help-17

[oak magnet]

Using matrices here is just a way to aufgehellt the writing

[sour thistle]

Okay that makes sense.
So when one of the x, in this case x_3, is a free variable it's linearly dependent? Could you maybe help me understand why that makes it linearly dependent if we have a free variable

[oak magnet]

Ok

Cuz having a free variable make that we have more than the trivial solution a1 = a2 = a3 = 0

[sour thistle]

Ahh right so if more than one solution we are independent cause to be dependent we need only one solution for X1 X2 X3 =0. Thanks man. Should I open another ticket for the next task as you already helped plenty or what should I do?

[oak magnet]

You can keep asking here

[oak magnet]

Switch dependant and independant

But yeah this is it

[sour thistle]

Ah I mixed them up two times in a row. Ok then next question:
Choose Lambda = 0. Thus the three vectors (as seen in my image above, only v_3 being different) form the basis of a three-dimensional
subvector space of the ℝ_4. Calculate the components of
vector b=
(-12//
-24//
8 //
-120)

I hope I didn't translate too badly, don't even know what that means to be honest.

[oak magnet]

Hmm

So how i understand it is

(Your three vectors | vector b)

Solve

Is that right ?

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[arctic mango]

I want to know more about the Ramanujan Theorem

[bronze osprey]

Google

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[gilded oriole]

can someone explain the first step i have no idea whats going on

[bitter pilot]

this is how sqrt(x²) is defined

[gilded oriole]

why would a = f(x)/x

theres like no point in doing that

[bitter pilot]

a is the slope of ax+b and the slope is known to be the quotient of some difference of y and some difference of x like Dy/Dx

So they consider something analogous y/x = f(x)/x

[gilded oriole]

but inf/inf doesnt make any sense at all

why would a be the same for horizontal asymptote and slant

[bitter pilot]

if the asymptote was horizontal then a would be 0

[twin meteor]

the bratwurst differential

[bitter pilot]

And again since b/a is just as a constant might as well just consider f(x)/x

[gilded oriole]

damn

makes sense

[bitter pilot]

But this procedure works of course if a != 0 else there is no slant but horizontal asymptote

[gilded oriole]

yeah

thanks man

i think i got it

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[robust atlas]

A friend has linked me to this server cause I need some help with this question. Heres the diagram, and I need to find the vector from O to D, and im not gonna lie I dont have any idea how.

[robust atlas]

I apologize for the low quality

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[steep moth]

can anyone help me please🙏

[hushed pewter]

Can you think of any ways to write FE in terms of FC CD and/or DE?

[steep moth]

i already got FD

it 2a

[hushed pewter]

Okay consider what FD and DE add to

[hushed pewter]

Btw FD is not 2a

[steep moth]

sorry i meant FE

typing error

[steep moth]

wdym

[hushed pewter]

Irrelevant. It was based on your typing error

What have you done for part b so far?

[steep moth]

i have every length other than FX and FM and CXE

@hushed pewter you tehre?

[hushed pewter]

What did you get for FM?

[steep moth]

2a - 1/2 b

@hushed pewter

[hushed pewter]

hm okay. This requires a good think

[steep moth]

for me or you @hushed pewter

[hushed pewter]

both

[steep moth]

do you know it ?

@hushed pewter

[hushed pewter]

No

I am thinking

[steep moth]

<@&286206848099549185>

nvm

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[marble marsh]

Hi. if we have the modular inverse from eea(m, r) where m is odd, greater than 1 and less than r, and r is some power 2. can this modular inverse ever be 1. just need someone smarter than me to sanity check before I try to prove this

thank you!

[marble marsh]

i am fairly confident it can never be 1 under these constraints

[nova mica]

i can help u with this if you want

these questions come up all the time in gcse

[steep moth]

icl i need to sleep rn sorry

[nova mica]

dont be

[steep moth]

but thank you tho

[marble marsh]

nevermind I just proved it, that was incredibly trivial

sorry for wasting time lol

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[sudden harbor]

I was using Taylor to solve this limit but I think there's a major mistake here, or multiple. It just doesn't look right

[flat whale]

,w lim x to 0 (cos(x)-cosh(x))/(pi/2-arctan(1/x^2))

[twin meteor]

Results provided by WolframAlpha

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

Requested by rie.mann

[flat whale]

Taylor expansion of arctan(1/x^2) doesn't work because it diverges to infinity as x goes to 0.

You could probably use some identity to cancel the pi/2

[sudden harbor]

I mean arctan(1/x^2) approaches pi/2 as x approaches 0 but that doesnt really do anything for me

I still end up with 0/0

I cant thing of anything else

I tried Hopital too but with no luck

[flat whale]

Look for identities involving arctan

[sudden harbor]

I cant think of anything besides arctan is the inverse of tan

[flat whale]

That's a definition

https://en.m.wikipedia.org/wiki/List_of_trigonometric_identities

[sudden harbor]

maybe this but I cant think of anything to do with it?

[flat whale]

Yea try that and Taylor expand arccot

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[sand bobcat]

Hi, I’m somewhat of a dumb dumb, but I’m trying to figure out how the equation in 5.1 b is used to refer to 5.1 a .
I don’t know which numbers goes where?

