help-17

[spring gate]

why does it = 4

is it just an arbatrariy number?

obviously not but i dont see the logic

[mild flower]

well, du/dx is the derivative of u with respect to x

[last creek]

do u know how to differentiate or find the derivative ?

[flat whale]

Is what an arbitrary number?

[spring gate]

yes

[last creek]

what's the derivative of 4x-3

[spring gate]

4

oh

huh

you know until you said it i just REALLY didnt read "du/dx" as "the dirivitive of U"

and i dont know how

,-,

[last creek]

let u = 4x-3
differentiate both sides you get du/dx = 4
now we can say that du/4 = dx

[spring gate]

i think dy/dx and f' was too in my head

[spring gate]

ye

[vocal sleet]

@spring gate Has your question been resolved?

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[vast shale]

Hi

[vast shale]

Afte the red arrow, if I suppose N terms take value k and (73-N) term take value (k+1) then why does it make a difference ..
Cuz

(k+1)(73-N)+kN = 54
73k+73-N=546
73k-N = 473

I get different 'k'

Yep

I'm dumb

73×7-38

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[strange maple]

Hi can i please get some help I’ve been asking for help since Thursday a few people have helped me a bit but it’s still not all the questions I need help with

[strange maple]

Yesterday when I got to Question A: for b I got the following…

But then idk what to do next I’ve solved did b but they are still other questions that I need help with

These are the questions I need help with

[vocal sleet]

@strange maple Has your question been resolved?

[strange maple]

<@&286206848099549185>

Please can I stop being ignored 💀

[dusk mantle]

lol dont talk like that

[edgy carbon]

soz

[dusk mantle]

I SAW THAT

I SAW

but i got no proof

[civic drift]

I saw that too

[tame pasture]

In a math discord, that means it isn't true ;)

I saw it too btw

[dusk mantle]

THEN

BRUUUH

[tame pasture]

ask a mod?

[dusk mantle]

yea

<@&268886789983436800>

[strange maple]

Well could I please get help I beg 🙏😭💔

[edgy carbon]

[strange maple]

Bruv please 💀

[sterile forum]

Yeah, don't be rude in the way of those now deleted posts.

[edgy carbon]

😭

sorry

[sterile forum]

Help here is given by volunteers who hang out in their own spare time and engage with people when they feel like it.
Instead of begging, you would increase your chances that someone wants to answer you if you spend your time writing a self-contained and focused question about some actual point that confuses you, rather than just dump an entire (hand-written!) homework problem on people and saying "help!" without any description or comments.

[strange maple]

Very firm and straightforward the problem is that I don’t understand much on the activity hence why I sent the activity through

I got to this part but I’m still confused when I asked for help on Thursday I sort of managed to solve b and q but it’s still incomplete..

Hence why I’ve been asking for help because I’m not sure what to do next

okay I’ve made a conclusion I’ve read this clearly and I think if I get assistance to solve Part A of the activity I’ll be able to do the rest so thank you for letting me know I can’t dump the whole activity and expect to get help for the whole activity

.close

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[trim walrus]

By observation, I can see that m=11, n=8,

But is there any other way to do it..

m,n are positive integers ...

Sry

[paper depot]

did you really mean ${m, n} \in \bZ$ exactly as written or did you mean that $m$ and $n$ should both be integers

[twin meteor]

Ann

[paper depot]

the {} are inappropriate here

[trim walrus]

Alr

I mean they're individually an integr

What does {} mean tho 🤔

[paper depot]

using a symbol without knowing what it means

{} the way you wrote it would be notation for a set

with the stuff inside being either a comma-separated list of elements or part of a set-builder notation

anyway!

m and n have to be positive (or at least nonnegative) integers, one would think. (otherwise we would get fractional values for 2^m or 2^n, which would do us no good here)

moreover, 2^m - 2^n has to be positive, therefore m > n

therefore 0 ≤ n ≤ m-1

do you follow thus far

[trim walrus]

No

[trim walrus]

You lost me here

Or I get it

[paper depot]

n has to be less than m

and n has to be nonnegative

[trim walrus]

Yes I get it now

[paper depot]

i mean ok like we can exclude the cases of either m or n being 0

since thatd give us 0 or an odd number

while 1792 is even

but it's whatever

since n ≤ m-1, we get 2^n ≤ 2^(m-1)

thus $2^m - 2^n \geq 2^m - 2^{m-1} = 2^{m-1}$

[twin meteor]

Ann

[paper depot]

do you agree or disagree

[trim walrus]

Agree

[paper depot]

ok

on the other hand 2^n > 0 clearly so 2^m - 2^n < 2^m

so we have $2^{m-1} \leq 1792 < 2^m$

[twin meteor]

Ann

[paper depot]

with m ∈ Z, this lets us find m unambiguously

[trim walrus]

I see that, m can be 11

[merry python]

are you again just guessing?

[trim walrus]

I mean 2^10 <= 1792 <= 2^11
1024 <= 1792 <= 2048

[merry python]

so a calculated guess

but actually you do not need guess

(your answer is correct tho)

[trim walrus]

Alr

[merry python]

You can proceed with a substitution
m=k+n

where k is any real number

[trim walrus]

Alr

Or I just plug the value of here

[paper depot]

@merry python dunno what you're suggesting here but that's at best very longwinded

at this point you can just take log_2 of that inequality and get m-1 ≤ log_2(1792) < m

[merry python]

I was thining to compare powers

[paper depot]

thus m = floor(log_2(1792)) + 1

[merry python]

$$2^{\left(k+n\right)}-2^{n}=1792$$
$$2^{k}\cdot2^{n}-2^{n}=1792$$
$$2^{n}\left(2^{k}-1\right)=2^{8}\cdot7$$
$$n=8,\ k=3,\ m=11$$

[twin meteor]

beard420

[trim walrus]

Alr thanks I get it now

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[verbal merlin]

okay i was thinking what if i use accelaration to prove that one is constant and one is not

cus rate of change of speed = acceleration rite

so acceleration = 0 = constant speed

but how do i show they move along the same path? is it related to positioning or smth?

