help-43

[compact pewter]

@vale perch Has your question been resolved?

[vale perch]

Yup ur writing is clean

Got it

Now how we carry this forward

[kind viper]

carrying means replacing the ten 1's with one 10 like i showed

so instead of sixteen ones we will be left with six

so the ones digit is 6

and the tens will be one more than what they were, so from 5 up to 6

thus, 666

[vale perch]

so much clear now

[vale perch]

We considered ten 1s and add to tens place where five tens are already present so it becomes 6 and that becomes 60

And six 1's are left as it is in one's place

Hence 666

Tysm

How carrying forward works in subtraction and multiplication

[compact pewter]

@vale perch Has your question been resolved?

[kind viper]

for subtraction you don't carry

you borrow, as it's usually called

but it's the reverse of the stuff i said about addition

and instead of combining lower denominations into higher, you make change by breaking a denomination into ten of the next lower one

ie break a 1000 into ten 100s, or a 100 into ten 10s, or a 10 into ten 1s etc.

multiplication is much trickier to look at in terms of money, but the algorithm essentially consists of multiplication by single digits and then following that with addition (which i already explained)

[compact pewter]

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[quasi wind]

How do i even approach this question??

[carmine garden]

I'd start by realising that $(a+2b)^2=c^2$

[boreal girder]

What a wonderful world !

[quasi wind]

Ok

my doubt is

We can find the common point with this relation

so

All the lines pass through that point

so shouldnt the distance be zero (shortest)??

common point is A(-1,-2) right

[iron stag]

i think question is poorly worded
You get two families of lines, one passing through A(-1,-2) and another through (1,2)

and from that u can find maximum distance between two lines from the families

||so all the options would be correct, as the distances range from 0 to root(20)||

[quasi wind]

how can u find max distance?

is there a formula?

[quasi wind]

wait

oh yes

[iron stag]

no try to visualize it? first obviously the lines need to be parallel

now think of different possible parallel lines

[quasi wind]

if u expand it from (a^2-b^2)=(a+b)(a-b) u get two conditions

im jus dumb

[quasi wind]

yes if they are not parallel the shortest distance would be zero yes?

[iron stag]

yes

note that they said "can be" so all possible distances are correct

[quasi wind]

Wouldn't the distance between the points be the largest distance needed (since apart from one pair of line all the other lines will have shortest distance less than the distance)

so all of the options r correct

[iron stag]

yeah

[quasi wind]

ty gng

[iron stag]

[quasi wind]

https://tenor.com/view/praying-cat-please-thank-you-thanks-gif-16119732

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[unreal moon]

Hi just trying to find this mystery function.

Can someone tell me what that math is called or a link to where i can find resources

[kind viper]

,rcw

[unreal moon]

maybe just some subject related to the mystery function anything anyone?

[compact pewter]

@unreal moon Has your question been resolved?

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[trim path]

Could someone explain how to do this please

[trim path]

Please ping

[compact pewter]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
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5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[trim path]

I got the answer by manually counting since it's easy in this case

I wrote all subsets (16 including null) and then counted them

But I can't think of a formal solution

[iron stag]

suppose some A and B are the disjoint subsets of S that u want
now each element in S can go in A or B or neither

[trim path]

Yeah

[iron stag]

what do u get from that

[trim path]

So there'll be 3^n possibilities in total, and since they're asking unordered, it should be 3^n/2

[iron stag]

almost

u need to subtract 1 from 3^n. the case where both A and B are empty sets

[trim path]

Oh yeah

But that doesn't give the right answer

[iron stag]

(3^n - 1)/2

[trim path]

The answer key says (3^n+1)/2 tho

[iron stag]

[trim path]

Why the +1 😅

Wait aren't we subtracting the null case twice?

Nvm

[iron stag]

wait i think we dont need to subtract this case

[trim path]

Why not

[iron stag]

two null sets are disjoint and both subsets of S

mb

[trim path]

Why are they considered disjoint tho

[iron stag]

well their intersection is also a null set

[trim path]

Mm yeah

I get it

[iron stag]

ok so we are still missing one case

[trim path]

So (3^n-1)/2+1

[iron stag]

oh wait

[trim path]

No we need to count the null pair only once right

[iron stag]

nvm yeah correct

thats it

[trim path]

Yep i got it, thanks a lot

[iron stag]

[trim path]

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[young raft]

i do not get this definition at all

[young raft]

why are there two S??

he never defines what S is (the regular one)

the only one it can be i assume is the S here and below in the page but i only see them in context of dS

never on its own

[regal storm]

They're all the same S. it's poorly written

[young raft]

what about here

[regal storm]

Yes

The same

A set $S$ is a smooth surface if for any $P$ you can find an open ball $N$ around $P$ such taht inside the ball there is a smooth function $g(x,y,z)$ that has a nonzero gradient near $P$ and near $P$ the surface is the set of points where $g(x,y,z)$ is nonzero

[boreal girder]

Oliver

[young raft]

where g is non zero?

it say $N \cap S = { Q \in N : g(Q)=0 }$

[boreal girder]

artemetra

[young raft]

should it be ≠ instead?