[sand bobcat]

Don’t even know if I did this right, first time I’m here asking…

[spare cobalt]

you can look up the internet :v

tho i dont think the formula 5.1b have anything related to 5.1a

r will be interest rate

n is how long in the bank

and b is the beggining amount in the bank

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[sand bobcat]

Thank you 🙏

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[tribal latch]

Whats 10x10

[steep crater]

idk

[tribal latch]

whats 10x10-30x3434343434.23423-2343

[steep crater]

1000 i think

[tribal latch]

naw 10000000

[steep crater]

No trolling btw

[tribal latch]

naw you being fr

[steep crater]

You can evaluate the expression using a calculator

[tribal latch]

I only have a 1 dollar budget

and I bought a pen

[steep crater]

Google dude

[tribal latch]

no couculator

[calm light]

desmos

[tribal latch]

Thats right demos

[steep crater]

Google calculator will work

Stop trolling pls ty

[tribal latch]

Im dumb I forgot

[tribal latch]

me?

[steep crater]

Yes you

[kind light]

please do not troll in help channels

[worn folio]

evaluate by drawing 10 lines 10 times, then count the resulting number of lines ofc

[tribal latch]

google calculator is dumb it helps you cheat

I dont cheat

[steep crater]

Brother

[tribal latch]

CHEATING SHOULD BE BANNED

[steep crater]

We have better use of time

[tribal latch]

SO SHOULD VAPES

[oak magnet]

<@&268886789983436800> sorry

[steep crater]

I will not reason

[tribal latch]

(I bet you use one)

[kind light]

was gonna ask if this is modpingable :p

[steep crater]

Its faking banned in singapore

[kind light]

fucking

[versed bane]

.close

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[unkempt spindle]

help

[unkempt spindle]

hello

how do i prove

i need it asap midterms soon

🙏

[gritty sage]

Do you know sech in terms of powers of e and so forth?

[unkempt spindle]

i do

its 2/(e^x+e^-x)

[gritty sage]

Right.

Do you know how to do an inverse function like f^-1 from earlier in algebra?

How to find one, I mean.

[unkempt spindle]

like switching x with y and y with x and solve for y ?

[gritty sage]

Right.

[unkempt spindle]

now what

[gritty sage]

x = 2/(e^y + e^-y)
xe^y + xe^-y = 2

You can do like that.

Then, you can multiply both sides by e^y.

Then, you can get it in the form of a quadratic polynomial (the "variable" being e^y) equal to zero.

[gritty sage]

.

[unkempt spindle]

like when i finishsolving it

what do i do

its confusing

it doesnt say anythign in the question

it just gave me this with a domain

[gritty sage]

Oh, you basically turn it into y = log((1 + sqrt(1 + x^2))/x) using valid algebra rules.

[gritty sage]

And since you used valid algebra rules and the formula you started with was correct according to the definition of sech(x), the result must be correct.

[unkempt spindle]

lemme solve it rq

[gritty sage]

OK.

[unkempt spindle]

i got lost after the 2/x=e^y+e^-y

[gritty sage]

[x = \frac2{e^y + e^{-y}}]
[x(e^y + e^{-y}) = 2]
[xe^y + xe^{-y} = 2]
[xe^y - 2 + xe^{-y} = 0]
[xe^{2y} - 2e^y + x = 0]

[twin meteor]

Chai T. Rex

[gritty sage]

Does it make sense so far?

[unkempt spindle]

where did we get 2e^y

[gritty sage]

We multiplied both sides by e^y.

[unkempt spindle]

ohhhh

okokok

to get rid of the ^-y no?

[gritty sage]

Right, and to get it ready to use the quadratic formula or whatever.

[unkempt spindle]

makes sense so far

[gritty sage]

OK, so solve for e^y using the quadratic formula or other method.

[unkempt spindle]

ive not done math in so long i forgot how i even get the a b c

Whats a whats b and c in this

😭

im cooked

[gritty sage]

OK, so to make it easier, you can substitute u = e^y.

So, like (xu^2 - 2u + x = 0).

[twin meteor]

Chai T. Rex

[unkempt spindle]

mhm

[gritty sage]

Then, u is our variable.

[unkempt spindle]

mhm

[gritty sage]

So, a is the thing in front of u^2, b is the thing in front of u, and c is the constant term.

a = x, b = -2, c = x.

[unkempt spindle]

so x is a

and -2 is b

and c is also x

ok

[gritty sage]

Right.

[unkempt spindle]

2 -+ sqrt(4-4x^2)/2x

[gritty sage]

Right, and you can simplify a little further:
[e^y = \frac{2\pm\sqrt{(-2)^2 - 4(x)(x)}}{2x}]
[e^y = \frac{2\pm\sqrt{4 - 4x^2}}{2x}]
[e^y = \frac{2\pm2\sqrt{1 - x^2}}{2x}]
[e^y = \frac{1\pm\sqrt{1 - x^2}}{x}]

[twin meteor]

Chai T. Rex

[unkempt spindle]

and that would be it no?

[gritty sage]

Nope, not yet.

We don't have the desired result.

[unkempt spindle]

by using ln

we eliminate the e

[gritty sage]

Well, there's a slight difference still.

The plus-minus symbol.

[unkempt spindle]

we wont use the -

because the answer would be negative

[gritty sage]

Right.

[unkempt spindle]

and ln cant be negative

so therefore we use +

no?

[gritty sage]

Yes, unless x is negative.

[unkempt spindle]

yup

[gritty sage]

But we're in like (0, 1].

So, we want the +.

[unkempt spindle]

exactly

so we only take the +

and ditch the -

and then use ln

[gritty sage]

Make sure to explain why you throw away the - in your proof.