[vast shale]

show that the two loci are the same

ie both r_1 and r_2 sketch out the same circle

[vocal sleet]

@verbal merlin Has your question been resolved?

[verbal merlin]

Oh wait ah lemme try

[verbal merlin]

Wait i dun really understand

[vast shale]

points on r_1 are of the form (2cos(3t), 2sin(3t))

(for all t, which is the parameterisation variable)

[vast shale]

can you show that this is a circle? what is the radius and centre?

[verbal merlin]

i dunt think i do TvT

[vast shale]

what is x^2 + y^2?

where (x,y) = (2cos(3t), 2sin(3t))

[verbal merlin]

wait ah brb in 10 mins

my discord has some issues TvT

i will keep the server first TVT

[verbal merlin]

apologies im back haha

[verbal merlin]

1

@vast shale apologies for long wait

[vast shale]

no it isn't 1

[verbal merlin]

o.0

oh shit its 2

my bad

[verbal merlin]

2?

not 1

and the 2nd one wld also be the same right?

@vast shale sorry for delay i managed to fix my discord finalllllyyy

it shld be 2 for both rite!

[vast shale]

still no

remember to square everything

x^2 = (2 cos(3t))^2 = 4 cos^2(3t)

[vocal sleet]

@verbal merlin Has your question been resolved?

[verbal merlin]

oh oop yes yes

sorry my parents called me

[verbal merlin]

so its 4

but whats the point in doing dis?

@vast shale

nvm i shall close this and open again

my discord got problems fr

.close

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[plain parcel]

hi

[sudden compass]

.close

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[fluid fjord]

im sory if i pose a simple problem but i struggle with it for some time and i would appreciate any help

Consider fractions $\frac{a}{b}$ where $a$ and $b$ are positive integers.
(a) Prove that for every positive integer $n$, there exists such a fraction $\frac{a}{b}$ such that $\sqrt{n} \le \frac{a}{b} \le \sqrt{n+1}$ and $b \le \sqrt{n}+1$.
(b) Show that there are infinitely many positive integers $n$ such that no such fraction $\frac{a}{b}$ satisfies $\sqrt{n} \le \frac{a}{b} \le \sqrt{n+1}$ and $b \le \sqrt{n}$.

[twin meteor]

ifunkinghatetransphobes

[quick fjord]

Please open up another channel

This one is going to auto close in a bit

[fluid fjord]

how?

[quick fjord]

Can you see the available help channels

[fluid fjord]

help 12 21 22 23?

[quick fjord]

#help-12 is available for example

[fluid fjord]

ok thanks

[quick fjord]

Yes

You can take one of those

[paper depot]

and now it is not

oh wait no

it's claimed by beep

[quick fjord]

Yea

[vocal sleet]

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[woeful bough]

LINEAR ALGEBRA - MATRICES

[woeful bough]

Can someone tell me what is the point of this step?

I understand
if c = -1 we will have no solution
if c /= -1 we will have unique solution
but what is the point of dividing the R3 like we did in the red box and having the matrix look like that?

__

Actually I don't even get the correct method to accurately find the values of C for no solution/unique solution/infinitely many solutions, any help plz?

[vocal sleet]

@woeful bough Has your question been resolved?

[mild flower]

remember that implicitly we're adding and subtracting equations here, just without the variable names and = signs

so the red box matrix represents three equations, the last of which is $z = \frac{c-2}{-c-1}$

[twin meteor]

Hayley

[mild flower]

you could leave it in the previous form I guess... and that might be easier to see the solution (and that's totally valid for solving the problem!) but having it in echelon form is nice and consistent

[woeful bough]

But what does having it in that form (red box) show us here? And why didn't we do it earlier? (Before the "the system will have no solution if c=-1" step)

[mild flower]

to get to that box we had to divide row 3 by -c-1, which is 0 if c=-1

[woeful bough]

If c=-1
R3 = [0 0 1 | 0] ?

Like what if I wanted to do this red box step here

Then try to find the values of C for which the system has
A) no solution
B) unique solution
C) infinitely many

What issue would that cause me?

[mild flower]

if you substitute c=-1 into that box you'll get [0 0 1 | 0/0]

[vocal sleet]

@woeful bough Has your question been resolved?

[woeful bough]

I see, okay thanks

[vocal sleet]

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[scarlet girder]

how do i find where the function has the slope of 0?

I would assume that I would have to do something like this

$0=2x^2-16x+8$

[twin meteor]

LilliPi

[scarlet girder]

and then solve for x

but i can see on desmos that the answer is 4

[flat whale]

nah

those are the roots

the slope is given by the derivative

[hasty pulsar]

4 is not zero value

[scarlet girder]

the derivative is $f'(x)=4x-16$

[twin meteor]

LilliPi

[scarlet girder]

do i just plug in 0 and -4?

