help-36

[forest pilot]

yea

[storm haven]

That's why I told u to use <

U randomly changed it to >

Just look at this dude.. I told u to use blah < y < blah

😭

[forest pilot]

OH

[storm haven]

Done bro?

[forest pilot]

yeah XDD

[storm haven]

Yeash

🤦

[forest pilot]

yeash indeed

[storm haven]

Ok you can type ".close"

[forest pilot]

sry for annoying ill prob be back in an hour or so ;p

.close

[final saddle]

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[tranquil pine]

Gaurav placed a light bulb at a point O on the ceiling and directly below it placed a table.
He cuts a polygon, say a quadrilateral PQRS, from a plane cardboard and place this
cardboard parallel to the ground between the lighted bulb and the table. Then a shadow
of PQRS is cast on the table as P'Q'R'S'. Quadrilateral P'Q'R'S' is an enlargement of the
quadrilateral PQRS with scale factor 1: 3. Given that PQ = 2.5 cm,
QR 3.5 cm. RS 3.4 cm and PS = 3.1 cm;
<P = 115°, <Q = 95°, <R = 65° and <S = 85°.
Based on the given information, solve the following questions:
Q1. The length of R'S' is:
a. 3.4 cm
b. 10.2 cm
c. 6.8 cm
d. 9.5 cm
Q 2. The ratio of sides P'Q' and Q'R' is:
a. 5:7
b. 7:5
c. 7:2
d. 2:7
Q3. The measurement of <Q' is:
a. 115°
b. 95°
c. 65°
d. 85°
Q4. The sum of the lengths Q'R' and P'S' is:
a. 12.3 cm
b. 6.7 cm
c. 19.8 cm
d. 9 cm
Q5. The sum of angles of quadrilateral P'Q'R'S' is:
a. 180°
b. 270°
c. 300°
d. 360°

[tired walrus]

!status

[final saddle]

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.
4. I got an answer and would like my work checked.
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

[tranquil pine]

1

[tired walrus]

ok, have you made a sketch of PQRS?

it doesn't even need to be to-scale but it can be a nice way to organize the info you've got for it

[tired walrus]

oh is that the picture the book gave you

[tranquil pine]

this is the image attached to the question

[tired walrus]

ok

[tranquil pine]

yes

[tired walrus]

right, that's good but i'd still like you to make a sketch of just PQRS alone

for organization purposes

and label all of its sides and angles

[tranquil pine]

idk how do i draw it

[tired walrus]

you should also recall some things about similar shapes (similar triangles primarily, but a lot of things carry over to arbitrary shapes -- such as preservation of length ratios as well as angles)

[tranquil pine]

that is my doubt

i know similarity

[tired walrus]

just draw a vaguely lopsided quadrilateral

[tranquil pine]

ok

[tired walrus]

MS Paint has a text tool

[tranquil pine]

my bad

[tired walrus]

anyway you should also label all sides and angles here

cause they're all known

and then refer to the diagram instead of having to dig through the text every single time

[tranquil pine]

ok

[tired walrus]

ok yeah good

so

the shadow P'Q'R'S' is an enlargement of PQRS with scale factor 3

do you know what this means for the side lengths

[tranquil pine]

I know basic proportuonality theorem and similarity of triangles

but basic proportionality doesnt apply here

This belongs to the case study of triangles chapter for me

[tired walrus]

hold on

what do you mean by "basic proportionality theorem" and why is it inapplicable here

[tranquil pine]

Thales

theorem

[tired walrus]

well you can apply it if you think three-dimensionally

in triangle OR'S' for example

[tranquil pine]

a line parellel to one side of triangles intersecting the other two sides , the other two sides are divideed in the same ration

*ratio

[tired walrus]

but really all you need to know is

P'Q'R'S' is an enlargement of PQRS with scale factor 3

which means that all sides of P'Q'R'S' are 3 times longer than the same-named sides of PQRS

while all of its angles are just the same as those of PQRS

capisce?

[tranquil pine]

idont undersran

d

[tired walrus]

mmm

which one are you?
(A) "WTF did you even say?"
(B) "I get what you said but why's it true?"
(C) "Secret third option."

[tranquil pine]

hmm

i got it

is the first one 10.2

R'S' =3 x 3.4 = 10.2

is it correcr

t

[tired walrus]

missing equals sign right after R'S' but yes

[tranquil pine]

Ty

sm

.close

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[marsh glacier]

what is the sum of n terms in an A G P and how to derive it?

[marsh glacier]

agp is arithmetico geometric progession

[high mural]

1. get 1 term
2. get n terms
3. sum

[marsh glacier]

bruh

theres gotta be a faster way

[tired walrus]

there's always going to be some crunch

[haughty prairie]

If √x + 1/√x=a then find the value of x^2 + 1/x^2

[terse crypt]

open ur channel

square both sides and u're done

[marsh glacier]

(a^2-2)^2-2

ezpz

[dapper hull]

there are specific formulas for it and you could derive them

[marsh glacier]

got the answer guys so nvm

.close

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Channel closed

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[serene panther]

Write it as S

Then multiply by the common factor of the gp(say r)
And then subtract it from S

[marsh glacier]

and shift subtract right?

[haughty prairie]

.

[serene panther]

Yeah

[marsh glacier]

i gave u the answer

[dapper hull]

!occupied

[final saddle]

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).

[serene panther]

Open your own channel pls

[marsh glacier]

its ok

[serene panther]

So you understood how to solve agp sums right?