[unkempt spindle]

Ofc

Yk im sorry my mind was a bit forzen cuz of winter break

i know those stuff i just forgot ab em

cuz of teh break

[gritty sage]

Oh, no problem. With some practice, it'll be back.

[unkempt spindle]

the hyperbolic functions are the worst

lemme see if i have any other problem with any other function

[gritty sage]

OK.

[unkempt spindle]

i had trouble simplyfing this

[gritty sage]

So, you need to derive arctanh?

[unkempt spindle]

yes

[gritty sage]

OK, so you can get rid of the denominator by moving it to the other side.

[unkempt spindle]

mhm

but my problem is always me questioning wether i should move the other side to the other or keep em

i always get confused

[gritty sage]

Does it give you the result you're aiming for?

What arctanh is defined as?

[unkempt spindle]

i have to get

[gritty sage]

OK, so, the ln tells you that you'll have e^something with y in it on the left side by itself.

That'll probably happen after the quadratic formula.

So, we need a quadratic equation with e^something with y in it as the "variable".

So, first we get rid of the denominator, then we can move everything to the left side to make a quadratic equation.

[gritty sage]

Like:
[x = \frac{e^{2y} - 1}{e^{2y} + 1}]
[x(e^{2y} + 1) = e^{2y} - 1]
[xe^{2y} + x = e^{2y} - 1]
[xe^{2y} - e^{2y} + x + 1 = 0]

[twin meteor]

Chai T. Rex

[gritty sage]

We're moving toward getting a quadratic equation like that.

[gritty sage]

Right.

[unkempt spindle]

now to make i easy

to make it easy

e^2 can be z

[gritty sage]

Wait.

You have (e^2y), but it should be (e^{2y}).

[twin meteor]

Chai T. Rex

[gritty sage]

You have e^2y when it should be e^{2y} in the LaTeX you type in.

[unkempt spindle]

oh latex render

yeah yeah

[gritty sage]

But I noticed something.

Factor out the e^(2y).

[xe^{2y} - e^{2y} + x + 1 = 0]
[e^{2y}(x - 1) + x + 1 = 0]

[twin meteor]

Chai T. Rex

[gritty sage]

It's starting to look like the answer.

Do you see how?

[unkempt spindle]

yes

how do u do the latex so quick

[gritty sage]

Practice.

[unkempt spindle]

what do we do after that tho

[gritty sage]

Well, we move the x + 1 to the other side.

[e^{2y}(x - 1) = -x - 1]

[twin meteor]

Chai T. Rex

[gritty sage]

Then we move the x - 1 to the other side.

[unkempt spindle]

divide

yes

[gritty sage]

[e^{2y} = \frac{-x - 1}{x - 1}]

[twin meteor]

Chai T. Rex

[unkempt spindle]

then we use ln

[gritty sage]

Well, first you can make it nicer.

Multiply the top and bottom by -1.

[unkempt spindle]

2y=ln((-(x+1)/(x-1))

[gritty sage]

[e^{2y} = \frac{1 + x}{1 - x}]

[twin meteor]

Chai T. Rex

[gritty sage]

Then, the ln.

[unkempt spindle]

but whyd we do that

[gritty sage]

Oh, to make it look like the answer.

Plus it reduces the number of minus signs.

[unkempt spindle]

i just dont get math sometiems

😭 😭 😭 😭 😭 😭

[gritty sage]

[\ln\qty(e^{2y}) = \ln\qty(\frac{1 + x}{1 - x})]

[twin meteor]

Chai T. Rex

[gritty sage]

And then it's easy from there.

[unkempt spindle]

yea

[gritty sage]

But to make fractions look nicer, you should minimize the number of minus signs.

[unkempt spindle]

just divide by 2

i gotchu

THANK U

[gritty sage]

You're welcome.

[unkempt spindle]

ur saving me rn

midterms tmr

ihate trig

i have another question

[gritty sage]

OK.

[unkempt spindle]

what does this benefit me

how do i apply it

[gritty sage]

Well, it changes an addition into a product.

Not only that, but you can use the angle addition formulas on the right side stuff.

Like sin((x + y)/2) has an addition in it.

And you can use the half angle formulas as well.

[unkempt spindle]

do i memorize?

[gritty sage]

Probably.

[unkempt spindle]

howd i prove these

[gritty sage]

You can use the double angle formulas, I think.

[unkempt spindle]

how tho

[gritty sage]

sin(2x) = 2 sin(x) cos(x)
sin(x) = 2 sin(x/2) cos(x/2) [by substituting x with x/2]
sin(x) = 2 sin(x/2) sqrt(1 - sin^2(x/2)) [Pythagorean identity]

And so forth.

Notice that sin(A/2) has 1 - cos(A) in it.

[unkempt spindle]

yes

[gritty sage]

That looks a lot like the Pythagorean identity.

[unkempt spindle]

hhow can u tell so quick

[gritty sage]

Tell what?

[unkempt spindle]

that its a pythagorean identity

[gritty sage]

Oh, it just looks like it. You have sin^2 + cos^2 = 1, so sin^2 = 1 - cos^2, and 1 - cos looks like that a bit.

[flat sigil]

the better question is how you write so fast

[unkempt spindle]

EXACTLY

I SAID THAT TOO

[gritty sage]

Hmm, it says that the half angle formula for sine uses cosine in it.

[unkempt spindle]

^

[gritty sage]

So, I was a bit off.

Sorry, I need to take a break.

