help-27

[zenith vortex]

it doesnt factorize into this

[meager terrace]

So like that?

[zenith vortex]

yes

i believe

[meager terrace]

Yeah i think its right as well, thankyou appreciate the help.

.close

[devout snow]

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[devout snow]

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[willow charm]

what are the steps in solving this equation?

[devout snow]

@willow charm Has your question been resolved?

[zenith vortex]

the integral, right?

[willow charm]

the equation

dx=integral sth dy

[zenith vortex]

x is function of y

[willow charm]

i dont know, i just saw the equation online and wanted to know how its solved

[devout snow]

@willow charm Has your question been resolved?

[devout snow]

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[gentle crow]

How to plot graph between f(x) and x?

[gentle crow]

.close

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@void knot Has your question been resolved?

[devout snow]

@void knot Has your question been resolved?

[devout snow]

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[viral lynx]

This is one section of the proof of a question

Im confused about the left integral

[eager nova]

Which one

[viral lynx]

How did they get from the first step to the second

[viral lynx]

Im just confused about this one step thats all

[mighty knoll]

They split the fraction into two integrals

[eager nova]

Exactly

[viral lynx]

But how did they integrate the -cos(2u)/32 to -sin(2u)/64

[eager nova]

Sub v=2u

[viral lynx]

Ahhhh yeah

Forgot I could do that haha

Thank you everyone!

❤️

.close

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[stray snow]

Let g be the inverse of the continuous function f. Let there be a point (α , β), where α ≠ β , is such that it satisfies each of y = f(x) and y = g(x) then:
a) the equation f(x) = g(x) has infinitely many solutions
b) the equation f(x) = g(x) has at least 3 solutions
c) f must be a decreasing function of x
d) g can be an increasing function of x

[stray snow]

ah i solved it

how do i end this session?

.close

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[mild sorrel]

prove that the perimeter of a triangle is greater than the sum of the medians on the three sides.

[mild sorrel]

Please only give hints.

[fleet moth]

https://www.vedantu.com/question-answer/prove-that-the-perimeter-of-a-triangle-is-class-9-maths-cbse-5f5b85d76e663a29cc4506c4

full solution

[mild sorrel]

I don't want as such.

[fleet moth]

then what do you want bro

[mild sorrel]

.close

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[short hare]

!nosols

[devout snow]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[fleet moth]

oh mb

[devout snow]

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[lost crag]

R,S are equivalence relations. why is the marked line a valid step

[thin fern]

Commutativity

S * R <=> R * S

[lost crag]

why can you use commutativity here?

[thin fern]

Because commutatitivity is a property of equivalence relations

or I don't think it's usually called commutativity

symmetry I think

[lost crag]

okay ty

.close

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[polar chasm]

How would I start thinking about something like this?

[grizzled yew]

Think mod 6

[polar chasm]

1 - winning iff 1 can be removed
2 - winning
3 - winning iff 3 can be removed
4 - winning
5 - winning iff 5 can be removed
6 - winning iff 3 can be removed
7 - losing

this is my observsation for first 7

[grizzled yew]

Ok nice

[polar chasm]

hmm okay, lemme try

[grizzled yew]

For example, if they play 1, you play 5

[polar chasm]

that way the # of coins will remain same modulo 6

unless the other player plays 3

[grizzled yew]

Oh oops

Yeah that won’t work

Hm let me take another look

[stone stump]

well you said for example 1 is winning if 1 can be removed. but if it cant, then it had to be removed previously at 2, but thats clearly not a play the other player would have made

[polar chasm]

what if the other player couldnt have made any other play

[stone stump]

then something would have happend at 4

[polar chasm]

yes

4 is 100% winning position, as is 2

[violet wind]

start with 8, remove 4

now the other player can only win by removing 2, which forces the situation in question

[polar chasm]

yeah

hmm

but nobody would want to remove 4 from 8

unless it was 9 previously

[violet wind]

so 9 is winning if u can remove 1

[polar chasm]

1 - winning iff 1 can be removed
2 - winning
3 - winning iff 3 can be removed
4 - winning
5 - winning iff 5 can be removed
6 - winning iff 3 can be removed
7 - losing
8 - winning if 1 can be removed

[polar chasm]

no, 9 should be losing i think

7 is losing, so 8 is winning if 1 can be removed

oh

[violet wind]

Let's play

I remove 1

now 8

[polar chasm]

oh i see

let me think if I can save this

[violet wind]

Oh wait

[polar chasm]

yeah, 9 is probably winning

[violet wind]

8, 4, 2, 1 and now it's stalemate

how is that handled

[polar chasm]

im not entirely sure

[grizzled yew]

It’s not winning

?

[polar chasm]

8, 4, 2, 1
Now the player can't remove 1 but also cant remove anything else

that's the problem

[grizzled yew]

Yeah

[polar chasm]

I'd consider this a win for the other player though

[grizzled yew]

It says “they may remove”, does this imply you can skip your turn?

