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[devout spruce]

can somebody help me? i will be glad for this

[lone heart]

@devout spruce Has your question been resolved?

[worn fox]

Probably write out its matrix and diagonalize

[devout spruce]

i wrote out

but eigenvalues is float

<@&286206848099549185>

[worn fox]

Show your work

[tacit arch]

^

[devout spruce]

okay

these are eigenvalues lol

[lone heart]

@devout spruce Has your question been resolved?

[devout spruce]

<@&286206848099549185>

[tacit arch]

Show how you got the matrix

[devout spruce]

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[devout spruce]

.reopen

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[violet bear]

first you should cover the 25.4 meters into centimeters, then subtract diameter of the spool plus the diameter of the tape by the diameter of the spool, then divide the length of the tape by the difference of the diameters

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@mellow crane Has your question been resolved?

[mellow crane]

so 25.4m is 2540cm

and then I do 3.6-(3.6+5.4)=5.4?

and then 2540/5.4= about 470?

but that's not a multiple choice answer?

[naive valley]

you need to work out how much tape is in each layer

for example, the first layer is at radius 3.6/2 = 1.8

what's the circumference of a circle with that radius?

[mellow crane]

3.6pi?

[naive valley]

yep

and the outermost layer is at radius 5.4/2 = 2.7, what's that circumference?

[mellow crane]

5.4pi?

[naive valley]

yep

now perhaps you could argue that on average, each layer has a circumference that is the average of those two

(it's not quite exact but maybe close enough to get the right multiple choice answer)

[mellow crane]

wait so do I do (3.6pi+5.4pi)/2?

[naive valley]

yea

that gives you the length of the average layer

(the ones closer to the spool are shorter and the ones farther out are longer)

[mellow crane]

wait but that average is 9/2pi which is also not in the multiple choice answers

[naive valley]

well that's because the question is asking how many layers, not what is the length of the average layer

but do you see how to calculate the number of layers, given the total length of the tape and the length of the average layer?

[mellow crane]

do I divide the total length by the number of layers?

[naive valley]

the number of layers is what you want to calculate, you don't know it yet

try: total length = (number of layers)(length of average layer)

[mellow crane]

so 2540=9.2pi*# of layers?

[naive valley]

where did 9.2 come from

i calculate 4.5

[mellow crane]

sorry, 9/2

[naive valley]

oh yea

[mellow crane]

I typed it wrong

[naive valley]

yep that equation is right

[mellow crane]

so then 2540/(9/2pi)=about 179.67 which is closest to 180?

[naive valley]

that's what i get

[mellow crane]

Ohh, ok, tysm

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[vague cloak]

here's my working, but, idk why i get two answers

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@vague cloak Has your question been resolved?

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@vague cloak Has your question been resolved?

[vague cloak]

nvm

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[warped topaz]

https://gyazo.com/70252a7e83e5bf6f6012f51903caadfc.png

[warped topaz]

I need the coordinates of P and Q

[last ether]

What are your circle equations

[warped topaz]

(x-a)^2 - (x-b)^2 = r

[last ether]

Well sure. You have two of those though

(Also it's r^2)

[warped topaz]

oh

[last ether]

$$(x-1)^2 + y^2 = 2$$
$$(x+2)^2 + (y-1)^2 = 4$$

[ocean seal]

Umbraleviathan

[warped topaz]

hmm

[last ether]

So it's an annoying system of equations

[warped topaz]

let me try something really quick

Cause maybe not squaring r was my mistake

[last ether]

Probably

I usually use a calculator for this kinda algebra tho because

Well frankly algebra kills everybody, regardless of skill, ultimately

[warped topaz]

I will close for a moment an possibly reopen

.close

[lone heart]

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[warped topaz]

.reopen

[lone heart]

✅

[warped topaz]

Still stuck lol

I thought I knew what I was doing, i got both equations in the form (x-a)^2 + (x-b)^2 - r^2 then set them equal to eachother but it didnt work

I got y value for P = 0 which is clearly wrong

[radiant jasper]

Did you solve It as a system of equations?

[warped topaz]

What do you mean

Im not sure what a system of equations is

[radiant jasper]

Solve for y or x on one of the equations and then plug It In on the other

[warped topaz]

well I just tried for P first since I knew the x was 0

but yeah I got y = 0 which is incorrect obv

[radiant jasper]

Y=1

[warped topaz]

how did you get that

[radiant jasper]

$$(0-1)^2 + y^2 = 2$$
$$ 1+y^2= 2$$

[ocean seal]

oxil764

[radiant jasper]

Then y=+-1

In This case clearly 1

[warped topaz]

Ok I get it

then for Q?

[radiant jasper]

This

[warped topaz]

I dont understand

plug which into which

[radiant jasper]

The other equation

[warped topaz]

(x-a)^2 + (y-b)^2 = 4

theres 4 unknowns

[radiant jasper]

This

[orchid parrot]

nedd steps for this

[warped topaz]

you nedd a different channel

theres 3 open bro

[warped topaz]

go #help-3

[elfin breach]

@warped topaz both the equations represent circles

plot it on a paper and find the poi

[warped topaz]

that feels like cheating a little bit, I need to know how to do it mathematically

[elfin breach]

that's not cheating dude

that's a graphical way of solving

but sure you can solve the quadratic in x by substituting y with x

the former would have been faster

[warped topaz]

I will try this, I think I understand you

[elfin breach]

bet

just for your reference

[warped topaz]

I now have (x-1)^2 + (x-1)^2 - 2 = y

am I doing this right

[elfin breach]

you need everything in terms of x

it's a tedious process

[warped topaz]

why x?

