help-28

[glossy valve]

Axe

[dry arch]

(holds for p>0)

[rugged raft]

ooh shoot its gamma?😭 I didn't recognize that

[dry arch]

yes offset by 1

oh technically the gamma function is the analytic continuation of the integral

do you need a proof or just to know the range of p?

the range should be p>-1

[rugged raft]

Proof😭

Also I went to learn gamma and I wonder why is this equal to 1? @dry arch

[dry arch]

i guess because that's required to make each term of the bottom sum look like x^n?

i don't fully understand that

you can show the same thing with integration by parts and induction

i wonder if you can argue that the area under the curve, for p>-1 and x >= 1, is bounded by the area under the curve with ceiling of p

and claim it converges that way

i have to go, hopefully someone else can help

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[static ore]

How do i do this?

[nova topaz]

What is the condition that makes a square matrix invertible?

[static ore]

yeah

how do i solve it

[nova topaz]

calculate the determinant of A and simplify. Then set that equal to 0 and solve for x

[static ore]

wait what

[sacred sparrow]

which condition are you familiar with then

you must have been given a def for invertability

[static ore]

I havent done this in awhile so I kinda forgot

[sacred sparrow]

probably via determinant

A is not invertible <=> det(A) = 0

so as Luca mentioned find x such that det(A) = 0

which is just a linear equation

[static ore]

wait what is det(a)

[sacred sparrow]

the determinant of A

[static ore]

interesting

[nova topaz]

the determinant is a measure of how the matrix stretches or squishes space

if the determinant is 0, space is squished down into a lower dimension

so its impossible to recover the information to invert that transformation

[static ore]

@nova topaz @sacred sparrow do i just use this formula and im chilling?

[nova topaz]

yeah. And then once you calculate it you will get an expression in terms of x. Then you need to solve for x when the expression equals 0

[static ore]

thanks bosses

[sacred sparrow]

yeah but keep in mind that using any old llm such as GPT3.5< is just asking for hallucation answers

[static ore]

im paying for it 😔

[sacred sparrow]

kk

[static ore]

its still lowkey dumb tho

[sacred sparrow]

yeah and not among the newest

if you take a current model and feed it with GB or PB of math book data it's more consistent

and also the way Anthropic etc achieve silver/gold medal scores on international math competitions

[static ore]

@sacred sparrow so im doing this question but im getting the wrong answer, do you know what i did wrong

[sacred sparrow]

sry gtg but there'll likely come another wanderer to help 🦇

[open hearth]

is c=23?

[static ore]

no c=35

[static ore]

@nova topaz wanna take a dabble at this question 🫡

[open hearth]

i messed up in a very dumb way 💀

[nova topaz]

ill take a look 1 sec

[static ore]

i mean chatgpt got 29 so you're as smart as chatgpt

[open hearth]

here you go

[open hearth]

chat gpt is always wrong

[static ore]

yeah

[static ore]

are you able to do it without deviding both?

[open hearth]

yeah do you want me to?

[static ore]

I dont really get this line

[open hearth]

its integration by parts

[static ore]

using this formula right?

[open hearth]

yup

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@static ore Has your question been resolved?

[static ore]

for this question, why did they put 0.5 on the 3rd one and not the first one?

@open hearth @nova topaz

sorry for ping 😭 the exam in a few hours

[open hearth]

its given in the question

read the first part again

[static ore]

im so confused

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[open hearth]

i don’t have time to do the whole thing

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[static ore]

bro im not asking you to 😭

like look at this, how did the go from this to what they wrote

[open hearth]

they are just writing the amt consumed and the const part is amt exp or imp

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[ember summit]

is PQ=Q-P not valid in any affince space ?

[limpid moat]

usually PQ=OQ-OP

where O is a generic point

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[ember summit]

thanks

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[west plume]

2.3 Painville
The debt of bread
There is a small village lost in the middle of nowhere, named Painville, where the
locals love bread. In fact, not only do they consume it regularly,
but their whole economy is built around bread. Every year, the bailiff of the
bread visits Painville in order to update a report that evaluates how many loaves
belong to every villager. However, if one of them finds himself owing du
bread to the state, the bailiff kidnaps him and forces him to work in a bakery to
pay his debt. Even if not everyone knows each other in the commune, everyone has seen
the psychological damage that such an experience causes, and so the village has vowed to
never have villagers in debt again. You are a renowned mathematician, and they
so they hired you to help them (you get paid with a dozen chopsticks).
Mathematical model
As a good mathematician, it is first of all about modeling the problem. Painville
forms a multigraph G = (V, E) whose vertices V represent the villagers and the
edges E represent the friendship relations (several edges between two villagers
mean that they are very close friends). We will always assume that the graph G is
related.
Each villager automatically has a certain balance of bread: it is whole and can be
positive or negative.
The villagers intend to redistribute the bread so that no villagers will find themselves
with a strictly negative balance. However, the villagers do not do things to
half and think of all their friends. Thus, a villager can choose to redistribute
bread to all his friends by giving them each one of his best baguettes. The
villagers can very well get into debt by such a redistribution (from where the bread comes out in
these cases there, it remains a mystery). In a similar way, a villager can ask
a baguette of bread to each of his friends, at the risk of putting them in debt.