[polar chasm]

I hope it doesnt

i'd say that turn can be skipped iff no move can be made

[grizzled yew]

Ok

[polar chasm]

okay so I know 8 is winning if 1 can be removed

idk if thats iff yet

yeah it is

1 - winning iff 1 can be removed
2 - winning
3 - winning iff 3 can be removed
4 - winning
5 - winning iff 5 can be removed
6 - winning iff 3 can be removed
7 - losing
8 - winning iff 1 can be removed

9 - winning

thats weird

almost every position is winning under almost every circumstances

10 - winning

10 is third fully winning

11 - winning

12 - winning iff 5 can be removed

it feels more or less random

13 - losing

losing repeat after approximately 6

usually

which makes some sense

there can be a period

1 - winning iff 1 can be removed
2 - winning
3 - winning iff 3 can be removed
4 - winning
5 - winning iff 5 can be removed
6 - winning iff 3 can be removed
7 - losing
8 - winning iff 1 can be removed
9 - winning
10 - winning
11 - winning
12 - winning iff 5 can be removed
13 - losing
14 - winning iff 1 can be removed
15 - winning
16 - winning iff 3 can be removed
17 - winning
18 - winning iff 5 can be removed
...

modulo 13

each condition is dependent only on the 5 preceeding it

so this far is enough to conclude that it's periodic with period 13

now this is interesting

2000 cards

2000 modulo 13

11

winning

that means player A wins

okay this was interesting

.close

[devout snow]

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[devout snow]

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[golden kestrel]

Let a,b,c € [0,1] so that ab+bc+ca =1

[golden kestrel]

I have to prove that a^2024 + b^2024 +c^2024 <= 2

So first, it doesnt really specify if a b and c are integers or real numbers

[frigid jetty]

one solution is a=1 b=1 c=0

Another a=0 b=1 c=1

and a=1 b=0 c=1

Idk

i just guessed

[stone stump]

well it says a,b,c in the interval [0,1]

[fair minnow]

the first part is actually a problem i think i've seen before

and it was prosed as 'find all the possible solutions'

trying to remember the method, but there is a way you can prove the only ones you need to consider as triples for (a,b,c) as the various combinations of (1 1 0) and also (sqrt(3)/3, sqrt(3)/3, sqrt(3)/3)

if you can do that you can just plug it into your proof and show it via exhaustion

not very elegant, but

[fair minnow]

for the obvious 3 i imagine you just split it into the two cases where a,b,c € [0,1] allows for the use of = 0 which gives you the three main solutions where you set either a or b or c to be 0 and the others to be 1, which of course satisfies the below expression

would you then just have to consider the inverse and manipulate some algebra with the assumption of a,b,c € (0,1) ?

something like that seems familiar

you'd either arrive at a contradiction if none such solutions exist or you would find the one im pretty sure exists

[violet wind]

Euro sign for set inclusion 💀

[fair minnow]

hey i'm just copying the thing...

im too lazy to alt code it i n

[violet wind]

fair enough

[devout snow]

@golden kestrel Has your question been resolved?

[golden kestrel]

i tried my best

[golden kestrel]

Right, yet thats the thing

[violet wind]

I wonder if you can solve this with Lagrange multipliers

[golden kestrel]

If they were integers

[fair minnow]

there is 100% a non interger solution

[violet wind]

Solution to what

[fair minnow]

mlem

i just cant remember how you prove it

ive been messing around with the algebra to little avail

[violet wind]

well yeah obviously there are values of a, b, and c which satisfy that equation

[golden kestrel]

Whats that?

[violet wind]

that doesn't really help solve the problem though

[golden kestrel]

Yeah ive found plenty

Indeed

[fair minnow]

there's only 1 of interest though

there's 4 in total afaik

[golden kestrel]

You can find a lot of them by just playing with 1s and 0s

[violet wind]

Surely this is about real numbers, otherwise the problem would be trivial

[fair minnow]

(a b c) = (1 1 0) , (1 0 1), (0 1 1) are the given ones

[golden kestrel]

Yeah

[fair minnow]

but proving (a b c) = ( sqrt(3)/3 , sqrt(3)/3 , sqrt(3)/3 ) backwards is a pain

no clue how you do it

[golden kestrel]

Or maybe titu's lemm

[violet wind]

I conjecture a+b+c ≤ 2 for this problem

[golden kestrel]

We have to work on the inequality

Maybe a factorization in the end

Im gonna apply titu's and see where i get to

[violet wind]

I doubt we're factoring a^2024+b^2024+c^2024

I think I have an approach

[golden kestrel]

Since its 2024 and we live in 2024 there must be a generalization

Or im just assuming stuff

[violet wind]

well too bad that polynomial doesn't factor generally

[golden kestrel]

Go ahead

[violet wind]

a+b+c ≥ a^2024 + b^2024 + c^2024

so it's enough to show a+b+c ≤ 2

[golden kestrel]

Right, yet if theyre all 1 it doesnt stand

[violet wind]

If they're all 1 then ab+ac+bc isn't 1

[golden kestrel]