[flat whale]

no

that would be finding the slope at those x values

you want to solve for what x value gives the slope = 0

[scarlet girder]

so $0=4x-16$

[twin meteor]

LilliPi

[lilac sparrow]

😎

[scarlet girder]

i got 4

and for the other

$-4=4x-16$

[twin meteor]

LilliPi

[scarlet girder]

I got 3

[flat whale]

yes

[flat whale]

why are you setting the derivative to be -4?

[scarlet girder]

because im trying to find where the slope is -4

[flat whale]

oh part b. other one. yes

[scarlet girder]

ok

thanks guys 😄

.close

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[empty linden]

?

[vast shale]

there is only really one thing you have to do to "simplify" this

[boreal remnant]

Just divide the top and bottom by e^(-x)

[empty linden]

the only lead i have is to use the negative exp rule

but thats about it

[vast shale]

yea

u got it if the negative exp rule is what i think you mean

[empty linden]

i first brought the e^-x over to the other side

since i made it = 0

[vast shale]

okaaaaaaay

pause

pause

[empty linden]

2 = 1/e^-x

but this is wrong

[vast shale]

what "other side"? this is not an equation, it is just an expression

[empty linden]

i thought i could make it = 0

[vast shale]

no...

this is not an equation, there is nothing to set it equal to 0 for

unless there is missing context to your question

[empty linden]

but since 1/e^-x is the negative exponent rule

howd it get to 2e^x?(the answer)

besides, theres already a numerator, which is the 2

[vast shale]

okay can you state the "negative exponent rule" so we are on the same page here?

[empty linden]

yes

same page

[vast shale]

what

[empty linden]

wait what

ok ok yk what

ill just say what i think

so, i know i should use the negative exp rule

thats it

i now want to know how it led to this

wait a minute

i think i know

since e^-x is 1/e^x

id have to divide 2 by 1/e^x

and in division, you gotta flip the divisor. as in, multiply it by its reciprocal. so itd be 2/1 multiplied by e^x/1

= 2*e^x = 2e^x

is there anything wrong with this method? was there an error?

[vast shale]

i think you got to the answer with a more roundabout approach, but sure, that works

[empty linden]

👍

not that bad i guess. it only turned out to be around 4 steps when i wrote it down on paper

[vast shale]

i'd advise to follow Mr. Gamer's approach for clarity's sake

[glossy maple]

how does that help

you get 2/e^(-x) on the numerator

same thing lol

[vast shale]

oh i thought that said multiply by e^(-x)

no wait thats still not helping i am tripping sorry

[glossy maple]

lol

multiply by e^x

[boreal remnant]

that's literally the exact same thing but ok

[glossy maple]

well if you're saying dividing by e^(-x) is the same as multiplying by e^x

just say its 2e^x

ur doing extra work

but it is the same yes

[boreal remnant]

lol

try khan academy, they have a lot of good lessons on exponential rules

[vocal sleet]

@empty linden Has your question been resolved?

[empty linden]

wait

why isnt the answer for this e^2 - 5e^6x? instead, its e^2x - 5e^5x(2nd img is the answer)

man i feel dumb. they didnt teach any of this in school

what i did was distribute the e^2x

and got e^2x - 5e^6x

thats it

[glossy maple]

yes and how did you get 6x

$e^x \cdot e^y = e^{x + y}$

[twin meteor]

redstoneplayz09

[empty linden]

yeah but

theres the property that said if theres brackets, we multiply

wait, ill find it

the 3rd one

which is why i was confused

[glossy maple]

the third one is

(a^m)^n

thats not what u have here

[empty linden]

i see

[glossy maple]

you dont have a base to an exponent, and then to another exponent

$e^{2x} \cdot \parens{1 - 5e^{3x}} = e^{2x} \cdot 1 - e^{2x} \cdot 5e^{3x}$

[twin meteor]

redstoneplayz09

[glossy maple]

you can also write it like this

$e^{2x} - 5 \cdot e^{3x} \cdot e^{2x}$

[twin meteor]

redstoneplayz09

[glossy maple]

order of multiplication doesn't matter

[empty linden]

ok

[glossy maple]

and $e^{3x} \cdot e^{2x} = e^{3x + 2x} = e^{5x}$

[twin meteor]

redstoneplayz09

[empty linden]

as for this, this one is 0 because:

ln(2 x 3^2) - ln 18
ln(18) - ln 18
ln(0)
0

i multiplied the 2 and the 3^2 because of the product property

[glossy maple]

not ln(0)

just 0

[empty linden]

yeah

[glossy maple]

but yes thats correct

🫡

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[obtuse nest]

Hey math legendaries, a quick question regarding latin squares. So my textbook explains a kind of weird method to find an orthogonal latin square of any odd-sized latin square. The pattern looks like rotating the original square 90 deg to the right(like top row, becomes last column), just asking if I got it right? Or maybe it works with a specific placements of items in original square. Thanks

[vocal sleet]

@obtuse nest Has your question been resolved?

[obtuse nest]

Alright, it's a little weird, and I'm sorry. I'm going to find another reference regarding that. Sorry for occupying the channel. Math bless us all

.close

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[viral badger]

13x^4 sqrt 35x^6

[vocal sleet]

Show your work, and if possible, explain where you are stuck.

[shy gulch]

what does this even mean

are you trying to simplify it?

[viral badger]

yes

[viral badger]

yes

[paper depot]

$13x^4 \cdot \sqrt{35} \cdot x^6$?

[twin meteor]

Ann

[vocal sleet]

@viral badger Has your question been resolved?