[west plume]

we wanna know whe the problem is solvable

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[pulsar bough]

HELP ME

[full forum]

Please don't occupy multiple help channels.

[pulsar bough]

OH MY BAD

[remote zealot]

factor out 3⁶

from under the root

[pulsar bough]

wdym factor out

[remote zealot]

like

[pulsar bough]

IS THE ANSWER B

[remote zealot]

yk how ab + ac = a(b+c)

[remote zealot]

no

[pulsar bough]

I DONT KNOW MATH

[pulsar bough]

what

[remote zealot]

damn

[pulsar bough]

is it d

[neat basin]

Yes

[pulsar bough]

PMG IM SO STUPID

[neat basin]

Just find the value of 3^6 and 3^7 , then add them

At last find the square root of the number you get by adding them

[pulsar bough]

can u help again

[neat basin]

Show your work

[pulsar bough]

wai

t

[neat basin]

Are you allowed to use a calculator if not use the method mentioned above by youssef

[pulsar bough]

damn im stupid as fuck

[neat basin]

Highly relatable

Send your work

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[cobalt atlas]

can anyone explain the second part ?

[cobalt atlas]

like why he is adding the factors everywhere but not next to the fractions

a good example of what I mean is at the end why not just add (x-4) to the 5/x+4 and make it equal

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[grim coral]

How did we simplify ? Is there Any rule applied?

[gritty rose]

,tex .exp rules

[glossy valve]

riemann

[grim coral]

Oh I see Thanks

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[boreal goblet]

I'm struggling with this sos please 😵‍💫

[wild sleet]

this is sum of two events

exactly 1 package or 0 packages

[keen vector]

what's the probability that none one of them have less than 36% fat

[boreal goblet]

Uhh waitt

[boreal goblet]

I don't really remember how you do thatt

[boreal goblet]

I need to start from the start I think 😭

[keen vector]

whats the prob that the first packagae doesn't have less than 36% fat

[wild sleet]

you get 0 packages if you roll 0.99 ten times

[keen vector]

let's abbreviate less than 36% fat to lt36f

[boreal goblet]

0.99 yes?

[keen vector]

yes

[boreal goblet]

And multiply that by the 2nd package..?

And so on?

[keen vector]

wb prob that first 2 boxes are not lt36f

[wild sleet]

yes

[keen vector]

yes

[boreal goblet]

Thenn

0.9801?

Or like

[keen vector]

that was a question

yes

0.99^2

[keen vector]

so what's this

[boreal goblet]

0.99^10 rightt?

[keen vector]

yea

so that's prob of 0 packages haveing l36f

now u want prob of exactly 1 package

there are 10 to choose from

this is binomial distribution btw if ur familiar

[boreal goblet]

I'm not sure what it's called in Englishh but I was gonna propose this like

1 over 10 but it's not division it's like

Let me try to find itt

[keen vector]

$$\binom{10}{1}?$$

[glossy valve]

christoffel

[boreal goblet]

Yes yes

Thattt

I was gonna say that

[keen vector]

ok

so 10 scenarios

[keen vector]

this equals 10 btw

[boreal goblet]

Oki

[keen vector]

they all have the same prob

u wanna find it

eg first is l36f but rest 9 are not

[boreal goblet]

Yesss I'm listening

[keen vector]

or second one is l36f etc

[keen vector]

ok what's the prob of this

[boreal goblet]

I haven't had this in a while let me think for a secondd 😭

[keen vector]

k

[keen vector]

u might wanna reason like this

step by step

[boreal goblet]

l36f means that it has less than 36% fat yes?

And the rest 9 have more

[keen vector]

yes

could just treat it like an abstract event

hence the abbreviation to not think about the details

but it could be helpful to not do that

[boreal goblet]

I'll tryy

Maybe likeeeee

0.99^9 * 0.01

😭 ❓

No I don't think that makes sense

[keen vector]

yes

[keen vector]

why not

[boreal goblet]

I don't know I just started doubtingg

[keen vector]

k so that's the prob of first l36f and rest not

what's the prob of 2nd l36f and rest not

[boreal goblet]

Like of the first 2 l36f and the other 8 not?

[keen vector]

no

1st not l36f, 2nd l36f, rest not l36f

so only #2

[boreal goblet]

Maybe 0.99^8 * 0.01?

[keen vector]

0.99 * 0.01 * 0.99^8

first no, second yes, rest 8 no

0.99 = no

0.01 = yes

[boreal goblet]

Oh so it's like here?

[keen vector]

yea it equals the same thing

[boreal goblet]

I wanted to say that agh

[keen vector]

same for 3rd 4th etc

[boreal goblet]

Okayy

[keen vector]

so 10 disjoint scenarios all with probability 0.99^9 * 0.01

what's the prob that any of them happen

[boreal goblet]

Nowww we use the binomial distribution?

[keen vector]

if u want

the whole thing is binom distbn

we are actually deriving it

[boreal goblet]

Oh woww

[keen vector]

for a special case

but same idea extends

[boreal goblet]

Soo like this? 😭

[keen vector]

umm

no that doesn't make sense

u can only have integers in that

[boreal goblet]

Ohh

[keen vector]

don't guess, think about the situation

[boreal goblet]

Well if we do (0.99^9 * 0.01)^10 that's the probability that none of them happen rightt?

