help-27

[wind root]

WHY IS SUCH A THING THERE

[formal delta]

Mathematics

[wind root]

😭 tf am i supposed to do w that

No wonder I saw that in my poisson table thingy

How I get given for lambada

Lambda

[formal delta]

$$P(X\leq 2)=e^{-3}(\frac{2+6+9}{2})=e^{-3}\cdot 17/2$$

[woven radish]

Benjamin

[formal delta]

That is 0.423

Which is 42.3%

Cuh

[wind root]

Howd we get the 9

[wind root]

Where get nine huzz

[formal delta]

You ask too many questions

[wind root]

😞😔

Sorry blud

[formal delta]

It comes from 3^2

[wind root]

Why we do that

I can fr tell ur frustrated at me fr

[formal delta]

Give me a sec buddy

Im jk im not mad

[wind root]

Oh ok that makes sense since fine shyt never gets mad at me fr

[formal delta]

The Poisson probability mass function (PMF) is:

$$
P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}
$$

where ( \lambda = 3 ). Buddy, calculate:

1. For ( X = 0 ):
   $$
   P(X = 0) = \frac{e^{-3} 3^0}{0!} = e^{-3}
   $$

2. For ( X = 1 ):
   $$
   P(X = 1) = \frac{e^{-3} 3^1}{1!} = 3e^{-3}
   $$

3. For ( X = 2 ):
   $$
   P(X = 2) = \frac{e^{-3} 3^2}{2!} = \frac{9}{2} e^{-3}
   $$

Cuh sum them:

$$
P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)
$$

$$
P(X \leq 2) = e^{-3} + 3e^{-3} + \frac{9}{2} e^{-3}
$$

Factor out ( e^{-3} ):

$$
P(X \leq 2) = e^{-3} \left(1 + 3 + \frac{9}{2} \right)
$$

$$
P(X \leq 2) = e^{-3} \times \frac{17}{2}
$$

Approximating using ( e^{-3} \approx 0.0498 ):

$$
P(X \leq 2) \approx 0.0498 \times 8.5 = 0.423
$$

Thus, the final probability is:

$$
P(X \leq 2) \approx 0.423
$$

[woven radish]

Benjamin

[formal delta]

Buddy there u go

💕🤭

[wind root]

Wtf is the squigle thingy

AHAHAHAHHSHDBDB

The double ~

Wtf bru I aint even using math in a farm

[formal delta]

Tilde

That’s what it is

[wind root]

Alright

[formal delta]

How does it feel to have the smartest helper help u

[wind root]

Thank you fine huzz 😔🫡

[wind root]

Um

[formal delta]

Np jit

[wind root]

Special fr 🙏🙏🙏

[formal delta]

Im glad you recognize it

Any more questions

[wind root]

Ofc I have a keen eye for things

[wind root]

Yah

Wanna see the frog I dissected? 🥰 It's giving max level 10000 cutie fr

[formal delta]

$$\mathfrak{yes}$$

[woven radish]

Benjamin

[formal delta]

Also the tilde means “is distributed as”

[wind root]

Thank you 🫡

[formal delta]

Yes

[wind root]

I sent it in your dms I'd send it here but I might get banned 😔

[formal delta]

Okay

[wind root]

Oki have a nice day huzz

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[wide ember]

Help

[final osprey]

this is an ellipse, make the right hand side 1 first

[wide ember]

Yea I did till that

[final osprey]

latus rectum is the line that passes through the focus of an ellipse

[wide ember]

2bsq/a?

But what is a and b here

[final osprey]

are you required to show your working? there is a direct result that'll let you find the length directly

yea this

x^2/a^2 + y^2/b^2 = 1

[wide ember]

No working just the use of formula

[final osprey]

is the form of an ellipse

[wide ember]

Root 12 is b ?

[final osprey]

yep

[wide ember]

Then the answer is coming to be 12

Which is not one of the options

Nvm got the answer tysmmm for the help tho I was using the wrong formula

❤️

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[true vapor]

Which one is the pivot?

[devout snow]

@true vapor Has your question been resolved?

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[desert girder]

I need help with part a and b

[devout snow]

@desert girder Has your question been resolved?

[eager agate]

You can use Polynomial Long Division to yield that exact format.

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[twin canyon]

Can someone please help me with this?

[pseudo basin]

@twin canyon

[twin canyon]

I tried finding sum of first 20 perfect squares

and got 2870

[pseudo basin]

right

[twin canyon]

but they said it is 3857

[pseudo basin]

so we know the squares in question can't have been the first 20 perfect squares

[twin canyon]

yeah

[pseudo basin]

what if the first 19 were as low as possible? what upper bound does that place on the highest square?

[twin canyon]

if first 19 perfect squares as low as possible then their sum would be 1885. so the last one would be 3857-1885? idk

[pseudo basin]

,w sum[k=1, 19] k^2

[woven radish]

Results provided by WolframAlpha

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

Requested by ann_dec

[pseudo basin]

2470 surely

[twin canyon]

o

yeah i guess i calculated wrong

[pseudo basin]

this leaves 1387

[twin canyon]

hm but that isnt a perfect square

it ends with 7

[pseudo basin]

so the largest square can't be bigger than 37

[twin canyon]

yeah

[pseudo basin]

yeah but it means we cannot have any square bigger than this, ever

so at least this narrows down the search space

[nova glen]

must they be distinct?

