help-17

[paper bridge]

$a^{2n+1}+b^{2n+1}=(a+b)(a^{2n}-a^{2n-1}b+\ldots b^{2n})$

[twin meteor]

vengeance

[vocal sleet]

@sudden cloud Has your question been resolved?

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[unborn owl]

2^m+2^n=520, how do i solve for m and n?

[paper bridge]

Find pairs of 2^m and 2^n such that their last digits add to 10

so tell me the pattern

[unborn owl]

pattern?

[paper bridge]

for last digit of powers of 2

[unborn owl]

the question on my paper was If 520=2^m+2^n, where m and n are integers, then m + n is equal to

im meh at math so i dont really understand what you're talking about

[paper bridge]

ok

so 2^1=2

2^2=4

2^3=8

2^4=16

2^5=32

so on

so we can say the last digits have a pattern of 2, 4, 8, 6, 2

let me bold it for you to see

[unborn owl]

ohh

i get what you mean now

[paper bridge]

oh ok

[unborn owl]

but how do i use that to solve for the question

[paper bridge]

well, which pairs add up to 0

[unborn owl]

uh

4 and 6?

[paper bridge]

and

[unborn owl]

2 and 8

[paper bridge]

bet

so we can say that a solution must be in form 2^(4k+1)+2^(4j+3) for k, j natural numbers due to our generalization

[half imp]

its probably easier to note that either 2^m or 2^n must be at least half of 520

which powers of 2 are between 520/2 and 520

[paper bridge]

yes yes

i guess small m and n

[half imp]

?

this would work for any number

[paper bridge]

yes yes

but read

[half imp]

i read

i just don't know what you are thinking

[paper bridge]

i am le thinker

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[carmine ether]

struggling to understand how to know what directions to plot arrows in for phase portrait

the null clines is making sense to me

just dont know how you can know which direction the arrows go in with th 3 different null clines

[carmine ether]

why do the phase portrait arrows circulate around the fixed points (+-1,0) but tend away from the fixed points (0,+-1)?

[white palm]

what’s your vector field?

[carmine ether]

2D in x,y

[white palm]

yeah sure but do you have an explicit formula for it like x’ = f_1(x,y) and y’=f_2(x,y)

[white palm]

yeah but what’s the vector field you are plotting now is it a) c) of g)?

[carmine ether]

I am doing 3b

so f=2xy g=1-x^2-y^2

srry not really sure what youre asking

new to this topic

[white palm]

It’s C^1 at least so you can linearize your system around the different fixed point

then the resulting linear system Ax=x’ can be studied and will have the same behavior as the nonlinear one around the different fixed points

if you’re not familiar with that yeah you have to get a feel for the vector field and plot it like you did

[carmine ether]

Oh is A the jacobin matrix here

[white palm]

yes

[carmine ether]

Jacobian*

[white palm]

the jacobian evaluated at the fixed point you want to study

[carmine ether]

So that way I find where the FPs are stable/ unstable?

[white palm]

yeah, if it’s a sink attractive or repulsive

[carmine ether]

Okay I see, thank you

@white palm how would I know the directions of the arrows in the quadrants outside the circle like where I highlighted below

[white palm]

you could go quadrant by quadrant pick x and y big in this quadrant and get a feel for the direction this way because like polynomials from R to R from my little experience playing with these vector fields the interesting behavior is around the origin.

[carmine ether]

Ah okay I see thank you

Been a great help

[white palm]

this empirical conjecture applies when your vector field is defined by polynomials, this won’t do if there is a sinx in your vector field

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[amber jolt]

Anyone know how to do this question?

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[finite maple]

can sopmeon explain
i get up to the 3rd line
and after i dont get it

[paper depot]

do you understand up to here?

[calm pecan]

the fourth line is literally just transposition

[finite maple]

yes

[calm pecan]

literally just transposition then mate

[finite maple]

oh yea

ok well what about after?

[calm pecan]

well the left hand has an annoying m there which you dont want

so youd divide by $m$

and thats it

[twin meteor]

Percy

[finite maple]

ohh ok

cool ty

.close

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[calm pecan]

I can't figure out why we have the $+ \pi$ in the second piece.

Graphically I get it, algebraically I don't see it.

[twin meteor]

Percy

[calm pecan]

I have like
\begin{align*}
y&=\arccot(x) \
\cot(y)&=x \
\tan(y)&=\frac1{x} \
y&=\arctan \left(\frac1{x} \right)
\end{align*}

and I don't see why we need the piecewise function

[twin meteor]

Percy

[calm pecan]

i mean, $\arctan(x)$ has a domain of $\mathbb{R}$ so I don't see why we need anything here-?

[twin meteor]

Percy

[calm pecan]

im at my limit now lol

[spiral turtle]

@calm pecan the reason it is there is because of the way we have done the branch cut

[calm pecan]

uh huh?

[spiral turtle]

So tan and cot are not one to one, right?

So when we take the inverse we need to decide which part actually gets mapped back as the inverse

[calm pecan]

I see, how do I do that?

[spiral turtle]

It's just convention

Someone somewhere decided that the function would be more useful if it inverted one region vs the other

To be clear, having tan^-1(1/x) does make an inverse of cot

[calm pecan]

Okay yeah well, how do I introduce the + pi though?