[elfin breach]

how you going to solve it?

manually plugging values?

[warped topaz]

I dont really know what im doing

No matter what I have 2 unknowns

[elfin breach]

give me a sec

odie do you know the basics of circles?

like in coordinate geometry

[warped topaz]

Im learning it right now

so kinda

[elfin breach]

are u aware of the equations?

[warped topaz]

https://cdn.discordapp.com/attachments/490557019623915520/1056795645471633519/349354952961032192.png

like this ^^ ?

[elfin breach]

yes

there's a method of finding poi of circles via the equations

but idk if your assignment allows you to do that

[warped topaz]

I think thats what im supposed to do

https://gyazo.com/70252a7e83e5bf6f6012f51903caadfc.png

I already got P since I knew the x of P is 0

but I dont know any points of Q so idk how to get the coordinates

[elfin breach]

you need to find the eqn of common chord of the circle

and from there you'll get ur two points

[warped topaz]

whats that

[elfin breach]

a chord that's common to both the circles

you'll probably learn it soon

[warped topaz]

I just googled it, I dont think ive covered it yet

weird that I would get this question before ive covered it

[elfin breach]

just ask your teacher for assistance then

[warped topaz]

Dont have one, I will ask youtube instead haha

Ok thank you for your help I will go watch some tutorials!!

[elfin breach]

👍

[warped topaz]

.close

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[warped topaz]

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[warped topaz]

is it possible to solve this without using the common chord of the circle? Cause I didnt cover it yet so there must be another way...

https://i.gyazo.com/70252a7e83e5bf6f6012f51903caadfc.png

I know that P is (0,1)

but I dont have x or y of Q so idk how to get it using any of the circle equations

[alpine sable]

In solving for P you had to solve a quadratic right?

[warped topaz]

no...

[alpine sable]

Couldnt you use the equations of both circles to get a system of quadratic equations?

[warped topaz]

I dont follow sry

what do you mean a system of quadratic equations

[alpine sable]

from this

[warped topaz]

but I still have 4 unknowns?

[alpine sable]

Well we know from the diagram that there are two such points that these equations are satisfied from

[pliant cedar]

ur equation shall have 2 pairs of solutions, but 2 unknowns

[alpine sable]

Both of these are accounted for as each quadratic has 2 solutions

[warped topaz]

So (x-0)^2 + (y-1)^2 = 2 turns into (x^2) + (y^2 - 2y + 1) = 2 , is that what you mean?

[pliant cedar]

how do u know that P is 0,1

[alpine sable]

It lies on the y axis so its x coord is 0 solve for y

[pliant cedar]

is this given

[alpine sable]

(from either of the equations of the circles)

its implied by the diagram

but nowhere explicitly stated

[pliant cedar]

well we dont know that exactly

maybe its like 0.0001, 1.0002

anyway if u know that P is (0,1), then u can just find the reflection of P about the midpoint of the centers of the circles

to find Q

[warped topaz]

Is the problem possible to solve if we assume that P2 might not be 0?

[pliant cedar]

assuming P is (0,1), u can do what i said above

[pliant cedar]

it is

can u write the equations of the two circles

[alpine sable]

It is 0, plugging into both circle’s equations give 1 for both the y coords

[pliant cedar]

right so it just happens to work in this case

[warped topaz]

Im not sure how to do this

[pliant cedar]

do u know how to find the midpoint

of two points

[warped topaz]

Are we looking for the midpoint of P and Q?

[pliant cedar]

yes

i mean no

[alpine sable]

I feel you shouldn’t use that method here as generally the coords of one point wont be given

[pliant cedar]

we want the midpoint of P and Q to also be the midpoint of the centers of the circles

[pliant cedar]

yes

[pliant cedar]

can u do this tho

[warped topaz]

For P (x-0)^2 + (y-1)^2 = 2 for Q (x-Q1)^2 + (y-Q2)^2 = 4

is that what you mean

[pliant cedar]

no..

like (x-a)^2 + (y-b)^2 = r^2 is the equation of a circle with center (a,b) and radius r

[warped topaz]

yes

[pliant cedar]

so one of ur circles has center (1,0) and radius root(2)

what is its equation then

oh

[warped topaz]

(x-0)^2 + (y-1)^2 = 2

[pliant cedar]

(x-1)^2 + (y-0)^2 = 2

u r mixing up the coords

[warped topaz]

is this for P?

or for 1,0

ahhh the centre

Sry I missread

[pliant cedar]

its not "for" anything, it is the equation of the circle

[alpine sable]

These are the equations

[warped topaz]

yes sorry

i get you

[pliant cedar]

right now u must solve these two simultaneously

can u do that

[warped topaz]

Im not sure how cause I have 2 unknowns

[pliant cedar]

have u learnt how to solve eqatuions with 2 unknowns?

like equations like 2x + y = 5, and x - 3y = 8

[warped topaz]

so like isolate x or y

[pliant cedar]

yes

[warped topaz]

then put in the new value

etc

I tried that with this

[pliant cedar]

u can try subtracting the two equations

[warped topaz]

I got (x-1)^2 + (x-1)^2 - 2 = y

is this right?