[marsh glacier]

yes

[serene panther]

Alright

[marsh glacier]

i closed

he can ask his question now

@haughty prairie

[serene panther]

I didn’t see anybody help you

But it’s fine

[marsh glacier]

btw answer was (a^2-2)^2-2

@haughty prairie

.close

[haughty prairie]

The books answer is a^4-4a^2+2

[haughty prairie]

Can you give me the steps

[jagged flare]

its not really how it works, you closed it but its still in your name, wait abit and the bot will automatically send the channel back to the available list

[marsh glacier]

if u expand u get the same answer

[marsh glacier]

spam square both sides

do it twice

[jagged flare]

!nosols sorry :(

[final saddle]

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

[marsh glacier]

the answer plops out

[haughty prairie]

Let me try

[marsh glacier]

?

what is !nosols

[tired walrus]

it's a command which makes the bot print this canned message

a reminder of server policy

[marsh glacier]

who was the msg for?

[tired walrus]

also do not encourage channel hogging

[tired walrus]

you, ostensibly

[marsh glacier]

not sure what u mean by that😅

[tired walrus]

this channel has your name on it

[marsh glacier]

im very new here

[tired walrus]

it is yours and you should not give it up for anybody else

other people who need help can and should open their own channels

[marsh glacier]

but i was done with my queation

[tired walrus]

then type .close

[desert mantle]

the channel is about to close anyway

[marsh glacier]

i did

[tired walrus]

oh is it

ok

still

[final saddle]

Available help channel!

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• Show your work, and if possible, explain where you are stuck.
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[gentle zephyr]

G(e1) = (0,0,0)
G(e2) = (2,3,0)
G(e3) = (-1,0,0)

[tiny gorge]

probably start by finding G o G right?

[gentle zephyr]

,w nullspace ({{0,2,-1},{0,3,0},{0,0,0}}^2)

[soft zealot]

Results provided by WolframAlpha

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

Requested by renatoxbz

[gentle zephyr]

Im(T) = <(1,0,0),(0,0,1)>

[gentle zephyr]

is it possible Nu(T o T) = Nu(T)?

we have to make choices here

if dim(Im(T)) = 2, then dim(Nu(T)) = 1
because dim(R3) = dim(Im(T)) + dim(Nu(T))

so its impossible Nu(T o T) = Nu(T)

T(x1) = (1,0,0)
T(x2) = (0,0,1)
T(x3) = (0,0,0)

T(T(x3)) = T(0) = 0
x3 € Nu(T o T)

@tiny gorge

we need to define T on a basis of his domain

[tiny gorge]

the wolfram result doesn't look right

[gentle zephyr]

?

did you check with sagemath?

[tiny gorge]

oh wait no it's good, i mistyped something

[gentle zephyr]

ok

T(x1) = (1,0,0)
T(x2) = (0,0,1)
T(0,0,1) = (0,0,0)
T(T(x2)) = T(0,0,1) = (0,0,0)
x2 € Nu(T o T)

[tiny gorge]

why is this impossible? consider a diagonal T with one nonzero entry on the diagonal and two zeros, it will satisfy this

[gentle zephyr]

Im(T) = Nu(GoG)

dim(Nu(GoG)) = 2 = dim(Im(T))

dim(Im(T)) = 2 <=> dim(Nu(T)) = 1

T : R3 -> R3

[tiny gorge]

ah right i have to keep scrolling up to remember what's equal to what, lemme paste it again

[gentle zephyr]

ok

[tiny gorge]

ok well you know what Im(T) has to be, you could just do the naive thing and set the first two columns of T to basis vectors for Im(T) and set the third column to 0 and see if that satisfies all the requirements

[gentle zephyr]

yes but I need to give the preimage of the image vectors aswell, I mean like this

T(x1) = (1,0,0)
T(x2) = (0,0,1)
T(0,0,1) = (0,0,0)
x2 € Nu(T o T)

but I think x1 = e1, x2 = e2, x3 = e3 satisfies all the requirements

T(1 0,0) = (1,0,0)
T(0,1,0) = (0,0,1)
T(0,0,1) = (0,0,0)

[tiny gorge]

yea, T^2 will send e1 to itself and it will send both e2 and e3 to 0

[gentle zephyr]

ok

hopefully this is good enough

we can also find the general form. T(x,1,x2,x3) = ...

but for my purposes is not needed

we can leave it defined like this, is fine

.close

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[blissful zenith]

help

[blissful zenith]

Here’s the question

My solution is here, have I done rigorously for this one

[tiny gorge]

should be nb + a in the denominator at the end?

[blissful zenith]

Yes, I think, the (74) has a typo I think should’ve been 1-t^b in the denominator

It was a question I did a couple days ago

[final saddle]

@blissful zenith Has your question been resolved?

[final saddle]

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[spare moth]

can someone explain this its asking for the period for this, i dont understand why dont we add the values of the periods of both what does it mean take the lcm

[tiny gorge]

well what if it were cos(x) + cos(x)?

the period of cos is 2pi

the period of 2cos is also 2pi

not 4pi

so adding periods is not the answer

[spare moth]

so like for this it will be 2pi/4 = π/2 for both so the ans is π/2 instead of adding them?