And like, justify its necessity

[calm pecan]

Just not the principal one, okay

fair enough

[spiral turtle]

But it doesn't make the function arccot, because arccot is defined as the inverse of the domain (0, pi) not (-pi/2, pi/2)

[spiral turtle]

That's just what we need to do to make the numbers work

[calm pecan]

How do I show that my graph is shifted down by pi compared to the principal arccot

[spiral turtle]

Well, the period is pi

[calm pecan]

okay and here we have a negative value for 1/x which doesn't work

i follow i think

[spiral turtle]

Negative values of 1/x are not a problem

[calm pecan]

no? it'd be outside the domain for the principal arccot though?

wait nvm

tfw i get confused between range and domain

So I get that, like, $\arctan( \text{some negative value})$ would be a negative value, which wouldn't be the principal range of the arccot function, and shifting it up by $\pi$ solves the issue

[twin meteor]

Percy

[spiral turtle]

To be fair, talking about the inverse you pun them a lot

[spiral turtle]

Exactly

[calm pecan]

Can I just write 'adding pi gives the principal range so ill do that', or is there a more rigorous way to like, i dunno, derive that we need to shift by pi

[spiral turtle]

So if we want to talk about rigor, your definition of arccot is incomplete

Because it is undefined at x = 0

[calm pecan]

That's also fair

[spiral turtle]

So I think maybe mathematical rigor isn't terribly important for this problem.

[calm pecan]

Fair enough

but out of curiosity, how would one show this rigorously

[spiral turtle]

(this isn't accusatory, just knowing when to pick your battles)

[calm pecan]

(yes i get it, going overkill isn't always the best idea)

[spiral turtle]

So you would need to define arccot as the inverse of cot limited to the domain (0, pi)

And also use the fact that cot is periodic with period pi

[calm pecan]

okaaayyy

im listening

[spiral turtle]

Then if you can show some function, such as arctan(1/x) maps to some other part of the cot domain, then you can use the fact that cot is periodic to show a vertical shift of kpi where k is an integer also maps some other part of cot

And then you show for k = 1 and k = 0 it maps the parts that you want

Then you have to cover the case where x = 0

But like

There's no rigor in why we choose the part of cot to inverse

It's just what we have decided is most useful to us for most applications.

So in a sense the choice is arbitrary, so the selection of k is also arbitrary.

This applies to arctan as well

Anyway, the above is rigorous enough for a proof

[calm pecan]

this works, yeah?

i dont like the way it's presented but i think it's okay

[spiral turtle]

Yeah, there's no problem here, other than the point wise discontinuity at x = 0

[calm pecan]

okayy thanks

.close

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[torpid goblet]

If I understood it correct it, deposited means that she loans money to a bank.

Meaning that she will earn money from the bank

Do I do 300x1,05squared

and 200x1,05powered to three

and calculate how much they earned and see who earned the most

and the girl in the green earns the most?

since she earns 31.525 and the other 30.75

[ruby jungle]

Woah, the last time I’ve touched interests was like 8 years ago.

I’m assuming this is Compounding.

$A = 200(1+5/100)^3$ and $A = 300(1+5/100)^2$

[twin meteor]

Maddie

[ruby jungle]

,calc 200(1+5/100)^3

[twin meteor]

Result:

231.525

[ruby jungle]

Yeps

The girl who’ve compounded for 3 years would earn more.

[torpid goblet]

Thank you brother or sister @ruby jungle

I just got this as a new unit

Im new to it too

In a solution there are 50 bacteria per ml. The number of bacteria increases by 15% each minute. How many bacteria are there in 5 ml after 4 minutes?

@ruby jungle

[ruby jungle]

So basically initially you have 50bacteria per mL.

[torpid goblet]

Mhm

I started with 50x4

[ruby jungle]

So, 5 ml would have errrr 250 bacteria.

[torpid goblet]

oh

[ruby jungle]

Bacteria*

[torpid goblet]

I calculated it wrong

is it 50x5 and then that times 1,15 powered to 4

[ruby jungle]

No, you’re fine.

wAit

Okay no more trolling.

So initially you’d have 250 bacteria.

[torpid goblet]

Yes

[ruby jungle]

250(1+15/100)^4

,calc 250(1+15/100)^4

[twin meteor]

Result:

437.2515625

[torpid goblet]

yes

[ruby jungle]

437 Bacteria

[torpid goblet]

I belive thats correct

Ty

[ruby jungle]

Also

Scientifically…

Bacteria shouldn’t be in decimals.. So i’d round to the nearest whole:)

[torpid goblet]

Ty

this question too

@ruby jungle

You are making a sauce and the recipe calls for vegetable stock. 1 litre sauce must contain 4% of stock. How much sauce can you make out of 180 ml stock?

Its so confusing, I dont understand it at all

[ruby jungle]

Okays, so 1 Litre of sauce -> 4% Stock

Means that per Litre, 40 mL is stock.

Do you understand?

4% =0.04
0.04 x 1L =0.04 Litres or rather 40 mL.

So, 180/40 will give u how many times you can make the sauce.

[torpid goblet]

Yes

[ruby jungle]

Do you understand up to here?