Oh right

[violet wind]

WLOG let a≥b≥c

we will show that if a+b+c > 2 then ab + ac + bc > 1

[crisp zealot]

We also have abc <= 1/3

Not sure how helpful that is

[golden kestrel]

Alright so titu just gets us in a lot of trouble

[violet wind]

we can assume 0<c≤b<1

[golden kestrel]

not worth it

[violet wind]

I have this people just keep on distracting me IRL while I'm trying to type it out

[golden kestrel]

thats unfortunate

[violet wind]

The case otherwise (b=1) is easy to check so that's why

anyways ab+ac+bc = bc + a(b+c)

one can check that decreasing the value of c and increasing the value of b by the same amount will strictly decrease this value

so we do that until b hits 1, only decreasing the possible value of ab+ac+bc

if a isn't 1, we decrease c and increase a until a hits 1 the same way

now a=b=1, c>0, so ab+ac+bc > 1 but we have only decreased it's value so it had to also be >1 before, QED

[golden kestrel]

Right, let me get a mathematical way to write it down lol

[golden kestrel]

I dont know how to explain this part

[violet wind]

let b' = 1, c' = c+b-1

now a, b', c' are a new triple

[golden kestrel]

Right

[devout snow]

@golden kestrel Has your question been resolved?

[golden kestrel]

Alright, then what?

Or I could write that in words actually

[frigid hare]

here's another approach: write the expression like this where 0 <= a,b <= 1 and then you take the derivative and get this, and notice it's positive on our interval so our function is increasing so our biggest value for that expression is at a = b = end of the domain = 1 so max(expression in 1st photo) = 1 + 1 + (1 - 1) / ( 1 + 1) = 2

[frigid hare]

and then you have this

[golden kestrel]

The expression is the first photo?

I might go on your solve tbh

[frigid hare]

yeah

you get c in terms of a and b then plug it in a + b + c

[golden kestrel]

Ill think about it tmr and ill ask you about it if i dont understand

[devout snow]

@golden kestrel Has your question been resolved?

[devout snow]

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[limpid bridge]

I need help with finding the domain of the expression, I have a couple questions under this. I know that you have to first set up restrictions such as dividing by 0 or negative square root #s but after that im slightly lost; the one im working on right now is 1 over 6 - x

[devout snow]

@limpid bridge Has your question been resolved?

[limpid bridge]

<@&286206848099549185>

[tame nest]

It’s 1/6-x

So basically
1/-x+6

So just -1/x-6

So basically here x cannot be 6 first of all.
X E R I x can’t be 6
Y E R I y can’t be 0

I think so

[limpid bridge]

can you explain the process? the next ones are more complicated and I wanan try and fully grasp it incase of confusing myself

[tame nest]

Sure, give me a second I’ll write it

Make sense kinda? Or shud I explain more

[limpid bridge]

yeah it makes more sense that way thanks

imma keep the channel open if you dont mind bc im pretty sure ill have questions abt these next ones lol

[tame nest]

No worries lmao dw

[limpid bridge]

the multiple choice awnsers are verbal btw idk if that makes a difference

all real numbers x such that x ≠ −6
all real numbers x such that x ≠ 0 and x ≠ 6
all real numbers
all real numbers x such that x ≠ 6
all non-negative real numbers

^ those were the choices for the previous question

which was all real numbers x such as x does no equal 6

[tame nest]

Oh okay, yeah

[devout snow]

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[bold wren]

How did he get the 15?

[royal laurel]

Show the original

It lacks of context

Batman!

[bold wren]

there, sorry

[royal laurel]

<@&286206848099549185> help him!

[glad seal]

He first calculates how "wide" each rectangle is. This is given by (7-1)/9 = 2/3

Then, to calculate the x-component of each rectangle, given by x_i, we do 1 + 2i/3 (The 1 is our minimum in our domain, and 2/3 is the width of each rectangle before). You can verify this if you want to, it may help

They choose to convert this to a single fraction $\frac{3+2i}{3}$

[woven radish]

X13warzone

[glad seal]

Then, they substitute this $x_i component into f(x_i) to find the "height" of each rectangle. They skip a step here and multiply 5x3 to get 15, since we have a 3 in the denominator of x_i$

You can write out that step as $f(x_{i})=\frac{5}{x_{i}}+2=\frac{5}{\frac{3+2i}{3}}+2=\frac{5\times 3}{3+2i}+2$

[woven radish]

X13warzone

[glad seal]

@bold wren Does this help?

[devout snow]

@bold wren Has your question been resolved?

[bold wren]

what is the x_i in this case? I get confused with xdelta and x_i

[glad seal]

xdelta is how wide each rectangle is, x_i is what x value should be inputted into f(x) to find the height of the i-th rectangle

So xdelta = 2/3 (each rectangle is 2/3 units wide), x_i = 1 + 2i/3:
i = 1: x_1 = 1 + 2/3 (to find the the height of the first rectangle, we use f(1+2/3), and not f(1), since we want to find the area under the graph)

[bold wren]

okay but I did put the answer down and its wrong. Why is the answer C?

where is the 2/3 from?