[pliant cedar]

how did u get that

[warped topaz]

Just isolating y

[pliant cedar]

can u like show ur work

[warped topaz]

Ok 1 moment

really messy sorry

[elfin breach]

I'll tell you something

common chord eqn is pretty simple

S1-S2 = 0

just subtract the two equations of the given circles

so the second degree x and y are eliminated

[pliant cedar]

yes

[warped topaz]

I will learn this way too once I reach this chapter, but I want to learn this way first

[elfin breach]

alright

its basically the same thing

[pliant cedar]

u dont need to know its called the "common chord eqn"

[elfin breach]

^

[warped topaz]

I have to take a break from this 😅 I will come back to it

Thank you guys for all of your help

you are so kind and patient

❤️

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[reef iris]

So I'm learning discrete math atm and I know what the reverse A and E symbols mean, but I saw something about their rules and I can't really interpret it well (even with a translator)

[reef iris]

For example this is what they use for (12)

From what I understand, if a function has a valid input for all values of X, it's valid for all inputs of Y?

And if someone has a link to this topic in English, I'll gladly devour it

[pliant cedar]

idk what the chinese is saying here

$\forall x, A(x)$

[ocean seal]

SilverSoldier

[pliant cedar]

might mean for all values of x, A(x) is true

[reef iris]

Wait

Lemme just translate it

[pliant cedar]

can u translate what the chinese is saying

yeah

[reef iris]

I can't keep making AB questions like this lol

Let a range be defined for all of Z

• Where P(X): X is an even number
• And Q(X): X is divisible by 2
• Let P(X) -> Q(X) be true

"Prove"
(1) Already stipulated
(2) Already stipulated
(3) Can be found by combining (1)(2) P(X)^(P(X)->Q(X)
(4) Here's the wrong step, but I do not understand this

It's wrong because clearly you can't assume all numbers are divisble by two, but what does that symbol on the lower right corner even mean?

Is there a rule or anything related to this I could perhaps search in English and read up on it?

The reason this confuses me so much is because in the previous powerpoint slide we had this rule explained, which states: "If A(X) is true, then all values of X in A(X) is true; when X is a independent variable and not a free variable"

/explanation

Oh I got, it was a mistranslation, abort abort. It was "If A(X) is true, then all values of X in A(X) is true; when X is a independent variable and does not appear in the front of the equation"

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[ivory dew]

sorry for the quality

help me please

[orchid valley]

hi

[ivory dew]

Hello

[tacit arch]

This translation doesn't make much sense in English

[lone heart]

@ivory dew Has your question been resolved?

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[sick smelt]

I solved the same question same times thrice, each time differently, only 3rd time I got the answer right.
I would like to know what I did wrong in the 1st two times I solved it.

[sick smelt]

Question. (√6+√8)^2

[rocky grove]

8 = 2^3

Also

[sick smelt]

I feel dumb 😩

[rocky grove]

$(a+b)^2 = a^2 + 2ab + b^2$

[ocean seal]

VulcanOne

[sick smelt]

yea that's what I used, It just goes a bit off page

[rocky grove]

Not $(a+b)^2 = a^2 + 2(a+b) + b^2$

[ocean seal]

VulcanOne

[sick smelt]

oh

If I hadn't done those mistakes would the answer have been right ?

[rocky grove]

Aww don't worry. You're pretty smart for asking for help. Others don't. When others point out your mistakes, you will learn and avoid them next time :)

[rocky grove]

Well try doing the question again but this time follow this rule: $(a+b)^2 = a^2 + 2ab + b^2$

[ocean seal]

VulcanOne

[sick smelt]

if I don't follow it, the result will still be same right ?

[rocky grove]

Nope

[sick smelt]

y ?

[molten pivot]

Because 2+1 is not the same as 2×1

Sadly.

[sick smelt]

It would just become 6+8=14

...

[tacit arch]

Should probably follow rules of math for math classes and math grades

[sick smelt]

where are the rules written/typed ?

[tacit arch]

Books

Your class should have one

[sick smelt]

like literally I'm using khan academy, text book and neither of them say what are the rules specifically

[rocky grove]

They're found in books, and you can find them by geometric intuition

[sick smelt]

my school book sux

[tacit arch]

Maybe a previous class's book's

[sick smelt]

nope

[tacit arch]

Doesn't mean you shouldn't read it

[rocky grove]

Maybe search YouTube for a good proof so that you get convinced :)

[sick smelt]

ig

alsoo rq another doubt

[rocky grove]

Sure

[sick smelt]

=> 14+2√(2^5*3)
=> 14+5+2√(3)

why can't I do that ?

[rocky grove]

Hmm

[tacit arch]

You shouldn't make up your own rules when it comes to arithmetic

[sick smelt]

the number extracted (2^5) can't be added to 14

[rocky grove]

Rules of square roots are missing

[tacit arch]

.

[sick smelt]

My book seriously had no rules written anywhere and neither are they on khan academy

[rocky grove]

You're using logarithm rules in place of square roots

[sick smelt]

I have no idea what logs are

[rocky grove]

They're on Khan Academy but you should search for them

[tacit arch]

You just implied you didn't read your book

[hard patio]

I mean sure if there’s another way to get there, but that one’s pretty basic

[rocky grove]

Decimals right?

[sick smelt]

i'm not joking when I saw my book is literally the worst book for math when it comes to the entire grade, it will give like 3 examples and throw 15-20 questions which are so much more complex

[vale wigeon]

what is the original problem

or rather

the one being discussed rn

[sick smelt]

u sure ?