[torpid goblet]

yes

Instead I used king henry does usually ..

converted 180ml to liter

and I got 0.18

[ruby jungle]

40 mL Stock -> 1 L sauce
180 mL Stock = 180/40 x 1

[torpid goblet]

Then that divided by 0.04 gave me 4.5

[ruby jungle]

Yes. That works too

[torpid goblet]

Yes

[ruby jungle]

What I did is valid because of unit consistency.

180 mL/40 mL would result in a dimensionless value. Which can be multiplied by a unit to give you the final unit.

^ Additional information.

so with 180 mL of stock, you can make 4.5 TIMES of 1 Litre sauce.

So, the volume of Sauce you can make with 180 mL is?

[torpid goblet]

Hmm

4.5 litres?

[ruby jungle]

Precisely

[torpid goblet]

Ty

The local lake in a town started to freeze one Sunday during the winter. The thickness increased by 2 cm each day. On a Sunday a couple of weeks later, the ice thickness has increased by 25% compared to the Sunday before. How thick was the ice then?

@ruby jungle the last questions I had some kind of idea how to do them but this just made my brain freeze

[ruby jungle]

Hm…

So the difference is 25% increase in thickness… And we know that 2 cm each day, so 7 x 2 = 14cm.

Interesting

[torpid goblet]

Wtf

[ruby jungle]

Ok nevermind..

[torpid goblet]

No

It seemed correct

But I was to lost in your calculations

sorry

[ruby jungle]

Nah, something was wrong

[torpid goblet]

No, no

[ruby jungle]

So percentage change is $\frac{(x+14)-x}{x} = \frac{25}{100}$ … Soo… x is the initial height and x+14 is the final height. The difference would be 14 cm as we know…

[torpid goblet]

Yes

[ruby jungle]

So.. unrealistically the height of ice is 70 cm 😂😂

[twin meteor]

Maddie

[torpid goblet]

The equation?

[ruby jungle]

Ok im trolling 😂😂 the change is simply 14/x

So 14/x actly the difference between one sunday and the sunday after

[torpid goblet]

Yes

[ruby jungle]

So the initial is x, final is x + 14.
The difference 14/x * 100% MUST be equals to 25% change

So solve for x.

My equation was 14/x = 25/100.:)

Solving for x will give you x = 56cm.

X+14 (Final height) is 70 cm.

You understand?

[torpid goblet]

I dont understand at alll

[ruby jungle]

😂😂 I figured

Are you familiar with equations?

% Change is = Final - Initial/Initial * 100%

[torpid goblet]

Oh

I get it now

So you basically did

X+14=1.25x

[ruby jungle]

Oo yes

Thats fine

[torpid goblet]

But then I dont understand why you add another 14

When you have already added 14 to the equation

[ruby jungle]

Oh no, because I was looking for difference using the full formula 😂😂

Then i forgot 14 is the difference

But irregardless, x+14 is the Final height and x is the INITIAL height.

Let me write it properly

\begin{align*} \frac{Final - Initial}{Initial} &= \frac{25}{100}\ \frac{14}{x} &= \frac{25}{100}\ \therefore \text{ x = 56 cm}\end{align*}

So basically, Final - Initial is simply 14 Because for each day, it increased by 2 cm. After 7 days the difference between first and 7th day is 14cm Hence Final - Initial can be represented as 14.

[twin meteor]

Maddie

[ruby jungle]

I’m in mobile

But as you mentioned, you can use x+14 = 1.25x:)
What you’re doing is you’re finding the 25% increase of x. 0.25x = 14 and x = 56 same thing.

Understand?

The initial is x = 56 and final is x + 14 = 70 🙂

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[fierce rampart]

Do i ask for help here?

[tidal dock]

yes

[jagged cargo]

yeah

[fierce rampart]

Do yall understand a bit of french to help me?

For something like this?

[oak magnet]

quelle question?

[fierce rampart]

Exercice 1?

Les deux?

[oak magnet]

ok

[fierce rampart]

You can also write in english its only that the tasks are in french

[oak magnet]

la première il faut regarder que la somme des proba est égale à 1

[fierce rampart]

Okay merci

[oak magnet]

et aussi que toute les proba sont entre 0 et 1

[fierce rampart]

Donc si c’est p.ex -0.5 c‘est impossivle?

[oak magnet]

une proba n'est pas négative

[fierce rampart]

Okay

[oak magnet]

pour la deuxième il suffit d'ajouter les probas qui vérifient la condition

donc pour P(X <= 1) tu as que c'est la somme des probas pour X valant -3,-2, -1, 0, 1.

[fierce rampart]

Donc 0,75?

Et l‘autre 0,80

?

[oak magnet]

oui 0.80 pour celui que j'ai dis

[oak magnet]

oui

[fierce rampart]

Okay merci

[oak magnet]

interpreter n'a pas vraiment de sens comme il n'y a pas de contexte

[vocal sleet]

If you are done with this channel, please mark your problem as solved by typing .close

[fierce rampart]

Comment est ce que je peux faire ca?

[oak magnet]

polynome du second degre

tu sais que la somme c'est 1

donc p² + p + 0.44 = 1

résoud pour p

il devrait y'avoir qu'un seul p dans [0, 1]

parmis les deux que tu trouves

[fierce rampart]

Mais comment est ce que je calcule ca?

