistake

[opaque hearth]

naw dawg you're helping me a lot

no offense, but it's proven that more information is retained when you teach others

and i kinda taught you just now

no offense btw

[junior ledge]

Yeah definitely

[junior ledge]

Non taken

[opaque hearth]

l.u.b = least upper bound
g.l.b = greatest lower bound

my answers:
a) l.u.b = 2 and g.l.b = 0
b) l.u.b = 1 and g.l.b = -1

[junior ledge]

So basically the interval of convergence?

[opaque hearth]

ye im not really sure

[junior ledge]

I’ve never seen one of those problems before

[opaque hearth]

exactly me too lol

i think it's asking for the bounds

like look up monotonic sequence bounded

the graph helps

[opaque hearth]

more like what it converges to i feel like

[junior ledge]

I see

[opaque hearth]

ima post the question again so ppl can easily see it

PLEASE CHECK

l.u.b = least upper bound
g.l.b = greatest lower bound

my answers:
a) l.u.b = 2 and g.l.b = 0
b) l.u.b = 1 and g.l.b = -1

METHOD (may be a bad method so please critique it)

just plug in 0 or 1 for g.l.b
find limit as it goes to infinity for l.u.b

[junior ledge]

Wait so, for (a), wouldn’t the glb be just 1? Since if you look at the graph, it doesn’t go below 1, and as the limit goes to infinity it approaches 2 which is what you got

And same thing for (b) which is also what the graph shows

But to be honest I’m not sure how to get the glb from not looking at the graph

[opaque hearth]

wait let me see the graph

.end

.close

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[shrewd trellis]

With this, I thought they give us the derivative

To find the primitive function

Or do we first differentiate

Then find it

[remote quartz]

f(x) is the derivative of F(x), which you'll have to find F(x).

[shrewd trellis]

i thought it would be

f’(x)

[remote quartz]

The difference is one is lower case f, the other is upper case.

[shrewd trellis]

I’m talking about in general

if we go back to calc

[merry finch]

Sometimes we use capitals to show the primitive, otherwise we’d be getting functions named f’ left and right

[shrewd trellis]

we used f(x) for a general function

[merry finch]

The ‘ is nothing more than a symbol to say that f’ is related to f in some way

[remote quartz]

Yeah, we use different notations sometimes. If you want to think of things using prime notation, you have F'(x)=f(x).

[merry finch]

In the case of calculus “related” means “is the derivative of”

[remote quartz]

It's like how we usually use x for an arbitrary variable, but nothing stops us from using a or b instead.

[merry finch]

It doesn’t mean f’ can only and must be used for derivatives

So whilst it makes sense to say here’s f’(x), find f(x)

It makes equally as much sense to say here’s f(x), find its primitive F(x)

[shrewd trellis]

Isn’t capital F(x)

[shrewd trellis]

No what im tryna say is

[merry finch]

If f(x) is some given function, we conventionally say that F(x) is the primitive of f(x)

[shrewd trellis]

here, you are given f’(x), find F(x)

[merry finch]

No you’re not

You’re given f(x)

[remote quartz]

Here you are given f(x)=F'(x), again they're using different notation than you're used to.

[amber veldt]

should say "a" primitive function

[shrewd trellis]

Yeah so that’s not the derivative right, we need to differentiate that to get f’(x)

[remote quartz]

The question isn't looking for f'(x) in this case, careful.

[merry finch]

But we don’t want to find the derivative of f(x)

[shrewd trellis]

wait so in this question they havnt gave us the derivative, correct?

[remote quartz]

To reiterate, you're given F'(x), now you need to find F(x) by integrating F'(x) (aka f(x), as they've notated).

F'(x) is the derivative.

[merry finch]

Don’t worry about this

[shrewd trellis]

Y/N

[merry finch]

They’ve given you some function f(x)

You need to find some other function(s) F(x) such that when you differentiate F(x) you get f(x)

[remote quartz]

Yes, f(x) is the derivative. It's equivalent to F'(x), which is the derivative you're thinking about.

[amber veldt]

the notation is irrelevant

imo

they give you a function and are asking for a primitive, that's the exercise

[merry finch]

You can’t think like this because in general every function is the derivative of some other function

(For the most part)

[amber veldt]

@shrewd trellis they gvie you a function and say "find the (should be find "a") primitive of this function, that's the question, maybe you're overthinking it?