[rocky grove]

$(\sqrt{6} + \sqrt{8})^2$

[ocean seal]

VulcanOne

[vale wigeon]

yeah and what about this?

[rocky grove]

Applying rules that shouldn't be applied in regular arithmetic

[rocky grove]

Yep I'm pretty positive about it

[vale wigeon]

are we discussing whether it is possible to simplify this while being forbidden from using certain methods?

[sick smelt]

alr imma look for em

[rocky grove]

We're discussing why the method used to simplify was incorrect due to some misuses of certain rules.

https://cdn.discordapp.com/attachments/490557019623915520/1056836625402957844/image.png

[vale wigeon]

well 8 isn't 2^4 for a start

[rocky grove]

Yep

And (a+b)^2 doesn't give a 2(a+b)

[tacit arch]

Sounds like most books tbh. Either way, you need to stop lying about having read your book

[sick smelt]

😔 oki, but im not lying

[sick smelt]

also this chapter isn't in my book

[rocky grove]

And $\sqrt{a^b} \neq a^{b-1}$

[ocean seal]

VulcanOne

[sick smelt]

where did I do this?

[rocky grove]

Lots of rules

[rocky grove]

3rd step in your 2nd column

[vale wigeon]

this work is full of mistakes

[sick smelt]

specificy more

[sick smelt]

why do u think I solved it 3 times :|

I should prob stop studying this chapter till my book gets delivered or this'll be torture

or I'll break thru with sheer willpower

[sick smelt]

I seriously don't see it

[rocky grove]

[tacit arch]

Vulcan re-uploaded sim's picture for some reason

[sick smelt]

oh wait nvm

[vale wigeon]

going from line 2 to line 3 in the top-right work, i see sqrt(8) being replaced with 2^3 for no good reason.

2^3 is 8 itself, not sqrt(8).

[sick smelt]

√8=√2^3, I directly removed the root, it should be 2√2

[vale wigeon]

yes, you stripped away the root

and in the next step you apparently rewrote 2 sqrt(6) as sqrt(12), which is also wrong

[sick smelt]

I did 2(√6) to get √12

[vale wigeon]

which is also wrong

$2 \sqrt{x} \neq \sqrt{2x}$, generally.

[ocean seal]

Ann

[sick smelt]

:O

ok

[vale wigeon]

i mean, think about it -- by your logic, sqrt(2) would be equal to 2.

$2 = 2 \sqrt{1} \wtfeq \sqrt{2}$

[ocean seal]

Ann

[sick smelt]

Imma just watch organic chem's vid on simplifying square roots and what ever radicals are

thx everyone for help

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[west wagon]

x1, x2 and x3 are distinct non collinear points in a plane. Define an infinite sequence of points as follows: Let xN for N>=4 be the reflection of x(N-3) over x(N-1)
Show that there exists a point A such that the triangles A, xk, x(k+1), for all positive integers
k, have the same area.

[lone heart]

@west wagon Has your question been resolved?

[lone heart]

@west wagon Has your question been resolved?

[molten pivot]

So the magnitudes of cross products must be the same

Which means det must be same

[lone heart]

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[spare fern]

could someone help me out

[spare fern]

I am trying to solve this quadratic: 5x^2 -17x + 6 = 0

[gilded vessel]

what have you tried

[spare fern]

I'm not sure how to write the answer to this in fraction form

my calculator won't do it for some reason

I'm not sure if I should round the answer

[gilded vessel]

you gotta put the fraction under the sqrt in one fraction first so you can take the sqrt of it

[naive valley]

shouldn't there be a 4ac term somewhere in that square root?

[spare fern]

no I'm using Poh Shen Loh's method

[naive valley]

this method somehow removes any dependence on the leading coefficient 5?

[spare fern]

yes it's basically the sreedhar acharya formula rearranged

[gilded vessel]

this method is for quadratic in the form x^2 + Bx + C = 0

you dont have a quadratic in that form

[reef grove]

its a way of using the quadratic formula without actually using the quadratic formula

[gilded vessel]

if you want to use this method, first reduce your quadratic to this form

[tacit arch]

That doesn't apply because of your 5

[naive valley]

you could put it in that form by dividing the whole equation by 5 if you want

seems uglier than just using the ordinary quadratic formula tho

[spare fern]

Oh I see

Alright I’ll try the regular sreedhar acharya formula

[gilded vessel]

its literally just plug in a = 1 to quad formula, boom "new" method

[reef grove]

in Germany we actually call it p/q formula and they want us to learn this instead of the quadratic formula

[spare fern]

So this is correct?

[tacit arch]

Fraction like needs to be longer

[spare fern]

What do you mean?

Oh I see

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[latent gulch]

why this is true? what property are we using here?

[latent gulch]

P_sigma is orthogonal (I don't know if it is relevant)

[rocky grove]

Well the result of det(P) and det(inverse of P) cancels out and leaves you with a net result of det(C-λI)
Edit: corrected

Wait

[vale wigeon]

matrix multiplication isn't commutative

however det(AB) = det(A)det(B)

[rocky grove]

[latent gulch]

Thanks a lot!