[oak magnet]

tu ne connais pas les polynomes du second degré avec delta = b²-4ac ?

[fierce rampart]

Ah oui delta

Donc je fais
5- 4 • 0 • 10?

5*2

[oak magnet]

?

donc p² + p + 0.44 = 1

est ce que tu vois comment j'ai eu ça?

[fierce rampart]

Oui

[oak magnet]

c'est sur celui ci qu'il faut faire delta

[fierce rampart]

Mais comment est ce que j‘utilise le delta la?

[oak magnet]

p² + p + 0.44 = 1
p² + p - 0.56 = 0

D = 1 + 4*0.56 = 3.24

[fierce rampart]

Mais c‘est plus que 1?

[oak magnet]

qu'est ce qui est plus grand que 1?

delta ?

[fierce rampart]

Mais je fais quoi avec la 3.24?

[oak magnet]

tu prends la racine

et tu calcule deux valeurs possibles pour p

[fierce rampart]

Bro I reallllyyy dont understand anything in math i just want to solve the fw excercices dont you wanna do them quickly?

.close

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[scarlet bison]

Hello! Quick differential equations question

[scarlet bison]

How do I know where to put the + C

vs.

[silk osprey]

it doesn't matter

[scarlet bison]

oh okay good to know

[silk osprey]

generally you put it on the right though

[scarlet bison]

but what is the reason for not doing a + C with both integrals?

oh wait

[silk osprey]

we only write one because the +C for both integrals can be combined into one constant

[scarlet bison]

i guess you could do + C with both integral?

but then you would subtract the costant

which would still be a constant

because the constants wouldn't cancel out?

[scarlet bison]

gotcha exactly

so the constants get combined

[silk osprey]

not necessarily and 0 is also a constant btw

yea

[scarlet bison]

okay that's a lot clearer

thank you so much

that's it

[silk osprey]

and you just write it as one

you're welcome

[scarlet bison]

.close

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[marsh robin]

Pulling my hair out

[marsh robin]

This was the first dy/dx

[silk osprey]

$\dv{y}{x} = -\frac{2xy^2 - 18}{2x^2y} = \frac{9 - xy^2}{x^2y}$

[twin meteor]

knief

[silk osprey]

yea you got that

[silk osprey]

i’m not going through this

it’s hard to read brother

[marsh robin]

Tried finding the second

But I can’t

I found the first

[silk osprey]

so

$\frac{x^2y(-2xy y’ - y^2) - (9-xy^2)(x^2y’ + 2xy)}{x^4y^2}$

[twin meteor]

knief

[silk osprey]

yuck

now

$\frac{x^2y(-2xy \cdot \frac{9-xy^2}{x^2y} - y^2) - (9-xy^2)(x^2\cdot \frac{9-xy^2}{x^2y} + 2xy)}{x^4y^2}$

[twin meteor]

knief

[silk osprey]

you had all this so far right

[marsh robin]

Yes but my two was in the front

[silk osprey]

$\frac{-2xy(9-xy^2) - x^2y^3 - \frac{(9-xy^2)^2}{y} - 2xy(9-xy^2)}{x^4y^2}$

[twin meteor]

knief

[silk osprey]

man these people are evil fr

let’s cancel xy

oh nevermind

that term doesn’t have it

[marsh robin]

Have been working on this for over a hour 🤣

[silk osprey]

you had this?

[marsh robin]

It’s a pretty evil problem for sure

[silk osprey]

$\frac{-4xy(9-xy^2) - x^2y^3 - \frac{(9-xy^2)^2}{y}}{x^4y^2}$

[twin meteor]

knief

[marsh robin]

I still had a 4 in the denominator since I’m not sure where it canceled out

[silk osprey]

i canceled the factors of 2 here^

$\frac{-4xy^2(9-xy^2) - x^2y^4 - (9-xy^2)^2}{x^4y^3}$

[twin meteor]

knief

[silk osprey]

ugh

then distribute i guess

$\frac{-36xy^2 + 4x^2y^4 - x^2y^4 - (9-xy^2)^2}{x^4y^3}$

[twin meteor]

knief

[marsh robin]

Thank you for helping me btw I greatly appreciate it

[silk osprey]

$\frac{-36xy^2 + 3x^2y^4 - 81 + 18xy^2 - x^2y^4}{x^4y^3}$

[twin meteor]

knief

[silk osprey]

$\frac{-18xy^2 + 2x^2y^4 - 81}{x^4y^3}$

[twin meteor]

knief

[silk osprey]

,w factor 2x^2 - 18x - 81

[twin meteor]

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Requested by kniefy

[silk osprey]

bruh

[silk osprey]

this should be simplified enough

@marsh robin

[marsh robin]

Checking now

I see the green check

[silk osprey]

https://tenor.com/view/sid-the-sloth-gif-27236411

[marsh robin]

I shed a tear

[silk osprey]

lmao

[marsh robin]

I thank you🤣🤣

[silk osprey]

i got you bro

🤝🏻

[marsh robin]

Wow what a problem, I’m giving him hell when I see him Monday

[silk osprey]

yea that’s just tedious work

[marsh robin]

Thank you so much

[silk osprey]

you’re welcome sir

[marsh robin]

I just type .close right?