[twin canyon]

yes

[pseudo basin]

try to assemble a sum of 3857 using 37^2 as your largest?

i don't see anything cleverer here

[wary maple]

When I get the answer for cos do I have to on calculator put the answer and than shift to get cos-¹ to get the answer

[nova glen]

!occupied

[devout snow]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[twin canyon]

maybe we can replace a perfect square in the first 20?

oh I got it

a^2 - b^2 = 987 => (a-b)(a+b) = 21 x 47 a = 34 b = 13. so largest is 34^2 right? or did i do something wrong @pseudo basin

[pseudo basin]

hmm where are you getting a^2-b^2=987 from

[twin canyon]

i did 3857-2870

[pseudo basin]

mmm

why

[twin canyon]

so i need to get a new perfect squre excluding the first 20, replace one of them

[pseudo basin]

oh hm.

[twin canyon]

the new perfect square should be 3857-2870 more than the existing one

[pseudo basin]

oh so you swap out 13^2 for 34^2?

[twin canyon]

yeah

does that work

[pseudo basin]

well it tells us we can make 34^2 as the highest square

but can we go higher?

[twin canyon]

idk

[pseudo basin]

,calc 35^2 - 34^2

[woven radish]

Result:

69

[twin canyon]

because in a^2 - b^2 = 987, u dont get any other prime factors such that one of them is <=20 and other > 20

3x7x47 yields 21, 47, 141,7 or 3,329

oh wait does 3,329 work then

no thats too big ig

so it IS 34? @pseudo basin

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[pseudo basin]

i... guess?

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[frail spruce]

whats the difference between concave up and convex down? or are they the same?

[pseudo basin]

never heard "convex down"

[frail spruce]

oh

[acoustic leaf]

usually it's "concave up or concave down" or "convex or concave"

[feral agate]

it's either concave and convex or concave up and concave down

[frail spruce]

so theres no convex up or down

[feral agate]

I mean if you want

[frail spruce]

its just concave (up) and concave (down)/convex

[acoustic leaf]

you could define it, but it's not standard so you would be less understood

[frail spruce]

but generally

[acoustic leaf]

concave up = convex, concave down = concave

[frail spruce]

oh mb

okok ty good to know

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[urban badge]

guys what is 2+2

[devout snow]

@urban badge Has your question been resolved?

[mystic scarab]

Are you trolling?

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[gray bluff]

Help

[gray bluff]

.close

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[gray bluff]

.reopen

[devout snow]

✅

[gray bluff]

?

[final osprey]

substitute $u=e^x$

[woven radish]

strizz

[gray bluff]

and what i change the x into?

[restive river]

strizz u lowk ate and devoured with that pfp

[final osprey]

ln u

[final osprey]

we love paul

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[compact hawk]

Can someone help me with both of these question?

[lavish nimbus]

,rotate

[compact hawk]

Wait nvm for no 1 I got the anwser

So I need help with no 2

:v

[frozen aurora]

by long division you get that $\frac{n^2+2023}{n+1} = n-1 + \frac{2024}{n+1}$

[woven radish]

artemetra

[frozen aurora]

so find all factors of 2024

[compact hawk]

._.

[frozen aurora]

is there something that confuses you?

[compact hawk]

I'm trying to process this information 💀

[graceful cipher]

to put it down, n-1 is an integer, because n is an integer. we need $\frac{2024}{n+1}$ to be an integer, in order for $\frac{n^2+2023}{n+1}$ to be an integer, so that $n+1$ divides $n^2 + 2023$

[woven radish]

therealtdp

[graceful cipher]

for $\frac{2024}{n+1}$ to be an integer, we need n+1 to be a factor of 2024

[woven radish]

therealtdp

[compact hawk]

So uh n+1 is an integer and we need factor of 2024 so it can be divided by n+1

So then factors are 2,4,8,11,23,88,184,253,506,1012

[compact hawk]

Got it

But now what

?

[hot steeple]

prime factors

2024 = 1012 * 2

etc

[fossil locust]

n + 1 = 2, 4, 8........

well it just asks you to find the number of positive integers

so you just needed 2024 = 2^3 * 11^1 * 23^1

hence there are (3 + 1)(1 + 1)(1 + 1) = 16 positive factors

but this includes n + 1 = 1, so we exclude that and we have 15

[fossil locust]

you missed quite a few

[compact hawk]

Oh. ._.

Oh there's 16 total right

So uh

Missing 3

Uhh

2³ x 11 x 23

Let's see

[fossil locust]

1, 2, 4, 8, 11, 22, 23, 44, 46, 88, 92, 184, 253, 506, 1012, 2024

[compact hawk]

Ok so uh n = 1,3,7,10,21,22,43,87,91,183,252,505,1011,2023?

One of these?

[fossil locust]

yes

read the question again, it asks you how many n there are

not what the n actually are

[compact hawk]

Ohhh

So we could have just found all the factors of 2024

• 1 cus of 1

[fossil locust]

just do this from the prime factorisation

you are choosing from
powers of 2: 0, 1, 2, 3
powers of 11: 0, 1
powers of 23: 0, 1

by the multiplication principle there are hence 4 * 2 * 2 = 16 choices

each choice gives a different positive integer, of course

[compact hawk]

Ohh honestly I knew this thing but I never knew why it worked xd

Ty

[fossil locust]

np!!

[compact hawk]

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[compact hawk]

I need help with 3 and 5

[compact hawk]

Also could someone check if this is correct for number 4?

This is a clearer picture of 5

[crude spruce]

can you show that $\frac{u^4 + u^2 + 1}{u^3 -1}$ simplifies to $u + \frac{1}{u-1}$?