[stuck juniper]

Even powers

[stuck juniper]

No

Simplify the function first

I mean yes

U can literally see

The period is pi/2

[spare moth]

i dont understand what u mean

if u dont mind explaininr

[stuck juniper]

Identities

[spare moth]

but isnt this a period question

[stuck juniper]

Yes

But it helps when u just have one function

Instead of two

Atleast i solve it like that

[magic coyote]

Simplify power 4?

[stuck juniper]

Yes

Its very simple

Sin^2x + cos^2x whole square

Then u will just have sin2x

[magic coyote]

Ah oops

[spare moth]

isnt that =1^2 = 1

[stuck juniper]

Expand it

[spare moth]

?

[stuck juniper]

U can literally see it

[spare moth]

see what...

[severe verge]

what is sinx*cosx

[stuck juniper]

[spare moth]

😞

.close

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• Show your work, and if possible, explain where you are stuck.
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[cedar obsidian]

If A is the sum of the digits of 4444^4444 and B is the sum of the digits of A then find the sum of the digits of B?

[cedar obsidian]

What is the apporach here?

[desert mantle]

find an upper bound on the result

then mod 9

[cedar obsidian]

first of all why do you think of upper bound ( what does it mean though?) and why do you think of mod 9?

[stuck juniper]

Binomial

Just divide by 9

[desert mantle]

sum of digits is clearly mod 9

and upper bound cause I hope that the end result has to be single digit already

[onyx peak]

i wouldnt be so sure of that

[cedar obsidian]

4+4+4+4 = 16

[desert mantle]

well, hope

[stuck juniper]

💀

[cedar obsidian]

16/9

[desert mantle]

I suppose more likely low two digits or something

[cedar obsidian]

16 == 0 mod 9

when?

[stuck juniper]

There in a upper bound

[cedar obsidian]

Why are you bothering me? I'm just not able to understand anything.

how 16 = 0 mod 9

[desert mantle]

its not

who said that it is

[cedar obsidian]

sum of digits

is mod 9

you said

[desert mantle]

sum of digits is clearly connected to mod 9

[stuck juniper]

Bro this is some imo level problem istg

[cedar obsidian]

maybe of 40 years back

[desert mantle]

thats just the basic divisibility rule

[cedar obsidian]

it is basic then please tell me how to think on this

[stuck juniper]

I think 7

Should be it

[cedar obsidian]

already been 15 min on this and i have to solve enitre exercise consisting of 34 questons

yes asnwer is 7

[stuck juniper]

Is this some jee worksheet

[onyx peak]

A number is divisible by 9 iff its digit sum is divisible by 9

do you know that rule?

[cedar obsidian]

yes

[onyx peak]

That rule is why you should connect digit sums to mod 9

more generally, we can even say that the digit sum of n === n (mod 9)

[cedar obsidian]

how do i think it is mod 9, i didnt get thought process of mod 9

[desert mantle]

we told you

there is nothing to "get"

sum of digits means mod 9 is relevant

thats literally the thought process

[cedar obsidian]

,calc 4444^4444

[soft zealot]

Result:

Infinity

[scarlet sequoia]

computers can't usually compute stuff like that

[onyx peak]

it has like 16,000 digits

[scarlet sequoia]

maybe WA can, but it's not how it's intended to be solved

The reason why we're talking about mod 9 being relevant

is because X has the same remainder mod 9 as the sum of its digits

[cedar obsidian]

where did X come from?

[scarlet sequoia]

X is ANY number (non negative integer)

so this can be applied to X = A

X = B

etc...

[cedar obsidian]

got this X point

[scarlet sequoia]

so

if by chance we prove that the sum of the digits of B, call it C, is small enough

[cedar obsidian]

yes

[scarlet sequoia]

maybe small enough to be less than 9

then do we have a lot of choices for the value of C?

[cedar obsidian]

lets see

[cedar obsidian]

i am still digesting this

so sad

[cedar obsidian]

for this did you think that what power of 4444 is, that is close to smallest multiple of 9

[scarlet sequoia]

uh I didn't quite get that

[stuck juniper]

This is def olympiad question

[scarlet sequoia]

yes this is an IMO problem

[cedar obsidian]

is this is IMO probelm

really

[onyx peak]

really

[cedar obsidian]

Now I’m feeling even more dizzy.

😦

[scarlet sequoia]

do you at least know how to compute remainders mod 9

[cedar obsidian]

yessss

[scarlet sequoia]

ok

try computing 4444^4444 mod 9 first

and we'll explain more from there

[cedar obsidian]

i am doing

but still there is why mod 9 in back of the mind

[scarlet sequoia]

this is an IMO problem, so there's at least a small part of it that relies on intuition

as we said, 9 has the property that "the sum of a digits of a number has the same remainder mod 9 as the original number"

that's because 10^m mod 9 = 1

so S(n) ≡ n mod 9 (S is the sum of digits function)

so it's clear that A ≡ 4444^4444 mod 9

and B ≡ A mod 9

and C ≡ B mod 9

so now, we hope that by doing the "sum of digits" operations a lot of times, the number gets small enough

to have only one option remaining for what C could be

(if you know what C mod 9 is and that C <= 9...)

[cedar obsidian]

I have done proof of divisibility of 9 before altho this is not standard proof but yes

[desert mantle]

that is the standard proof

(except only 5 digits of course)

[cedar obsidian]

i heard that standard proof is by letting a_n, a_n-1, a_n-2 ........ a_2, a_1 , a_0 this an...a_0 is all under the bar as they are digits

[scarlet sequoia]

you write the number "a_n....a_0" as the sum of 10^n * a_n

so under sum notation everything is fine

[desert mantle]

anyway, your proof also shows that the number abcde and a+b+c+d+e are the same mod 9

which is the point of the entire thing

[cedar obsidian]

tell me what the person who created this question intended, what exactly do they expect a student to think in order to solve it?