[silk osprey]

yea

[marsh robin]

.close

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[visual coral]

lil bit of a dumb question, but how do you simplify cos(arctan(x)?

[edgy kayak]

try drawing a right angle triangle

label an angle theta

tan(theta) = x/1
so the opposite of that angle is x and the adjacent of that angle is 1

i hope u can carry from there

[visual coral]

ykw that's extremely less convoluted than the unholy method i was trying 😭

i was trying to reverse a silly thing i did to get the derivative of arcsin(x) kind of similar ot logarithmic differentiation but w sin and then itnegrate to somehow get f(x) back

but

uhhh

that wasnt working at all 💀

tysm for your help!

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[novel wharf]

How to simplify?

[viral copper]

might be sophie-germain

[novel wharf]

What's that

[jagged cargo]

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

[jagged cargo]

it is indeed sophie germain

[river kettle]

hi fungus

[novel wharf]

Oh ok, I'll try using it

[jagged cargo]

just an intuition, i would try only examining a small part of the fraction to apply sophie germain

the first term in each tesseract follows a linear pattern, i.e. 6x + 4

$\frac{(6(x + 1) + 4)^4 + 4\cdot3^4}{(6x + 4)^4 + 4\cdot3^4}$

[jagged cargo]

try applying the identity on this first. see what you get

[novel wharf]

Ok

[novel wharf]

Actually wait, I don't get the tesseract part

Uh, just explain to me in highschool terms

[patent nymph]

another name for fourth power

[novel wharf]

Oh

[patent nymph]

like, squared, cubed, tesseracted

[novel wharf]

I see

[patent nymph]

since a tesseract is a 4d cube

[novel wharf]

Ah, make sense

Well, after playing around with the sophie-germain formula, I think I have an idea now

Anyways, imma take it from here myself, I'll close the channel now

.close

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[bleak prawn]

Part (b)

Very rough diagram

So we know that m and n have slopes of 2

And if we get where the slops of the tangent to the circle are equal to 2 we get their locations

But I dont really know how to find where the slopes of the tangents are equal to 2

Do I just take the derivative of the circle and set x = 2? (im just guessing)

[willow pike]

take y = 2x + c

and then find distance of (3, 0) from it

[bleak prawn]

Ahhh

It should be equal to sqrt(20) right?

[willow pike]

yes

plus or minus

if algebraic distance is found

if you put modulus in the distance formula then it's already sorted

[bleak prawn]

1 moment working things out

[bleak prawn]

How do I do this with the c being unknown though

0 = 3(-1/2) + c
c = 3/2
2x + c = -x/2 + 3/2

c = -5x/2 + 3/2

x = 3/5 - 2c/5

[willow pike]

do you know the formula for distance of a line from a point?

[bleak prawn]

No I didnt know there was a formula

[bleak prawn]

I always did it this way

[willow pike]

oh

[willow pike]

then that was a given method in the book you followed?

[bleak prawn]

Im not following a book 😅

I dont have a teacher you guys are my teachers hehehe

[bleak prawn]

But I would like to know this formula that sounds useful

WAIT I do know this one I just googled it

|ax + by + c|/sqrt(a^2 + b^2)

[willow pike]

yes

this is the one

basically

take all the coefficients of the equation on one side

[bleak prawn]

So x = 3, y = 0

[willow pike]

yes

[bleak prawn]

a = 0, b = 2, c = c

[willow pike]

wait hold on

[bleak prawn]

And the distance is all equal to sqrt(20)

[willow pike]

what did you say the slopes of the tangents were?

[bleak prawn]

2

[willow pike]

2 or -1/2?

okk

[bleak prawn]

-1/2 is the red line slope

[willow pike]

then your line equation should be y - 2x - c = 0

[bleak prawn]

Ahh yeye I forgot about that step

a = 1, b = -2, c = c

Right?

[willow pike]

a is -2

[bleak prawn]

Oh

[willow pike]

and b is 1

[bleak prawn]

Okok

[willow pike]

a is coefficient of x

-2x + y - c

[bleak prawn]

|(-2)(3) + (1)(0) + c|/sqrt((-2)^2 + (1)^2) = 2sqrt(5)

[willow pike]

-2, 1, -c

are the respective terms

[bleak prawn]

|-6 - c|/sqrt(4 + 1) = 2sqrt(5)

[willow pike]

just substitute 3, 0 here

yes

[bleak prawn]

|-6 - c|/sqrt(5) = 2sqrt(5)

[willow pike]

yes

[bleak prawn]

|-6 - c| = 2(5)

[willow pike]

yes

[bleak prawn]

|-6 - c| = 10

How do I get rid of the ||?