[west raptor]

I think he means $13x^4\sqrt{35x^6}$

[twin meteor]

math_is_fun

[vocal sleet]

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[dark sleet]

I have to write this complex number in standard form and I already know that the first part is 5 - but I assumed the second part was 3.5 but that doesn't make sense $5-\sqrt{-12}$

[twin meteor]

Iucas

[dark sleet]

i just dont really understand how to get the second part

if its not 3.5 or 3.4

[paper depot]

what's wrong with keeping it as sqrt(12), or perhaps as 2*sqrt(3)?

[dark sleet]

$5-2\sqrt{3i}$ i thought this was wrong for some reason but this is correct thank you

[twin meteor]

Iucas

[paper depot]

$2 \sqrt{3} i$ not $2 \sqrt{3i}$

[twin meteor]

Ann

[dark sleet]

ah

[vocal sleet]

@dark sleet Has your question been resolved?

[viral badger]

yeah

[vocal sleet]

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[regal bane]

Complete the square for x, and complete the square for y, separately

Oh wait no, I didn't see the xy term

[winter raft]

yeah

[paper depot]

hold up

this is sus actually

oh wait fuck typo sorry

3xy not 2xy

this changes everything actually

theres apparently only one point that satisfies this eq

[winter raft]

yeah

[winter raft]

how do you solve the equation anyways?

[paper depot]

can't think of any non painful ways to do it

[winter raft]

oh ok

[empty frigate]

it's $(\frac{1}{2}x + 3y + 3)^2 + (4y+2)^2$

[twin meteor]

bee [it/its]

[empty frigate]

i don't really know how they expected you to deduce that, it's not at all obvious

1/4 is 1/2^2 so that suggests you're going to have 1/2 x
13 is 4 + 9 which is 2^2 + 3^2
the 3 going with x/2 is a good way to generate 3x
34 is weird so it looks like a sum of two y terms which makes sense because we know there's an xy term
25 could be 5^2 but it's also 3^2 + 4^2, and 3y going with x/2 would generate that 3xy term
^ this is basically roughly what i did

[winter raft]

so did you just guess till here?

[empty frigate]

yeah pretty much

[winter raft]

or was there any way of rearranging it into that form

[empty frigate]

it was smart guessing, i didn't just literally try random numbers until it worked, but i don't know of a mechanical way to do what i did

[winter raft]

wait so how do u solve for x and y after that?

[empty frigate]

well we know that this expression is equal to zero

and both of the squares are >= 0, which means they must both be zero (if one is larger than zero the other can't be negative to cancel it out)

so then you get 4y + 2 = 0, and x/2 + 3y + 3 = 0

[winter raft]

ahhh, that makes ssnese

thanks dude

[empty frigate]

(i'm not a dude)

[winter raft]

thanks*

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[steep musk]

how is it approaching cos y

[merry python]

what have you tried?

[steep musk]

I’ve tried taking derivative

but idk how it’s gonna work cuz of y

I tried close the sum to product formula

on sin

[paper depot]

y is just y

it's a constant

[steep musk]

ohhhhh

ok

thankyou

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[floral ledge]

qn 19 well i understand nothig

[thin vale]

question 19 isn't in this image

[floral ledge]

mb 14

[dreamy viper]

lol almost helped with 14

ah

just means the sphere has a radius the same has the cylinder

go from there

[floral ledge]

ok i did that alreeadly

but i dont understand what the qn wants by suface area of can in contact with water when the sphere is inside it

[dreamy viper]

yyou said you understood nothing

[floral ledge]

er 🙂 well i dont sort uf understand the qn

or what they want you to find

[dreamy viper]

okay

[floral ledge]

whether the sphere is inside it shoud have the same surface area in contact with the water i thought?

[dreamy viper]

you can figure out how deep the water is since you know the radius of the sphere is the same as the radius of the cylinder

since a sphere is well.. a sphere

[floral ledge]

how?

[dreamy viper]

then you can use the SA formula for a cylinder to figure it it out

but has to be for a cylinder with no top

[floral ledge]

i dont understand how u find how deep the water is from this

[dreamy viper]

how deep do you think the water is

[floral ledge]

i have no idea

[dreamy viper]

whats the radius of the sphere

[floral ledge]

3.4cm

[dreamy viper]

whats the diameter

[floral ledge]

6.8

[dreamy viper]

how deep is the water in relation to the sphere

"just covers"

[floral ledge]

but isint the diameter horizontal the height is vertical tho?

[dreamy viper]

spheres are spherical

they arent ovals

[floral ledge]

so their diameter and height are the same?

[dreamy viper]

not an ovoid

yes

[floral ledge]

oh

that solves how to do part 2

but im still not sure how do u do part 1

[dreamy viper]

do you know the formula for the surface area of a cylinder

[floral ledge]

4pieR^2

[dreamy viper]

what

[floral ledge]

no?

[dreamy viper]

nope

[floral ledge]

eh?

[dreamy viper]

surface area not volume

what is surface area

[floral ledge]

1/2 x4pieR^2?

[dreamy viper]

what is the definition of surface area

[floral ledge]

huh

isint

volume

4/3pieR^3

[dreamy viper]

thats the volume a sphere

surface area of the cylinder

[floral ledge]

isint this the surface area of a sphere

oh

pieR^2

[dreamy viper]

cylinder is not a sphere

[floral ledge]

no you asked for the formula for the surface area of a sphere

which i replied 4pieR^2

[dreamy viper]

my bad

but regardless

the question asked for the cylinder no?