[glad seal]

Remember the formula to calculate the area of a rectangle is height x base, in this case height = 15/(3+2i) + 2, base = 2/3

[bold wren]

got it, thank you

[devout snow]

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[hybrid moss]

How do I go about this question?

[ashen stirrup]

since its a parallelogram, opposite sides are equivalent and so are the vectors you could say

you want to find the combination of vectors that would have a resultant equal to vector D

vector D starts from a the bottom left corner and goes up to the top right, you can try to follow a path of vectors that starts and ends at the same positions

[devout snow]

@hybrid moss Has your question been resolved?

[hybrid moss]

so A - B?

[ashen stirrup]

nah

[hybrid moss]

oh

its C + B?

right @ashen stirrup

[ashen stirrup]

nah

[hybrid moss]

sorry can you reexplain it

@ashen stirrup

im still confused

is it A + B?

i think it is but idk why

[ashen stirrup]

it is, take a quick a peek at this sick diagram

the vectors must be translated be tip to tail

[hybrid moss]

omg

[ashen stirrup]

tip being the arrow head

[hybrid moss]

thats helps like crazy

wtf

ahhh

okok

tyvm

[devout snow]

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[gentle urchin]

can anyone help

the question was what intervals of p make the integrals divergent

[devout snow]

@gentle urchin Has your question been resolved?

[devout snow]

@gentle urchin Has your question been resolved?

[gentle urchin]

.close

[devout snow]

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[rare zealot]

helpp idk how to solve this proof

[solid osprey]

is that the original question?

[rare zealot]

yep

[hoary glen]

I believe that for it u have to use the AA criterion

[rare zealot]

ummm ok but

before we even get there

what do i do

if i use reflexive then ad = ad then can i get somewhere

<@&286206848099549185>

[devout snow]

@rare zealot Has your question been resolved?

[rare zealot]

@serene rampart pls help

[static horizon]

well since AB II CD then <EAB = <CDE and <EBA = <ECD as transversal angles

[rare zealot]

do i have to prove if like AD exists

before i write that

[lapis locust]

Yeah and <AEB = <CED

So then there's AAA

Which proves similarity

[rare zealot]

wait but

dont i have to prove those lines in the middle exist first

or no

[lapis locust]

I don't think so, no

[rare zealot]

so it would look something like

Statement |Reason

AB // CD |Given

<EAB = <CDE | def of traversal angles?

[lapis locust]

Ig you could say that yeah

No sorry, those are vertically opposite

<EAB = <CDE would be alternate angles, that's how I learnt it

[rare zealot]

now im confused bc we havent proven AD as a bisector yet

[lapis locust]

You don't need to: When AB is parallel to CD, AD is a transversal and that means that the angle are alternate and equal

[rare zealot]

ohh

wait

[lapis locust]

Yeah?

[rare zealot]

i have to mention AD as a transversal

so do i do it like

AD = AD | def of transversal

[lapis locust]

Probably not to that extent, unless your professor is extremely strict, you could probably get away with just mentioning the alternate angles on the parallel line

[rare zealot]

umm no hes really strict

[lapis locust]

Ok, mention it then

[rare zealot]

but when you say something is transversal do you say its equal to itself

or like

AE = ED | def of transversal

[lapis locust]

Yeah, so you could say (The angles are equivalent)| AD is a transversal and the angles are alternate

[rare zealot]

wait um i have to do it in this kind of style

[lapis locust]

Ok, I guess you could do that

[rare zealot]

wait but idk the tranvsersal symbol this is an example of how he writes stuff

in this case how do i write it

[devout snow]

@rare zealot Has your question been resolved?

[devout snow]

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[mental wagon]

Woops

Wrong image

PLs help me

[junior chasm]

which

[mental wagon]

The one I underlined

[junior chasm]

oh thats underline

i thought u were censoring smth

[mental wagon]

Yo?

Anyone here

@junior chasm

<@&286206848099549185>

[static meadow]

we have to put the value of y after differentiating?

[mental wagon]

Maybe

I have to separate the y on one side I think

And then use the y(45⁰) = 1

But how do I separate y

[static meadow]

I did the differentiation

[boreal helm]

Here for the help

[static meadow]

dy/dx = 1/(cos(xy) + 1)

[mental wagon]

Ok I believe you with that

Now what

Integrate?

[boreal helm]

Which one in the pic.

[mental wagon]

I UNDErLINED THE DAMN QUESTION

[boreal helm]

Okok

[mental wagon]

Literally two person asked me that

Isn't it clear

[boreal helm]

sin(xy)+y=c?

[mental wagon]

Yup

[boreal helm]

Ok what we have to find particular solution?