[woven radish]

kizzyyy

[crude spruce]

[compact hawk]

Uhhhhhh

No idea

😭

I was thinking of u⁴ + u² = u³ + 1 but that's just untrue cus I was thinking of (n + 1)(n-1) = n² - 1

[compact hawk]

Oh wait it's u from last time

[crude spruce]

mmm when all hope is lost

you can always do polynomial division

[compact hawk]

Oh I finally got it after so many tries 💀

[crude spruce]

nice

so you’re done

u + 1/(u -1) - u

where u = 2023

[compact hawk]

Oh so 1/2022

[crude spruce]

smth like that

[compact hawk]

👍

.close

Ty

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[cerulean badge]

Can someone calculate and tell me what they're getting? I wanna check if I got the answer right or not

[pseudo basin]

yours first

[cerulean badge]

Well I'm actually getting two different answers 😭

[pseudo basin]

show both of your answers along with working

[cerulean badge]

I'm probably doing it wong

Like, very weong

[pseudo basin]

very wrong indeed...

[cerulean badge]

[strange arch]

is that a new variance formula from outer space?

[pseudo basin]

first off you should probably be working with the numbers 2, 3, ..., 11 unless you have enough foresight to tell that adding 1 won't affect the variance

[cerulean badge]

Ann, I'm going to fail maths 😭

[pseudo basin]

secondly you should find the mean of all these things

uh.

do you want to bitch/whine/vent/etc. or do you want to know how to solve this problem properly

[strange arch]

at least look up the formula you've been taught to solve the problem

[cerulean badge]

I want to know how to solve this

[pseudo basin]

ok right

first off you should probably be working with the numbers 2, 3, ..., 11 unless you have enough foresight to tell that adding 1 won't affect the variance
secondly you should find the mean of all these things

these 2 points stand

[cerulean badge]

I did, I watched a video too I'm so lost

[pseudo basin]

can you write down the formula you attempted to use?

just the formula in general, nothing plugged in yet.

[cerulean badge]

Yes one sec

[pseudo basin]

ok right...

so, what's your mean here?

you calculated $\bar{x}$, right?

[woven radish]

Ann

[cerulean badge]

Yeah, I got 5.5 however that's a big calculation, I don't wanna go through that so

Assumed mean

[pseudo basin]

wdym "assumed mean"?

[cerulean badge]

...subtracting and adding a value from the middle of the series that is closest to the mean?

[pseudo basin]

i mean... ok?

but shifting down by that doesn't make the new mean 0.

you cannot just pretend that your mean is 5 instead of 5.5

that's not how it works

[cerulean badge]

Okay

So I just...put myself through the long calculation?

[pseudo basin]

well you will have to deal with it anyway

[cerulean badge]

I mean, I could do that but this is like an 80 mark paper with 3 hours to finish it

[pseudo basin]

the fractions don't get any denominators worse than 4

you can sweeten the pill somewhat

your values of x - mean are -4.5, -3.5, -2.5, ..., 2.5, 3.5, 4.5

or -9/2, -7/2, -5/2, ... up to 9/2

which is not THAT bad

[cerulean badge]

Yeah, I suppose it's not... 😭

Sighs okay I have like 23 more questions to go

So I'm going back to hell

See ya

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[sly badger]

Can someone please tell me what type of function this is?

[supple knot]

try using change of base formula

,tex .log base change

[woven radish]

riemann

[sly badger]

i get this

[supple knot]

and maybe uhhh

,tex .log power

[woven radish]

riemann

[sly badger]

but what i meant is like how would i classify this function? like how we say that y=x^2 is a quadratic function, how would we call this one?

[stone stump]

not all function types have a name

[sly badger]

so this one just isn't classified?

[stone stump]

those classifications you are thinking of only work for very basic examples

its way too easy to come up with functions which wont fit

this one being one of those

you might call it sublinear but thats more a property instead of a general "classification"

[sly badger]

alright thank you 🙂

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[round talon]

i tried to help someone but i kinda explained it badly

[round talon]

why does the same angle make them pararell?

angle BAD and DCB

[placid rover]

its one of the geometry rules

[round talon]

also i realized my mistake on the angles AED and CEB so just ignore that

[round talon]

what rule is it again

[placid rover]

do these things have consistent names

idk, take this

[round talon]

hmm ok i think i get why

its like reversed

same coresponding angle, parrarel lines cuz pararell lines = same corresponding angles

ok

thx

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[cerulean quarry]

anyone able to help with part 3 i dont get what it means

[pseudo basin]

the damping term is $\gamma\dot{x}(t)$

[woven radish]

Ann

[pseudo basin]

compare the period with damping to the period without

[devout snow]

@cerulean quarry Has your question been resolved?

[cerulean quarry]

Because T_0 is 1 is omega then 2pi and I use that to get a value for T_d

Along with my gamma from the 2nd part

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[slim thistle]

I don’t understand why my method (on the top) doesnt work with the actual answer (on the bottom)

[stone stump]

its the same

[lusty sapphire]

Your solution is correct

[stone stump]

due to all the trig identities, the same things can look very different

your expression is equal to 1/6-1/6cos^6(x)

[spring salmon]

It varies by a constant

[stone stump]

so the same up to the constant of integration

[slim thistle]

Oh okay thank you !

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[sudden roost]

How do I prove x=10?

[sudden roost]

Also sorry for bad handwriting

Let me rephrase- how do I prove the question has only one solution?

Sorry there's a plus somewhere instead of minus

.close

[devout snow]

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[sudden roost]

How do I prove x=10 is the only solution for this?

Nevermind it's not even 10

.close

.close

[silk prawn]

i was about to say

[sudden roost]

Why uh

Isn't it closing

[silk prawn]

it's already closed

[sudden roost]

Oh

[silk prawn]

takes time to become free againn

[sudden roost]

Wait no it is 10 I just wrote the equation wrong twice

[silk prawn]

[sudden roost]

😭

Let me open a new one

[devout snow]

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[velvet coral]

im kinda confused on part i. Im not really sure how I should use the hitn

[upper schooner]

(how do you have being (not/) connected defined, btw?)