[regal storm]

Shit. my bad. I meant $g(x,y,z)$ should be zero. the gradient should be nonzero. @young raft

[boreal girder]

Oliver

[regal storm]

but yea in all images the $\mathcal{S}=S$

[boreal girder]

Oliver

[young raft]

this is the part that confuses me

how can there be a neighbourhood where the function is only zero but the gradient isn't

wouldn't it be a constant

[regal storm]

Becuase we're only thinking of surfaces, not all of space

Take $g(x,y,z)=x^2+y^2+z^2+c$ for any constant $c$. The surface defined by $g(x,y,z)=0$ is a sphere with radius $\sqrt{c}$. Which works. Also $\nabla (g) = (2x,2y,2z)$ which is not 0 anywhere on the surface - it's 0 only at the origin

[boreal girder]

Oliver

[young raft]

z^2 - c btw

hmmmmm

oh i think i get it

[calm trout]

the condition that grad g != 0 is the condition that the surface has a tangent plane at each point

this ensures that S locally is actually a surface, rather than some singular looking thing like the union of two planes

[young raft]

okay i get it now

[young raft]

i am still unsure why they use two S's and keep them clearly distinct

[calm trout]

its definitely a typo

don't worry about it

[young raft]

$\mathcal{S}$ seems to be the surface of integration while $dS$ is an area element

[boreal girder]

artemetra

[calm trout]

that is unrelated

[young raft]

ok

thank you!!!

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[heady hawk]

someone help this was in my school olympiad

i guessed the answer in the test it was right some how

[bleak dock]

search up the formula for the inradius of a right triangle

then you can set up an equation for sides ST, TR and also don't forget Pythagoras

so you will know what sides ST, TR are

[heady hawk]

oh

[bleak dock]

yes then this is the next step

[compact pewter]

@heady hawk Has your question been resolved?

[bleak dock]

the idea is that you let the radius of the big circle = r

[heady hawk]

i clicked x by accident

[bleak dock]

oh rly?

[heady hawk]

oh

[heady hawk]

i think ill get it

im solving it

ty

[bleak dock]

and then using similar triangles you can get an expression for the red length in terms of x

npnp!!

like side-chasing okok I'll leave you to it!

[heady hawk]

tyy

[bleak dock]

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[compact pewter]

@deft tangle Has your question been resolved?

[bleak dock]

well if $S = 0 \cdot 1 + 1 \cdot nC1 + \cdots + n \cdot nCn$

[deft tangle]

Yeah

[boreal girder]

south

[deft tangle]

Sun r=0 to n
rncr

[bleak dock]

then you also have $S = 0 \cdot nCn + 1 \cdot nC(n - 1) + \cdots n \cdot nC0$

[boreal girder]

south

[bleak dock]

so you can add the two sums together

[deft tangle]

But how can I add them?

[bleak dock]

0 * 1 + n * 1

• (1 * nC1 + (n - 1) * nC(n - 1))
• ...

and nC1 = nC(n - 1) ofc

[deft tangle]

Can you type this step in latex?

[bleak dock]

$0 \cdot nC0 + n \cdot nCn = 0 \cdot nC0 + n \cdot nC0 = n \cdot nC0$

[boreal girder]

south

[bleak dock]

$1 \cdot nC1 + (n - 1) \cdot nC(n - 1) = 1 \cdot nC1 + (n - 1) \cdot nC1 = (1 + n - 1) \cdot nC1$

[boreal girder]

south

[deft tangle]

You are adding one from left end and other from right end

[bleak dock]

yep!

[deft tangle]

Now i understood

[bleak dock]

great so now you should be able to continue

bear in mind the AM = S/2^n

[deft tangle]

So n will taken out

n(nc1+nc2+....ncn)=n(2^n-n

[compact pewter]

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[deft tangle]

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[pine osprey]

Got anything ?

[copper lion]

-2(sinx + cos2x) = 0

cant solve it

[potent berry]

i dont think this is even right

,w derivative 2cosx -sin(2x)

[boreal girder]

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[potent berry]

okay

guess im stupid

[potent berry]

well

from here either $-2=0$ or $\sin(x) + \cos(2x)=0$

[boreal girder]

Percy

[pine osprey]

Cos(2x) = ... in terms of sin²

= 1 - 2sin²(x)

Now its 2nd degree polynomial in sin(x)

u = sin(x) if you want to visualise it

[copper lion]

ok so -2sin^2x + sinx + 1 = 0

[pine osprey]

Seems like

Yeah

[copper lion]

so -2x^2 + x + 1
(x+2)(x-1) = 0

if i let x sin

[granite forge]

take sin(x) = u not x

[potent berry]

$u \not x$?

[boreal girder]

Percy

[potent berry]

im assuming you mean $u_0 x$?

[boreal girder]

Percy

[copper lion]

you get what i mean bruh

[eternal pulsar]

The second step does not directly follow.

Factorize it again.

[granite forge]

no

i meant to say take sin(x) = u

[compact pewter]

@copper lion Has your question been resolved?