So we're doing kinda like the opposite of thatt

Or no that's the probability that all of them happen

Uhh

[keen vector]

none of them

[keen vector]

the opposite would be atleast 1

we r looking for exactly 1

[boreal goblet]

Ohh we haven't accounted for that have we..?

[keen vector]

we have

[boreal goblet]

Oh yes yes

[boreal goblet]

Am I going in the right direction here at leastt? 😭

[keen vector]

yea ig

ok what's prob of 0 happening

[boreal goblet]

It wasssss

0.99 ^ 10

[keen vector]

yes

lemme know if u got specific questions

[boreal goblet]

If it's specific okay but rn I'm kinda questioning everything 😭

[keen vector]

k i mean

[boreal goblet]

Well maybe

[keen vector]

u just gotta think about it then

[boreal goblet]

(0.99^9 * 0.01)^-10

Because it's like

The opposite

[keen vector]

and not rush it

idk what ur confused about tho

[keen vector]

what 💀

no

[boreal goblet]

Oh 😭

I'll go over it again wait

Does it matter a lot that the scenarios are disjoint you're emphasizing it sm 😭

I don't knoww help mee @keen vector

[keen vector]

yes a bit

[keen vector]

idk how

u gotta ask smth specific

[boreal goblet]

Well this I guesss you can tell me

And we'll go over the answer

[keen vector]

k so it's 10 * (0.99)^9 * 0.01

prob of exactly 1 happening

add this to prob of exactly 0 happening

which is (0.99)^10

that's the answer

[boreal goblet]

Ohh

Could we go over the whole thing again rq though pleasee 😁

[keen vector]

no

[boreal goblet]

Just like

Each point

[keen vector]

maybe someone else will 🤷 💀

this is a big server

[boreal goblet]

Well okay I guess I kinda understood it 😭

Thank youu

[keen vector]

i bet someone's reading these messages rn

[keen vector]

welcome

[boreal goblet]

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[white cradle]

Calculate the area of ​​a figure bounded by the graphs of functions:

[glacial pasture]

what has one tried?

[rotund birch]

!show

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[white cradle]

i am bad at this

graphic

i dont understand this

[viral jasper]

,rccw

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[summer python]

Not really math but physics

I got 3.46 for the answer

Several AI’s also got 3.46

The answer key in brackets says 2.1 tho

Can someone do it and see if they also get 3.46 or 2.1?

[gritty rose]

did you do any calculations yourself

[summer python]

Yes

One sec

@gritty rose apologies for ping but any update?

[gritty rose]

why is this -205sin(29) ?

how do you interpret "with a force along the handle" ?

[summer python]

FN = MG - 205 sin 29
Sin 29 is the Y component

[summer python]

Handle doesn’t mean anything it just a word used so that we know that the force that is applied is on an angle and connected to the box

[gritty rose]

why - and not +

[summer python]

Different directions

[gritty rose]

"with a force along the handle" means the force has a y component in the same direction as gravity

not opposite

the force is applied down and left

what you interpret is up and left

[summer python]

I still don’t get it

What should I have done instead

[gritty rose]

.

[gritty rose]

mg + 205 sin(29) here

[summer python]

ahhh

I get it

[gritty rose]

if you draw the force diagram, the force of gravity and the y component of the force of 205N along the handle are in the same direction, not opposite

[summer python]

Like as in free body diagram?

[gritty rose]

yes that. forgot the name

[summer python]

Ahhh I see

[summer python]

Also since it’s pushing you do + instead of minus right?

On all previous questions and the next one I used minus and got the correct answer

[gritty rose]

Luck

You really should just be doing the free body diagram each time

[summer python]

Bro it’s working tho

For every single question

Thanks again for the help

@gritty rose

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[winter ember]

can y'all help with this

are the eigenvalues +- sqrt(7)i?

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[wicked zodiac]

how do i do this

[craggy tapir]

💿

[wicked zodiac]

disk

i see

[wicked zodiac]

how would i know if it starts from where tho?

ik the end bc u can set the two to eah other

and find the end point (where they intercept)

[formal scarab]

and when you set them equal, don't you get two solutions?

that'll tell you where they intersect both times

[wicked zodiac]

so same value but one is -

[brisk obsidian]

Yup

[wicked zodiac]

i see

thanks

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[torn jolt]

hi

[torn jolt]

i need to show that Un is > 0

for all n

i reached a point where i got that Un = (Un+1)/3

idk what to do

<@&286206848099549185>

BRO IT CANT BE THIS HARD

[cloud creek]

Just put in the values and show that it's increasing

[torn jolt]

what

i dont have the sequence

i just have those 2 indices

[cloud creek]

You have u0 and u n+1 find u1

Put n=0 first

[torn jolt]

...

did you read my question

put n = 0 where? i dont have the formula

[cloud creek]

It literally the second line of your question

[torn jolt]

so just replace 1 and -1 with U0?