[floral ledge]

would the surface area of the can in contact with the water be the surface area of the cylinder-curved surface area of sphere?

[floral ledge]

yes

[dreamy viper]

just think about the can not the sphere

[floral ledge]

but pieR^2 is not the ans tho

[dreamy viper]

sphere is irrelevant now we know our height

[floral ledge]

i have the ans but i have no idea how to get it

its 182cm^2 but idk how

[dreamy viper]

it would have to be a numerical answer

ok

so im going to give you the formula of a cylinder and explain why

[floral ledge]

oh

ok

[dreamy viper]

its pi(r^2) + 2pi(r*h) for a cylinder with no top

we can lay out the pieces of a cylinder has the top and the round walls

the area of a circle is pi(r^2)

[floral ledge]

oh i didint know about this forumla

[dreamy viper]

the walls are going to be a rectangle

the length will be the circumference of the circle

the height will just be the height, which we found

make sense?

[floral ledge]

ok yes

[dreamy viper]

cool

so now we can plug in numbers

we have the radius which is given and the height which we figured out by finding the diameter of the sphere

[floral ledge]

ok

im still stuck-_-

[dreamy viper]

ask questions

[floral ledge]

still not sure how to find part (i)

although you know the water is 7cm high but thats not the total volume of the cylinder right?

height*

[dreamy viper]

it is the height that you want

because it asks for the SA of only the cylinder that is touching the water\

[floral ledge]

the SA in total is only 38.485

SA of cylinder

but the ans is more than that which i dont understand how

putting a sphere inside the cylinder doesnt increase surface area? doesnt it decrese it

[dreamy viper]

you did pi(3.4^2)+2*pi*3.4*6.8?

[floral ledge]

how did you get the 6.8?

[floral ledge]

and this is the volume or? total area?

[dreamy viper]

surface area

[floral ledge]

oh

wait what

isint the surface area just the circle

at the bottom

[dreamy viper]

no

the rectangle and the circle

[floral ledge]

oh

6.8 is from?

[dreamy viper]

the diameter of the sphere

[floral ledge]

ohhh

i understand it now i see

thanks for the help!

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[teal radish]

In this diagram from the following video: https://www.youtube.com/watch?v=gC6xHzgm28k&list=PLgPbN3w-ia_MzatWGASfuPg9hon_Fsz1V&index=17&ab_channel=ProfessorBryce around 8:00, i dont quite get how the max flow is 3 because the value of flow is the flow(in) = flow(out) right? But flow(out) of s is 5, but flow(in) of t is 3

[mild flower]

max flow across the network needs to consider the network as a whole

sure, max flow out of s is 5 if you just look at that node but the flow "has nowhere to go", so to speak

[teal radish]

what do you mean by "nowhere to go"?

[civic pendant]

@teal radish a flow function needs to satisfy certain conditions. For instance, the total flow into a vertex must be equal to the total flow out of the vertex (except for s and t)

[regal bane]

If you pump 5 units/sec out of s, where's some of that going to go?

Only 3 units/sec can enter t, so...

Your system will be gaining 2 units/sec, and that's not okay

[civic pendant]

There is a difference between flow and capacity @teal radish

[regal bane]

So you can't use the max flow rate of the two pipes coming out of s

[civic pendant]

To back up what Kaynex is saying, if you set the flow in s->b to 3, it cannot be transfered outside of b, because the total capacity of the outgoing arcs of b is 2.

[teal radish]

Oh so why is it okay to go from the path s->a->b->t?

[civic pendant]

It's ok if you set the flow to 2,

because you never exceed the capacities

you can transfer 2 units of flow through this path s->a->b->t

but this is not the max flow

[teal radish]

But if you look at a here, we only have flow of 2 incoming, but outgoing of 4. How does that satisfy the flow conservation?

[mild flower]

those arrows are capacity not flow

[civic pendant]

@teal radish again, you should make a distinction between flow and capacities

It would be helpful for you to add another column with c(e)

[teal radish]

In the video he's denoting them as flow I think so I was confused

[civic pendant]

c(e) is not the flow, this is the maximum amount of flow that can be transfered through this edge

[teal radish]

because he has e, and f(e) only in the table

[civic pendant]

f(e) is what you need to find

that is, the output of the algorithm

c(e) are given as input as part of of the flow network.

[mild flower]

(that's why the table is currently blank)

[civic pendant]

Yep, so the c(e)'s, i.e., the capcaities, are written on the edges

[teal radish]

Oh I see, so the numbers currently depicted on the graph are just capacities.

[civic pendant]

Yes

[teal radish]

And my job is to determine what level of flow is optimal

[civic pendant]

Yes

Go back and revisit the definition of the problem, now that you get it

Before you jump in to understand the algorithm, make sure you understand the problem

[teal radish]

Ahh that makes so much more sense now

Thank you guys

[civic pendant]

Yw

[teal radish]

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[prime spear]

these are the two diagrams ihave come up with

and idk which one is the right one

im assuming the 1st one tho

[merry python]

makes 26 degrees WITH the WALL

[prime spear]

becaise it says ladder agaisnt wall makes the degree

so it's the first diagram?

[merry python]

yes

[prime spear]

alright

thanks mate

have a good day

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[prime spear]

.reopen

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✅

[prime spear]

so i've completed 7a

and for b)

when it says

height "it" reaches above the ground

is that referring to the

ladder

[merry python]

This length

[prime spear]

thanks pal

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[karmic iron]

Is there a more elegant way to write that I'm adding up some vectors and then taking the mean of the components?