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[slate sequoia]

can someone help me with this

[slate sequoia]

just 4 and 5 i guess

but 4 definitely

[pseudo rapids]

use sum from n = 1 to infinty of x^n = 1/(1-x)

!nogpt

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Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[atomic ledge]

Lol

[trim marsh]

Yo

Is this still open

[potent berry]

!occupied, not for you

[compact pewter]

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).

[trim marsh]

Okay

I just found the question interesting

[compact pewter]

@slate sequoia Has your question been resolved?

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[trim marsh]

I got the infinite sum as 25

[glacial mortar]

could you share your work?

[slate sequoia]

.reopen

please guys

[slate sequoia]

wait can i dm u

[trim marsh]

Oh my bad

Should i send it here

I didn't derive the n^2/2^n summation because it was kinda lengthy and i found it online

[acoustic stump]

i got it as 2/27

not sure if i did it correct

[trim marsh]

Can you share your work

We will check each other's

[acoustic stump]

i dont have phone to send my work but what i did was

just like how you calculate sum of an AGP

[trim marsh]

But this is not a simple agp right?

[acoustic stump]

i converted into one

i remember doing similar question before

[trim marsh]

Can you send me the steps or is it too lengthy

[acoustic stump]

write the series S = -1/2 + 2^3/4 - 3^3/8 +......

divide by 2 and shift one term to right

S/2 = -1/4 +2^3/8 -3^3/16 + ....

now add them

3S/2 = -1/2 + 7/4 -19/8 + 37/16 - ......

do the same step again

3S/4 = -1/4 + 7/8 -19/16 +....

shift one term and add

9S/4 = -1/2 + 6/4 -12/8 + 18/16 - .....

add 1/2 both sides to get

9S/4 + 1/2 = 6/4 -12/8 +18/16

now you can see an AGP forming

@trim marsh

@slate sequoia@glacial mortar

[glacial mortar]

yeahh you are going correct @acoustic stump

[acoustic stump]

the final S should come out to be 2/27

[glacial mortar]

uhh yeah it might be, not sure

[compact pewter]

@trim marsh Has your question been resolved?

[compact pewter]

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[scenic holly]

Need help here i think i messed up rounding ill show work below

[scenic holly]

P(x=0) c (10,0)=1 p^0=1

(0.71)^8 0.0646 * .71^2 = 0.0646 * .0541 ~ 0.0326

P (X=1) C(10,1) p^1 = .29
q^9 = (0.71)^9 = (0.71) ^8 *.71 ~ 0.0459
P (x=1) = 10 .29 *.0.0459 ~ 10 * 0.0133 -= .1330

0.0326+0.1330+0.2445= 0.4101

[compact pewter]

@scenic holly Has your question been resolved?

[scenic holly]

<@&286206848099549185>

an1 :?

[scenic holly]

^

bump if someone sees

[kind crane]

!noping

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[scenic holly]

apologizes

[kind crane]

,w binomialcdf(10,0.29,3)

[boreal girder]

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[kind crane]

,calc 0.0325524 + 0.377346

[boreal girder]

Result:

0.4098984

[kind crane]

yea rounding error

keep all digits and don't round until the very end

[scenic holly]

Preciate 💯 it was correct

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[quartz yoke]

@floral fjord

[floral fjord]

lol

took u so long

@quartz yoke ok

[compact pewter]

@quartz yoke Has your question been resolved?

[small mason]

writing log_e

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[stark ridge]

A and B throw a pair of dice one after another. A wins if the sum on the dice is 5. B wins if the sum on the dice is 8. Find the probability of A winning given that A starts first

[stark ridge]

for A to win sum can be 1+4;2+3;3+2;4+1
for B to win sum can be 2+6;3+5;4+4;5+3;6+2
so P(A)= 4/36
and P(B)= 5/36
also P(A')=32/36
and P(B')=31/36

what we are looking for is
= P(A)+P(A')P(B')P(A)+P(A')P(B')P(A')P(B')P(A)....

right?

[lone quartz]

I think I have done this one before. Jee?

[lone quartz]

Yes

[stark ridge]

yeah

[lone quartz]

Figured

[stark ridge]

ok nevermind i figured what my mistake was

i wrote 3 instead of 36 in P(A) in the final sum

got it, thanks

have a nice day

[lone quartz]

You too

[stark ridge]

.close

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[primal ledge]

oh

[compact pewter]

Please don't occupy multiple help channels.

[primal ledge]

.close

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[fast frigate]

hello can someone explain why the integral value is like this?

[fast frigate]

my working is like this

[floral fjord]

(-1)^n is constant right?

[primal ledge]

no it is not a constant

[torpid pier]

In the Integral it

Is

Not in the sigma

[primal ledge]

ohhh

but generally it is not a constant

i wanted to say that

[kind viper]

don't drop the brackets

[floral fjord]

my bad

[fast frigate]

ohh wait so (-1)^n is a constant in the integral so we take it out first and then integrate the rest?

okk thanks

.close

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[keen glacier]

how to find the curve of intersection between the plane z = 0 and the cylinder x^2 + y^2 = 4

[azure vault]

you want the intersection curve of those two surfaces, you will then have both equalities verified

so z = 0 and x^2 + y^2 = 4

[keen glacier]

how do u get the curve of intersection without graphing

[azure vault]

Well let's think about it

z = 0 means that the freedom of movement in z is removed

so it's kinda like we removed that dimension

from there, what does x^2+y^2 = 4 represent if there were no "z dimension"?