[unkempt spindle]

OKOKOK

[gritty sage]

You can see some derivations for half angle formulas in the "Using Half-Angle Formulas to Find Exact Values" section of https://math.libretexts.org/Bookshelves/Algebra/Algebra_and_Trigonometry_1e_(OpenStax)/09%3A_Trigonometric_Identities_and_Equations/9.03%3A_Double-Angle_Half-Angle_and_Reduction_Formulas.

[unkempt spindle]

imma see

i actually understand now

😭 😭 😭 😭 😭 😭 😭 😭 😭

i got the sin (x/2)

and cos and tan

well that was easy ..

now i need to understand the sum product thing

[gloomy goblet]

eventually these rules u will me memorising them

[gloomy goblet]

can you send it

[unkempt spindle]

for sure

[gloomy goblet]

do you memorize all of your rules ?

[unkempt spindle]

i didnt

im doing it rn

[gloomy goblet]

these are the rules

u want me to show you proof of them

or what

[unkempt spindle]

mhm

[gloomy goblet]

what is ur question then.

[unkempt spindle]

one sec

i dont get this kinda

[gloomy goblet]

you studied them ?

the csc and sec

[unkempt spindle]

yeah

[gloomy goblet]

u dont understand periodicity?

[unkempt spindle]

mhm

how does the answer = to sintheta

is that just a rule i memorize or what

[gloomy goblet]

which one

[unkempt spindle]

like any sin cos tan (x+2pi) = sinx cosx tanx

[gloomy goblet]

in math we have what we call a periodic functions for exemple if you take cosin or sin when u study them u use the circle right?

[unkempt spindle]

yup

[gloomy goblet]

if you make a 2 pi turn

don't you return to the same point?

[unkempt spindle]

yes

360

1 rev

[gloomy goblet]

so going around or backwards with 2pi makes no difference?

[unkempt spindle]

yes

[gloomy goblet]

and no matter how many times you turn

with 2pi

5 ,10,6556 times

u always stay on the same point

[unkempt spindle]

u still end up at the same point

cuz its 1 rev

[gloomy goblet]

exactly

that is what we call periodicity

when you add something to x

and you get the same result

sin cos and tan are called 2pi periodic

[unkempt spindle]

ohhh I GET ITT

[gloomy goblet]

which means if you or substract 2 * n * pi ( n is an integer can be 1 ,2 ,3 ,4.....612516) you get the same answer

in general we type cos(x)=cos(x+2npi)

[unkempt spindle]

YEAHH

[gloomy goblet]

because no matter how many n times we turn

we get the cos of x

or sin or tan

[unkempt spindle]

but what if its 3pi

[gloomy goblet]

okay okay you're playing with fire i like it 😄

[unkempt spindle]

it wouldnt be the same

[gloomy goblet]

yes

how we solve that ?

[unkempt spindle]

itll be 1.5 rev

[gloomy goblet]

pretty easy

we use the first thing we know

always whn you have x+deta

try to make that deta between -pi,pi or 0,2pi whatever interval u're using that is 2pi long

let is work with 0,2pi if u want as an interval

in our case 3pi is out of this interval

[unkempt spindle]

mhm mhm

[gloomy goblet]

so what we can do to get it back in this interval is substract 2pi which means get a -2pi there since going back with 2pi won't change the result

right?

[unkempt spindle]

yes

[gloomy goblet]

so it becomes x+pi right?

[unkempt spindle]

correct

[gloomy goblet]

okay in this case we get different result depending on what functions you're using

let is start with cos

try to get a point in a circle

[unkempt spindle]

hmm

∏/2

[gloomy goblet]

what can you notice here

[unkempt spindle]

its symmetrical by the origin point

[unkempt spindle]

yes

[gloomy goblet]

exxactly

so what we can notice is

that cos(x)=-cos(x+pi)

[unkempt spindle]

odd

function

[gloomy goblet]

no no

that is something else

[unkempt spindle]

i thought cos is always even

[gloomy goblet]

forget about odd

and even

for now

[unkempt spindle]

okok

can u call

[gloomy goblet]

the conclusion is

[unkempt spindle]

that'd be cookinn

[gloomy goblet]

when u add pi to cos x

you get a minus out

[gloomy goblet]

wanna go on a call?

[unkempt spindle]

yea

[vocal sleet]

@unkempt spindle Has your question been resolved?

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[raven bronze]

So I have to find AF/DE

[vocal sleet]

@raven bronze Has your question been resolved?

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@raven bronze Has your question been resolved?

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@raven bronze Has your question been resolved?

[raven bronze]

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[valid sleet]

Hey need some help on this question

[valid sleet]

I understand how to find the values in which cos is equal to 1/2 and in which sin is equal to radical2/2

but from there im lost

[coral chasm]

Ok tell us the possible values of theta

[valid sleet]

cos could be pi/3 or 5pi/3

pi/4 or 3pi/4 for sin

[coral chasm]

Ok focus first on cos

Since we have an inequality, what values of thete from 0 to 2pi is the inequality true?

The answer must be an interval

[valid sleet]

would it not be true from (0,pi/4), (5/3,2pi) ?

[coral chasm]

Are you referring to the answer of the whole question?

[valid sleet]

No just for cos

[coral chasm]

In this interval, pi/3 is not included but if we solve it 2cos pi/3 is 1 which is greater than of equal to 1

[valid sleet]

oh sorry i apolgoize, i meant pi/3 and not pi/4

im just a little lost overall

[coral chasm]

Its okay, youre on the right track

To make it clear the interval for the cosine part is [0,pi/3]U[5pi/3,2pi]

Right?