[karmic iron]

(The context is this is a KL divergence derived loss function for a machine learning model)

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@karmic iron Has your question been resolved?

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@karmic iron Has your question been resolved?

[patent nymph]

which ones are the vectors?

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[mossy geode]

Can somebody please help me find the inverse of this function?

[lime gorge]

What have u tried

@mossy geode

[mossy geode]

I started by multiplying by the denominator, then distributing x on the left side...

then making the natural logs' multiples their arguments' exponents

Then move all y variables to one side.

Not sure how to proceed after this.

I've also tried raising to base e after step 2, still no luck.

I just don't have much experience with logs, so the process doesn't really pop out to me.

[vocal sleet]

@mossy geode Has your question been resolved?

[mild flower]

solve for lnx first

then raise to base e

[mossy geode]

Sorry, still don't get it.

[harsh canopy]

try to think of it as y = .....

then rearange to make x the subject

after doing that just plug back in x where you had a y

[mossy geode]

I get that much, I just can't figure out how to isolate the variable.

[harsh canopy]

multiply by the denominator first

[mild flower]

treat lnx as one unit, replace it with the letter L if that makes it easier

[mossy geode]

But there are two of them, how do i bring down the variable in their arguments without getting them to divide or multiply so I can isolate the variable?

[harsh canopy]

from this you can collect the lnx terms

and factor out the lnx

[mossy geode]

Oh, didn't know you could factor out logs.

[harsh canopy]

yeah you can

[mossy geode]

neat

[harsh canopy]

after that it should become easy to isolate the lnx

[mossy geode]

Thanks friend

[harsh canopy]

np

[mossy geode]

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[void ledge]

Why can't I say that the limit of floor(1/x) as x approaches 0 doesn't exist because from right the limit is infinity and from the left the limit is -infinity

[half steppe]

you can say that. your picture is considering x*floor(1/x), however

[void ledge]

but I can't use the limit laws unless both the function are continous

[half steppe]

what limit laws?

[void ledge]

aha I got confused now I get it

sry

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[vast shale]

I need help with this

[vast shale]

• Justify that YZ is the bisector of AF⃡ , then write what AŶZ and ZŶF are like.

Justify that YZ is the bisector of AF⃡ , then write what AŶZ and ZŶF are like.

it is the napoleon theorem

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[old juniper]

How would the outlier be described for this particular graph

[vast shale]

Let me study what outliers are and I'll come back

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@old juniper Has your question been resolved?

[old juniper]

Ok

[vast shale]

bruh

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@old juniper Has your question been resolved?

[calm light]

well the typical method for determining if something is an outlier is using the formulas:
mean - 1.5(Q1)
mean + 1.5(Q3)
any values less then the value of the first equation or greater than the value of the second equation is an outlier

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@old juniper Has your question been resolved?

[old juniper]

Ok

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[undone aurora]

I saw a question like this on youtube

[undone aurora]

but idk how to come up with an equation

in the video the dude just started with the equation 2^n - 1

how would I come up with that

[fossil ginkgo]

induction maybe

[undone aurora]

ye but like

how would i come up with the equation with the help of induction

[fossil ginkgo]

hmmm

well

starting with ur base case

[undone aurora]

for 1 it would be just 1

[fossil ginkgo]

1 disk takes one move

right

and then in order to solve a tower of n+1 disks, you need to solve a tower of n disks first

and then use like 2 more moves to shift it over

idk something like that

i’ll think more abt this

[undone aurora]

wth i dont get it at all

[fossil ginkgo]

ah yes

[undone aurora]

how would i solve a tower of n disks witout knowin the equation doe

[fossil ginkgo]

well you assume the equation in the induction step and then prove that it works

[undone aurora]

oh so I start with the claim that the equation is 2^n - 1

and then prove it usin induction

[fossil ginkgo]

yes

[undone aurora]

but lets say

a question like this popped up on my test

how the hell

would i even come up with a claim like that

[fossil ginkgo]

i guess just try it out and see if u can find a pattern

[undone aurora]

because right now i searched it up and saw it was 2^n - 1

[lyric fossil]

the idea is that you can view this problem as moving all n-1 disks to the middle

[fossil ginkgo]

i feel like that’d be too cruel of a test question

[lyric fossil]

then moving the bottom n disk to the end

[fossil ginkgo]

yes

and then solving again

[lyric fossil]

then moving the n-1 disks to the end

[fossil ginkgo]

to move the rest onto the last

so it’s 2(2^(n-1)-1)+1

[undone aurora]

oh so

its

[fossil ginkgo]

idk i feel like they wouldn’t ask u this on a test that feels mean😭

[civic pendant]

Let's denote it by T(n)

[undone aurora]

f(n-1) + 1 + f(n-1) ?

[fossil ginkgo]

yes

[civic pendant]

you get T(n) = 2T(n-1) +1 yes

solve this

The idea is that in order to solve this, you have to move all n-1 discs to the mid rod in order to move the last disc to the target rod

And then you have to move them again to the target rod

So you have to make 2T(n-1)+1 steps

And actually, this is also the recursive solution

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@undone aurora Has your question been resolved?

[undone aurora]

so

would this be correct

[undone aurora]

wait how would solving that help

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@undone aurora Has your question been resolved?

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[stuck harness]

How can I find B

[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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[stuck harness]

1

[glossy maple]

Notice that you have a few triangles: 1 big one, and 2 small ones

and notice that they all have very similar angles

what does this tell you?