[velvet coral]

not connected is there eixsts open sets O1, O2 such that O1 U O2 supset K and O1 cap O2 = empty st and K not subset of O1 and K not subset of O2

[chilly hornet]

The definition of connectedness involves two open sets, so it would be natural to try and use that
Suppose K is not connected. Then there exist open sets U and V such that U∩V is empty and K ⊆ U∪V. now consider the sets K'_n = U^c ∩ V^c ∩ K_n. Clearly each K'_n is compact, K'_n ⊆ K'_(n-1) ⊆ ... K'_1, and the intersection of the K'_n are empty; This contradicts a property of compactness(which I assume you have already proven)

[velvet coral]

but how do we know that K'_n = U^c ...

ah are you constructing something new?

[chilly hornet]

yep
basically as the hint said

[velvet coral]

ahhh wait okay. I think i mistinerpeted the hint and was trying to do some funny stuff with sets

[velvet coral]

each K'_n is compact since intersection of compact set ic compact right?

or intersection of compact with closed set is compact

and then we can apply that indutively

to as many closed sets + comapct set

oh wait also arbtiary intersection of closed set is closed

[chilly hornet]

yes, because a closed subset of a compact set is compact

[velvet coral]

so here we actually dont need th eintersection to be countable (sry kinda going on tangent)

or i mean how we index

because it can be arbtirary

and in any metric space compact => closed + bounded

[chilly hornet]

yep

[velvet coral]

so actually any intersection of compact + closed => compact

not just countbale index thing

[chilly hornet]

that's actually only for euclidean spaces, not all metric spaces

[upper schooner]

All metric spaces, it's the converse you're thinking of

[chilly hornet]

oh mb

lmao

[velvet coral]

but how is their intersection empty?

[chilly hornet]

misread it

[velvet coral]

isnt their inntersection oh wait

their intersectino is

O_1^c cap O_2^c cap K

but that isnt empty i think

ah shoo

wait

we have htat O_1 U O_2 supsetof K

[chilly hornet]

this is empty since K ⊆ U∪V was assumed at the start; in words, If K is included in U plus V, and you take away U and V from it, you're left with the empty set

[velvet coral]

so O_1^c cap O_2^c subset of K^c, thus O_1^c cap O_2^c cap K subset of emptyset

[velvet coral]

for part ii, using the hint and the fact it is in R^2, i think i need to find an unbounded and probably not closed set

im thinkin gof the set ${(x,y): x \in [0,1], y \in [0, n), (x,y) = (\frac{1}{n}, n) }$

[woven radish]

LXDL

[velvet coral]

where the "connected" point that goes unbounded is 1/n, n

[chilly hornet]

There's a lot of approaches for this one, just make sure to satisfy the 'nested sequence' part
an unbounded U-shape like R^2 minus [-1, 1] * [n, inf), with n getting larger, is one simple example

[velvet coral]

ah okay, but here that set is connected?

because i think it looks disconnected since once cannot reach inf

or ig i would try a proof bu contradiction approach

maybe

[chilly hornet]

Oh it's supposed to be [-1, 1]*[-n, inf), my bad

[velvet coral]

wait but then its no logner nested

[chilly hornet]

the way I originally wrote it, it wouldn't even be a nested sequence

[chilly hornet]

the sets are R^2 - [-1, 1] * [-n, inf)

[velvet coral]

because we have $K_1 \supset K_2 \supset K_3 \ldots$

[woven radish]

LXDL

[velvet coral]

but for the [-n, inf)

[-2, inf) is not subset of [-1, inf)

[chilly hornet]

yes, in fact, [-1, inf) is a subset of [-2, inf)
but I'm not using those sets but rather their complement with respect to R^2, and the complement of [-2, inf) is a subset of the complement of [-1, inf)

think of something like this, where the square is R^2 and the shaded area represents each set K_n

(excuse the horrible drawing)

[velvet coral]

ahh okay

[velvet coral]

ohhhh

waut

and its like

the line divicing the middle

makes it eventually disconenxcted

like split into two

[velvet coral]

so bascially, (-infity, -n)?

[velvet coral]

connected sets do not have to be closed/open right?

[chilly hornet]

yep, but obviously you'd need at least some other elements in the set since otherwise the intersection is empty

[chilly hornet]

yes

[velvet coral]

i thnk ill go with the set $S_n = {(x,y): x \neq 0, y \in (-\infty,-n) }$

[woven radish]

LXDL

[velvet coral]

thus the intersection will be everything except the y axis

if i wanted to prove that each $S_n$ is connected, should i use contradiction?

[woven radish]

LXDL

[velvet coral]

or would just going with the nomral defn be better

[wicked rover]

its disconnected

x!=0

[velvet coral]

oh yeah

[velvet coral]

maybe i change the set to $S_n = {(x,y): (x \neq 0 \text{ OR } y \notin [-n, \infty) }$?

[woven radish]

LXDL

[wicked rover]

ye connected

i didnt read anything before this so idk if u need other properties

[devout snow]

@velvet coral Has your question been resolved?

[velvet coral]

ye ive ben trying to show its connected but am having trouble

[wicked rover]

id show path connected

[velvet coral]

how would i do that? it seems to be quite messy/ many paths

[devout snow]

@velvet coral Has your question been resolved?

[wicked rover]

drawing the set helps u see the path

u will see two cases, the points lie on opposite sides of x=0 or they lie on the same side

[devout snow]

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[short depot]

It's my physics que can anyone answer it?

[short depot]

Que in which wire phrase or netural is electric switch connected?

If not then I will colse

[devout snow]

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[eager agate]

Correct. Now all you need are your justifications.