[keen glacier]

a circle

[azure vault]

exactly

[keen glacier]

but we can't solve it directly

we have to visualize it in our head

[azure vault]

?

Didn't I tell you how you should visualize it in your head?

you look at your cylinder

and you remove the freedom of the z dimension

it's like you squashed your cylinder in 2D

to where you cut it with the xy plane

[keen glacier]

i know but we can't get it from an equation

one is a function of z, the other is a cylinder

i can't set them "equal" to each other

[azure vault]

where did I mention equations in this

[keen glacier]

I know you didn't mention any equation

[azure vault]

so apply that

[keen glacier]

I was just saying that there's no way to find it using an equation

[azure vault]

wdym

do you want to find it using equations or not?

[keen glacier]

if u have z = 4-y, and z = 0, u can set them equal to each other and find the intersection

here, u can't do that

[keen glacier]

I do

[azure vault]

you can tho

z = 0

and x,y independent from that equation

and x^2+y^2 = 4

z independent from that equation

in that case

we can do cylindrical coordinates for example

(x,y,z) = (rcos(t),rsin(t),z)

->

z = 0

r = 2

there's not much else to think about

and so if you want the formula for the path that describes your curve

f(t) = (path for a circle for x,y ; z=0)

so f(t) = (2cos(t),2sin(t),0)

and that's it

[keen glacier]

ok thanks

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[drowsy tusk]

Why can I only divide top and bottom by a polynomial when the limit is of the form 0/0

[kind viper]

the reason you can factor out (x-2), and then cancel it out, is because num and denom are both polynomials and have x=2 as a root

[drowsy tusk]

ah right ok

makes sense

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[astral cipher]

can anyone help with physics help or no

[proven latch]

what level?

[humble mortar]

!da2a

[compact pewter]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[proven latch]

sorry

[astral cipher]

ok np

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[opal stirrup]

ho do i calc this without calculator? r = 1000 X 0.125 X 1/12

[potent berry]

can you write 0.125 as a fraction

[opal stirrup]

its percentage

[subtle zealot]

Personally, I would do it (1000*0.125)/12

[short ferry]

calculate 0.125 * 1000 first

keep in mind that you're just multiplying by 1000, so just shifting the decimal point

[opal stirrup]

yeah i get 125

then i divide 125 by 0.08?

no i mean 125 x 0.08

[subtle zealot]

using one significant digit is very inaccurate, and it would add extra steps to calculate 1/12 by hand then multiply by hand

you can put it together in one step by dividing 125 by 1/12

[opal stirrup]

how do i calc that on paper

[subtle zealot]

whoops i meant dividing 125 by 12

do you remember long division?

[opal stirrup]

oh yea

ty

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[lament token]

it was a tricks video for matrices where he mentioned if we put r and n as 1 we get the determinant value as zero which i understood how , but i couldnt understand how that would also mean that the summation of all determinants would be 0

[kind viper]

that sounds quite suspicious, can you link the video

[kind crane]

entire determinant where r would go from 1 to n would also be zero
do you mean sum of the determinants?

[lament token]

https://www.youtube.com/watch?v=HIVZT-_E4ok

[lament token]

yess

🫠 hello?

[lone quartz]

Are you sure it's sum of determinants and not sum of matrix and then determinant?

[lament token]

yes its jus the sum pf determinants

[lone quartz]

Okayyy

Let me check that for higher numbers like 2 or 3

[kind viper]

ok let me watch this

@lament token it looks like the guy just arbitrarily decides to see what happens if n were 1 and the summation had only 1 term

in the name of speedrunning, not math

[lone quartz]

Okay I checked, and it does come 0 for all of it. It's like you take a sequence of sum but the sequence is just one term and that term is 0.

[kind viper]

$

if we abandon speedrun strats, i think we can have $$\vmqty{r-1&n&6 \ (r-1)^2 & 2n^2 & 4n-2 \ (r-1)^3 & 3n^2 & 3n^2-3n} = n(r-1) \vmqty{1&1&6 \ r-1 & 2n & 4n-2 \ (r-1)^2 & 3n & 3n^2-3n}$$

[boreal girder]

Ann

[kind viper]

does that help us any?

[lament token]

Ohh ikay

[kind viper]

at least the first row is now constant

[torpid pier]

Math on go ahh 😔

[kind viper]

maybe subtract 6*col2 from col3

that might cook

[torpid pier]

Ek week pehel hi dekha tha ye lmao

[lament token]

Okay so that's mentioned wrong in the video

[kind viper]

but still i guess we have to place trust in the question that there even exists a correct answer option

[torpid pier]

Ye tabhi kaam karega agar options mai constants hain

Nahi toh cooked

[lone quartz]

We can just apply summation formulas for a n, n^2 and n^3

Natural numbers

[kind viper]

$$n(r-1) \vmqty{1&1&6 \ r-1 & 2n & 4n-2 \ (r-1)^2 & 3n & 3n^2-3n} = n(r-1) \vmqty{1&1&0 \ r-1 & 2n & -8n-2 \ (r-1)^2 & 3n & 3n^2-21n}$$