[willow pike]

-6-c can be 10 or -10

first check the solution for 10

that gives one line solution

[bleak prawn]

Wait how sry

I get that |-6 - c| cant be negative

But how does it give us 2 solutions

[willow pike]

there are two tangents to a circle with the same slope

like a sandwich

|x| = 10 means x can be at -10 or 10

[bleak prawn]

Ahhh yeyeye

[willow pike]

for example

if we needed to equate it to 5

for EXAMPLE

|-6+1| = |-6+11| = 5

so c can be -1 or -11

same way for 10

a normal line equation is a straight line

but the modulus for a single variable is a V

so one solution in the y axis cuts through two possible values in the x axis

[bleak prawn]

Ok I thiiiink I get it

So -6 - c = 10 and -6 - c = -10
-c = 16 and - c = -4
c = -16 and c = 4

So the equations of the line are y = 2x + 4 and y = 2x - 16

(x - 3)^2 + y^2 = 20 is the circle equation

[willow pike]

i think

you can graph it to check

[bleak prawn]

So sub in (x - 3)^2 + (2x + 4)^2 = 20 and (x - 3)^2 + (2x - 16)^2 = 20

And that will give us the x values of the coordinates

Then get the y values using those x values

[willow pike]

and you should get a perfect square

[bleak prawn]

then done?

[willow pike]

(ax + b)^2 = 0 form for both

[willow pike]

yes

[bleak prawn]

Okokok I get it now 😄

Thank you so much for explaining!! You are a really good teacher

❤️

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[finite hatch]

How do i differentiate the end part

2pi^2r^3

Do i take both powers down by 1?

Ohhh im doing derivative with respect to r

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[hallow shadow]

can anyone help me with part a & b

[hallow shadow]

how do i find the value of a? in the right angled triangle?

[marble elk]

length l is half of 42

since O is midpoint

then use pythagoras

[hallow shadow]

ohh alright i get it now

and for part b do I just find the area of the rectanglular prism and add it to the triangular prism?

[marble elk]

surface area of rectangular prism + surface area of triangular prism - the area of common face

[hallow shadow]

what do you mean by the area of common face?

[marble elk]

there's a face common to both the prisms

that wont be included in surface area to giftwrap

[marble elk]

the 42 x 15 face

[hallow shadow]

ohh okay I get it

thank you

[marble elk]

so ur net surface area is 4 trianhles + 5 rectangles

[hallow shadow]

alright thanks

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[native kestrel]

take combinatorics

[slow patio]

??

i was thinking newton leibniz formula

<@&286206848099549185>

[paper depot]

does looking at f'(x) give us anything

[slow patio]

we can find maxima and minima by looking at f'(x)

i cannot simplify f'

@paper depot

[paper depot]

FTC and chain rule?

the derivative of $\int_0^{b(x)} g(t)\dd{t}$ is $g(b(x))b'(x)$

[twin meteor]

Ann

[slow patio]

??

i mean i obviously know newton leibniz formula

[winter hawk]

you have f(x) defined in this form

[slow patio]

but the resulting f ' i cannot simplify it

yep

[winter hawk]

ignore simplifying for a sec

how does the min relate to f'?

[slow patio]

by taking the derivative we can check at what points is the function f(x) minimum

or maximum

[paper depot]

and how do we do that

how do we find those points

[slow patio]

we have f(x)

[paper depot]

what do we do with f'

what do we do AFTER finding f'(x)?

[slow patio]

i mean i assume you guys know application of derivatives / monotonicity

[paper depot]

yes but do YOU know them

[slow patio]

yes ofcourse !

[paper depot]

and are YOU able to state clearly what needs to be done

then do it

[winter hawk]

exactly what computation with f' lets us find the min?

[slow patio]

THATS THE POINT AFTER DIFFERNTIATING I AM NOT ABLE TO MOVE FORWARD !

[winter hawk]

f'=0 at the min

[slow patio]

yes then

[paper depot]

show us what your f' is

[slow patio]

wait a min i ll send a photo of my working

[paper depot]

you could have sent that at the beginning and saved us some effort

[slow patio]

@paper depot

[paper depot]

ok right

so

again

you need to find when this equals zero, yes?

[slow patio]

yep

[paper depot]

note that x tan^-1(x) is always >= 0

and also e^(anything) is always > 0

[slow patio]

yep

[paper depot]

this can let you get rid of a lot of the complexity of the equation f'(x)=0

[slow patio]

i ll try it wait a min

can you just tell why arctanx is always greater than equal to zero

[paper depot]

no

i mean x*arctan(x)

not arctan(x) itself

[slow patio]

for x less than zero it will be ...?

[paper depot]

if x is negative then arctan(x) is negative also.

[slow patio]

oh i see

so denominator is always >=0

and also e to the power everything

@paper depot i have got a smaller term but cant simplify it further

<@&286206848099549185>

[paper depot]

show

[slow patio]

@paper depot

[paper depot]

bad transformation

arctan(x) + x/(1+x^2) = 0

note that you have here the sum of two numbers which are always of the same sign

either both positive or both negative

how can it ever happen that their sum is zero?

[slow patio]

yes!

does that mean no solution ?

@paper depot

[paper depot]

no, it means the only solution is x=0.

[slow patio]

oh i get it

.close

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[flat whale]

no fractions don't turn from addition to multiplication

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[paper panther]

hello, a friend sent me this because she couldn't find the mistake, and its supposed to be solved without substitutions, so idk what to do

[willow pike]

how did tan²x become sin²x in the 4th step?

oh

never mind

misread

[paper panther]

its cos⁴x in the denominator

[willow pike]

yeah

there's no mistake

[spiral turtle]

There's no mistake

[peak matrix]

he had basically undone what he did tho

[spiral turtle]

you just get back to F = F

[paper panther]

yeah thats what I thought

[willow pike]

you just get F(x) = F(x)

[paper panther]

but that's useless

[spiral turtle]

Why not simplify sin^2 / cos^4 to sec^2 tan^2 ?