[valid sleet]

Yessir

[coral chasm]

Okay how about the sine part

Tell me its interval

[valid sleet]

well i know it is [0,pi/4]

and then would it also be [3pi/4, pi]?

[coral chasm]

Yes but 2pi

[valid sleet]

yes my mistake, typo

[coral chasm]

Its okay,

Now you just have to find the intersection of those two interval

Then you can have the answer

I suggest you try to sketch it in a number lime so u wont get confused

[valid sleet]

okay

[valid sleet]

yup

been doing that

[coral chasm]

Okay nice

[valid sleet]

okay so

one of the intersections would be [0,pi/4] right?

[coral chasm]

Correct

[valid sleet]

okay i got that part

now the 2nd interval confuses me

as to how id find the connection from 5pi/3 and 3pi/4

[coral chasm]

Okay try to sketch in the number line 5pi/3 up to 2pi

[valid sleet]

got it

[coral chasm]

And that goes the same for 3pi/4 up to 2pi

And whatever interval they intersect will be the answer

[valid sleet]

3 pi/4 right ?

[coral chasm]

Ah yes sorry typo

[valid sleet]

no worries

well i made the numberlines yet I am still a little ocnfused

[coral chasm]

Can I see?

[valid sleet]

yea one secod

[coral chasm]

Try to sketch them in a single number line

[valid sleet]

done

[coral chasm]

Can I see

[coral chasm]

Okay can you see the intersection?

[valid sleet]

I am still a little confused on that sorry

[coral chasm]

Okay, try to sketch or like draw another line from 3pi/4 to 2pi

And to the same to for 5pi/6 to 2pi

So that you can see the intersection better

[valid sleet]

got it

[coral chasm]

Can I see

[valid sleet]

you mean two seperate lines correct?

[coral chasm]

Yes but still on this number line

[valid sleet]

wdym?

[coral chasm]

So that you can visually see the interval of each

Okay Ill draw it for you

[valid sleet]

haha thanks man sorry

[coral chasm]

Like this

Can you understand the lines?

Its the intervals of each

[valid sleet]

yes okay got it

[coral chasm]

Now can you tell me the interval where they intersect

[valid sleet]

well they intersect from 5pi/6 to 2pi no?

[coral chasm]

Yesss

Now you have the answer

Can you tell me the full answer

[valid sleet]

well would it be [0, pi/4] U [5/6, 2pi]?

as thats where they intersect

[coral chasm]

Good job just fix the typos hahahaha

[coral chasm]

5pi/6

[valid sleet]

got it i see

but the issue is

none of the answer choices have that

i assume the answer is A

[coral chasm]

Wait

[valid sleet]

but like, for the 2nd interval on A i do not understand where they got 5pi/3

[coral chasm]

We made an error

It was 5pi/3 from the start

Idk where we got the 5pi/6 hahaahha

[valid sleet]

oh yeah

oops

hahaha

ahhhh

okay that is actually pretty easy

[coral chasm]

Hahahaha but do you understand it now?

[valid sleet]

Yessir i got it now, thank you man

[coral chasm]

Nice, youre welcome

Goodluck bro

[valid sleet]

is there any like part that requires testing regions or smething like that?

[valid sleet]

thank you man 🙏

[coral chasm]

I guess its needed when you want to be sure where the interval is, for example in the cosine part we know its 1 when it is pi/3

Then you can test region for example values for theta then its less than pi/3 or greater than pi/3

Then whatever region that makes the inequality true, then thats the region that u include it

[valid sleet]

i see, perfect makes sense

i appreicate the help bro you were great at explaining it haha

[coral chasm]

Im glad I was able to help you hahaha

[valid sleet]

have a great rest of your day/night 🙏

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[ruby kestrel]

can someone help me with this

anyone?

[arctic oar]

Im lookin

[ruby kestrel]

ok

anything

hello

[arctic oar]

ya thats hard

[ruby kestrel]

thats why im asking for help

[arctic oar]

lol

[visual marsh]

you giving jee this year?

[ruby kestrel]

yep

[visual marsh]

ah me 2

[ruby kestrel]

i found this in my book

[visual marsh]

bro there are like tons of solutions inyt check them out

2016 paper 💀

there are two deadly roational questions in the 2016 paper

[ruby kestrel]

ik

[visual marsh]

but why preparing for adv when mains is in 15 days?

[vocal sleet]

@ruby kestrel Has your question been resolved?

[ruby kestrel]

no

already did my mains

.close

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[jagged cargo]

i wanna check my proof that composition satisfies additivity

$T(f+g)(x) = (f+g)(q(x)) \stackrel{*}{=} f(q(x)) + g(q(x)) = (Tf)(x) + (Tg)(x)$

[jagged cargo]

where (*) makes sense because q is fixed

am i correct here?

[cobalt crypt]

uh

yes?

[jagged cargo]

is there anything with it?

i was unsure

[cobalt crypt]

with it?

[jagged cargo]

yeah

[cobalt crypt]

what does that mean?

[jagged cargo]

actually, does it even matter if q is fixed?

either way, thanks for confirming

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[cobalt crypt]

well (f + g)(q(x)) wont equal f(q(x)) + g(r(x)) ig

if you uh

change q to r

[jagged cargo]

oh i see

[cobalt crypt]

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[exotic ravine]

i dont get how to start

ping me please

[fiery elbow]

work = computer times days

[vestal meteor]

I need help in this

fomula d=rt

[vocal sleet]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[exotic ravine]

17M?