[boreal girder]

Ann

[kind viper]

,rccw

[lament token]

Okay then do u jus multiply them or can we simply further

[lament token]

Ohhhh

[lone quartz]

Well the first and last column are same soo

[torpid pier]

It's like saying ki sin^2x+cos^2x ki value batao

Would only work substituting the value

Agar koi identity hai

[lament token]

😭also I can't seem to understand why we use n(n-1) here instead of n(n+1)

[lone quartz]

Cause you have r-1 and not r

Summation for natural no. starts from 1, here it will start from 0 if you put r=1

[kind viper]

i don't know how to proceed further without going through some rather unpleasant bivariate polynomial simplification

[lament token]

Ohhhh okayy got it thankyou

[lament token]

🥲 right so i guess ill go ahead with the summation method

[lone quartz]

The determinant will be 0

[kind viper]

you mean the dirty trick

it is a trick. it is trickery.

[lament token]

When i use the summation method i understood how when we get two columns we get it zero but i didn't understand how here u called two columns common

[lone quartz]

Let them use it lol, it would help them good in exams (if you are giving jee)

[kind viper]

,w det{{r-1, n, 6},{(r-1)^2,2n^2,4n-2},{(r-1)^3,3n^2,3n^2-3n}}

[lament token]

No mean

[boreal girder]

Results provided by WolframAlpha

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

Requested by ann_dec

[kind viper]

oh that is so ugly.

[lament token]

😭gosh

[torpid pier]

It is a mcqs based exam wouldn't it be time saving if we just substitute some values (assuming the options are constants)

[kind viper]

sure

but again it is MCQ exam trickery.

[torpid pier]

plus that question was never meant to be solved using conventional methods

[lone quartz]

Agreed

[kind viper]

yes, it was meant to be solved using trickery.

ok i am done tooting this horn sorry

[torpid pier]

You get 2mins per question

Jus sub and move on

[lament token]

Yes yes understood these two are the same

[lone quartz]

So the determinant will always be zero

[potent berry]

whats going on

[lament token]

😅yea i know that part i actually meant it for the one Ann sent above mine , but thanks

[lone quartz]

Oh ok

[potent berry]

what even is the question.

[lament token]

🤡

[lament token]

.

[potent berry]

that's not a question.

i mean the orig ques

[lament token]

Yea so i shouldn't use the a=n= 1 part as it doesn't always work?

[lament token]

Its on the board

[potent berry]

okay.

[lone quartz]

No, substitution is a great trick to use, and it works. Just check whether all options are constant.

[potent berry]

i dont like the word 'trick'.

[lone quartz]

Why? "Logic" is better?

[torpid pier]

Advance ki prep kar rahi hai toh solution dekh agar mains ka kar rahi hai then you are wasting time

[lament token]

Okay but like as shown when we did a and n as 1 we got delta 1 as zero

But its a summation from 1 to n right

So when we add the parts like delta 2+ delta 3.....+delta n how do i know its still zero

[lament token]

Going to 12th 🤡9 months hai - what do i say

[potent berry]

kar 'rahi' directed to who

[torpid pier]

@lament token

Her name sounds feminine

[lament token]

🤡🙏

[torpid pier]

Dekhle solution

[potent berry]

you're gonna get banned soon

[lament token]

Koi batayega

[fair frigate]

you dont know that, you just know its not any of the other options

[lament token]

Ohhhh

[fair frigate]

which is why substitution doesnt even work when its not mcq

[torpid pier]

chill bruh

[potent berry]

No. Try not to assume pronouns.

[fair frigate]

it could be some complex polynomial in n you just dont know but if u put values and see it might match with only 1 option

[torpid pier]

Vehencho 😭🙏🏻

[lament token]

Okayyy

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[sleek sapphire]

hm so how do i apply this to the problem

the only thing i can say is that "when x is less than 1 it converges"

[eternal pulsar]

What would your r be?

[sleek sapphire]

x

i dont know what x is though

how do i communicate that in interval notation

[kind viper]

no, your r isn't x

[sleek sapphire]

huh how

[kind viper]

look at the series more closely

it is not $\sum x^n$ and it is also not $\sum \text{(constant)} \cdot x^n$

[boreal girder]

Ann

[sleek sapphire]

oh i was confusing it with my a

my a is x

my r is 1/3

[kind viper]

wrong again

[sleek sapphire]

or, (1/3)^n

wat

x^n/3^n

[kind viper]

the series is $\sum_{n=0}^{\infty} \paren{\frac{x}{3}}^n$; if it were as you say and $a=x$ and $r=\frac{1}{3}$, the series would have been $\sum_{n=0}^{\infty} \frac{x}{3^n}$.

[boreal girder]

Ann

[kind viper]

you see that those two series i wrote aren't the same at all, yes?