[wanton aspen]

Thanku : )

.close

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[brisk ocean]

is th5e second integral wrong, shouldnt' the cubic be negative or smthn

[drifting mauve]

it will automatically be negative because uhh that's how the function works

just plot the integral graph you'll see that it is negative in that interval

[brisk ocean]

is it becuase of the abs value

[drifting mauve]

just lemme plot the integral gimme a minute

and you'd see

the integral function is

<0 for x=1 to x=4

so you won't have to worry about what values would be -ive or +ive

since the integral will automatically give you -ive and +ive values

[devout snow]

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[hushed wave]

Area and Perimeter of this Algebra Tiles

[hushed wave]

<@&286206848099549185>

[solid osprey]

any info at all?

[solid osprey]

whats x? can you screenshot the whole problem

[drifting mauve]

uhm, it should be pretty obvious with the area

three whole x^2 squares and three x^2/2 half squares

[hushed wave]

Idk but people said it 14 but I am confused on how.

Perimeter.

[solid osprey]

!original

[devout snow]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[drifting mauve]

it's 7...

[hushed wave]

Only look at B.

[hushed wave]

!original

[devout snow]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[drifting mauve]

an easier method would be

making an x * x square

then subtracting or adding to fit the shape

that will give you an easier time with area and perimeter

[hushed wave]

I don't get how they got 14cm as the perimeter.

[drifting mauve]

btw because of how projections work, this will have perimeter equal to x*x square + twice the length of the small boxes

[drifting mauve]

dw the answer is 7

i think

[hushed wave]

5+1+3+2+1+1+1=14..

Someone said that.

I just don't get how.

[drifting mauve]

because of how projections and perimeters work, this will have perimeter equal to a rectangle of dimensions 5*

so the perimeter becomes 2 * 5x + 2 * 2x

= 7 * 2x

which is just 7

[drifting mauve]

the answer's wrong

if the value of x is 1/2

[devout snow]

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[green haven]

i'm preparing for an integration bee in which we get a "cheat sheet" of sorts, one of the facts on this sheet is (a), but I just kinda don't get it

What is y in this situation? where does this come from? is this like a special case of (b)? How can I use this for definite vs. indefinite integrals?

thanks to whoever helps me :)

[dense lynx]

$\int f^{-1}(y), \mathrm{d}y = \int x, \mathrm{d}\left(f(x)\right) = \int x, f'(x), \mathrm{d}x$

[woven radish]

Mqnic_

[dense lynx]

IBP: $\int x, f'(x), \mathrm{d}x = x, f(x) - \int f(x), \mathrm{d}x$

[woven radish]

Mqnic_

[dense lynx]

^ (a)

(b) observe such diagram for the geometric interpretation of (a) under suitable intervals

[green haven]

I see, do you have an example in which we can apply (a)?

[dense lynx]

i have literally never seen this used in my life

[green haven]

figured

[dense lynx]

and i have seen a fair number of integration bees

b is important, however

know how to recognize tricky sums related to of inverse functions

[green haven]

good to know

thx :)

[dense lynx]

yw

[green haven]

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[normal lagoon]

I know I should be using a theorems based on primes to answer these.

No clue where to start from... or which one is an easier answer.

@upper schooner Is it not based on primes? Or am I tripping cause this was given to us on the chapter about primes?

[upper schooner]

You don't need prime factorisation to show the statement true, at least

I'm presuming that maybe they might use that somewhere else, though it would be quite weird imo if they're expecting you to show it from prime factorisation

[normal lagoon]

I'll go down the chapter this is in and see if we don't need to. I'm actually kinda of confused this chapter not gonna lie.

[upper schooner]

Awwww

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[devout snow]

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[normal lagoon]

.reopen

[devout snow]

✅

[normal lagoon]

Still confused.

I should reopen this once I have gone down the chapter.

.close

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[compact hawk]

What's the probability of rolling a sum of 12 with 4 six-sided dices

[compact hawk]

I just gotten to a+x+b+z+c+y+d+q = 12 and minimum value is 1 for a,b,c,d so it's x+y+z+q = 8

So js using stars and bars I gotten 165 total combinations

But this includes the fact that a,b,c or d can be greater then 6 so the max value is uh 9

Idk how do I calculate how many like restricted combinations are there

I was thinking of doing js if a = 9 so b+c+d = 3 = 5C2 which is 30 but after calculating it it's wrong compare to the awnser sheet

[pseudo basin]

what are all these letters?

[devout snow]

@compact hawk Has your question been resolved?

[pseudo basin]

the count can be rawdogged like this

[crude spruce]

[pseudo basin]

probably easier than trying some stars and bars shit cause the corrections for upper bounds on dice scores might get quite nightmarish

[compact hawk]

💀

[pseudo basin]

also btw http://anydice.com/

[woven radish]

Results provided by WolframAlpha

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

Requested by kizzyyyyyyyyy

[compact hawk]

💀

[pseudo basin]

why skulls

both times

[compact hawk]

Cus

Why not

[pseudo basin]

💀

☠️

[compact hawk]

Also doesn't this take a long time

Like in the test I get 2 minutes per question

And this is the easy part of the test

[pseudo basin]

i didn't time myself

took me just about 2 mins in all

[compact hawk]

How would you even calculate it like this then 💀 cus I can't think of a way without spending like 5-10mins

[pseudo basin]

dunno

i am not the kind of person who knows how to speedrun tests

[compact hawk]

No like how did you like calculate this quickly

[pseudo basin]

which part

finding the patterns?

finding how many permutations each one gives?

adding up the permutation counts?