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[thick lynx]

$\lim_{x \to \infty} \sqrt{x}(\sqrt{x + 3} - \sqrt{x - 2}) = \lim_{x \to \infty} (\sqrt{x^2 + 3x} - \sqrt{x^2 - 2x}) = \lim_{x \to \infty} \frac{x^2 + 3x - (x^2 - 2x)}{\sqrt{x^2 + 3x} + \sqrt{x^2 - 2x}} = \lim_{x \to \infty} \frac{5x}{\sqrt{x^2 + 3x} + \sqrt{x^2 - 2x}}$

How would you proceed from here?

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[thick lynx]

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[hollow ruin]

How would you do this?

[lone heart]

Please don't occupy multiple help channels.

[remote heron]

[hollow ruin]

I am sorry but yeh these are the 2 questions that i have

[hollow ruin]

Yeeeh hi there mate

[alpine sable]

what do you need in order to calculate the area of a triangle?

[hollow ruin]

The 2 sides right?

See the problem here is that this is a non calculator question

[alpine sable]

what's the formula for the area of a triangle?

[hollow ruin]

1/2 ab sin theta?

[alpine sable]

ok I didn't even know there was that formula

so you know what that is equal to

[hollow ruin]

Yeh there is that one too, for like an irregular triangle

[alpine sable]

and you know two sides

this leaves you with one unknown quantity in one equation, which you can solve

[hollow ruin]

Yup got that that ones easy

How bout the second part?

[alpine sable]

ah ok

are you familiar with the law of cosines?

[hollow ruin]

I think so yeh?

OH

YEH

I am

[alpine sable]

so do you know how to apply that

to find the length of the third side?

[hollow ruin]

Ah yeh yeh i forgot that

But its non calculator based tho

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[viral pine]

Good mornin USAAA

[viral pine]

Q11i)

may I know what went wrong in my answer

[rocky grove]

You forgot the 1/2 for -1/(1-x)

[vale wigeon]

you misread the function being differentiated

it's $\log\paren{\sqrt{\frac{2x-1}{1-x}}}$ not $\log\paren{\frac{\sqrt{2x-1}}{1-x}}$

[ocean seal]

Ann

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@viral pine Has your question been resolved?

[small kernel]

Is there any one who study in university??

[lone heart]

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[viral pine]

.reopen

[lone heart]

✅

[viral pine]

another question

how do I go about doing 11ii b

im completely lost

[pseudo ice]

...so you want two points x that give you a gradient of what they've said, and a quadratic equation?

[viral pine]

the gradient as in

the dy/dx?

[pseudo ice]

Yeah, in part (a), what do the solutions of that quadratic give you?

[viral pine]

(2x+1)/2(x^2+x+1)

wait

am I supposed to like factories it

factorise

[pseudo ice]

No not quite

[viral pine]

hm

[pseudo ice]

I was asking for a description of what those solutions would be, if that's clear?

In other words, how did you show part a?

[viral pine]

ohh wait I was reading the wrong question

accidentally read 12iii

hold on

alright so I compared the dy/dx to the 1/2k gradient

as they’re the same thing

[pseudo ice]

I seem to be unable to read myself tbh 😂

[pseudo ice]

Yeah, so the solutions x to that are the points where you have the gradient 1/2k

And it's a quadratic, and they're asking for two solutions

[viral pine]

o

okay one sec

alright

i am still lost

could you perhaps give me the workings to obtain the answer

I think it’d be easier to understand that way

@pseudo ice

[pseudo ice]

I'll just say this and it should be fine from there: work out the discriminant of that quadratic

Then remember the conditions on the discriminant for the number of [real] solutions you'd want

[viral pine]

ahhhh

discriminants, roots, quadratic and whatnot

i forgot about them

i did too much differentiation itself that I forgot about the previous topics

oops

thanks anyway

ill make sure to revisit these things

[pseudo ice]

Yep, always a good idea to go over stuff when you can (as I am learning the hard way )

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[ornate ginkgo]

How do i differentiate
$2\sqrt{cot(x^2)}$

[ocean seal]

ColdTee

[ornate ginkgo]

$\frac{d}{dx}2\sqrt{cot(x^2)}$

[ocean seal]

ColdTee

[alpine sable]

use the chain rule twice

[ornate ginkgo]

Well that's what im confused about do i take $\sqrt{cot(x^2)}$ as x the first time then im left with 2?

[ocean seal]

ColdTee

[alpine sable]

2 is a constant, so you can take it outside the derivative

yes, you could use chain rule there, but nothing happens

[alpine sable]

take 2 out of derivative cuz its a constant

[ornate ginkgo]

Then im left with
${(cot(x^2))}^\frac{1}{2}$

[ocean seal]

ColdTee

[ornate ginkgo]

So first time

[alpine sable]

the derivative of ONLY cot(x^2) is -2 cosec(x^2) . cot(x)

[ornate ginkgo]

The first time i will solve it in the form nx^n-1 right

$\frac{cot(x^2)}{2}$

[ocean seal]

ColdTee

[ornate ginkgo]

Like this?

$\frac{1}{2\sqrt{cot(x^2)}} × \ - cosec^2(x^2) × 2x$

Is this right

[ocean seal]

ColdTee

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[wild raven]

Hi

[wild raven]

i need help for qn 29a

[wild raven]

lol

[viscid pawn]

ahahah 😭

[ivory igloo]

you dont need that?

[wild raven]

ya thats what i thought

i was kinda confused

[desert tendon]

He'll need to take the log to the num so he'll do chage of base

[wild raven]

is this correct?