[desert mantle]

mod 9

in a competition setting it couldnt be more telegraphed that this is about mod 9

they might aswell have written mod 9 in the question

[cedar obsidian]

Please explain how to approach thinking in mod 9, the explanation above was unclear and felt more like a riddle than a proper solution.

im sorry for repeating the same thing

[sonic jacinth]

idk who's the author of this q anyway
it's ||IMO 1975 p4 of 6||

[cedar obsidian]

but i cant

[scarlet sequoia]

I'll try again

[cedar obsidian]

Ask me guiding questions like “What would happen if this were the case?”—so I can gradually work my way toward the answer.

[desert mantle]

mod 9 isnt something you reach

[scarlet sequoia]

Three part reasoning:

1. Sum of digits has same remainder as original number mod 9. That is a property people should think of immediately when they hear "sum of digits" (competition setting).
   Applying it to our problem, this means sum of digits of sum of digits of ... of 4444^4444 has same remainder as 4444^4444 mod 9

[desert mantle]

mod 9 is something you have to know immediately

sum of digits being connected to mod 9 is something that is expected to be known

[scarlet sequoia]

2. Sum of digits of a number makes it smaller, MUCH smaller. So taking sum of digits 3 times, the number is expected to be VERY small

[desert mantle]

well maybe back in 1975 it wasnt cause this question is too easy otherwise I feel. but nowadays it is

[sonic jacinth]

I try it, and after that I noticed I just done same thing as the original key

Bruh

[cedar obsidian]

idk how you feel it too easy, do i not know the concepts?

[desert mantle]

clearly

[scarlet sequoia]

3. The number we seek, call it C, we know its remainder mod 9 and it's SMALL. It's then expected that C is THE remainder mod 9

[cedar obsidian]

How am I supposed to just dream up the idea that the sum of the digits is divisible by mod 9? If I were a mathematician, maybe it would come by intuition right?

[desert mantle]

this isnt a question that is asked in a vacuum

people are expected to know things

[cedar obsidian]

nvm ill accept sum of digits is divisivle by mod 9

[desert mantle]

but I mean, a smart third grader can manage to find that sum of digits is connected to divisibility by 9

sure they cant prove it

but finding the connection is easy

[scarlet sequoia]

if you enter a math competition in 7th grade, it would put you at a disadvantage to not know Pythagoras's theorem (trying to come up with an analogy)

[cedar obsidian]

i am in 11th grade

but its if

[pale niche]

why are you trying imo problems if you don't know sum of digits' mod 9 property

or was this an assignment

[desert mantle]

being in 11th grade and doing a competition problem but not knowing that the sum of digits is connected to mod 9 is just genuinely a skill issue. that is expected knowledge

you need to know a lot of stuff for competition problems

thats just how it works

[cedar obsidian]

I have covered (took lectures from crash course) these topics in number theory which is required for this type of competition, then should I not be able to think sum of digits is connected mod 9 ?

Diophantine equation -02 is remaining only

[desert mantle]

first lesson, divisibility tests

[cedar obsidian]

But I know that I didn't practice alot with easier problems in the particular topic

[desert mantle]

you didnt make the connection
sum of digits
divisibility rule 9
mod 9

[severe verge]

mod 9 is like the most basic thing you think about when talking about sum of digits

[desert mantle]

I just dont know what to tell you

[scarlet sequoia]

well not practicing "easy problems" is what led you to not know this easy thing

[cedar obsidian]

I did even this type of basic problems aswell

,rotate

[cedar obsidian]

with this basic knowledge i should be able to connect mod 9

what to do now?

[desert mantle]

more practice

[cedar obsidian]

yes, not going to give up definitely

[final saddle]

@cedar obsidian Has your question been resolved?

[final saddle]

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[flint ibex]

weird help

[flint ibex]

but

does anyone do a maths degree

i wanted to ask

how hard is it

if its like this

im gonna do econ

like

how good do u have to be at maths

to do maths at degree level

[shadow aspen]

You should know high school math. You do not need to be "talented" if you are willing to put in some work

[flint ibex]

u defo do

i want to be the best

im in a british ciriculum

so i need like

top uni

then top degree

wtv

like rn

[shadow aspen]

well that is a completely different hope than just getting a math degree

[flint ibex]

im getting A*s to maths and fm

but thats AS

not like final yr

[flint ibex]

yh

like

best maths degree

or at least up there

with one of the best

[trim jasper]

if you're getting A*A* at AS level maths & fm and intend to continue them to A level then you can definitely do it

[flint ibex]

really

bro these proof questions look insane

apparentaly they make up majority of the stuff

[tranquil pine]

most of the courses you will take in undergrad will be proof based yes

assuming you do a math degree

[flint ibex]

yh

it looks insane

is the jump

that big

from A level to uni

in terms of workload

and

difficuitly

[tranquil pine]

in the UK yeah I'd say so

[flint ibex]

oh finished

might just do econ+maths

[final saddle]

@flint ibex Has your question been resolved?