[cloud creek]

Replace n as 0 so u1= cuberoot(3+1) -1

Cause we know u0

[torn jolt]

ohhhh shit

[viral jasper]

I think the series is decreasing though

[cloud creek]

Is it I haven't ried it yet

[torn jolt]

thanks a lot

[viral jasper]

Neither have I, but the -1 on the outside feels dominating. I could be wrong though.

[torn jolt]

yeah

i jsut need to prove that the entries are positive

[cloud creek]

Makes sense

The cube root is always greater than one

Hence always positive

[torn jolt]

thanks for the help

[viral jasper]

@torn jolt it sounds like you understand it better now, but I want to point out to you -for future reference- that this kind of definition for a function is, in fact, well-defined, It is known as a recursive definition. You have a definition for some specific value, then you recursively define each sequential value based on the previous value(s)

[torn jolt]

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[torn jolt]

thank you

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[static pewter]

help understanding why it diverges

[lime ether]

write out what k^k means and what k! means

[cloud creek]

Expand them

[lime ether]

great idea

[static pewter]

what does that mean

[lime ether]

do you know what k! is

how is it defined

[static pewter]

yeah k factorial

[lime ether]

sure but what is it defined as

mathematically

[static pewter]

k * k-1 ...

[cloud creek]

Yes and what is k^k defined as

[static pewter]

k * k * k up to k

[cloud creek]

So write it out

[trail ivy]

the easiest way to show it diverges in this case is by using the ratio test tbh

[static pewter]

i tried that but i got = 1

and doesnt that mean the test is inconclusive

[trail ivy]

oh yeah i guess so

sorry i didnt actually try it out 😅

[lime ether]

did you write this out

[static pewter]

i mean yeah but idk how that helps

[lime ether]

what did you get

[static pewter]

oh you wanmt me to write the series out

[trail ivy]

okay actually i did try out the ratio test and i didnt get 1, so the ratio test does work in this case

[lime ether]

no just k^k / k! will suffice

[static pewter]

what how, i got lim of k / k+1

[trail ivy]

i got a different limit, maybe you did the algebra wrong ?

[lime ether]

$\lim_{k \to \infty} \frac{(k+1)^{k+1}}{(k+1)!} \cdot \frac{k!}{k^k} = \lim_{k \to \infty} \left(\frac{k+1}{k}\right)^k\cdot \frac{k+1}{k+1}$

[glossy valve]

knief

[trail ivy]

yes

[static pewter]

ohhhh i forgot about the exponent

[lime ether]

i think it’s best to just understand it intuitively

[trail ivy]

i mean this limit works out to be the definition of e

[lime ether]

that’s true yes

but

what i was getting at is

[static pewter]

i just didnt know where i was going with that idea

[lime ether]

$\frac{k^k}{k!} = \frac{k\cdot k \dots \cdot k \cdot k}{k(k-1)\dots(2)(1)}$

[glossy valve]

knief

[lime ether]

you’re multiplying k by it self k times

whereas in the denominator the terms decrease to 1

[trail ivy]

it suffices just to show that each term will always be at least 1

[lime ether]

$\frac{k}{k} \cdot \frac{k}{k-1} \cdot \frac{k}{k-2} \dots \frac{k}{2} \cdot \frac{k}{1}$

[glossy valve]

knief

[trail ivy]

which should be made obvious by the expansion

[lime ether]

each term is >= 1

[static pewter]

ohh okay

so it just approaches infinity right?

[lime ether]

i think doing it this way is better for understanding

[cloud creek]

Yes

[lime ether]

instead of just using the ratio test which is just pure computation

[static pewter]

yeah it makes sense but my brain immeditley goes to using one of the tests

[lime ether]

despite still being valid

[lime ether]

math education

[static pewter]

lol

[trail ivy]

i think the ratio test has a level of intuition to it

[static pewter]

the ratio test woujld have been so much more work

should the first step of solving series be writitng the terms out?

[lime ether]

it does but unfortunately i think most people don’t understand it

[trail ivy]

i wouldnt immediately discard it as just a mechanism of Pure Computation; the ratio test makes sense why it works, and doing the work just some algebraic manipulations and is probably the intended way of going about solving this problem

[lime ether]

also requires more algebra

[static pewter]

you guys knew instatnly that it diverges which is crazy to me

[lime ether]

but the ways in which it’s used for intro calc from what i’ve seen is exclusively just "plug and chug fuck it it works, if you say so!"

[trail ivy]

like all the ratio test says is "okay, in the long run, this series is going to end up looking like the geometric series, and we know if B>1 then it is going to blow up to infinity and if |B|<1 then it's going to shrink"

[lime ether]

it’s commonly known that x^x grows faster than x!

[static pewter]

yeah i got that much later lol

[lime ether]

right but from what i’ve seen no one even thinks about this for intro calc

again it’s perfectly valid

it’s also longer though tbf

and easier to make a mistake

[trail ivy]

it is more generalizable and useful for more problems, this is like one of the only problems with this specific intuition to it

i would say that it's best to use one of the tests whenever possible and only resort to this kind of stuff if you don't think the tests will work

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@static pewter Has your question been resolved?

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[spark ivy]

got this problem from elsewhere.. I am getting 25/12 after solving it

idk if am wrong or the options are :(

[leaden ermine]

options

[spark ivy]

are wrong right..?