[mental wagon]

@static meadow how am I gon integrate cos(xy) 💀

[mental wagon]

Ig so

[boreal helm]

take xy=t

xdy/dx+y=dt/dx

[mental wagon]

😭

That's too complex guys

I need an easier way

[boreal helm]

Listen 1+cos(xy)=2cos^2(xy/2) which is sec^2(xy/2)/2

Now take xy=t

[static meadow]

It will be sin(xy)/y + c

[boreal helm]

I hope this will make it

[static meadow]

But there we have 1/cos

so

[mental wagon]

💀

Howd u do that miss

[boreal helm]

wait imma integrate wait

U want me to find derivative of this sin(xy)+y=c?

Imma not getting what u want

[static meadow]

by taking xy = z

then differentiate

then substitute

[static meadow]

Yes

[boreal helm]

Wait

[static meadow]

First differentiation

[boreal helm]

Understood

[mental wagon]

Wait

Hol up

I got it I think

[static meadow]

No wait I did the differentiation wrong 😭

silly mistake

[boreal helm]

Bro

Imma gettin

[mental wagon]

I trusted you with the differentiation

[boreal helm]

-ycos(xy)/[1+xcos(xy)] = dy/dx

This is correct

[static meadow]

the rhs is gonna be zero

[boreal helm]

I took dy/dx other side

Listen simply tell now what u want me to integrate this

Bro but it will simply revert us back to sin(xy)+y=c what u all want guyz 💀

[mental wagon]

Bruh what r u guys doing

@static meadow @boreal helm

[boreal helm]

Bro u are not telling me

What to do

[static meadow]

😭

[boreal helm]

I did the derivative correctly

[static meadow]

is that the ans?

[boreal helm]

Bro fr

I get it know

[static meadow]

we needed to calculate the c?

[mental wagon]

I got the damn answer

[static meadow]

😭

huu

Huh

[boreal helm]

Ques wants the particular solution out of this general solution

Just put in x and y guyz

To eliminate c

[static meadow]

I thought we had to calculate the y

[mental wagon]

I don't even understand the statement I just found c

[boreal helm]

Bro

Listen

U already have the function

Now there are only two things u can do

1. U find c and get the particular solution
   2.get the differential equation by differentiating

U were not telling me what to do

[mental wagon]

Cuz I DONT KNOW MYSELF

what to do

[boreal helm]

Lmao bro

Wait imma read

Listen

They asking 1. Find a particular solutin

Solution*

[mental wagon]

Ok

[boreal helm]

Just find c in each

[mental wagon]

Got it

[boreal helm]

And u are done

[mental wagon]

But how does that work

[boreal helm]

Just put x and y for which they have given a definite value

like y(π/4)=1

sin(π/4)+1=c

Done

[mental wagon]

The one in green is the general solution

[boreal helm]

Yes

[mental wagon]

And the one in yellow is the initial condition?

[boreal helm]

Yes

[mental wagon]

Am I right

Oh

[boreal helm]

Yes

Yes

[mental wagon]

You smart

Thanks

[boreal helm]

You too

Np

[mental wagon]

Blushes in 8k

[boreal helm]

Fr

[mental wagon]

Ok

Thanks

[boreal helm]

Bye imma go

Duty calls

[static meadow]

thanks guys

[mental wagon]

Call of duty

[boreal helm]

Lol

[static meadow]

bye tek care

[boreal helm]

Bye

[mental wagon]

Are you solving differentials too

[static meadow]

Yep

[mental wagon]

Fun fact: this is my first ever differential equation

Thats why I got no idea

[static meadow]

😭oh

we have different kinda differentiation

we don't have the general solutions in our question

We find for y

so that's what I was doing here

[static meadow]

I'm down with differentiation and started integration

Ig I'm halfway thru it

[boreal helm]

Here for the help

[mental wagon]

☠️

[boreal helm]

Fr

[static meadow]

😭hyh the question

[mental wagon]

I hate word probs

[static meadow]

me too

I sux

at those

[boreal helm]

Ye differential fr

Wait let's read

[mental wagon]

Then how did you even attempt a differential equation

Woah

[boreal helm]

Fr bro

Not possible

[static meadow]

the word probs have a different chapter called "application of derivative" and I'm left with that, I've to learn maxima and minima for then

[mental wagon]

phonk music plays

*Troll face on @static meadow *

[static meadow]

ouch

[boreal helm]

dy/dx=x(1000-x)k I think this will be the equation

[mental wagon]

Hehe 👀

[boreal helm]

x is infected 1000-x not infected

[mental wagon]

I only know the answer

It's 276

[mental wagon]

What's up with the k

[static meadow]

Yep

[boreal helm]

Bro to remove proportionality I made it

[mental wagon]

💀

[boreal helm]

k

[mental wagon]

How are you tryna solving this

You said you just started integration

[mental wagon]

Oh

Dumb me

[boreal helm]

Bro x(1000-x) can be integrated ezily

[mental wagon]

Soru

[boreal helm]

No prob.

[static meadow]

😭😭bro is this normal equation

[mental wagon]

Should I do 1000x -x²

[static meadow]

I can solve equation

[boreal helm]

Bro

Yes

[mental wagon]

Lmao

Ok

[static meadow]

this sum is kinda ez

[static meadow]

Yes

[mental wagon]

Now integrat itM

?