[ivory igloo]

$\implies \frac{8}{2log_3{x}} - log_3{x} = 3\$
$say~g=log_3{x}\$
$\implies \frac{8}{2g} - g = 3\$
$\implies \frac{8}{2} - g^2 = 3g\$
$\implies 4 - g^2 = 3g\$
$\implies g^2 + 3g - 4 = 0\$
$\implies (g+4)(g-1) = 0\$
$\implies g = 1\$

[ocean seal]

GameSwitch

[desert tendon]

Correct till herwy

And then as above did

[ivory igloo]

$log_3{x} = 1 \implies x = 3$

[ocean seal]

GameSwitch

[wild raven]

wait how did 1/logx to the power of 3 become just log3x?

[desert tendon]

Where

[ivory igloo]

yes and?

$\frac{1}{log_x{3}} = log_3{x}$

[ocean seal]

GameSwitch

[desert tendon]

It's not to power 3

[wild raven]

oh wait nvm sorry i get it now

when solving log do we always have to use “let something be somethint”? or are there other ways?

[ivory igloo]

the assumption is used so that the equation is easier to view and solve

we can also solve without assuming

[desert tendon]

Sometimes you can combine all logs into one term and solve but can't do that here

[wild raven]

ohhhh okay thank you sm

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[quartz cave]

Hello

[quartz cave]

I have a question about shapley values combined with monte carlo sampling

When we use this intuition

One would say that with n = 10 and every step between each point = 1

Then the value of the 10th agent would be 1 + 1/2 + 1/3 + 1/4 ... + 1/10

This would be a value of 2.93

However

When I try to experiment with this through monte carlo sampling I get way higher values for the 10th agent

What am I missing in this problem?

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[quartz cave]

<@&286206848099549185>

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[humble heron]

Why does no one wanna help me 🥲

[alpine sable]

cuz u didnt ask a question ig?

[humble heron]

Maybe cause it’s stats related

[alpine sable]

are those decimals?

[humble heron]

3.9,

[alpine sable]

3.9,3.7... etc ryt?

[humble heron]

Yes

That’s a set

[alpine sable]

mean u can do ryt?

[humble heron]

This is the method and I don’t get it

[alpine sable]

wait u did the method urself and dont get it?

[alpine sable]

the median is 2.7

[alpine sable]

mode is 3.7

[alpine sable]

they are correct

gud job

:>

@humble heron

the median should be 2.7

u wrote 3.7

count upto the 8th term

which is 2.7

[humble heron]

Don’t we have to arrange?

The data first

Also this is how we were supposed to do it XD

[alpine sable]

ohh yes im sry my bad forgot

[alpine sable]

u want an explanation of wts happening here?

[humble heron]

Yes 🐣

The median is the value of x in the middle of the interval.
That is, when x=x1+(4.4-2.6)/2=2.6+0.9=3.5. The median is me=3.5.

Especially this part

[alpine sable]

is frequency same as xi+1 in the chart?

[alpine sable]

here to make the calculation easier they have divided the data into 6 intervals with a difference of 0.3 if u notice carefully and then for example if we take 3.5 and 3.8 then find avg between them and see what values in the original data are within that average range like (3.5 + 3.8)/2 is 3.65. Now the original data which u arranged in ascending order ( 2.7...etc) see there u have to see the numbers that lie between 3.5 and 3.8 which are 3.7, 3.7, 3.7 so since there are three numbers lying between so we write the m1 as 3

[rocky grove]

Yep

Also this way of arranging data is a lot less cumbersome since you deal with ranges of data and you can represent a single range by the range's mean, which helps you calculate things a lot faster

Btw quick answer to a previous question you had about computers: your MacBook is good for productivity stuff. It will handle most tasks nicely but don't expect it to do a lot of hardcore processing. But I think that if you use it for school works, then you will be good to go (Unless you wanna game on it xd)

Anyways

You get the number of intervals by taking the range (biggest number minus smallest number) then divide it by a suitable interval length. Since your range is 4.4-2.6 = 1.8, and the numbers have a tenth of a difference between them, then 0.3 should be good. 1.8/0.3 = 6.

Now you use the 0.3 interval length to create your intervals

2.6+0.3 = 2.9. So your first interval is from 2.6~2.9.

Hmm

@humble heron Did your teacher tell you to add an additional interval to the number of intervals?

Like instead of 6, you use 7 intervals

And your start will be 2.6-0.3/2 = 2.45

And the end will be 4.4+0.3/2 = 4.55

For the last interval so that you don't have any errors with the bounds

Like the points that lie exactly on the borders

But I don't think you do that with your teacher

Hmm

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[slender crater]

where do i go from here

[gaunt sparrow]

it's not visible

[slender crater]

oh wait

i substituted x for a

nvm

.close

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[steady basin]

am i bugging or is this incorrect?

should c be -pi/2?

[vale wigeon]

no, c is +pi/2 exactly as is written

[steady basin]

but if u rearrange, -arccos of 0 - arcsin of 0 is -pi/2

[vale wigeon]

oh

[steady basin]

its minus arccos not + arccos

[vale wigeon]

wait shit

okay, yeah

-arccos(x) = arcsin(x) - pi/2

this way it will be correct

[steady basin]

yh so its wrong

in the pic i sent c should be -pi/2

heres the whole thing for context

[vale wigeon]

well they made two sign errors which cancelled each other out yeah

-arccos(x) = arcsin(x) + c => arcsin(x)+arccos(x) = -c

= -(-pi/2) = +pi/2

if done properly

[steady basin]

yep kl

thx

[limpid spade]

? Does this contribute to SA's question?