[shrewd trellis]

I understand now but wdym find a

Oh u mean by adding constant of integration

[narrow crypt]

Do u know power rule for integration

[remote quartz]

Because an integral has a +C factor, which makes it not unique (unless you think of the solution as an element of an additive torsor).

[shrewd trellis]

Yes add 1 to the exponent and divividr by the same number

And add a c

[narrow crypt]

Okie

[shrewd trellis]

What’s an integral

[remote quartz]

Integral = Anti-derivative.

[shrewd trellis]

oh there’s many names i assume

[remote quartz]

I'm from America, we've never called the integral of a function it's primative (I honestly had to Google it).

[shrewd trellis]

damn I see

[narrow crypt]

Ye😭

Never used primitive as well

Anyhow could u find the primitive of the given function here?

Using the power rule

@shrewd trellis

[odd edge]

@shrewd trellis Has your question been resolved?

[shrewd trellis]

wdym

[merry finch]

Can you find a function, when you differentiate it, gives f(x)

Such a function is called a primitive of f

[shrewd trellis]

Hmmm

Yes I could but where’s the q im solving

[shrewd trellis]

oh this?

3x^5/5 + 2x^4/4 + C

[north sparrow]

that’s most probably what they want for F(x) here

[remote quartz]

What happened to the constant 2?

[shrewd trellis]

oh 2x

[remote quartz]

Yep.

[shrewd trellis]

3x^5/5 + 2x^4/4 + 2x + C

thanks

.close

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[fallow shuttle]

i may sound dumb, but can someone help me find the range of 4x - xy = 8

[rose coral]

Can you try isolating y?

[tribal ingot]

you're never dumb for asking a question btw

everyone starts somewhere asking questions is part of the learning process

[fallow shuttle]

ty for this 😭

[fallow shuttle]

y = -8/x + 4

[rose coral]

Now, do you know how to find the asymptotes of this?

Alternative method: isolate x instead.

[fallow shuttle]

no idk

[rose coral]

With most rational functions, you want to think of the process BOB0, BOTS, EATSDC.

[digital shore]

y=-8(1/x)+4
so vertical asymptote at x=1 i think

oops 0

typo

[rose coral]

When you take the highest degree of the numerator and denominator into account:

• Bigger on bottom: 0
• Bigger on top: slant
• Exponents are the same: divide coefficients

[rose coral]

@fallow shuttle can you convert $y = -\frac8x + 4$ such that it is one big fraction?

[clever fjord]

@rose coral

[rose coral]

why is the bot female

[fallow shuttle]

Oh

i get it now

thanks

[rose coral]

!done

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[silent solar]

I'm struggling with the process for D

[silent solar]

eig(D) = -2,-2

Null(D+2eye(2)) = [1 1/3; 0 0] so an eigenvector is [-1;3]

Now ker(D+2I)^2 = [0 0; 0 0] = C2

Choose any vector in C2 not in ker(D+2I) as a generalised eigenvector e.g. [0;1]

P = [0 1; 1 -3]
J = exp(-2t) * [1 t; 0 1]

Typing PJP^-1 into matlab

I get a different result to the answer

[odd edge]

@silent solar Has your question been resolved?

[odd edge]

@silent solar Has your question been resolved?

[odd edge]

@silent solar Has your question been resolved?

[amber veldt]

you got t=0 in matlabs answer

[odd edge]

@silent solar Has your question been resolved?

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[gloomy mason]

no clue

what to do

[wooden python]

do you remember how i took you through reading a plot like this?

just earlier today though it was some hours ago

(and potentially yesterday for you bc i dont know what timezone you are in)

@gloomy mason

[gloomy mason]

solved this one

[wooden python]

k

[gloomy mason]

cant imagine how dum i get sometimes

[wooden python]

you can .close if you're done then

[gloomy mason]

.close

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[meager raft]

I need help with like this whole, problem

I dont understand what its asking me, how to find the answer, or explain the signigicance of blue and red

[indigo dirge]

blue (t) and red(t) are just arbitrary functions

[meager raft]

meaning...

[indigo dirge]

same thing as f(t) and g(t)

just plug the ts into each function

so for instance blue (0) would be 80(0)+20

[meager raft]

the gears are turning slightly

[indigo dirge]

and blue(1)=80(1)+20

so what would blue(2) be?

[meager raft]

b(2)=80(2)+20

?

[indigo dirge]

yes, that's correct

[meager raft]

then it keeps going

input output

sort of thing

[indigo dirge]

you do the same thing for red

yes

[meager raft]

okay so what would be the significance that im looking for

do i need to find what one is traveling faster?