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[steady wraith]

this is probably dumb but

[steady wraith]

what the heck does degree 3 mean

[tranquil pine]

Hello 👋

[steady wraith]

hey

[trim jasper]

The highest power of x is x^3

[vital crag]

a + bx + cx^2 + dx^3

[steady wraith]

https://tenor.com/view/of-course-gif-21373815

ok got it

wait so repeated root with cubics

let me think about this

[tranquil pine]

Yaa

[steady wraith]

for quadratics usually its pretty easy to work out repeated roots

but its not quadratic

actually it gives me major clues

[tranquil pine]

Polynomial

[steady wraith]

a + b + c + d = 1
a + 2b + 4c + 8d = 2

[trim jasper]

Have you heard of factor theorem?

[steady wraith]

and we also know when x is 0 is a root

so a = 0

[tranquil pine]

Yaa me to

[steady wraith]

so its actually just

x(a+bx+cx^2) = 0

cx^2

sorry

[severe canyon]

P(x) = ax(x - M)²

[steady wraith]

yeah

i was about to write that

well not the a part thanks for reminding me to write a

[severe canyon]

Because the repeated root is at M

And a is just a scaling coefficient

[steady wraith]

okay so a(1-m)^2 = 1
2a(2-m)^2 = 2

[severe canyon]

x is the factor (x - 0), since 0 is a root (because you're given P(0) = 0)

[steady wraith]

a(2-m)^2 = a(1-m)^2 🤔

[severe canyon]

Yep

[steady wraith]

huh

[steady wraith]

(2-m)^2 = (1-m)^2

[severe canyon]

Exactly

[steady wraith]

the a can cancel out it cant be 0

[severe canyon]

(because a ≠ 0 clearly)

[severe canyon]

Perfect

[steady wraith]

4 - 4m + m^2 = 1 - 2m + m^2

3 - 2m = 0?

did i make a mistake somewhere or not yet

because i dont usually type my working out

[severe canyon]

All good

Even if you could have used a shorter way

[steady wraith]

wait hold up

b^2 > 4ac

but like

-2 is b

3 is a

3 is c

uhhhhh

[severe canyon]

Oh wait

[steady wraith]

somethings wrong

[severe canyon]

On the left it should be m², not 4m²

[steady wraith]

oh yeah

thats why

[severe canyon]

Yeah sorry, I didn't see it at first

[steady wraith]

oh

that makes things so simple ow then

now*

m = 3/2

the answer is (e)

[severe canyon]

Awesome, good job 💪

[final saddle]

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[sacred night]

I feel like I’m missing something here

[sacred night]

Mb

I was missing the area under the curve

Aigh thx

.close

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[blissful zenith]

Help

[blissful zenith]

I am taking multi variable calculus soon

I wonder if there is a neat proof for green and stokes theorem in measure theoretic approach

I know it’s a bit weird but I passed measure theory adapting folland’s and I am curious I asked chat gpt it gives me some steps I barely understand

I really want a human understandable scratch for proving stokes theorem in measure language

[vital crag]

!nogpt

[final saddle]

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).

[blissful zenith]

I was asking it but what it generated passed my understanding significantly

Well, stokes theorem is a theorem of integrals and I think there must be a proof for it in measure language but I don’t know how to convert boundary conditions to integration, and I just passed measure theory I feel quite overpowered

I am thinking using radon nikodym derivative though, however radon nikodym derivative can’t change the integrating domain E to partial E

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[civic furnace]

how do i do this

[final saddle]

Please don't occupy multiple help channels.

[elfin osprey]

I love learning math

[civic furnace]

.close

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[hushed fable]

I was wondering if anyone could help me figure out what I did wrong

[formal trail]

what makes you believe your answer is wrong?

[sage meteor]

just differentiate the answer if you think your answer is wrong

[hushed fable]

I’m looking online to find a solution and everything else is letting u = sec

[formal trail]

there can be potentially more than one way of integrating it

[hushed fable]

or does that not matter?

Oh Yeha that’s what I thought

[ripe jewel]

$\int \sin x \cos x , \mathrm dx$ has at least 5 ways to solve it

[soft zealot]

gfauxpas

[ripe jewel]

as an example

[hushed fable]

Yeha there was a lot of ways to do when I was solving itttt

Ok tank u everyone

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[spare moth]

i only know that sin inverse a + cos inverse b = pi/2 when a^2 + b^2 = 1 but how would that work here when theyre both theta

[loud sundial]

i only know that sin inverse a + cos inverse b = pi/2 when a^2 + b^2 = 1

,w true or false arcsin(1/sqrt3)+arccos(sqrt(2/3))=pi/2

[soft zealot]

Results provided by WolframAlpha

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[loud sundial]

It's actually a well known fact that $\arcsin x+\arccos x=\frac{\pi}{2}$ such that the expression is defined

[soft zealot]

Civil Service Pigeon

[loud sundial]

I'll leave that to you to justify

so you just need to figure out what x are in the domain

[spare moth]

sorry im confused i dont get wht ure trying to say with this

[loud sundial]

Counterexample.

oh wait wolfram didn't even say if it's true or false

,w arcsin(1/sqrt3)+arccos(sqrt(2/3))

[soft zealot]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by civil_service_pigeon

[loud sundial]

,w pi/2

[soft zealot]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by civil_service_pigeon

[spare moth]

english not my first language so i dont know what ure trying to say if u can say it in simpler terms.. 😅

[loud sundial]

Counterexample: An example of when a general statement is false

So I'm giving an example of

i only know that sin inverse a + cos inverse b = pi/2 when a^2 + b^2 = 1
being false

[spare moth]

ohh wait i was using the wrong identity its sin^-1 a + sin^-1 b = π/2 for a^2 + b^2 = 1

i still dont get how to solve this

[loud sundial]

It's actually a well known fact that $\arcsin x+\arccos x=\frac{\pi}{2}$ for all $x$ such that the expression is defined

[soft zealot]

Civil Service Pigeon

[loud sundial]

I just blanked and forgot to write a fragment of the sentence

here you go

[spare moth]

.close

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[shrewd bobcat]

I know how to handle the bottom part (54m^3), but how do I find the volume of the top part?