[willow pike]

or convert using double angle?

[paper panther]

what's the point of that

[spiral turtle]

then you have u = tan x and du = sec^2 x dx

[peak matrix]

its supposed to be solved without subs

[paper panther]

no substitutions

[spiral turtle]

oh

[willow pike]

apparently no subs

[spiral turtle]

my bad

[willow pike]

):

[paper panther]

it's very easy with substitutions

but they werent taught substitution yet ☠️

[twin horizon]

Are you allowed to use the general form for \int sec^n x

[paper panther]

no 😭

[willow pike]

is that a thing

generally

[twin horizon]

I saw it somewhere

[paper panther]

do you think people who do not know about substitutions would know this formula lol

[spiral turtle]

would it be possible to reduce this using the double angle formula for cosine?

[willow pike]

trust me

i tried

[paper panther]

cos²x = (1+cos2x)/2 ?

[willow pike]

this behemoth though

[spiral turtle]

cos^4 x = 1/4 (1 + cos 2x)^2

[paper panther]

that lowkey seems harder to solve

[willow pike]

oh

🤔

[spiral turtle]

maybe you can then do a partial fractions?

[willow pike]

probably

[twin horizon]

Maybe (cos²x+sin²x)²=1

Idk if that helps

[willow pike]

hey hey HEY

i think

i did something similar

but was it with cos⁴

im not sure

brb

[twin horizon]

[willow pike]

yup

works like a charm

[twin horizon]

What works?

[willow pike]

the integral

it's smooth sailing from here

[twin horizon]

[willow pike]

@paper panther does the answer have tan³x somewhere?

[spiral turtle]

I mean, this is where I got to, but without a substitution you can't eval tan^2 sec^2

[willow pike]

oh yeah

[twin horizon]

I guess so

[willow pike]

my bad

well we can always reverse engineer

[paper panther]

,ask integral 1/(cosx)^4

[twin meteor]

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Requested by eigenmirai

[paper panther]

why does it write it like that

[willow pike]

no idea

my eyes burn

[paper panther]

that is tan³x yes

[willow pike]

https://tenor.com/view/tbh-creature-autism-creature-yipee-yippee-confetti-gif-26570203

now how do we get there

im getting $\tan x + \frac{1}{3} \tan ^3 x$

that's the same as this?

[peak matrix]

,w differentiate tanx + 1/3 tan^3(x)

[twin meteor]

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Requested by methisalwaysright

[willow pike]

nice

[peak matrix]

yep, the same thing

we could reverse engineer the solution

[willow pike]

hah

another supporter

what do we split it into though

[peak matrix]

sec^4(x) = sec^2(x) + sec^2(x)tan^2(x)

idk if this is any easier tho

oh its what we had before

nvm

[willow pike]

i suggest splitting tan³x

[peak matrix]

can we use IBP?

[twin horizon]

Idts 😭

[twin horizon]

@paper panther

[peak matrix]

That'd leave us with exactly 0 applicable rules

[twin horizon]

[paper panther]

i think so

but like atp just drop it im probably just wasting ur time

[vocal sleet]

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[pulsar dome]

why do we add two gradients in 2 constraints Lagrange optimization problems, what is the geometrical interpretation of it? I understand in the case of one constraint, level curves of two function touch at some point and in that point gradients are parallel, but cant wrap my head around when there are two constraints

[spiral turtle]

So imagine you have two spheres, the place where they intersect is a circle

This circle is a level curve where both constraints are satisfied.

Unfortunately, with more constraints and more dimensions of the underlying function, we rapidly stop being able to effectively visualize these things because they start requiring more than 3 dimensions.

@pulsar dome ^

I'm sorry if it's not entirely clear

[pulsar dome]

thank you for explaining, but it is still not entirely clear why do we add ,in this example, gradients to the circle

[pulsar dome]

I see

[vocal sleet]

@pulsar dome Has your question been resolved?

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[spiral turtle]

.reopen

[vocal sleet]

✅

[spiral turtle]

@pulsar dome sorry I went AFK

The idea I was trying to illustrate is, say for instance, you have f(x, y) and a constraint that makes a circle, like x^2 + y^2 = 4

your constraint reduces the number of dimensions of the valid output by one.

typically, this is true of any constraint.

now in the second case, you have f(x, y, z) and the two constraints each make a sphere.

let's say (x-2)^2 + y^2 + z^2 = 4, and (x-1)^2 + (y-1)^2 + z^2 = 9

these two constraint surfaces together have an intersection. And this intersection is the mutual constraint surface

This is typically (but not always) each reducing the number of dimensions of the possible solution space by one.

But the way they interact with the function is different.

so they take two different legrange constants.

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@pulsar dome Has your question been resolved?

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[astral echo]

even though the answer to the first box should be negative cube root of y-3 in the site my homework is wrong there is no cube root

have i done something wrong

??? where is my cube root

[oak magnet]

Can't you type a fraction with the x^... ?