[boreal helm]

Listen after integrating

U will have y=k(500x^2-x^3/3)+c now see y is no of days

Now after 4 days that is y=4 x=50 so u put this in thee equation

To get k

[mental wagon]

Got the c

[mental wagon]

Wai what

I had to find k

But I found the c

Lmao

[boreal helm]

c?

how without k

[mental wagon]

Constant if Integration

[mental wagon]

I ignored it

[boreal helm]

Ok check answer

Let's see

[mental wagon]

But I have to find the value of y

I mean x

After putting y 6

[boreal helm]

Yes listen u have considered I think y=(500x^2-x^3/3)+c right

Now put y=6 and find x

[mental wagon]

Yes

But now I'm thinking Abt the k

[boreal helm]

If it's correct congrats

If not find l

[mental wagon]

What Abt the k

[boreal helm]

K*

Listen bro I feel u have found the k

Bcz when y=0 x=0

[mental wagon]

How

[boreal helm]

Listen

[mental wagon]

.

[boreal helm]

if there were no days

Out of 1000 no one will get infected

So y=0 when x=0

So c=0

So equation becomes y=k(500x^2-x^3/3) ok now u put y=4 and x=50

U get k

Now after finding k put y=6 and solve for x

Try

[mental wagon]

How did you even integrate

The k is still thee

There

[boreal helm]

Listen bro

dy/dx=kx(1000-x) ok

I integrated then so we got +c also

But u can understand like that if 0 days passed then 0 person will be infected

So c=0

Ok

Now simply u have to find k

And put y=6 to get x

[mental wagon]

If 0 days have passed then 0 persons are infected

But how C BECOME ZERO CUZ OF THAT

Oh

Ok

Mb mb

[boreal helm]

Did u get it

It

I thought u did it

So I went

@mental wagon

[mental wagon]

I've done it @boreal helm

All thanks to you smarty pants 👖

[boreal helm]

Okok 👍

[mental wagon]

You snart

Smart

[boreal helm]

No all thanks to your calculation and hardwork

Np np

[mental wagon]

Hehe

.close

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[dense jay]

,rccw

[stable ginkgo]

Thx

How do I continue this

[dense jay]

aim?

[stable ginkgo]

Wdym

[stable ginkgo]

Are you asking why am I doing this?

[dense jay]

im asking what it is youre trying to do

solve for x?

[stable ginkgo]

I want to find a value for m and n so that ( 3x - m )^2 + ( 4x - n )^2 - ( 5x - 5 )^2 = 2x

[dense jay]

alright, you need to compare coefficients then

constants on each side and x coefficients on each side must be equal

[stable ginkgo]

I don't know how to do it yet

[dense jay]

say you have ax+b=cx+d with a,b,c,d being constants

for the equality to hold it must be true that a=c and b=d

[because you have (a-c)x+(b-d)=0]

[stable ginkgo]

Alright, so where I put de coefficients

[dense jay]

you just have to look at both sides

and equate them

[stable ginkgo]

,rccw

[stable ginkgo]

So this is equals to 0

[dense jay]

indeed

[stable ginkgo]

Great

M^2 + n^2 = 25

Awesome

[dense jay]

[stable ginkgo]

Thank you, have a great day

.close

[devout snow]

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[terse stone]

I need some help with this

[glacial wyvern]

Is your problem coming up with the supremum or justifying the supremum?

[terse stone]

Just finding it

[glacial wyvern]

Well can you come up with any upper bounds for that set?

[terse stone]

1?

[glacial wyvern]

yes, that's an upper bound for E

meaning every element of E is at most 1

[terse stone]

This is the solution but I don't understand it

[glacial wyvern]

What part of the solution don't you understand?

[terse stone]

How did they come up with M'?

[glacial wyvern]

Well by plugging in n=1, you find that 0 belongs to E

So therefore you can't have a negative upper bound

because 0 in E would contradict that

1 is an upper bound, but in order to be a supremum, every other upper bound needs to be greater or equal

So therefore we can attempt to derive a contradiction by assuming the existence of some M' that contradicts that requirement

i.e., if there exists some M' that is LESS than 1 (meaning "better" than 1)

(but still positive because no negative upper bound exists and 0 is not an upper bound)

then our supremum of 1 was not right

but since we derive a contradiction, we can conclude that 1 was in fact the supremum

[terse stone]

What is the contradiction?

Tbh I think I need to go practice some simpler exercises

Thanks for your help tho

[glacial wyvern]

np

[terse stone]

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[eternal sandal]

please help it's 4am and I've lost it

[torn vessel]

you're S_2N isn't correct, you've added a bunch of terms to the sum that aren't there.

[eternal sandal]

sorry I don't follow wdym that aren't there?