[sour egret]

mmm wait

what is SA?

sorry idk

[steady basin]

wdym

[steady basin]

me

[limpid spade]

dont worry

[steady basin]

why

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[distant plover]

if $f$ and $g$ are functions limited in the interval $I$, is it true that $\sup{|f(x)|g(x); x \in I } = \sup{|f(x)|; x \in I } \cdot \sup{g(x); x \in I }$ and $\inf{|f(x)|g(x); x \in I } = \inf{|f(x)|; x \in I } \cdot \inf{g(x); x \in I }$?

[ocean seal]

Amarinya

[mortal trellis]

No

Only <= and >=

Eg f(x)=x, g(x)=1-x on [0,1]

[distant plover]

ahh alright

so that means sup{f(x)g(x)} <= sup{|f(x)g(x)|} = sup{|f(x)| * |g(x)|} <= sup{|f(x)|} * sup{|g(x)|}?

[mortal trellis]

Yes

[distant plover]

alright so, i have f and g being integrable on [a,b], and i have to show that fg is integrable. i'm pretty sure |f| and |g| is integrable, so my idea was to show that sup{f(x)g(x)} <= sup{|f(x)|} * sup{|g(x)|} and the opposite for inf so that i could squeeze the upper area to the lower area and show that it's integrable. i'm not sure it'll work, but i'll try

i'll reopen if i have more questions

.close

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[past dagger]

not sure how to sketch this

[long axle]

Is there a part of the question that u didn’t show

[past dagger]

yeah

but i dont think its relevant

[long axle]

Send it anyways

[past dagger]

i dont care about how to do it for THIS question, i just dont how to sketch negativee quuadratics

[long axle]

Oh that’s it?

Alright so

The problem is you want to graph it, but don’t know how to?

So we have: y = 6 - (x-3)^2

See if u can put that into vertex form

[pseudo ice]

It is relevant (see the part (ii))

[past dagger]

nver heard of vertex form

[pseudo ice]

Unless you just want to sketch it in general?

[past dagger]

just in general

[long axle]

I think they just want this

https://www.chilimath.com/wp-content/uploads/2022/09/vertex-form-of-a-parabola.png

@past dagger

[pseudo ice]

Don’t think so - see the parametric equations that you have

[past dagger]

never seen this in my life

[past dagger]

.

[long axle]

Dimonoc do u just want to graph the quadratic

Is that all u want

[past dagger]

yes

thats literally al

all

[pseudo ice]

It’ll probably be some section of the y=6-(x-3)^2

[long axle]

Do u want to sketch the quadratic, independent of the parametric at the start? @past dagger

[long axle]

Thanks for making me clarify this, I forgot abt that

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[rugged dust]

Hi, I'm practicing proof by induction. I'm really struggling with this one because of the x..

[rugged dust]

Could anyone provide an approach to this?

[plain flame]

i wouldnt write x there

[rugged dust]

Why is that? c.c

[marsh rapids]

Just take the equation from the hypothesis and add 1/(n+1)(n+2) to both sides, then simplify the RHS

Usually with induction you just do the operation that turns one side into what you want, and computations on the other side to get the wanted result, starting from the result from the induction hypothesis

[rugged dust]

aaah i see now

yeah calling that whole part x was kind of pointless

[marsh rapids]

If there is no such operation, then induction might not be such a good approach (or maybe it is but you only notice it after some time looking, and do you just wrap your proof inside of an induction lol)

[rugged dust]

got it, thanks you two

.close

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[warped topaz]

https://gyazo.com/c51b80a97ab4d416fa5571c7f42798be.png

[warped topaz]

I hate geometry

[pseudo ice]

How did you find your answer?

[warped topaz]

I got the equation of the circle and isolated y, then did the same for the equation of the line, then set them equal and tried to solve

[gray isle]

can you show your work

[warped topaz]

its really messy but ok 1 moment

I couldnt find my work so I did it again and solved it lol

sry about this

.close

[lone heart]

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

Hey

[celest storm]

Have to determine the n-th derivative of x -> x^2n with 2 different ways

I got the first method i guess
I conjectured a k-th derivative

Then prove it ( by induction ) and replaced k by n

Then i tried using liebniz and i got to nowhere

[marsh rapids]

Formule de Leibniz ?

[celest storm]

Oui

[marsh rapids]

n-th derivative of a product ?

[celest storm]

Yes

X n x n

[marsh rapids]

Or induction on x^2n * x^2

Not sure though

[celest storm]

To be honest, i tried with Liebniz because the exercice got an indication saying that we could write x^2n = x^n × x^n

Post bac

[marsh rapids]

J'ai vérifié en regardant ton historique

[lone heart]

@celest storm Has your question been resolved?

[lone heart]

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[lone heart]

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[gilded marten]

can someone teach me how to get the percentiles

for d2 d6 and p32 p89

or just the formula on how to get it

[lone heart]

@gilded marten Has your question been resolved?

[modern topaz]

Je suis tres intelligent

[modern topaz]

Je aide toi

[lone heart]

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[spare fern]

how would I solve these types of quadratic equations

[spare fern]

using sreedhar acharya's formula

[woven plaza]

Just as any other question

It's just that b=0

[spare fern]

oh ok

[dim root]

wait

[spare fern]

so a^2 = x^2

and wouldn't B = -64

and C = 0?