If yes you can do ^(1/3)

Else your answer seems correct so i advice you to contact them to ask

[flat whale]

missing a - sign

[quasi timber]

I want to isolate I

[flat whale]

!occupied

[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).

[astral echo]

i can only check two times on the site

[oak magnet]

!occupied

[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).

[astral echo]

so next try i wont know if i got it wrong

[oak magnet]

Riemann point it, missing a minus

[astral echo]

that is true

should this be the proper formatting

[flat whale]

no

you did the algebra wrong

[astral echo]

how

it literally should be right

[flat whale]

you took the cube root before multiplying by -1

(-x)^n is not always the same as -x^n

[astral echo]

idk then because the google ai begs to differ

[flat whale]

then follow google ai and close your channel

[astral echo]

.close

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[sage yarrow]

i need visualizing -660 degrees. would appreciate to be taught how to draw them

[cunning plaza]

Think circle. Start at the rightmost end, a quarter turn clockwise is 90 degrees. After doing 4 quarter turns, you will end up back where you started and at -360 degrees. This shows that -360 degrees is = 0 degrees. Keep going for -660.

[sage yarrow]

and the opposite for positive?

[cunning plaza]

yup, anticlockwise

[sage yarrow]

simple enough

thanks!

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[pulsar dome]

thank you very much! now it is much more clear

[pulsar dome]

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[somber ore]

today we did graphing quadratics in factored form but i had a doctors appointment today and this is a new unit we started today

[sour night]

What exactly do you need help with?

[somber ore]

well since i missed todays lesson I need help with all of it

[sour night]

Alr, lets start with the x-ints first

do you know how to find them?

[somber ore]

no
its a new subject we started today

[sour night]

alr

do you know exactly what an x-int is?

[somber ore]

yeah its where y = 0 and the line touches the graph

[sour night]

yup, so when would f(x) equal 0 for the equation above

[somber ore]

when y = 0?
i dont think i get the question

[sour night]

do you know exactly what y means?

[somber ore]

up and down
vertical

[sour night]

yeah, but in this case it also means the Output

[somber ore]

huh

[sour night]

do you know what x is?

[somber ore]

horizontal

side to side

[sour night]

yeah, but it is also the input

so for example if you have y=2x+1

what is y when x is 2

[somber ore]

5

is f(x) = y?

[sour night]

yup!

now do you know when an equation equals 0?

[somber ore]

when F(X) = 0

[sour night]

yeah

now set the equation (x-2)(x+3) equal to 0

do you know for what input values (x) the equation equals 0?

[somber ore]

uhh

lets see

2 and - 3

[sour night]

yup!

those are your x-ints

do you need help for the rest of them too?

[somber ore]

yep

[sour night]

you wouldn't write the x-ints like that

it would be (2, 0) , (-3,0)

[somber ore]

oh wait its not a singular point hold on

btw ive got a answer key to check with just in case

[sour night]

do you know what AOS stand for?

[somber ore]

axis of symmetry

[sour night]

yeah, now do you know how to find it?

[somber ore]

no

I litterally now none of them for factored equasion

[sour night]

if you don't mind me asking what level math is this

[somber ore]

2

honors

[sour night]

alr

so do you know how to find an average?

[somber ore]

add all numbers and divide by amount of numbers

[sour night]

yup

just find the avg of the x ints

and thats your AOS

[somber ore]

oh that sounds easy

one sec

-1/2

[sour night]

yeah

so it would be x=-1/2

as the AOS

since it is meant to be a straight line

[somber ore]

ok vertex time

[sour night]

remember to set it equal to x

[somber ore]

its just X=-1/2

[sour night]

yup

the vertex should be pretty easy to find now with this new info

do you know what exactly the vertex is?

[somber ore]

the maximum or minimum

[sour night]

yup

and do you know what the x-value of the vertex is?

[somber ore]

its the AOS

[sour night]

yup

so u would just plug it in the equation

for the y-value

[somber ore]

oh so just solve for y

thats easy

[sour night]

yeah

[sour night]

yup

[somber ore]

so vertex is (-1/2 , -6.25)

[sour night]

yup

[somber ore]

ok then, open and domain are easy because domain is pretty much always all real numbers and its positive so it opens up

[sour night]

yup

[somber ore]

last thing is range

[sour night]

range is just the vertex

and whether it points up or down

[somber ore]

(-6.25, ∞) i believe

or was it the x

[sour night]

yup

no it's y

[somber ore]

ok

one last thing

could we do another problem but in this one A is greater than 1

[sour night]

its the same thing

its only different during the vertex and range

[somber ore]

elaborate pls

[sour night]

it only affects the y-values

[somber ore]

alright lets just start

x ints are 1 and 4

[sour night]

yup

[somber ore]

aos is 2.5

[sour night]

yup

[somber ore]

so whats up with the vertex

hows it different

[sour night]

you have to multiply it by 2 now

[somber ore]

so not much changes

[sour night]

yeah

[somber ore]

just same steps + multiplying

[sour night]

yup

[somber ore]

@sour night like this?

[sour night]

yup

[somber ore]

ta dah

alright thanks bubba
theres some other questions I might have trouble on a bit later in the work but i think ive got it

close

.close