[leaden ermine]

yup

[spark ivy]

thanks lol.. was thinking abt where i went wrong tbh 😭

.close

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[sacred shard]

Just wanna make sure i have these 3 questions correct

[viral jasper]

Ah hello again

[sacred shard]

Ay wassup man

[viral jasper]

Getting through the day

For 1), graph is good, range is incomplete

[sacred shard]

Would it be (-inf, -1] U [-1, -inf)?

Or +inf since e^x goes off to the right?

[viral jasper]

Yeah +inf

But [-1, is not correct

That Union would just give you numbers

[sacred shard]

Then i'm not too sure

0 is covered

Actually i kinda think it's (-inf, +inf)

[viral jasper]

Where in the graph do we have a y value of -0.5?

[sacred shard]

Nowhere

(-inf, 0] U [-1, inf)

Yay or nay

[viral jasper]

Almost

Where in the graph do we have a y value of 0?

[sacred shard]

At x = 0, no?

[viral jasper]

What is f(0)?

[sacred shard]

|0| = 0

[viral jasper]

But $f(x)=-e^x$ for $x\ge 0$. If $x=0$, then $x\ge 0$, so $f(0)=-e^0$

[glossy valve]

SWR

[sacred shard]

But if x = 0 then x < 1, so f(0) = |0|

Like we talked about yesterday, it's kinda fucked bc both parts of the piecewise overlap

[viral jasper]

And this is where i don't blame you for the confusion

I'm quite sure the problem meant x<0, not 1

[sacred shard]

Aight

If that's the case

[viral jasper]

x<1 would make f not a function

[sacred shard]

(-inf, 0) U [-1, inf)?

[viral jasper]

Yeah it's unfortunate, that typo

[viral jasper]

Looking good

2 is correct as well

[sacred shard]

2a and 2b?

[viral jasper]

Also, nice job correctly identifying domain for 2a. Very subtle, so good job

[sacred shard]

Thx

[viral jasper]

3 is good

[sacred shard]

So all 3 questions are good to go?

[full forum]

@sacred shard Has your question been resolved?

[viral jasper]

Yup

[sacred shard]

Ok cool

Lemme drop the rest of my assignment to double check everything

I'm 99% sure everything is correct, just playing it safe

Btw SWR thanks for all your help with domain and ranges. You're the goat, and it's fine if you dont wanna do all that double checking

[full forum]

@sacred shard Has your question been resolved?

[sacred shard]

Shiiiiiii not really but fuck it, save the rest for tomorrow

[full forum]

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[dusky walrus]

Hi could I check with anyone what are the steps to evaluate the contour integral of z conjugate defined as x - iy along the contour of a unit circle defined as w = e^(iθ) where θ ranges from 0 to 2pi?

[oak gorge]

parameterize Unit Circle
express dz
setup the Integral
evaluate the Integral

[dusky walrus]

Is it possible for me to do it with Cauchy's Residue Theorem?

[oak gorge]

yes

[dusky walrus]

Can I ask how do I find the pole for z conjugate?

.close

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[peak estuary]

yo so ive done this

and im not getting the sameresult

idk if im wrong

could someone double check for me

ive done it like 10 times

[left hazel]

wait

I got the same result

the problem is in you

[peak estuary]

could you show me ur steps ?

[left hazel]

wait

@peak estuary

[peak estuary]

what ?

[left hazel]

?

[peak estuary]

this isnt the question

[left hazel]

you are about (A) or (B)?

you need to see all previous simplifications I done?

[peak estuary]

b

[left hazel]

general solution for it: $C_1x+C_2x^4$

[glossy valve]

mathell

[left hazel]

I wrote it

$t=e^x$, thus we have $y(x)=C_1e^{4lnx}+C_2e^{lnx}$

[glossy valve]

mathell

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[sick crow]

hi

[sick crow]

step by step how would I do this

[quiet zodiac]

quadratic formula

with discriminant

[sick crow]

how

[reef moss]

factor

[quiet zodiac]

nah

[reef moss]

factor it and cancel out like terms

[quiet zodiac]

i mean yes factor but with quadratic formula

[reef moss]

????

[sick crow]

oh so i do it for the top and seperate for the bottom

[reef moss]

yes

[sick crow]

and then divide tem or leave it by itseld

[reef moss]

yep

[sick crow]

okay lemme have a go

whats a and what would be b and what would be c

[reef moss]

you don't need to use this if you know other factoring methods

[sick crow]

usually i do the factoring method

but for easier ones

how do i do it for this

[reef moss]

what factoring method do you use?

[quiet zodiac]

u do

if its 6 and 2 and 3 fine

[sick crow]

2 numbers that times to give the last one and add or munus to give the middle number

[quiet zodiac]

but its inconsistent as hell

[reef moss]

you do not NEED to use the quadratic formula to factor when there are much quicker methods

[sick crow]

and then put them into two x brackets

[reef moss]

yep

[sick crow]

this is for gcse content btw

[reef moss]

try using this method for the numerator and denominator separately

[sick crow]

ik but i cnt seem to figure how to get 2 numbers that times to 1 and add to give 3x

[reef moss]

for this question?