[torn vessel]

$\sum_{n=1}^{2N} n^3 - (n+1)^3 = (1^3 - 2^3) + (2^3 - 3^3)+(3^3-4^3)+\ldots ((2N-1)^3 - (2N)^3)\ \neq (1^3 - 2^3) + (3^3 -4^3) +\ldots + ((2N-1)^3-(2N)^3$

[woven radish]

Zybikron

[eternal sandal]

wait isn't it the same...

am I overlooking smth

[torn vessel]

just look at the first two parentheses

(1^3 - 2^3) + (2^3 - 3^3) + .... is the summation you wrote for S_2N, but the terms you have written out above are 1^3 - 2^3 + 3^3 - 4^3 +....

you have both 2^3 and -2^3, but 2^3 should only appear once

and the same with every term after this

[eternal sandal]

ohhhh

wait the terms are duplicate

is that what you mean

[torn vessel]

yes, in your S_2N formula

[eternal sandal]

hm I do realise we have to separate the negative and positive terms

and that we need to do that through the powers being odd or even

but I'm a bit confused as to what expression I should use

[devout snow]

@eternal sandal Has your question been resolved?

[devout snow]

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[chrome linden]

Can someone help me with this I have an answer but its wrong

[restive river]

what answer u get?

[chrome linden]

6sin(pi/15x)

[restive river]

this will make the amplitude to be

6 on top and bottom

but in the bottom its -2

[twilit comet]

hence, there must be a constant that is added

[restive river]

yep

[chrome linden]

Ahh i see

So where would i go from there?

[restive river]

rn its making the range to be [6, -6]

addition or subtraction will make it bad

so that means taking 6 in the firt

place was bad choice

[chrome linden]

Ohhh

[restive river]

6 = a*1 + c
-2 = a*-1 + c

c = 6 - a
-2 = -a + 6 -a
-2 = -2a + 6
-1 = -a + 6
a = 7
c = -1

y = 7sin(kx) - 1

[chrome linden]

ohh okay, got it

so after that, I would continue with finding the K correct?

[restive river]

yep

B = a*-1 + c```

these equations always gonna work to find the a and c just put the A and B to be the top height of the graph and bottom

respectively

[chrome linden]

okay, got it thank you

[restive river]

now

find the K

[chrome linden]

okay

would you mihd helping with that? sorry im a little lost

Do i use law of sines and cosines

[chrome linden]

I got pi/6 x as the K

is that correct?

[devout snow]

@chrome linden Has your question been resolved?

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[unique pasture]

Let's say the base probability of success is .03 and each failed attempt increases the probability by .002 so that the second attempt has a probability of .032, the third .034, and so on. How can I perform a calculation or write code to solve for the expected number of attempts?

How accurate is the statement below and if true, why?:

The expectation of the number of tries is equal to the sum over k of:

k * P(first k-1 tries all fail) * P(kth try succeeds

[unique pasture]

Grabbed this from stackoverflow and trying to understand the answer (if it’s true)

[devout snow]

@unique pasture Has your question been resolved?

[unique pasture]

<@&286206848099549185>

[devout snow]

@unique pasture Has your question been resolved?

[devout snow]

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[abstract sonnet]

Ill get the translated version i just want you to see the y=kx+b

its for my cousin

but does anyone know the correct answer

I solved it and got C

He got C but a teacher said the right answer is B

its basically asking which graph is y=kx+b

[jagged harbor]

the only graph whose y-component is negative only when the x-component is less than 2 is graph B

[abstract sonnet]

wdym

[jagged harbor]

it's given that kx+b < 0 only when x < 2

provided y = kx+b, we have y<0 only when x<2

[abstract sonnet]

ohhhhh

wait

[jagged harbor]

thus, check each graph for x=2 ; the correct graph will contain the point (2,0)

and have positive slope

[olive solstice]

5+5x5

[jagged harbor]

i.e. for x < 2, the graph's y-component (the vertical component) will be below the x-axis

and x>2, above

[abstract sonnet]

IM SO CONFUSED RN

ITS OKAY GUYS

WE'RE OVER THAT QS WE'RE MOVING ON TO ANOTHER QS

.close

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[restive river]

so there is this paragraph now the thing is im not getting like how can any function be symmetric around the y if u put -x ?!

like isn't thats just for even functions

[radiant hare]

If you have an even function, the reflection would be itself

[restive river]

but they didn't say the function is necessarily

even

for -x to make

symmetry

😭

[radiant hare]

They didn’t say the reflection would be itself

[restive river]

OH yeah but like

points are at equal

distance

in the diagram

[radiant hare]

Yeah they are 2 different functions red and blue

[restive river]

but they are assuming that the blue function is even right?