[shrewd bobcat]

Is the radius 1.5?

[pliant shore]

indeed!

[shrewd bobcat]

Okay! I think I can do this now.

How did I do?

I got 64.

Wait...

Okay, I fixed it. I got 118.

Did I make any mathematical errors?

... I'll take that as a no and input the answer.

... ah, it wasn't correct.

I'll close this.

.close

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[pine fractal]

How do I find all the real zeroes of this function aside from plugging in every factor of 84 and solving it
without it taking an absurdly long time? I have done rational root theorem and found the factors of 84 , but I wish to know if there are simpler, less absurd ways of finding the real zeroes including the rational ones

[tired walrus]

well you can start small right

like check 1, then -1, then 2, then -2, etc

when you hit a root, knock the degree down by one

itll only be absurd if there are no rational roots at all

[pine fractal]

I mean absurd by just meaninglessly long and tedious

.close

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[novel forum]

I bet python (The programming language) has some stuff to help, its used in data science, check out things like numpy

idk its good for a lot of number stuff

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[high canyon]

What do I further in question 6

[high canyon]

I think the answer will be 4 but in answer key it's 3

[steel talon]

,rccw

[steel talon]

Good luck

[high canyon]

Why isnt the answer 4 ? From 0 to 2pi each of the values of sin(2theta) will be repeated twice right ?

[amber holly]

The extraneous solution probably comes from the step where you squared

[high canyon]

But i did nothing mathematically wrong ?? What is the correct method then ?

[amber holly]

There are certain types of steps that cause extraneous solutions, squaring both sides is one of them. I can’t think of a better approach here, so the best you can do is verify which of the four values are actual solutions by plugging them into the original equation

[high canyon]

Ohkk tysm

Uh but here wont we have to go into inverse functions of -2/5, 2/3 how am I supposed to know that without a calculator..

[amber holly]

I don’t think it’s necessary, it is enough to determine the sign of (sin(theta) + cos(theta))/(sin(theta)cos(theta))

The reason squaring is an issue is that what you get from x = y is the same as squaring x = -y, meaning you end up with solutions for both of the equations

So the extraneous ones will just make the one side of the original equation have a different sign

[high canyon]

Ohh but then how can the answer be a odd number , like 3

[amber holly]

Good question, let’s find out together actually

[amber holly]

So what I would do is rewrite sin(theta) + cos(theta) as sqrt(2)sin(theta + pi/4)

The sign of sin(theta + pi/4) is easy to determine now as long as we can compare theta with 3pi/4

Okay so when sin(2theta) = 2/3, 2theta must be less than pi and theta is less than pi/2

Uh, wait, looks like we need to consider both values from sin(2theta) = 2/3 here

So for one of them 2theta between 0 and pi/4, the other is between 3pi/4 and pi

In other words, theta between 0 and pi/8 and between 3pi/8 and pi/2

In the first case sin(theta + pi/4) will be positive

[high canyon]

Uhh, Waitt i dont understand this step

[amber holly]

sin is an increasing function from 0 to pi/2 and 2/3 < 1/sqrt2 = sin(pi/4)

That means arcsin(2/3) < pi/4

[high canyon]

Okk ty got it

[amber holly]

And in the second case theta + pi/4 < 3pi/4 < pi, so that’s still positive

So both values that came from sin(2theta) = 2/3 are valid

The other case can be analyzed similarly

I suppose one of the values will make sin(theta + pi/4) obtain the wrong sign there

[high canyon]

Okk tysm, I will do the other one myself sorry for taking up soo much time

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@high canyon Has your question been resolved?

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[cedar obsidian]

can someone tell the proof of this prime number property

p|a^n => p|a => p^n|a^n

[cedar obsidian]

where n is Natural number

[sinful fiber]

whats the prime factorization of a^n in terms of a

[cedar obsidian]

aaa*a......n times

Oops

[severe verge]

prime factorization

[sinful fiber]

like write it out in p1^c1 * p2^c2 * p3^c3

[final saddle]

@cedar obsidian Has your question been resolved?

[desert mantle]

euclids lemma for the first step

[cedar obsidian]

Hmm

[tired walrus]

the second step is straightforward from definitions

[lime zodiac]

if p|a then a=pk so a^n = p^n k^n

[desert mantle]

!nosols

[final saddle]

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

[lime zodiac]

oh sorry i didn’t notice giving full solutions

[desert mantle]

you can still edit/delete/spoiler tag it

[cedar obsidian]

Now it's clear from there

P^n|a^n

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[brisk hemlock]

guys where can i learn calculus?