[sick crow]

yes

[reef moss]

you need 2 numbers that add to -8 and multiply to 15

[sick crow]

oh sorry i was looking at the wrong question mb

[reef moss]

and on the bottom two numbers that add up to 3.5 and multiply to -7.5

but not sure what methods you've learned

[sick crow]

why those numbers

since it says 15

and -7

[reef moss]

it would be -15/2

since we look at the coefficient of a as well

[sick crow]

ohhhh because of the 2 before the x?

[reef moss]

yep

[sick crow]

so even if it was + 15 and +7 i would still dived them

[reef moss]

yep

[sick crow]

so it would be x-3/2x+3

[reef moss]

wdym

[sick crow]

that would b the final answer no?

[reef moss]

you should get a simplified rational function as final answer

wait

yeah that's right sorry didn't realize cuz of no parentheses

[sick crow]

its fine dw

also for thsi one

would I do

multiply for 0.5 and add for 1.5

for the top hald

half

[reef moss]

add to -1.5

but yes

[sick crow]

but 3/2 =1.5 so how comes minus

[reef moss]

essentially this it probably hasn't been explicitly named though

[sick crow]

ngl i did not think it was this compolicated

istg u just factorise and simplify

and then thats it

[reef moss]

that's basically it

it pretty much depends on the factoring method you use since there's a lot of tricks people learn

[reef moss]

if youre looking for numbers that sum up to something and multiply to something it would be this though

[sick crow]

oh okay

[reef moss]

it's basically what you've been doing

[sick crow]

idky its not workinggggg

[reef moss]

for the top one?

[sick crow]

yes plz

[reef moss]

do you have any example problems where your teacher or whoever used the factoring method you learned?

[sick crow]

yes hang on

it worked their

and i understood but then i didnt get it in the previous question i sent

hello???

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[sick crow]

.close

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[hoary ember]

for prime $p$, find all positive integers $(a,b,c)$ such that
$$a^p+b^p=p^c$$

[glossy valve]

Skissue ping4response

[hoary ember]

i dont think fermats last theorem is given here so i cant use it i dont think :p

[umbral dome]

it doesn't apply, note the right side

[hoary ember]

?

[granite torrent]

left side has p exponents, right side has c exponent

[hoary ember]

oh right

[lilac oyster]

I suppose first thing I'd do is reduce mod p and play around a little

[hoary ember]

my only idea is to split into cases where p=2 so you only work with odd p

[hoary ember]

so a+b=0 mod p right (by fermats little theorem)

[lilac oyster]

I don't see a reason to split even/odd p at this stage yet, but haven't thought that far ahead either

[lilac oyster]

yeah

good

[hoary ember]

i was hoping there would be like something where one is odd and one is even and something can come out?

[lilac oyster]

maybe there's a more direct way though by splitting even/odd since you could directly factor a^p+b^p now that you mention it

[hoary ember]

does a^2+b^2=2^c have any solutions?

[lilac oyster]

also a good point

[hoary ember]

im p sure no but i cant think of why

[lilac oyster]

1^2+1^2 = 2^1

[hoary ember]

wlog a>=b well its either a,b odd or a,b even and if they are both even then you can factor 2 out of them so it either becomes odd+even or odd+odd

[lilac oyster]

in fact, when p=2 look at when a=b

[hoary ember]

it would be then a=b=2^n right

[lilac oyster]

yup and c=?

[hoary ember]

n+1

[lilac oyster]

close I think

[hoary ember]

what

is it not c=n+1

2^n+2^n=2^(n+1)

[lilac oyster]

that would be a+b=2^c

[hoary ember]

agh

2n+1

[lilac oyster]

nice

ok so we at least got infinitely many solutions but not necessarily all of them for p=2

or maybe we did and we can reason it out, I was starting to work the odd p case at the moment, just playing around still haha

[hoary ember]

lemme think about it aswell

[lilac oyster]

at least a=b can't happen for odd p right

[hoary ember]

if you say something pls ping cause im gonna be on my notes app

wait

how do you know a=b is the only solutions for p=2?

[lilac oyster]

I don't

[hoary ember]

ohg

[lilac oyster]

I had to take care of some chores and garbage and stuff around the house, so bit distracted but

for odd p I think factoring might be a good option

maybe focus on a^3+b^3 = 3^c to get the idea started if you're unsure what I mean by that

oh I guess another separate strategy that'll be handy to know

if (a,b,c) is a solution, then (pa, pb, c+2) is a solution

so now once we find one solution - we have infinitely many

I guess the p=2 case was a special case of that as well when we had (1,1,1) earlier and then got those infinitely many more from it

to just get it immediately for some parameter n, (ap^n, bp^n, c+2n) if (a,b,c) is a solution

you don't by chance know the lifting the exponent lemma do you? thinking that might also help here

where did this problem come from?

[hoary ember]

can 2^n -1 be a perfect square?