[stuck harness]

I'm not sure

[glossy maple]

what do you know about the relation between 2 triangles that have the same angles?

[stuck harness]

They are similar or congruent

[glossy maple]

right

so if they are similar, what do you know about their side lengths?

[stuck harness]

The side lengths are similar too?

[glossy maple]

what do you mean by "similar"

similarity is something between triangles

not lengths

[stuck harness]

all sides have the same ratio

[glossy maple]

be more specific

correct

now try taking two triangles here that you know the most info about

and construct an equation to find b

using the fact that ratio between sides is the same for similar triangles

[stuck harness]

But I would need the measure of the bottom measure for the smallest triangle

[glossy maple]

no you don't

you have everything already

look at the biggest triangle and second biggest one

[stuck harness]

I am still confused and the method I tried in my notes didnt work

[glossy maple]

alr let me draw it real quick

so

triangle ABC is similar to triangle ADB

and what I mean by that is:

<A = <A (angle is shared in both triangles)

<B = <D = 90 degrees

or more accurately i would say
<ABC = <ADB

including all three letters so u know what angles im talking about

and <ACB = <ABD

now let me draw these triangles again, but next to each other

[stuck harness]

What how

<acb is congruent to the 90 degree angle?

oh nvm

[glossy maple]

its way more clear if I draw it like this

make sure you understand how I got it

each side in triangle ABC has its "partner" side in triangle ABD

[stuck harness]

is there not a shortcut to solving this that doesnt require so much understanding because this is a summer class and we are just breezing through lessons with shortcuts

[glossy maple]

that we scaled up

[glossy maple]

math is about understanding

[stuck harness]

not my grade though

[glossy maple]

yes it is

your grade is supposed to understand ur level of understanding

this isn't too difficulty, just try to get it intuitively first

[stuck harness]

we are learning methods in the sense we dont get why we need to do those methods but they work and they are fast

[glossy maple]

all you have to do here, it notice which side is paired up with which side from the similar triangles

[glossy maple]

you dont have to draw it but im just doing it so you can see it more clearly

which sides are paired with which

now you can easily see the ratios

try to write down the ratios

[stuck harness]

22/b times b/16?

then solve algebraicly?

[glossy maple]

yes

[stuck harness]

Okay wait I think shortcut that we learned is probably harder than fundamentally understanding it

[glossy maple]

shortcuts only make things messy later

whenever you learn a concept, just try to understand the principle behind it first

similar triangles can be thought of as "scaling up/down"

so of course the ratios will be equal

[stuck harness]

its tough to do that at the speed we are learning lol because we have a years worth of lessons crammed into 6 weeks which isnt ideal

[glossy maple]

thats the education system for you

what can I say

[stuck harness]

yup

hopefully geometry doesnt carry on to algebra 2

thanks for the help

[glossy maple]

np!

[stuck harness]

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[dark sleet]

am i supposed to write this like this (x+243)(x-243)

[dreamy viper]

whats 729 cube rooted

also have to use SOAP

[dark sleet]

i dont know what that is

[dreamy viper]

sign opposite, always positive

[dark sleet]

oh 9

[dreamy viper]

alright

[dark sleet]

oops

[dreamy viper]

so when factoring a cube like this

[karmic imp]

It says sum or difference of cubes, key word is cubes

[dreamy viper]

it follows (cuberoot(a)+-cuberoot(b)) as the first term

[karmic imp]

Image is better fyi

[dreamy viper]

then we can denote these new values as something else because we have to use them to find the quadratic multiplied by it

i was just about to send an image

[dark sleet]

this kinda confused me more

uh

i dont really understand what ab is

[karmic imp]

You have (x^3 - 729), using the image above, what is a and what is b?

[dark sleet]

x^3 and 729

[dreamy viper]

a is x and b is 9

[dark sleet]

oops

yeah

[karmic imp]

Notice how the left side, in the image, is x^3 - b^3, so to find a, you would do a^3 = x^3, solve for a, and b^3 = 729, solve for b, that's how arctic got x and 9

[dark sleet]

so it would be x^3-9^3=(x-9)(x^2+x*9+9^2)

thats what the ab is right

(x-9)(x^2+9x+81)

thank you

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[wanton hound]

What is the difference between compound interest and annuity .?

[strange oak]

$A = P(1 + \frac{r}{n})^{nt}$

[twin meteor]

.differentialequations

[wanton hound]

I know difference in formula and i also know that annuity has something to do with geometric series sum but i don't get the concept behind annuity .

[strange oak]

Like what it is?

[dreamy viper]

annuity has constant deposits

[strange oak]

Yea

[dreamy viper]

compound interest is just one at the start

[strange oak]

Yea that’s it

[wanton hound]

Okay i want to know that only thx

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[fleet vessel]

How to answer 5

[worthy citrus]

Rewrite it in normal operations first

[livid grotto]

Out of curiosity, where are you learning this? Is this an undergrad course? I've never seen tropical geom sneak its way out of more advanced stuff before!

[mild flower]

i did a whole math major and some of a masters and had never heard of it until now

[worthy citrus]

Also they've said the word equation and then not written an equation

I would assume they want the root(s)

[livid grotto]

It is used in current alg geom research because it gives good combinatorial control of degenerations of varieties and birational maps.

[livid grotto]

Yeah I mean it will be a minimisation problem, it depends on x.

They may mean the tropical hypersurface defined by that tropical polynomial, which is the locus where the minimum is achieved at least twice (i.e. the boundaries of the regions).