[radiant hare]

No

,w Plot x, -x

[woven radish]

Results provided by WolframAlpha

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

Requested by deadtomato_

[frozen stump]

think of it like…
f(x) = f(-(-x))

[radiant hare]

Orange is the reflection of the blue

[restive river]

OH SHIT

OOOOOOOOOOO

bruh

that makes so much sense

negative x in the left of y will become positive

to give same graph

and also the fact that -x is not actually being put in the function

but -(-x)

its makign so much sense now

i was thinking like they actually put in the -x

and get a

f(x)

function

smth smth

oh well

like this says "replace"

this is different then

f(-x)

function

@frozen stump @radiant hare THANK YOU SO MUCH GUYS

f(-x) was function transformation with horizontal

stretching

this is different then function being symmetric around the axis

thing so so much clear

.close

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[opaque talon]

How do I do this?

[gusty nexus]

Is N number of Microbes and T is time in days or sth like that

[opaque talon]

yeah

[gusty nexus]

Are you allowed to use a calculator?

[opaque talon]

ye

[signal crag]

no need to use calc tho

just find the gradient of the line

that must be m

y-intercept is c

[opaque talon]

oh right thanks

[signal crag]

yeah the rest is just subbing in

but make sure to change your values

cause its in the log form

[gusty nexus]

This is why I asked for calculator

[opaque talon]

whats the y intercept here

is it just 1.7

[signal crag]

i mean you can be a bit more precise with it

[opaque talon]

so far I have logbase 10 N= 14/29 logbase 10 3 + logbase 10 A

[signal crag]

what is logbase 10 A

?

[opaque talon]

idk

[signal crag]

use the formula given to you

[opaque talon]

oh

Its c

the y intercept

[signal crag]

log10(N) = m * log10(T) + c

why is c, the y-intercept expressed as a log

[signal crag]

you can find a number like this

[opaque talon]

from this question

[signal crag]

ah

so you want to change it into a and/or b

[opaque talon]

ok got it

[boreal helm]

Yo

[opaque talon]

p sure I got the gradient function wrong

[boreal helm]

Imma back for the help

Explain me fast

What u need

Fr dr

Fr*

Deleted

What

[opaque talon]

we're on this

[boreal helm]

Ok

[signal crag]

ok ill explain this again, you can calculate the gradient from this graph

the gradient will be the value m

[boreal helm]

that is the slope of line

[signal crag]

to calculate the gradient, you can take two points on the line

[boreal helm]

∆y/∆x

[opaque talon]

I used the coordinates of (1.4,4.7) and (0,1.7)

[signal crag]

preferabbly on the ends

[boreal helm]

Use those points through which line passes exactly

[signal crag]

okay now use rise/run

[signal crag]

don't extrapolate points

interpolate them

[boreal helm]

Yes

[opaque talon]

I got 15/7

[signal crag]

for this, subtract the y values of the points and then divide that value by the difference of the x-values of your points

[boreal helm]

Bro giving lessons after discovering the neutron fr fr 🔥🔥🗣️

[signal crag]

okay if this is correct, this will be your m value

[boreal helm]

This is the slope of the line

:>

[signal crag]

now, your y-intercept

read that off the graph

[opaque talon]

is it not 1.7?

[boreal helm]

Now for y intercept

Put x =0

[signal crag]

1.7 is too rough

you can do better than that

ill just tell you

i believe its 1.85

[opaque talon]

yeah

I see that

now

[signal crag]

always use the scale given to you

[boreal helm]

1.85 seems good

[signal crag]

anyways thats your c

[boreal helm]

Yes now u just have to substitute T=3

And find N

[opaque talon]

N = (1.85)(3)^15/7

[boreal helm]

Fr

[opaque talon]

doesn't get me the right answer at all tho

[signal crag]

actually you really don't need a calc

[boreal helm]

Fr

[signal crag]

you should have a value for m given in terms of a and b

[boreal helm]

Bro be estimating from the graph for real

[signal crag]

then you can calc from your original function

[signal crag]

were you able to do this?

[opaque talon]

in terms of logs yea

[signal crag]

no like express m as either a or/and b?

[opaque talon]

oh

uh

ig m = b

[signal crag]

are you sure?

[opaque talon]

yea

[signal crag]

yup you are correct

okay if m = b, then we can sub whatever value you had for m into b which is in your original equation

we can also do the same thing for c

c = log10(a)

is this familiar

[opaque talon]

yes

a = 10^c

got it

[signal crag]

yup

now sub those two things into your original equation

sub T = 3, and solve for N

[opaque talon]

cool ok thanks

.close

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[lofty rose]

can someone help me find the area in exact form and decimal form

[untold ravine]

think of it as a square with four quadrants cut at its corners

so find the area of the square and subtract those of the quadrants

[devout snow]

@lofty rose Has your question been resolved?

[lofty rose]

so would you do 4.2 x 4.2 to findthe area of the square

i got 3.46 for the area of the corners

so would i minus 3.46 4 times from the squares area?

[untold ravine]

no

4.2mm as shown in the diagram is 1/3 of the side of the square

the actual area of the square would be (3*4.2)^2

and the area of each quadrant is pi/4 * (4.2)^2

[devout snow]

Channel closed

Closed due to timeout

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[lofty rose]

.close

[devout snow]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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

[inner ibex]

can someone pls help me?

[inner ibex]

when i do this i get 8.66

which is false