[sleek sapphire]

i see

i cant split the series either

[kind viper]

you do not need to do any splitting or other such surgery

[sleek sapphire]

so series x^n * series 1/3^n

[kind viper]

no

no

this is so far from the truth it does not even hold for finite summations

[sleek sapphire]

i genuinely have no clue then

[kind viper]

in the series $\sum_{n=0}^{\infty} \paren{\frac{x}{3}}^n$, the common ratio is $\frac{x}{3}$

[boreal girder]

Ann

[kind viper]

and the first term is just 1.

[sleek sapphire]

i didnt consider that because how can i evaluate it when i dont know my x

1/1-x/3

[kind viper]

no, you are still overthinking it

in math sometimes you do not immediately know the "value" of every single letter involved

and you should make yourself OK with leaving an answer in terms of one or more letters when it is appropriate to do so (and in this case it is).

x, right now, is just x.

the expression $\frac{x}{3}$ is not something that needs to be exorcized by ``evaluation''. it just is.

[boreal girder]

Ann

[sleek sapphire]

okay so how do i find out its divergence/convergence

with that in mind

it diverges when x is this and converges when x is that?

[kind viper]

well, yes, that's roughly what your conclusion would look like

now,

[kind viper]

thinking back to this

the condition under which a geometric series is convergent is |r| < 1.

here, the role of r is played by x/3.

write down the inequality in x that describes the convergence of your series.

do NOT attempt to solve, simplify, or otherwise manipulate it until i tell you otherwise.

[sleek sapphire]

hmmm
so it converges when |x / 3| < 1

[potent berry]

oh lord

what'd she say about not attempting to simplify

better

[sleek sapphire]

i just wrote down the inequality

i didnt isolate x

[potent berry]

no, you wrote down $\frac{x}3<1$.

[boreal girder]

Percy

[sleek sapphire]

is that not an inequality

[kind viper]

x/3 < 1 would be an inequality but it'd be the wrong one

|x/3| < 1 is correct

[sleek sapphire]

this is what i think of when i hear "inequality"

[kind viper]

those are properties of inequalities

[sleek sapphire]

right. this is less than that

[kind viper]

ie they are like rules of the game by which you can transition from one inequality to another

[potent berry]

an inequality is a relation between two values that is not an equality.

[kind viper]

anyway, now that we know the inequality we're looking for is $\absv{\frac{x}{3}} < 1$, now you need to solve it.

[boreal girder]

Ann

[potent berry]

okay yeah too handwavy fair enough

[sleek sapphire]

hmm i havent done inequalites in ages
do i do -x/3 < x/3 < 1 and then like, simplify something

something like that ik its wrong

[potent berry]

geq is also ineq

anyway

[potent berry]

no

[sleek sapphire]

ok....

[kind viper]

then you have to review those tbh]

[sleek sapphire]

yea probably but not rn. last time i did them was 2 years ago in pre calc algebra
can i do |x/3| < 1
|x| < 3

[kind viper]

like we can fumble our way to a solution here kinda but it is just objective fact that if you have not worked w/ inequalities in ages then you have to revisit them as a topic in their own right

can i do |x/3| < 3
bad notation

[sleek sapphire]

oops

i meant < 1

[kind viper]

but yes, you can multiply both sides by 3 and get |x| < 3.

that is a good step in fact.

[sleek sapphire]

well that tells me x is always between -3,3

[kind viper]

mmm don't like that comma.

but yes.

[sleek sapphire]

(-3,3)

[kind viper]

sure, that's the interval notation for it.

[sleek sapphire]

i never care about the importance of notation i probably should 😭

[kind viper]

do you understand why you need to use () and [] ?

[kind viper]

yes you really really REALLY should.

in math, notation is important.

[sleek sapphire]

yes i know notations

[sleek sapphire]

so the conclusion i should come to is "x is between -3 and 3, and only converges to the sum when |x| < 3)

diverges otherwise

[kind viper]

the conclusion you should come to is that the interval of convergence is, as you mentioned before, (-3, 3).

[sleek sapphire]

yea but if i wanna explain what its interval of convergence means here

[potent berry]

i dont like that

[sleek sapphire]

is that how i would word it

[potent berry]

x is between -3 and 3 exclusive

[eternal pulsar]

(Redundancy.)

[sleek sapphire]

hm okay

well thanks for all the help and corrections

.close

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[glacial slate]

This is an unacceptable thing to say. Particularly to helpers who are just volunteering their time to help other users. If I catch you saying anything like this to a helper again you will be banned. Further, for moderation purposes you need to use english.

Once the mute expires do not make this same mistake again

[compact pewter]

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[keen glacier]

How to draw a closed surface on 3D desmos?

I know how to draw a volume or an open surface. How do we draw a closed surface? I'm trying to draw a capped cylinder for example.

Here is my attempt. I'm getting an open surface, not a closed one.

[rigid perch]

add the caps separately

[keen glacier]

I want to draw the net overall surface.

how do I do that

i know I can easily write the equations for the caps on another line

can i use braces to draw the overall surface here

[kind crane]

this isn't a desmos server

[compact pewter]

@keen glacier Has your question been resolved?

[compact pewter]

@keen glacier Has your question been resolved?

[paper rune]

x^2 + y^2 <= 5? if thats what youre asking

[mortal scroll]

wdym net?

this server might be helpful https://discord.gg/unofficial-desmos-655972529030037504

[compact pewter]

@keen glacier Has your question been resolved?