[fiery elbow]

since the total work doesnt change, lets let total work (W) equal to 17M

[exotic ravine]

then?

i think i would do work done in 1 day right?

[fiery elbow]

at first you have M computers, then everyday they decrease by 4

so the work would be M + (M-4) + (M-8) …

[exotic ravine]

m,m-4,m-8... 25 terms?

[fiery elbow]

ueah

[exotic ravine]

cuz its 25 days

okayy its an AP sum

i get it

[fiery elbow]

then you set the two works equal

[exotic ravine]

yeah i got 150

[fiery elbow]

👍

[exotic ravine]

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[vast shale]

Here, I get what's going on with the product rule being applied to get from y to y'.

But then the 3x^2

I don't see why the chain rule is being applied to (2+x^3)^-1

I would have thought it would be just e^x(2+x^3)^-1 + e^x(-(2+x^3)^-2)

Like if the chain rule is d/dx[f(g(x))] = f'(g(x)) * g'(x) then with (2+x^3)^-1 <---- that's only the g(x) portion right? what am I not seeing here please?

[empty frigate]

f(x) = x^-1
g(x) = 2+x^3

[vast shale]

Oh okay I kind of see what you mean

so the f(x) = x^-1 is the ( )^-1 ?

[empty frigate]

yep

[vast shale]

Okay so gonna follow up with something that might seem obivous, I just can't see it.

e^x(2+x^3)^-1

I don't see an x there being multiplied to the (2+x^3)^-1...

Thanks for helping btw.

[empty frigate]

...?

why would (2+x^3)^-1 be multiplied by x

[vast shale]

It wouldn't by looking at it I guess but you said f(x) = x^-1 so I thought it was?

[empty frigate]

the x in f(x) = x^-1 isn't the x of the expression you're differentiating, it's what ends up being the 2+x^3

$f(\textcolor{red}{x}) = \textcolor{red}{x}^{-1}$, so $f(\textcolor{red}{2+x^3}) = (\textcolor{red}{2+x^3})^{-1}$

[twin meteor]

bee [it/its]

[vast shale]

Interesting, just gonna work it out, brb in 5min lol

thanks again.

I get it!

okay perfect yeah I just had to see how the rules relate and now it makes intuitive sense

thank you! @empty frigate

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[random ice]

Inverse cos and normal cos in derivatives, how does this work?

Specifically

[random ice]

from what I know f'g * g'

but on the second picture where is the g?

[strong grove]

also there shouldnt be a 3x in the 3rd line , d/dx 3x = 3

[random ice]

where is g'

[strong grove]

the 10x is g'(x)

g(x) = 5x^2
g'(x) = 10x

[random ice]

oh but then where is g(x)

inside the sqrt?

[strong grove]

do u know whats the derivative of arc cos (or cos inverse)

[random ice]

-1/sqrt 1-x^2

[strong grove]

yeah , and for g(x) those x would be change into g(x)

[random ice]

hmm im trying to process this as I weirdly didn't see this for trigonometric functions

cause in cos x, its -sin x

right

[strong grove]

derivative of simple trig was easy , cos turned into sin

[random ice]

yea

[strong grove]

yeah , but for inverse its different

[random ice]

lets say for sin(5x^3)

it would be

cos(15x^2) * (15x^2)

[strong grove]

no

[random ice]

its wroong?

[strong grove]

yes

u said
(f(g(x))' = f'(g(x)) * g'(x)
use that

[random ice]

oh bruh

my bad

cos(5x^3) * 15x^2

and y - sin^-1(2x-5) would be

1/sqrt (1-2x-5)^2 * 2

[strong grove]

if u meant
$$\frac{1}{\sqrt{1^2 - (2x-5)^2}} \cdot 2$$ then yeah

[twin meteor]

JustToPro

[random ice]

yea I kinda still can'

t wrap my head around it

but it kinda make sense since or else where would the g' go

g*

[strong grove]

easy way to learn about it

$$sin^{-1} (2x-5) = f(x)$$
let $u = 2x-5$
$$\implies sin^{-1} u $$

[twin meteor]

JustToPro

[strong grove]

differentiate with respect to x

u = 2x-5
du/dx = 2

f(u) = sin^-1 u
f'(u) = 1/(1-u^2) du/dx

swith u and du/dx into x

and this is chain rule

[random ice]

chain rule?

ik what is chain rule but what do you mrean

[strong grove]

all that i wrote is chain rule

[random ice]

wait is u = 2x-5 and du/dx = 2 related

no right?

srry im kinda confused cause im not that familiar with implicit differentiation yet, just learned it like awhile ago

[strong grove]

let me type it in texit so maybe its easier to notice

$$sin^{-1} (2x-5) = f(x)$$
let $u = 2x-5$
$\implies \frac{d}{dx}u = 2$
$$f(u)= sin^{-1} u $$
$$f'(u) = \frac{1}{\sqrt{1-u^2}} \cdot \frac{d}{dx}u$$
From above
$$f'(x) = \frac{1}{\sqrt{1-(2x-5)^2}} \cdot 2$$

[twin meteor]

JustToPro

[strong grove]

but yeah , they are related

[random ice]

oh

the derivative of u

I thought that was something else

ahhh

I see

I understand now

Thanks a lot for making it more understandable @strong grove

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[ashen veldt]

i need help with this😓😓

[ashen veldt]

its abt Multiplying and dividing rational expressions

[atomic jasper]

!status

[vocal sleet]

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

[ashen veldt]

1? 😓

[grim lotus]

,rotate

[grim lotus]

ur supposed to find the area?