Given they are asking for a sketch of a `tropical solution' later on I presume that's probably correct actually.

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@fleet vessel Has your question been resolved?

[livid grotto]

So to spell it out, we need to find the values of x for which $\operatorname{min} {0.001, 1000 + x, 100 + 2x, 3x }$ is achieved at least twice. To do this, write out the various inequalities and solve them, then take the boundary value of x. Its usually pretty long to do it this way, so there is a more technical thing called the discrete Legendre transform which gives a more systematic approach via the Newton polygon, but I doubt that is required here.

[twin meteor]

kimbo7

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[verbal turtle]

For (d), how do you know what L is. I mean obviously, 0 and 1 are in it because it is a finite field and in (c) it is shown that a^3+a^2. However, how is the last element found of L?

[vocal sleet]

@verbal turtle Has your question been resolved?

[livid grotto]

Since L is a ring it definitely contains a^2 + a^3 + 1=(a^2 + a^3) + 1. Remains to see that a^2 + a^3 + 1 is a distinct element of the field. It cannot be a^2 + a^3 since then 1=0, it cannot be 0 since (unravelling the definition of K and a) this would mean that x^4 + x^3 + x^2 + x + 1 divides x^2 + x^3 + 1, and it cannot be 1 since then x^4 + x^3 + x^2 + x + 1 divides x^2 + x^3. There may be a simpler way to see it though.

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[orchid matrix]

I tried to solve using the Quotient rule but my answer doesn't seem to match any options

[orchid matrix]

my answer:
$\dfrac{(2xy+1)(cos(xy²+y)}{x²+1}$

[twin meteor]

chegaro

[slender sky]

what did you do?

[slender sky]

thats wrong...

[orchid matrix]

how its in terms of variables?

[dense igloo]

what did you do here

[stiff pumice]

Y

[stiff pumice]

Ynot

[stiff pumice]

Derivative wrt y

Where x=Constant

[slender sky]

Its asking diffrentiation wrt y right?

[stiff pumice]

Partial differentiation

Not normal

[dense igloo]

that is not the correct quotient rule tho

[slender sky]

oh

[dense igloo]

ah

[stiff pumice]

You don't need quotient rule

You treat x as a constant

Chain rule

[orchid matrix]

ohh yes wait a min

I tried Quotient rule,
$\dfrac{(x²+1)(\dfrac{\partial{sin(xy²+y)}{\partial y})-(\dfrac{\partial{x²+1}}{\partial y}})(sin(xy²+y)}{(x+1)²}$

[twin meteor]

chegaro

[orchid matrix]

sorry I am bad with latex

[stiff pumice]

This is correct imo

[orchid matrix]

but its not in option

[slender sky]

check the options.

[stiff pumice]

Ikr

But like what ese

Else

[slender sky]

idk what is partial diffrentiation. lemme see what it is

[orchid matrix]

$\dfrac{(x²+1)\left(\dfrac{\partial{sin(xy²+y)}}{\partial y}\right)-\left(\dfrac{\partial{(x²+1)}}{{\partial y}}\right)(sin(xy²+y)}{(x+1)²}$

[twin meteor]

chegaro

[orchid matrix]

this is what I did but Ig its wrong

and if we consider x as constant then still its same with less mess

[dusk python]

don’t try to help people if you don’t understand it yourself

@orchid matrix your answer is right

[orchid matrix]

ok so there's something wrong with options then?

[dusk python]

its seems that they forgot to include what values of x and y it is evaluated at

[slender sky]

does it hurt anyone?

[dusk python]

it causes confusion

that can be easily avoided

[orchid matrix]

ok that seems more logical explanation at this point

[stiff pumice]

It's clearly not infinity ryt

Since denominator can never be 0

[orchid matrix]

yes

we can get the answer if they provide a point I guess then.
for now I assume it as x = 0, y = 0 so the answer as 1

anyways I will ask my teacher tomorrow for whether there is an error

[stiff pumice]

Let x=0

F=siny

[orchid matrix]

yeah

[stiff pumice]

Df/dy=cosy

Poinless

[orchid matrix]

D:

It is pointless to guess the x, y

I guess it's just the printing error

anw Thank you @stiff pumice 🙂

[stiff pumice]

Yea ryt

Wlc

[orchid matrix]

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[gray yarrow]

hey

[gray yarrow]

the top thing is the definition for a contraction

and in my case, phi is a function f and d is the 1-Norm

[wraith venture]

yes

[gray yarrow]

the context is that i have a function that goes from R^2 to R^2

but ||x,y| | is not really defined, is it? or do we mean | | (x,y) | | with this

so is the x,y a vector in R^2 or is x and y each a vector

because in the latter case, i would be confused
in the first case, i dont see why we dont use brackets then

however

if x,y is a vector, then f(x),f(y) is weird

it would be f(x,y) then

[hard atlas]

should be -

norm(f(x)-f(y)) <= lambda norm(x-y)

[gray yarrow]

hmmm, so i can just say d(x,y)=x-y?

[hard atlas]

they are being slightly imprecise. d(x,y) = norm(x-y) but they still use it as a function with two inputs

[gray yarrow]

ahhhh ok

hmm

ook that makes sense, thank you^^

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[uneven jay]

i just wanna check if i did this correctly:It is a linear system of equations and we are tasked to find alpha and beta so that it has 1, no, or infite solutions.

[uneven jay]

i found alpha = 2 for no solutions

but i do not know how to do the rest

[lyric fossil]

alpha = 2 for no solutions?