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[gray stirrup]

How do I do this question?

Let me know if I need to show more information

[pliant beacon]

the ratios of the area is the ratio of the number

[gray stirrup]

So u times 9(pi) by 378699/2599000
then u divide the answer of that by pi
Then square root it after?

@pliant beacon

[pliant beacon]

sounds good

[gray stirrup]

🙂

Thanks

[pliant beacon]

what do you get

[gray stirrup]

3.53

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[pliant beacon]

seems a little bit off

[gray stirrup]

I redid it and got 3.6cm

[pliant beacon]

3.62?

[gray stirrup]

Yeah

[pliant beacon]

cool

[gray stirrup]

👍

Made typo

Maybe

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[rough raft]

Hello I need help with this Poisson probability question. I used excel like the question asked and it’s just not correct can someone please let me know what I did wrong?

[compact pewter]

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

[rough raft]

This is my work

[wheat pasture]

I don't know, perhaps stupid question, but why 134 instead of 1.34?

[rough raft]

I thought since n is 100

I thought the 1.34 defects is for one vehicle

But ima try that rn

[wheat pasture]

Fairs, I'm not knowlegable enough (been a bit long since I've needed to do stats, and I forgot it ) but that sounds a little bit sus

[rough raft]

💔

<@&286206848099549185> hi I’m really stuck on this question

[torpid spear]

why would in excel the cumulative = 1?

[rough raft]

It’s just there as placement I think because 0 means false not cumulative

[torpid spear]

did you calculate the solutions to the other problems with excel's POISSON.DIST?

[rough raft]

No I used =NORM.S.DIST

[torpid spear]

honestly not sure why your mean is 134

because it says 100 random cars sampled?

[rough raft]

My friend has the same question but different numbers

Okay I just found out what I did wrong ugh 💔

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[proud coral]

why is this line incorrect?

[junior vault]

maybe because you put a random dot after the top 0

formatting issue

[proud coral]

that random dot is a multiplication sign

[junior vault]

but above that youre subtracting the tan

i dont see multiplication

[proud coral]

truee

this is still incorrect

[junior vault]

idk if youd call plugging in numbers algebra

[proud coral]

ohhh evaluation

now I see

thanks you

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[pure spear]

Hello

[pure spear]

I don't understand where the extra p is coming from

[kind viper]

rho, not p

also, from context it looks like they are going for cylindrical coordinates

does it say that anywhere earlier in the text

[pure spear]

Yes

[kind viper]

ok then yeah the jacobian is $\dd{x}\dd{y}\dd{z} = \rho \dd{\rho} \dd{\phi} \dd{z}$

[boreal girder]

Ann

[kind viper]

it is basically the same as the polar jacobian, but with a dz tacked on

[pure spear]

Is the dz the same as the original cartesian dz?

Am I supposed to use this?

[kind viper]

no, this is spherical

you're looking for CYLINDRICAL coords

[kind viper]

can think of it that way, why not.

[kind viper]

yeah

[pure spear]

So it's always r or rho when you use jacobian for clyindrical?

[kind viper]

...yes

[pure spear]

Okay thanks!

[kind viper]

it is basically the same as the polar jacobian, but with a dz tacked on

[pure spear]

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[stark terrace]

help

i took logarithm on both side and double diffrentiated it but i dont see where the e^e comes from

[latent ridge]

it asks you for the result of multiplying the second derivative with e^e

[stark terrace]

i was trying to prove it 💀

(dumbness)^2

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[stark terrace]

thanks

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[shell cairn]

I was thinking that since x* is an interior minimizer i can create a ball around it b(x*, \varepsilon) such that every feasible direction from x* has the following property gradient f(x*) times direction > 0

Now whenever x* is not an interior point and is on the boundry of omega, we cant take into account all fesible directions

[kind viper]

do we even know that f is differentiable here

i think you're overcomping it cause we do not need anything like a gradient or any of that fancy stuff here -- we do not even know that Omega is a subset of anything with more structure than a topological space

you are right in that an open ball centered at x* will come into play somehow

[shell cairn]

alright ill think about it thank you

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[pale ocean]

How do I solve for c here?

[pine osprey]

develop the b(x+c) as bx + bc and use the relations between graphs and expression of function

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Please don't occupy multiple help channels.

[pine osprey]

no we don't want your discord server

<@&268886789983436800>

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[trim oak]

could someone explain how the got the step func equation in B

[timid lodge]

start by graphing 2u(t). then you notice that between 1 and 2, you have a height of two.

you want a height of 0 at that point, so if you subtract off 2u(t-1), you get what you want.

now you look at 2u(t) - 2u(t-1) and notice that the height between 2 and 3 is 0, but you want it to be 2, so you add 2u(t-2).

and finally you look at 2u(t) - 2u(t-1) + 2u(t-2) and notice that the height after 3 is 2, but you want it to be 4, so you add 2u(t-3)

so you go one little interval at a time and add or subtract the step function (shifted to that point) that gets you the value you want

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[trail berry]

Working Laplace transforms, and I just need to know if I'm starting this problem correctly:

[trail berry]

Never seen one that doesn't simply start with F(s).. can I factor that F(s) out and just.. actually, that should be division..