[onyx peak]

khanacademy is pretty good

[tardy void]

Watch 3 blue 1 brown in the beginning

[onyx peak]

and 3b1b too, yeah

3b1b should only complement some other course / book tho

3b1b itself isnt enough

[tardy void]

Yeah

[brisk hemlock]

got it thanks

[tardy void]

Just for the visual stuff

[brisk hemlock]

.close

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[flint bobcat]

I wasnt getting answers so I come here.

[quiet zinc]

based

[flint bobcat]

I am trying to find the area of the larger figure and the distance between the length of one figure and the length of the other

I am stuck at not knowing how tofigure out the ratio of one axis to the other in the oval used to provide rounded corners to the larger figure.

Ok, still mostly wonky, but more clear on what I am looking for.

No takers?

[quiet zinc]

whats Z1.2

[flint bobcat]

the value of Z times 1.2, Z being the distance from the centroid to the corner and the difference being the corner to the arc

[quiet zinc]

oh alright

Z is also unknown? yeah

[flint bobcat]

Z is unknown but easy to calculate I guess... I just didn't put my mind to solve Z yet.

[quiet zinc]

yeah got z

[flint bobcat]

Is z 54368394.7163624 micrometers? cause if it is my math aren't that bad... But I cant fathom the distance gradient of the arc in the quarter circle or quarter oval at the corner.

[quiet zinc]

why are you measuring in micro meters

[flint bobcat]

ah... they were the best fit for conversions I was making... and that way I round up to 2 decimal on micrometers and get several detailed numbers, maybe not a good reason though.

[quiet zinc]

yes , it is 54.36 m

you can use significant digit aproximation to help the calculations

[flint bobcat]

I guessed that exact numbers are best until one gets to the answer, or the answer gets skewed.

[quiet zinc]

consider using significant digits , measuring m in micro metres is a nightmare read imo

[quiet zinc]

im sorry i didnt get that point

[flint bobcat]

oh ok, but I used a mix of both, I show the number in meters too.

[quiet zinc]

fair yeah

[flint bobcat]

but the numbers can become placeholders, the issue is more geometric algebra or algebraic geometry

[quiet zinc]

is this to scale , like is the other side 9.27m too?

or the other one is wider

[quiet zinc]

yeah fair enough 🥀

[flint bobcat]

this is to mock scale, so, you cant use rulers on the image to solve it, the other side is 9.27 m too

[quiet zinc]

yeah fair , i was asuming the other side was thiner

[flint bobcat]

yeah, its meant to be symmetric, I drew on power point and did it quickly...

[quiet zinc]

npnp got it , imma try

i mean if we consider this x/2 ig it might be solvable , i cant think of much here rn

[flint bobcat]

x/2?

What's the channel's rules on AI? I tried AI first, if you want I can show you what the AI told me, I didn't trust its answer... But you may understand it better. If there is no rule breaking going

[quiet zinc]

ai usually doesnt help , but if we dont see anything we can use it to get idea , but it requires proof reading fact checking

[quiet zinc]

yeah is it?

i can find the area and x if that is x/2

[severe hawk]

!nogpt

[final saddle]

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).

[flint bobcat]

I dont understand the font...

[severe hawk]

AI harms WAY more than helps

[flint bobcat]

Oh ok, no it was perplexity, but ok. I wot mention it then.

I wont mention it any more then.

[quiet zinc]

like is this x/2

i did some construction

[flint bobcat]

oh wait, I think I see your process! its brilliant! I will try it myself to see if I make it be even more sound to me cause I am still not sure I am following.

[severe hawk]

...why does that thing have 15 significant figures

[quiet zinc]

yeah fair

[flint bobcat]

calculated with excel. arithmogeometric means.

[severe hawk]

!original

[final saddle]

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.

[flint bobcat]

I made up this problem trying to find up a solution to stuff I was making, no one stated the original to me.

[severe hawk]

Okay. So state to us your original problem. "Here, have a shape" is not really helpful if all the relevant dimensions are not clearly labeled

[flint bobcat]

So the triangle whose hypethenuse is Z1.2 has one of its sides as (x+31.34 m) and the other (44.39+4.64 m)

Ok, I did that at first and got called a troll for doing a wall of text.

[severe hawk]

there's no triangle there with hypothenuse z1.2

[quiet zinc]

it looks like you are doing something with an unusually long pool table

[flint bobcat]

think football field

[severe hawk]

z1.2 is a segment, not a hypothenuse of a triangle

[quiet zinc]

right.

[flint bobcat]

a line that serves as the hypothenuse of a triangle you can break the rectangle into...

I thought that was what Weirdo meant.

[severe hawk]

z1.2 goes outside the rectangle and doesnt cover all the rectangle, so you cannot break the rectangle with it either

[quiet zinc]

this comes with some assumtions , so take it with a grain of salt ( mainly the x/2 part ig)
unless im imagining some wrong

[severe hawk]

okay, so i'm gonna state what i see on the original problem

[flint bobcat]

I am too dumb to figure out if you are on the right track... Can I put the wall of text now? No problems with that?

[severe hawk]

a) dark green rectangle.
b) Light green "rectangle", where the corners are, instead, a quarter of a circunference of unknown radius

[severe hawk]

i dont think that gives you the area, since you assume that the rounded edge of the light green triangle is an actual corner, not rounded

[quiet zinc]

oh yeah i see it now , i forgot to take that other x/2 part into account

[flint bobcat]

I have a rectangle 62.7792146497993 meters wide (x-axis) and 88.7832167928779 meters long (y-axis),

You have an indeterminate oval or circle, break it into 4 equal quarters. Where the radii of these quarters meet make them coincide with the rectangle's corners. Connect the oval quarters with straight lines.