[ashen veldt]

yes yes

[grim lotus]

area of rectangle = length * width

use that

u can split any shape into rectangles

[ashen veldt]

okay but how do i find the perimeter of the figures???

[grim lotus]

sum of side lengths

find the length of each side then add them all up

start with q48

[ashen veldt]

okok thabks for the advice

[vocal sleet]

@ashen veldt Has your question been resolved?

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[vast shale]

Completely forgot how to do this after winter break

[dull bear]

For b), can you work out what the lengths AC and OC are?

For c), you may find it easier to work out the area of the sector AOB, and the area of the triangle AOC

[vast shale]

I just got it thank u can u help me w another trigo problem

[dull bear]

For part (a), you may wanna “add in” a point on MP such that you break the trapezium into a rectangle and a triangle (the rectangle having height 5cm)

That should then hopefully make it easier to find the angle MPQ from the triangle, as you can figure out two of the side lengths “instantly”

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[vast shale]

I know how to get the exponential

Not sure what to do after that

[round plover]

in exp form, you can multiply the exponent easily [e.g. (Re^it)^8 = R^8e^(8it)]

and a real number has exponent a multiple of pi

since rotating by pi lands back on the real axis

similarly, a^m is imaginary whenever the exponent is a half pi past a multiple of pi

[vast shale]

Gotchu thanks

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[crude yoke]

Is there any way to simplify the Area function? https://www.desmos.com/calculator/cgyjpczpbe

[crude yoke]

I’m trying to find the analytical solution for the motion blur of a disk moving in a circular path. I’m integrating u(x, y, t) / (b - a) from t = a to t = b. My approach involves turning the inequality condition in u into equality, solving for t, and getting two sets of solutions: A(x, y) + B(x, y, k) and -A(x, y) + B(x, y, k) (the first is larger)

Then, I compute n1 as the smallest k where the larger solution is > a, and n2 as the largest k where the smaller solution is < b. After that, I sum the areas for all intervals in between, excluding n1 and n2, and add the partial areas at n1 and n2. Finally, I divide everything by (b - a) and simplify

It works beautifully, but the function is way too long. Is there a way to make it more elegant?

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@crude yoke Has your question been resolved?

[crude yoke]

<@&286206848099549185>

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@crude yoke Has your question been resolved?

[crude yoke]

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[gaunt zenith]

my problem is that I don't know how to use x^2 + bx = c, and i don't know if it is any different from ax^2 + bx = c.

[heady vine]

it is the same form just with a=1

[gaunt zenith]

but how to use x^2 + bx = c?

[heady vine]

do you have the original problem

[gaunt zenith]

it gave me the rule and x^2 + 14x = 15 and said solve by completing the square

[heady vine]

and you know what is completing the square ?

[gaunt zenith]

yes: when a trinomial is a perfect square, there is a relationship between the coefficient of the x-term and the constant

that is how we studied it

[heady vine]

so you'll add b/a to both sides

[gaunt zenith]

b/a?

[heady vine]

i'll visualize it

@gaunt zenith
So when someone says completing the square you are just literally complete the square by adding a constant term
In this example the left square (after we added 49) is now a complete square with side length (x+7)

[gaunt zenith]

thank you very much I made you tired that is so much easier

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[lilac plaza]

Hello

[lilac plaza]

Why is f_y(1, 2) not 1/e^2

Where is -1-(1)^2 coming from

[heavy yoke]

well we have -1 + x^2 + xy
x = 1, y = 2, so x^2 = 1 and xy = 2

[lilac plaza]

Why is there two additions?

Isn't this correct?

Oh wait

My bad

Thanks ahaha...

.close

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[lilac plaza]

.reopen

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✅

[lilac plaza]

Umm

I misread

I still stand correct

[heavy yoke]

your derivative at the top is incorrect, the partial derivative of e^(xy) wrt y is xe^(xy)

[lilac plaza]

Ohhh right

Which equals 0

Then

Ahh thank you so much

.close

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[tired sable]

Question 59, how did we get the sin and cos? shouldn't sin be -sqrt(3)/2?

[hushed pewter]

It's not asking you to find sin

[tired sable]

Wym?

We need it for the sin(x + phi) formula no?

[hushed pewter]

Oh you want it that way, I see. I thought it wanted just $\cos x=\sqrt{1-\sin^2 x}$

[twin meteor]

SWR

[hushed pewter]

Either $\sin\alpha$ or $\cos\alpha$ could be $-\frac{\sqrt3}2$. The choice is arbitrary and you will end up at the same result.

[twin meteor]

SWR

[tired sable]

Why can they be the same?

Oh the R is multiplied I see

[hushed pewter]

One would give you the cosine angle sum, but it would be easy to transform it to sine angle sum. But I see how I misled you. One moment

You want to rewrite your expression in the form $R\sin(x+\alpha)$. Use angle sum formula: $R\sin(x+\alpha)=R(\sin x\cos\alpha+\cos x\sin\alpha)$

[tired sable]

Yes this is what I'm trynna do

[twin meteor]

SWR

[hushed pewter]

Do you see from this how $(R\cos\alpha)\sin x+(R\sin\alpha)\cos x=-\sqrt3\sin x+\cos x$?

[twin meteor]

SWR

[tired sable]

Yes

I got it, thank u brotha

.solved

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[tired sable]

May sum1 confirm pls?

[flat whale]

just plot both in desmos

[tired sable]

.solved

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[tired sable]

.reopen

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✅

[tired sable]

Why am I wrong? Did I have to factor the 2x into the R when I find the sin and cos?