[compact hawk]

Yes

About the ss u sent

How did u calculate it quickly like that

[compact hawk]

Cus like for me I always js used stars and bars to calculate these types of questions so I got no idea how u did it like that quicjly

[pseudo basin]

so just to be clear: you understand the "pattern" column, but want to know how i did the "permutations" column specifically, correct?

in that case: in each pattern, i calculated the number of rearrangements of n objects with repetition

eg for the pattern 6321 all four dice are distinct so it's gonna give 4! permutations, but the pattern 5331 identifies the 2 threes so 4!/2! = 12 permutations, and 4422 identifies two pairs this way therefore 4!/(2!*2!) = 6

[compact hawk]

Oh alright

Ty

.close

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[wind tulip]

hi

[wind tulip]

i need help

hi i need help on this problem ive been working on: Consider the first 30 terms of the following integer sequence:

1,4,7,12,20,32,44,60,81,108,135,168,208,256,304,360,425,500,575,660,756,864,972,1092,1225,1372,1519,1680,1856,2048
I need to determine a pattern or formula for the
n-th term. I have already checked the OEIS database and attempted polynomial interpolation, but I was unsuccessful.
I found the pattern lies in the second difference, but its a periodic sequence plus a polynomial of somesort.

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[devout snow]

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[wind tulip]

.reopen

[devout snow]

✅

[wind tulip]

no

[devout snow]

@wind tulip Has your question been resolved?

[devout snow]

@wind tulip Has your question been resolved?

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[strong ocean]

I need to prove this using induction

[strong ocean]

I've don the base case

And now I'm proving this

But somehow the next step became this

(I'm looking at the answer book)

I don't understand how 1+(k+1)x became (1+x)(1+x)^k

wait no

[olive snow]

No its the left part that become this

[strong ocean]

wrong image

1

This

I don't understand how 1+(k+1)x became (1+x)(1+kx)

[olive snow]

Ok

See

Take your original inequality

And multiply by 1+x both sides

Don't flip inequality cuz 1+x >0

[strong ocean]

But why multiply 1+x on both sides?

[olive snow]

Cuz x >=0

[olive snow]

To have the lhs as you want

Now you need to show that rhs >= 1+(k+1)x

[strong ocean]

Yeah

But I didn't multiply the lhs?

I need to prove for k+1

so the k+1 didn't come from a multiplication of (1+x)

[olive snow]

It does for lhs here

[strong ocean]

OH

So in induction we don't substitue k for k+1

[olive snow]

Wdym

[strong ocean]

but we just want to transform the expression of k to something in the form of k+1?

[strong ocean]

Like I thought we need to replace all ks with k+1

[olive snow]

Oh yeah or either starting from k+1 assomption and use the fast that its true at n=k to prove your induction

[strong ocean]

ohhhhhh thanks

so we just transform the expression

no substituing right?

[olive snow]

Yeah transform what you have to what you want OR transform what you want with what you have to end something true

[strong ocean]

Okay! Thanks!

That clears a lot of stuff up <3

[olive snow]

You're welcome

[devout snow]

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[strong ocean]

.close

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[lavish folio]

james josh and judy are sight seeing in Egypt, James (A) is south of the great pyramid of giza, josh (C) is west of the pyramid, whilst judy (B) is in between them on the line AC. the angles of elevation from A, B and C to the top of the pyramid are 23 degrees 24 degrees and 25 degrees respectively, find the bearing of the great pyramid of giza from where judy is standing to the nearest degree

i was told by the person who gave me this question that the answer nice is ||North 3 degrees east|| however when i did it i got ||92.772|| so i think i added ||90 degrees|| somewhere

my working out was:
||let g be the base of the pyramid, let t be the top, let theta be the bearing from B to g, let h be the height of the pyramid aka gt tan24=h/GB GB=h/tan24 tan23=h/AG AG=h/tan23 tan25=h/CG CG=h/tan25 tan<CAG=CG/GA = (h/tan25)/(h/tan23) = tan(23)/tan(25) thus <CAG = arctan(tan23/tan25), angle <BGA = theta (alternate angle on parrel lines, B to its north and GA is north to south, hard to explain) sin<GBA/GA = sin<GAC/GB sin<GBA/(h/tan23) = sin<GAC/(h/tan24) tan23 x sin<GBA/h = tan24 x sin<GAC/h tan23 x sin<GBA = tan24 x sin<GAC sin<GBA=sin<GAC x tan24/tan23 <GBA = arcsin(sin<GAC x tan24/tan23) theta = 180-GAB-GBA (angles in triangle add up to 180) theta = 180 - (arctan(tan23/tan25)) - arcsin(sin(arctan(tan23/tan25)) x tan24/tan23) but that makes theta 92.7 when the answer is meant to be 3 degrees, i cant find my mistake T-T||

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@lavish folio Has your question been resolved?

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@lavish folio Has your question been resolved?

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[acoustic helm]

If vectors of some set A are a linear combination of vectors in B then B is necessarily found in A

[devout snow]

@light hazel Has your question been resolved?

[mystic scarab]

What do you mean? What two sets are you talking about?

But the first isn't a vector space

Or you mean the vector space spanned by (0,1) and (1, 0)?

[devout snow]

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✅

[acoustic helm]

I don’t think so. The closest thing would be that W = span{(1,1)} is a subspace of V = span({(1,0), (0,1)})

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[lone sphinx]

Could anyone guide me on this question? I'm not so sure how to appraoch 1.a

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[midnight totem]

well the most obvious method is to just find R and R'(in terms of m) and equate

but that will take too long

its supposed to be solved in under 5min

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[wooden cairn]

just looking at 'CE = EP'

does that mean the vectors are the same

or just their magnitudes?

[hollow pulsar]

just the magnitudes

if it was referring that the vectors were equal, then they wouldve added the arrows on CE and EP

[wooden cairn]

and them being on a straight line makes their directions the same right?

so overall they are the same vectors when u combine them two facts?