At any rate, is that allowed?

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[outer halo]

The inverse Laplace transform is a linear operator.

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[proper swan]

i need help with "factoring by grouping"

[proper swan]

i chose the pairs shown on the picture but im not sure what to do next

[old mica]

a good thing to do is to write it as (ab + a) + (-ub - u) c:

literally grouping the pairs by parentheses

now you can focus on each pair individually

and hopefully you‘ve seen how to factor each pair

[proper swan]

oh sorry it's not a u its A 4

[proper swan]

im not sure

[old mica]

ah, my mistake

but it won’t change anything

[proper swan]

okkk ty

il put the thingys

[old mica]

hm, if I gave you the expression 8x + 8, could you find factor a greatest common factor from it?

[proper swan]

8

[old mica]

yeah!

so you wanna do the same thing to each of your pairs

[proper swan]

ohh ok but what next c:

[old mica]

what do you have now? :o

[proper swan]

a and 4

[old mica]

right

after you find the GCF, you can factor it out of both terms in the expression

for example, 8x + 8 = 8(x + 1)

what we’re using here is just the distributive law

[old mica]

you wanna do this for both your pairs

[proper swan]

ok il try!

[old mica]

[proper swan]

i meant -4 whoops

[proper swan]

wait if i gather -4 from -4 do i get 1 like usual or -1?

do i keep the minus with b or

[proper swan]

what do i do in this case

[old mica]

you get 1!

[proper swan]

ty what about b

[old mica]

you can leave it c:

[old mica]

have you written down the factored expression yet?

[proper swan]

mhm

[old mica]

awesome!

you’re almost done!

now look closely at your two terms

what’s the greatest common factor they share?

[proper swan]

b?

[old mica]

unfortunately not

almost though!!

[proper swan]

b+1?

[old mica]

yes!!

[proper swan]

:D

[old mica]

so we can do the same thing one more time

and factor b + 1 out of both terms

can you write down what we get after we do that?

[proper swan]

i dont know how

[old mica]

here, let’s do a sneaky trick

you have a(b + 1) - 4(b + 1) right now, yes?

[proper swan]

mhm

[old mica]

we’re going to say “let b + 1 = c”

this is just renaming b + 1

[proper swan]

mhm

[old mica]

now our expression looks like ac - 4c

now, can you factor this?

same thing as you did earlier for each pair

[proper swan]

il try

[old mica]

write the expression here when you do c:

you did an excellent job earlier with this 🤍

[proper swan]

here (ignore the bottom thing

[proper swan]

ty

[old mica]

yes!

wonderful!

now, just remember that b + 1 = c, and we get…?

(just replace c with b + 1)

and write down the final expression c:

[proper swan]

ok!

[old mica]

excellent!

congratulations, you’ve solved the problem

[proper swan]

yay thank you so much

:).

[proper swan]

good byee thanks again for the help

!.close

[old mica]

have a great day!

[proper swan]

ty

!close

. close

[old mica]

[proper swan]

.close

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[proper swan]

lol

[old mica]

there you go c:

[proper swan]

yay

c:

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[proper swan]

i need to figure out A1 with taletes theorem but i dont know how

[quartz yoke]

proportionality

[proper swan]

i dont know what to do

i don't understand thr theorem

[quartz yoke]

since two parallel lines divide them

if a line is drawn parallel to one side of a triangle, intersecting the other two sides, then it divides those sides in the same rati

ratio

[quartz yoke]

the ratio of B to A will be same as the ratio of B1 to A1

[proper swan]

ok what next

[quartz yoke]

...

have you tried anything?

[proper swan]

i don't understand what I'm supposed to try

[quartz yoke]

you need to find A1

right?

[proper swan]

yeah

[quartz yoke]

by the basic proportionality theorem

[quartz yoke]

.

this holds true

5:3 = 6:A1

now you just solve it

you need an example?

@proper swan

[proper swan]

im trying to write it down il show you what i get to make sure it's like correct

[quartz yoke]

sure

[proper swan]

im stuck on that

i dont know the next step

[quartz yoke]

EXACTLY

dude

write x instead of that question mark

[proper swan]

ok

[quartz yoke]

multiply by x on both sides

yo?

[proper swan]

sorry i was writing it down

[quartz yoke]

its okie

[proper swan]

like this?

[quartz yoke]

nope

lemme send

[proper swan]

oh

ok

ty

[quartz yoke]

understood?

you just solve for x

[proper swan]

why did 5/3 turn into x

[quartz yoke]

i just multiplied by x on both sides

okay forget those

solve this

6+x=8

what is x

[proper swan]

8-6

2

[quartz yoke]

good

addition becomes subration when its taken to the other side of equality

and vice versa

right?

[proper swan]

yeah

[quartz yoke]

similarly

multiplication becomes division

solve this

5x=10

x=?

[proper swan]

10:5=2

[quartz yoke]

good

10/5=2

solve this

20/x = 4

[proper swan]

20•4

80

[quartz yoke]

nope

20:x = 4

this?

[proper swan]

20•4

[quartz yoke]

nahh