[dim root]

wait

ankit cant you solve it

using algebra

x=8

[spare fern]

?

oh you mean adding 64 to both sides

and taking the square root?

[wary stream]

Yes

[wary stream]

Don't give out answers

[last ether]

Well it's not just 8

So they didn't give out full answers

[wary stream]

It's 2 * 4

[last ether]

But yeah

:0

WHAT!!?(

REAL?!1!1!1!1!!!!???

[wary stream]

No, it's 5 + 5 − (5 + 5)/5

[last ether]

Vine boom sfx

[wary stream]

(1 + 1)^(1+1+1)

[ivory igloo]

facepalm moment

[last ether]

You don't even need quadratic formula

Just know your perfect squares

[spare fern]

how about this?

20-x = x^2

square root of 2-x?

[leaden igloo]

The easiest (or at least most standard) way to solve this is to make a quadratic formula equal 0.

So you take 20 from each side and get (x^2 + x -20)

=0

x^2 + x -20 = 0

Then you factor out or use quadratic formula.

[spare fern]

oh you mean x^2 + x = 0?

[wary stream]

No

[leaden igloo]

No the equation needs to stay the same

x^2 + x - 20 = 0

[wary stream]

Subtract 20 on both sides

[leaden igloo]

Subtract 20 on both sides

ye

lol

[spare fern]

Ax^2 + Bx + C = 0

[leaden igloo]

mhm

[dim root]

im back

[leaden igloo]

wb

[wary stream]

Yes that's the general form for a quadratic

[spare fern]

so x^2 + x -20 = 0

[dim root]

hm

[leaden igloo]

yes

[dim root]

yes

[leaden igloo]

Then factor or quadratic

[wary stream]

You can choose to factor or use the quadratic formula

[spare fern]

so B = x

[dim root]

a is 1 and b is also 1 and c is -20 rigfht

[spare fern]

and C = -20

[dim root]

?

[spare fern]

yeah

[leaden igloo]

b is the number next to x

[wary stream]

No

[leaden igloo]

ax^2 + bx + c

[dim root]

rember

always take the coefficient

for the formula

[wary stream]

You want the coefficients

[dim root]

u shouldnt take variable

[spare fern]

but isn't A = x^2?

[wary stream]

No

What's the coefficient with x^2?

[dim root]

its the coefficient of x^2

[spare fern]

1

[dim root]

yes

[spare fern]

so B = 1

as well

[dim root]

yes

always take the coefficients while using quadratic formula

[spare fern]

oh ok

[dim root]

and c is the constant

[spare fern]

so the solutions are -5 or 4

[wary stream]

Yes

[dim root]

yea

[wary stream]

To check, you can plug each one back into the original problem

[dim root]

mhm

[spare fern]

and one more

6x^2 = x

I tried dividing both sides by 6

and I got x^2 = 1/6x

[raven rover]

Is there a reason for that step?

[spare fern]

just tried to use algebra to isolate

x

[dim root]

it should be x/6

[wary stream]

A tip for you, most of the time, it's better to get all the terms on one side so it equals 0

[dim root]

hm'

[raven rover]

This

[spare fern]

so I should subtract x?

[wary stream]

Yes

[spare fern]

ok so 6x^2 - x = 0

[raven rover]

Yes

[dim root]

yhes

[raven rover]

Ideally, you factor

So you get (something 1) • (something 2) = 0

[spare fern]

but If I want to use the formula?

[wary stream]

Factor by GCF is a hint for you

[raven rover]

Setting each respective thing to be equal to 0 will give you your solutions

[raven rover]

You can, but it’s a lot more work

[wary stream]

You can but it's not needed

[spare fern]

honestly I'd rather not factor

formulas work better for me

[dim root]

factoring is quite easy

youll get the answer faster

but yes you can apply formula too

[spare fern]

how would I use the formula?

[wary stream]

Formulas, in my opinion, especially the quadratic formula is the last resort method. You should ideally factor using factoring methods, like GCF, seeing the patterns like difference of squares, since it's faster

[dim root]

hm

[wary stream]

If you were to use the formula, same process, identify A, B, and C

[spare fern]

ok so a is 6

and b is -1

don't know about C

[raven rover]

Is there a number just hanging out by itself in this equation?

[spare fern]

0

[wary stream]

That's on the same side as the other stuff

Recall that the general form is Ax^2 + Bx + C = 0

[spare fern]

hmm

x^2?

[wary stream]

You have 6x^2 - x = 0
Use the general form of Ax^2 + Bx + C = 0 to compare the two to determine what A, B, and C are

[lone heart]

@spare fern Has your question been resolved?

[spare fern]

not sure

I know that a = 6

and b = -1

really not sure about C

[wary stream]

.

Compare the two

6x^2 - x = 0
Is there a constant?

[lone heart]

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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[vale fjord]

I am having trouble finding the x coordinate?

I found that the equation for the sine graph is y = 5 sin(pix / 4) - 3

How do I solve for x because I keep getting weird numbers?

If I arcsin 0.6, I get a large number that definitely isn’t the x coordinate for P

[carmine reef]

yes

If y = sin(x)

Then arcsin(y) represents only one of the many possible values of x such that sin(x) = y

For example

,w arcsin(1)