[hollow pulsar]

yes, as a matter of fact

u have a theorem stating if E is the midpoint of [CP] then vector CE = vector EP

[wooden cairn]

kl

thanks

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[rocky carbon]

I need help

[rocky carbon]

Make an integral that shows B - A

i got this

why is the answer

the one that is c and b

[winter torrent]

the given answer is this?

[rocky carbon]

no

the whole thing

[rocky carbon]

this

i wrote cb integral - the ab integral

but i dont understand

why there is a minus on cb

integral

[onyx cedar]

because its below the x axis

[rocky carbon]

oh

[winter torrent]

this will be a negative number

[rocky carbon]

so the area is negative?

bro how can area be negative

[winter torrent]

the area is positive (we think of areas as positive) so if we want to talk about the positive area we need to multiply that by -1

[rocky carbon]

this is trippy

[rocky carbon]

alr

thanks

so everytime its under the x axis

i put a minus

sign

[onyx cedar]

if u want the area ya

[winter torrent]

i mean it depends on what you want to do

,w integrate[-1, 1] (x^2 - 1)

[rocky carbon]

oh

[woven radish]

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[rocky carbon]

alr

thanks

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[restive river]

anyone help pls

<@&286206848099549185>

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[gleaming rose]

Yes

[restive river]

i got the answer tnku bro

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[urban spindle]

$\Omega \subset \mathbb{R}^n$ is an open set, $A, B$ differentiable functions,

$A : \Omega \rightarrow M_{mn}(\mathbb{R})$,
$B : \Omega \rightarrow M_{nk}(\mathbb{R})$,
$u : \Omega \rightarrow \mathbb{R}^n$.

$\forall x \in \Omega, V \in T_x\mathbb{R}^n$, we define $V(A), V(u)$ as the differential of A and u along the vector field V.

I try to see if $V(AB)=V(A)B(x) +A(x)V(B)$.

Need some help please. One of first things I'd like to understand is, how does $V(A)$ even look like?

I know if $V=\Sigma_{i=1}^nV_i\partial_i, u=(u_1(t), ... u_n(t))$, then $V(u)=\Sigma_{i=1}^nV_i\partial_i(u_i(t))$. How would it work on a matrix though?

[urban spindle]

HI

[woven radish]

Wild123

[urban spindle]

Formatted the question.

I think I have an idea. I'll reopen later if I get lost.

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[jaunty mantle]

Why don’t you just write V \in T\Omega

[urban spindle]

I didn't know the command 😄
Ohh I was copying it as I had it

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[obsidian plank]

y=3x^3-6 hello I am supposed to determine if this is one to one

[obsidian plank]

right now I had 3(x^3-2) am i on the right track

have*

[misty crest]

do you know what the definition of one-to-one is

[obsidian plank]

nah

[misty crest]

🤔

[obsidian plank]

I think It means if two numbers add up to the same number or something

[restive river]

every value input value only has one unique output value

no two input values have the same output value

[misty crest]

why are you doing homework before knowing the material

[misty crest]

no

a function f is one-to-one if f(a) = f(b) implies a = b

[restive river]

for example if
f(x) = x^2
f(-1) = 1
but f(1) = 1

Since two input values give rise to the SAME output value, the function is not 1 - 1

[obsidian plank]

cause I need to and this isn't in the posted notes

[misty crest]

doubt it

[obsidian plank]

I could send you the notes

[restive river]

you have
$$f(x) = 3x^3 - 6$$

[woven radish]

Edmund Cloudsley

[restive river]

set f(x) to a and b. If this implies that a = b, the function is 1 to 1

[misty crest]

3a^3 - 6 = 3b^3 - 6

use this to show a = b

thus establishing f is injective/one-to-one

[restive river]

https://www.youtube.com/watch?v=abB9rWbOdUw

you could also watch this video

to strengthen your understanding

[obsidian plank]

uhh kinda confused on the a to b thing is 3x^3 a and -6 b or ?

[misty crest]

bruh

[obsidian plank]

what

oh its the whole thing

got it

[misty crest]

read what we’re saying to understand it not to figure out what you should copy on your paper

[restive river]

$$f(a) = 3a^3 - 6 = f(b) = 3b^3 - 6$$

[woven radish]

Edmund Cloudsley

[restive river]

@obsidian plank as @misty crest mentioned earlier, please do go through the material first

it would be hard to solve this question if you don't know what a 1-1, many-1 and 1-many mapping is

[misty crest]

the last two have to be some red coat shit bro wtf is 1-many

[obsidian plank]

dude I am just trying to understand what you guys are saying bro what's the attitude for you can't expect me to know something the first time you say something and have me not ask questions

[restive river]

like the graph of y = +-\sqrt{x}

1 input value maps to many output values

,w plot y = +- \sqrt{x}

[misty crest]

so not a function?

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[restive river]

sorry I mean just a mapping

I am tripping wtf

[misty crest]

if an input value maps to multiple outputs then it’s not a function

[restive river]

not a function

yup yup

[misty crest]

you didn’t even read my definition

[misty crest]

^

do you understand what that definition entails

[obsidian plank]

yes I did I just didn't understand it

I understand it now though

[misty crest]

an equivalent formulation (the contrapositive) is if two input values a and b are such that a ≠ b then f(a) ≠ f(b) for a one-to-one function

[obsidian plank]

trust me dude I am trying to understand this if I wanted to just get it done I would just use chatgpt and cheat on this shit but im not

I think I got it now though

[misty crest]

essentially it just means distinct inputs yield distinct outputs

[obsidian plank]

thank you

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[south fiber]

how did the max become 3? wouldn't it just be e

[stone stump]

well its kind of self-referential

the thing you are trying to work out in the first place is the range of what it could be

[valid vector]

$\max\qty{e^z}$ for $0\le z\le 1$ is just $e^1=e$.

[woven radish]

mathisfun

[valid vector]

Since e^x is increasing for all real x.