[lilac oyster]

consider it mod 4

[hoary ember]

that implies a^2-ab+b^2=3^(c-1)

[hoary ember]

oh

[lilac oyster]

I'm thinking it's kind of close but not quite

a+b = 3^m and a^2-ab+b^2 = 3^n with m+n=c

[hoary ember]

well doesent this mean p=2 the omly sols are a=b

[hoary ember]

apparently french tst 2012, i tried searching it online but got nothing

[lilac oyster]

gotcha, fun

[lilac oyster]

taking a look now one sec

ah yeah

I'm thinking even more immediately reducing that mod 4 forces 2=2^c mod 4 so c=1 is the only option

[hoary ember]

ok thats much nicer lmao

[lilac oyster]

but I like the square argument too, we don't know what kinds of tricks we'll need for odd p so good to cover different strats now I think haha

[hoary ember]

ive read about it a bit

[lilac oyster]

ok cool so that settles p=2 I believe

[lilac oyster]

nice yeah you said that too ok so we're good

let me make sure though having second thoughts that I missed something on it

we know a=b mod 2 so they're both even or both odd at least

if they're both even we can just pull the 2 out and go from (a,b,c) to (a/2, b/2, c-1)

so they're both odd

and we just did that, ok cool I feel better now

[lilac oyster]

ok I'm able to get $c=1+v_p(a+b)$ from LTE

[glossy valve]

Merosity

[lilac oyster]

that make sense?

need you to check me cause I'm distracted haha, so I could be wrong here

I think it's pretty straight forward though in this case

[hoary ember]

yes

[full forum]

@hoary ember Has your question been resolved?

[full forum]

@hoary ember Has your question been resolved?

[lilac oyster]

I would try asking in #competition-math or #elementary-number-theory not sure how to finish it

@hoary ember

[hoary ember]

sure

.close

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[cursive wagon]

yo help

[cursive wagon]

im stuck

[narrow ermine]

[cursive wagon]

CHALLENGE 1 (5 marks)
The pdf data you have been given is somewhat realistic but there are aspects of it that are artificial.
The probability density during the chip’s operating life would be much lower than 0.01 per year, the
probability density would not change discontinuously, and there would be some small probability of
the chip lasting a very long time. How might you modify your bathtub curve to take these into
account? Make a sketch by hand of your modified bathtub curve.

[cursive wagon]

I think this is right, but i have no clue

please someoneee

[full forum]

@cursive wagon Has your question been resolved?

[cursive wagon]

NO BRO

<@&286206848099549185> where r u

<@&268886789983436800> do ur job

before i do it for u

sorry pookies

[buoyant sierra]

Bro wht

[cursive wagon]

js vv stressed

[full knoll]

I can’t answer they keep timing me out

[cursive wagon]

what did u do wrong then

were u naughty

[buoyant sierra]

[cursive wagon]

@buoyant sierra luv u

can u two help me now 🙂 ❤️

[buoyant sierra]

You are here to ask questions just wait for help. As for me , I have no idea what that is so sorry I can't do anything. Moderators Don't control helpers so pinging them is useless

All you can do is just wait

[cursive wagon]

oh... just chatgpt then youll get it 🙂

then u can help me

:))))

[sage sparrow]

what?

[cursive wagon]

im vv stressed it was out of line.

I apologise 😦

[sage sparrow]

dont ping moderators for math help.

please read #❓how-to-get-help

[cursive wagon]

u are a very good mod, i adore ur help

u should get a raise @sage sparrow

<@&286206848099549185> HELP MEEEEE PWEASEEEE ill gift nitro.

[wooden siren]

Uhh

[dawn rock]

sigh alr

what is it that you need help with @cursive wagon

[cursive wagon]

i have this question...

[cursive wagon]

this is the data

[dawn rock]

Alr give me a second (it’s 4 am rn and I have school in a couple hrs)

[cursive wagon]

this is my graph output... i knows its not hand drawn lol

i am just wanting to verify if its correct

[dawn rock]

kk

So

[cursive wagon]

https://tenor.com/view/umm-i'm-not-nervous-nervousness-need-a-relax-help-anxious-gif-11511780752195456503

[dawn rock]

So the bottom graph is the data

and the top is what u did as a output?

[dawn rock]

Based on these requirements u seem correct

[cursive wagon]

0*yup

W?

[dawn rock]

1. Ur probability is a nonzero constant lower than .01 during usual life

2. The curve is smooth which is nice

And the tail at the end basically shows the small probability of it lasting longer

[cursive wagon]

https://tenor.com/view/cat-gif-17716769089925234109

[dawn rock]

yuh

drop the nitro 🙏

jk u don’t have to lol

[cursive wagon]

https://tenor.com/view/thanks-cat-thank-you-cute-adorable-gif-16960579

[dawn rock]

yuh ofc

Do .copse

.close

or wtv it is

[cursive wagon]

https://tenor.com/view/smurg-smurf-kittennn-discord-iheartu-gif-24765093

have a good night!!! i may send nitro 🙂

[dawn rock]

yippie

[cursive wagon]

add me i think

[dawn rock]

it’s 4 am and I have school at 7:45

Am I cooked

[dawn rock]

sent

[raven ingot]

What are the units on the vertical axis of your pdf graph?

[cursive wagon]

^ what this bum said

@dawn rock sorrryyy

[dawn rock]

Usually failures per year

(Or per unit of time)

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[minor turret]

Trying to cover over some stuff during the holidays that I didn't quite understand. Need some help with C. I don't know where I'm supposed to go from here, or if I even did it right. Thanks!