[flat whale]

show your full steps

[tired sable]

I did inverse sin to get pi/3

[flat whale]

that only applies when it's Acos(x) + Bsin(x)

you have 2x

[tired sable]

So

Did I compute these incorrectly?

[tired sable]

Should I have factored in the 2x?

[flat whale]

you can just set t = 2x and then solve for t

then to get back to x, re substitute x = t/2

[tired sable]

Sorry I'm confused wym?

What is t?

[flat whale]

instead of 2cos(2x) = 1. it's 2cos(t) = 1

[tired sable]

Ahh I see

Okay ty

.solved

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[surreal herald]

I do not understand how they get this

[surreal herald]

Or I guess I don't understand how they solve it here

I can sort of understand the X_3 bit

where it's -5 because it's solving for x_3 and the other variables are 0, so it's wanting some number that multiplies by -4 to get to 20

but the X_1 part has me totally confused

[stone gazelle]

x_3 is trivial, you have -4x_3 = 20 so x_3 = -5

[surreal herald]

yeah, that part I can understand

[stone gazelle]

As for x_1 and x_2, you have, from to top row of the augmented matrix:
-2x_1 - 8x_2 = 8

You noticed already that x_2 is free, so it can take on any number; it’s like any other variable. Which means you want x_1 in terms of x_2

Because you can’t know the x_1 that satisfies -2x_1 - 8x_2 = 8 without knowing x_2 first, but x_2 can be anything

[surreal herald]

So couldn't we just say x_2 is 0 and x_1 is -4?

where did they get the uh.. -4X_2?

-2 * -4 = 8

[stone gazelle]

-2x_1 = 8 + 8x_2
Divide by -2
x_1 = -4 - 4x_2

[stone gazelle]

That would only be one solution though

You are asked for the general solution

[surreal herald]

I understand this

let me try another one of the problems, and thank you for taking the time to explain it

the book I'm using, I dn't understand it the best or it feels like it doesn't explain it the best, because I'm getting to, what I think is the parametric form

[stone gazelle]

I don’t think i’ve come across a solution expressed like they did for ‘general solution’ before. I’ve mostly come across parametric form

[surreal herald]

I typed it wrong, but I was able to follow the steps you taught me

The book definitely feels strange or hard to follow at times

Yep, I understand it now

Thanks for teaching me and helping me

I feel like the way they explained it could've been better, or they could've elaborated more on the steps to take

Especially since when I took calculus we had to basically explain every step and every rule we used to get an answer, the textbooks should follow the same requirement to express every step so a learner or reader can go 'oh ok I see every step they did without the shortcuts'

[stone gazelle]

Yeah, they could have added a quick sentence before the solution explaining how they went from the system of equations to parametric form

[surreal herald]

on the other hand this helps me understand why we have to usually write out the rule we use and aren't allowed to just make shortcuts, it leads to things being confusing at times x.x

I'll close the ticket in a minute

[stone gazelle]

I think this seems good at surface level, but being able to extrapolate what is going on when some steps are skipped is definitely a skill that should be improved

[surreal herald]

I'll try working on that in the future

thank you again for helping

.close

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[wicked mist]

Question and answer format. I know how to expand it i just dont understand what answer it wants

[pale perch]

it wants you to enter the coefficients (and constant)

[wicked mist]

howd i get that?

[pale perch]

by expanding it

[wicked mist]

but like once expanded

[pale perch]

youll get x^3+bx^2+cx+d
the answers are b,c and d

in that order

[wicked mist]

ohh

thanks

yup i got it rights

thanks again

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[graceful scarab]

So do I plug the function f(x) into each x in the function g(x)? I've never had to plug in functions that are divided by one another so im quite confused

[atomic jasper]

yes, thats how it works

[graceful scarab]

alrighty thank you, I was just confused on the fac of putting a fraction in another fraction

[atomic jasper]

$g(f(x))=\frac{5-f(x)}{2f(x)-1}$

[twin meteor]

Bonk

[atomic jasper]

and then fill in f(x) in there ofc

[graceful scarab]

gotchu ty very much :3

Now how would I start upon simplifying this? TvT

[atomic jasper]

i think *(2x+1) would be a good start

[graceful scarab]

Oh thats a fraction, I write fairly big and messy so my apologies

[atomic jasper]

yes i know

multiplying the top and bottom by (2x+1) is a good start

[graceful scarab]

OHHH bc then that will cross out both 2x+1, correct?

$g(f(x))=\frac{5-(x+5)}{2(x+5)-1}$

[twin meteor]

Kirbynt

[atomic jasper]

almost

[graceful scarab]

oh wait i forgot abt the constants

[atomic jasper]

the 5 should be 5x(2x+1) and the 1 should be 1(2x+1)

[atomic jasper]

you forgot to distribute to the 5x

[graceful scarab]

Oh wait i messed up there was no 5x on g(x)

[atomic jasper]

ah, youre right

then you need to distribute to the 5

[graceful scarab]

I assume the 2 at the bottom as well?

[atomic jasper]

wdym? the 2?

[graceful scarab]

the 2 on the denominator beside the x+5

[atomic jasper]

i dont understand

[graceful scarab]

oh dear i just realized my 2's look like z's

[atomic jasper]

you are still missing the (2x+1) on the 5

[graceful scarab]

whoops ty

so then that leaves me with

$g(f(x))=\frac{(-5x+25)}{(2x+10)}$

[twin meteor]

Kirbynt