[stone stump]

they also arent claiming that the max equals 3

they only claim that its bounded by 3

[south fiber]

so would i be able to out down 2.8 instead

[valid vector]

Sure. Anything greater than e (reasonably, of course)

[south fiber]

thank you so much bro

[stone stump]

the question is whether you know at this point what e is

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[spiral rampart]

Hi, this is a picture of the truss problem I have been working on. I need to solve forthe forces in the members listed in the top right (BF, AF, and FG). The thing that is tripping me up is that both the supports are pins and therefore have horizontal reactions. I cannot solve for these reactions no matter how hard I try. Any guidance would be much appreciated, thank you.

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[spiral rampart]

<@&286206848099549185>

Please

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[inner ibex]

why did they word it like this?

[inner ibex]

wat do they mean?

[topaz axle]

i can't even guess

just missing half the question

[thorn perch]

yeah

[inner ibex]

this is the example

[inner ibex]

ohh u are right

i just checked the pdf textbook

what does In mean?

[winged bluff]

Do you know about logarithms?

[summer summit]

ln is natural log, its the same as saying log base e, where e is euler's number.

some basic proofs are that $\ln(e^x)=x$, $\ln(e)=1$, and $\ln1=0$.

[wispy ice]

im here to save the day

[inner ibex]

hi

[woven radish]

Krish

[inner ibex]

um

the only time ive heard of eulers number is when u differentiate e^x u get e^x

[summer summit]

yes

[summer summit]

ln also is all based around eulers number.

[inner ibex]

is there a graph 4 it

[summer summit]

there is a graph

[acoustic leaf]

,w graph ln(x)

[woven radish]

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[summer summit]

$\ln$ tells you what power you need to raise $e$ to in order to get a certain number.

[woven radish]

Krish

[inner ibex]

ok thanks

im going to pray this isnt in the test

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[junior heron]

struggling on number 3, have i gone about it correctly?

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[wispy ice]

<@&286206848099549185>

[cinder nova]

when you did the u sub, you haven't substituted the du

[junior heron]

thank you

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[brisk ocean]

Ik its prolly basic, but which areas do we substruct. Is it area under the linear - area under the parabola or smthn else

[brisk ocean]

for part 2 btw

[wispy oyster]

the top curve - the bottom curve

on the domain

of interest

so the linear - the parabola, yes

but tbh, the order does not matter. if u get a negative answer, just absolute value it and u get the right answer

[brisk ocean]

though

i was thinking

the area under the linear given there is no parabola will just be infinity

@wispy oyster could u solve it as well since i need to double check my answers coz there is no solutions available

I have gotten 9/2

[supple trench]

she means area between two curves

!showwork

[devout snow]

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

[brisk ocean]

yeh sending now

[supple trench]

after u've finished ur work u, u can self-check by

1. wolfram alpha
2. other CAS, say https://integral-calculator.com

,w int x-4-(x^2-4x), x=1..4

[woven radish]

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[brisk ocean]

@supple trench would the theory behind be correct

[wispy oyster]

your working / answer is correct, yes

[brisk ocean]

ight

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[slender galleon]

I'm confused of ChatGPT's answer about the vector matrix multiplication. I run it on my ide and I got a different solution than ChatGPT told me. It's even the same code. How is that possible? 😟

[wraith horizon]

!noai

chatgpt is nonsense

it doesn't actually do the calculation, it just finds the sentence that looks the most likely to follow

[slender galleon]

So is it just like the problem with the number of r's in the word strawberry, where ChatGPT thinks that it's 2, instead of 3?

[wraith horizon]

yup

[slender galleon]

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[rain solar]

Hey, first a bit of context (sorry its so long). I am working on an implementation of the bin packing problem (you have x items of a certain size, how do you pack them optimally in the least amount of bins of size y) but a small tweak: there are multiple possible bin sizes.

I have an implementation which is working and I'm happy with for now, but I now want to add progress reporting to the calculation. The algorithm itself is a simple branch and bound for now. I had the idea I could a crude upper bound for the total amount of possible (or impossible) packings: (bin_sizes + items)! / (bin_sizes - 1)!. Reasoning behind it is: first item you have bin_sizes choices (you put the item in a new bin of one of those sizes), next item has bin_sizes + 1 choices (add them to the existing bin or create a new one)... and so on. I know this also includes impossible packings since it does not take into account capacity of the bins and it always assumes all choices up until now have been the creation of new bins.
Now I want to also keep track of the explored packings, since the ratio would then ideally be the progress of the calculation.
When I reach a leaf node, I have a complete packing: increment one to the explored space.
When I bound a branch I intend to also very aggressively update the explored space to counteract the upper bound which is too large. I update with (bin_sizes + bins_in_current_packing + items_left - 1)! / (bin_sizes + bins_in_current_packing)!

I'm now wondering if this is a valid way of tracking the progress. I actually know it isn't since my implementation of this fails (never reaches 1% progress, or my implementation is wrong). Intuitively it seems to me like it should work, if I envision the decision tree of the total packings, then my updates just prune the subtrees. But I might be wrong in that it gets all the subtrees, which is where my intuition ends lol... I'm kind of stuck now on how to fix this, or if there's a better way to track the progress.

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[rain solar]

Hey <@&286206848099549185> can anyone help me? 🙂

[rain solar]

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[deep relic]

Prove that Ax=lambda x eigenvalue is real if A=A*

\begin{gather*}
A^* = A  \rightarrow a_{i,j}=\overline a_{j,i} \\
Ax = \lambda x \\ 
A = \lambda x \cdot x^{-1} \\
A = \lambda I \rightarrow a_{i,j}=\lambda \in \mathbb{R} \\
\end{gather*}

is this correct?