[rustic frigate]

It might be less ugly if you substitute u = x - 1, and and solve for u

[minor turret]

Got to here. I'm guessing the next move is to bring the u² to the left side?

[rustic frigate]

Yes

[minor turret]

So it would become u³ right?

Or wait is it u²+u

[rustic frigate]

It would be u^3

Since we’re multiplying both sides by u

[minor turret]

Yep got it

So I got u³=4

[rustic frigate]

Correct

[minor turret]

This is where I got stuck lol

[rustic frigate]

Now to get rid of ^3

We take its inverse, namely ^1/3 on both sides

Does it sound familiar?

[minor turret]

So it would look like this?

[rustic frigate]

Yea

[minor turret]

So, u would be = to

Or did I get that wrong

[rustic frigate]

that’s correct

[minor turret]

So would I now sub back in the x-1?

[rustic frigate]

Yes!

Now it’s more comfortable to do so

[minor turret]

So I tried subbing it in, but I'm not sure if that's the right thing to do

[rustic frigate]

Oh

That would be unpleasant

And also not really what you’re looking for

[minor turret]

Lol

Yea I had a feeling

[rustic frigate]

You know that u = x - 1

So 4^1/3 = x - 1

Makes sense?

[minor turret]

Yep

Im just not sure where to go from here unfortunately

Wait I think I've got it

Thank you!

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[tawny coyote]

2+2

[tawny coyote]

2+2??

[tawny coyote]

whats the root of 4

[lyric ember]

2?

[tawny coyote]

Just trying to figure out how this bot works 😔

im confused

[thick hedge]

You just ask your question here

an actual person will then try to help you

[tawny coyote]

Oooh!

That makes a lot of sense actually

thanks for telling me

sorry for bothering you by the way 😅

Have a nice day!

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[short ermine]

im curious about why we need to split the second integral

[short ermine]

why do we split it in 2 parts -infinity to 0 and then 0 to x

[green kite]

because f(t) behaves differently when t is negative

and positive

[short ermine]

ohh

just one more quick question, for the second its stated that x is nonnegative

cant we just do the lower bound 0 and upper x

does it perse have to go from -infinity to x, even though we know x is nonnegative

[torn jolt]

always

then you just specify what you have tof ill in for -infinity, since in many cases it isnt defined for some values

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[wise zephyr]

is it bueno?

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[autumn nest]

It says too: solve the systems of equations using the addition method. And like I rlly dont understand how it works, like ive been trying but it just turns pit silly

[brisk obsidian]

The idea is to cancel out either the x term or the y term by multiplying one equation by some factor and then add that result to the other equation. If done properly, the term will cancel out leaving the the sole other term for which you can solve for.

For a, you can multiply the first equation by -4/3 so that you end up with a -4/3 * (3x + 2y = 9). Do the math and you should end up with a -4x in the new equation. Add that result to the second equation which will cancel out the x terms.

[autumn nest]

Ahhhh

U pro like that

Thank you

[brisk obsidian]

yw

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[static bramble]

Looks like triangle

[mellow pike]

hmm I dont see triangle. For the last step i do:

or reverse, rather.

[pulsar kiln]

$( |\sqrt{a} - \sqrt{b}| \le \sqrt{|a - b|} )$

[glossy valve]

Quantie
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[mellow pike]

oh ok haha

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[livid iris]

how do you solve this or find which answers are the right ones

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[sudden bolt]

Hello, I want to ask something. In limit, is (0/0)=(0/0) still considered zero or indeterminate for it to determine if it is continuous or not? thank you

[mental ruin]

0/0 is indeterminate

[sudden bolt]

so it is not continous?

[mental ruin]

is what continuous?

oh the function which limit you are taking

[sudden bolt]

i'm studying basic calc limits

[mental ruin]

yeah

the function wouldnt be continuous at that point because the limit is nonexistent

[sudden bolt]

uh continuity on an interval

[sudden bolt]

oh thank youu

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[inner rover]

.close

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[vocal sage]

Find the locus of the point of intersection of a pair of tangents to the ellipse
[
\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1
]
if the difference of eccentric angles of their points of contact is
[
\frac{2\pi}{3}.
]

[glossy valve]

riddle

[vocal sage]

how is sintheta_1 + sintheta_2= 1 in this question?

ik u can do it by formula

[hollow herald]

Use SS_1 = T²

[vocal sage]

i want to solve by obtaining quadratic in sintheta

[hollow herald]

Hmm

[vocal sage]

for ellipse tangents are xcostheta/a + ysintheta/b= 1

we can satisfy (h,k) in it

[hollow herald]

Suppose one point is (a cos t, b sin t), another is (a cos(t + 2π/3), bsin(t + 2π/3))
Tangent to both points can be written

[hollow herald]

✓

[vocal sage]

so u get ksintheta/b = 1- xcostheta/a

now square ig

wait

keep xcostheta alone so it can be replaced by (1-sin^2theta)

[hollow herald]

? You're looking for intersection of (x cos t)/a + (ysin t)/b = 1 and (xcos (t + 2π/3))/a + (ysin(t+2π/3))/b = 1