[indigo dirge]

well

what units do the functions each output?

[meager raft]

what?

[indigo dirge]

units, as in seconds, km, grams, etc

[meager raft]

km?

[indigo dirge]

yes!

so what would be the interpretation of blue(1)=red(1)

[meager raft]

wait

this

?

[indigo dirge]

[meager raft]

the significance is that they have the same output

when t=1

[indigo dirge]

could you translate that into real-world terminology?

[meager raft]

when the travel time is equal to 1 both cars travel the same distance which is 100 km

[indigo dirge]

good!

[meager raft]

wait i did it omg

bless ur heart and soul

[indigo dirge]

great job :D

[indigo dirge]

thank you :)

[meager raft]

.close

[odd edge]

Channel closed

Closed by @meager raft

Use .reopen if this was a mistake.

[odd edge]

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[faint stump]

@rancid moat

[faint stump]

<@&286206848099549185>

[indigo dirge]

!15mins

[odd edge]

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.

[faint stump]

Can someone help me with this problem

[faint stump]

I got 11.25 but idk if that correct

[faint stump]

well basically

[indigo dirge]

what's the length of BA

[knotty kindle]

Probably not unless you rounded

[indigo dirge]

continue mb

[faint stump]

Alr

[faint stump]

Ba is

6

[ivory grove]

just a note, you've opened help channels with the @Helpers ping a lot of times already. please stop. use it only if people haven't answered your question for 15 mins.

[faint stump]

Alr

Mb

[indigo dirge]

mk, so now can you find AC?

[ivory grove]

thank you!

[faint stump]

But how can I

[indigo dirge]

ty marianne :)

[indigo dirge]

pythag theorem

[faint stump]

alr

Yea

I got ac is

3 radical 5

[indigo dirge]

yes

that's right

[faint stump]

I js did

3 radical 5/2

Since that’s the radius

then after that I use area of circle formula I got pie (3 radical 5/2)to the power 2

[indigo dirge]

yea

[faint stump]

I got 45/4

Whihc is 11.25

[indigo dirge]

did you multiply by pi?

[vivid wyvern]

Pie

[indigo dirge]

🥧

[vivid wyvern]

A lil treat

[indigo dirge]

now i want pie ;-;

[faint stump]

I think the ans key made a mistake

[indigo dirge]

yea lol

it should be D

[faint stump]

I need help with this pls

[indigo dirge]

find ac :)

[faint stump]

Alr

[indigo dirge]

either use pythag or memorize triples lmao

[faint stump]

Actually I’m kinda confused

Nvm

[indigo dirge]

mk so

what's 15^2?

[faint stump]

225

[indigo dirge]

and 12^2?

[faint stump]

144

[indigo dirge]

good

now what do we do with these two to get AC^2?

[faint stump]

Wait

Can u identify the legs and the hypotenuse for me

[indigo dirge]

well

where's the right angle?

ignore point D for a second

[faint stump]

Yes

[indigo dirge]

^^

[faint stump]

uhm

[indigo dirge]

yes

ok, so what does pythag say? just give side lengths, no numbers yet

[faint stump]

Alr

C^2=A^2+b^2

Wait is ac=9

[narrow crypt]

Compared it will the 3,4,5 triangle

[indigo dirge]

ok, so what's the area of the triangle?

[narrow crypt]

(Area can be represented in two ways)

[brittle plinth]

either that or use inverse Pythagoras

[indigo dirge]

or similar triangle

[faint stump]

Alr

[brittle plinth]

the area way is much more intuitive

[faint stump]

Alr now what do I do?

Ac=9

[indigo dirge]

^^

[faint stump]

Alr

I’m stuck

[indigo dirge]

what's the area of a right triangle?

[faint stump]

B*h/2

[indigo dirge]

yes

what's the (length of the) base

[faint stump]

15?

[indigo dirge]

uhm

what are the lengths of the legs

[faint stump]

alr

Ac=9

Bc=15

Ab=12

[indigo dirge]

ok

which two are the legs?

[faint stump]

alr

Ac and ab?

[indigo dirge]

yes

so what's the area

it would be the product of the legs over 2 right?

[faint stump]

54

Oh is it cause they r perpendicular

[indigo dirge]

yes

[faint stump]

Got it

Wha

Aren’t the 2 legs ab and ac?

[indigo dirge]

yes