Find X (width of larger figure - width of rectangle)/2
The major axis of the oval has a radius of 9.26857757727846 meters

Half a diagonal of the rectangle is Z
Z times 1.2 is a projection of Z into the arc of the quarter oval.

Much clearer now?

This is is an intersection of algebra and geometry.

This is the kind of stuff that may need to be drawn to be better understood?

I bet a drafting compass solves it.

but I dont have any drafting compass available.

I could draw it on a scale of 1 cm = 10 meters and it would become an easy challenge.

[severe hawk]

okay, first off, i'm gonna use 62.8m and 88.8m as dimensions of the rectangle

[flint bobcat]

Ok.

[severe hawk]

second, the second statement isnt clear. An oval or a circle dont have "a" radius, so you cannot use "the radius" meet as a condition

[flint bobcat]

I will explain it with another image, this one I hope made my power point so it wotn be wonky:

these are two quarter circles or quarter ovals, respectively.

[severe hawk]

what definition of oval are you using

[flint bobcat]

the straight lines are what I refer to as the radii and the arc is the round part...

What definitions of oval there are? Am I using the wrong word? I thought an oval was like a circle but elongated.

[severe hawk]

well, depends on the context

in technical drawing, an oval is made by joining four curved arcs, of which each pair has the same radius

in this case, the two black arcs left and right, and the two red arcs top and bottom

[flint bobcat]

I am not that much well informed to figure out how that works out... I know that drafting compasses can be used to make circles and (ellipse then?)

[severe hawk]

in projective geometry, a thing that i dont care or understand

in regular maths, a "kind-of-close-to-ellipse but we didnt bother to define it properly"

[flint bobcat]

I am not an architect, but this would need to be understnadable for an architectural project... I guess ellipse was the right term...

I thought ellipses were 3d ovals, so I ruled out the word, my bad.

[severe hawk]

okay, so an ellipse is a well defined curve. Are we to understand the corners are quarters of ellipse?

[flint bobcat]

Yes.

[severe hawk]

and no, an ellipse is a 2d curve, defined by having each point have the same sum of distances to two points

[flint bobcat]

All I know about ellipses is that they get two axis, the major and the minor.

[severe hawk]

the most important thing they got is the 2 focii

[flint bobcat]

with the major being analogous to a rectangle's length and the minor to its width.

[severe hawk]

thats... not correct, since you can rotate it

[flint bobcat]

and the focii determine the axis?

I am going well over my head here... Please bear me, I think ellipse was the right term all along.

[severe hawk]

kinda? you can get all the stuff from the ellipse from wikipedia

it isnt hard to understand. Basically, a circle would be an ellipse where both focii are in the same spot

[flint bobcat]

yep, where the two axis measure the same.

[severe hawk]

if you separate them, then you get an increasingly "not circle" ellipse

[flint bobcat]

so, great, we have this sorted out

[severe hawk]

okay, so the corners are quarters of ellipses, not circles

[flint bobcat]

but my intuition is that if you drew them you wouldn't tell at simple glance that they are quarters of ellipses not of circles.

[severe hawk]

that's why i mentioned here what i was seeing

and why i explicitly asked for more information

[flint bobcat]

thanks, I preferred the written question format but got called a troll for writing a long piece of text.

[severe hawk]

okay, you said the major axis of the oval was 9.26m. Does that mean HALF of the major axis (so distance from dark green rectangle to light green rectangle), or the actual major axis (so distance from dark-light green rectangle would be 4.63m) or a different thing?

[flint bobcat]

9.26 m is half of the major axis.

[severe hawk]

half of the major axis, are we're not given the minor axis

okay

[flint bobcat]

that's part of what has to be found... But we are indirectly given the value of a line that bisects this quarter ellipse. (I am not sure it bisects it in the mathematicla sense)

[severe hawk]

okay

so we also have the diagonal of the rectangle

and you called that 2z

projection in this case is not correct

[flint bobcat]

Yes because we have its sizes and know that Z + 20% of Z reaches the perimeter of the outer figure.

[severe hawk]

so you're saying that, if we extend the diagonal of the rectangle (not properly drawn in your diagram), it eventually intersects the corner ellipse, and it does so with a line 20% larger than the diagonal of the rectangle

[flint bobcat]

yes

[severe hawk]

okay

[flint bobcat]

I used to be good at math at high school, got ruined when university "professors" taught it to me horribly... and like Millhouse and Bart in an episode of the Simpsons I didnt only not learnt, but started unlearing what I knew. I intuit this isnt too complex but I get stuck on it.

You've been quick to get the ideas, so... I am grateful ya're helping me.

[severe hawk]

im drawing the thing, gimme a bit

[flint bobcat]

(I am also grateful to Weirdo, even though' he left the building already)

Oh ok, thanks.

[severe hawk]

BEHOLD, mspaint

i'm not drawing all 4 corners, fuck that

okay, that's all the data you've given me

[flint bobcat]

lol... I work with krita, ms paint and powerpoint, figured out power point was the best to use now, you prove me wrong.

[severe hawk]

1. Always use the right tool for the job
2. The right tool for the job is always a hammer
3. Everything can be used as a hammer

[flint bobcat]

4)Specially MC Hammer!

[severe hawk]

okay, so the problem i see here is that we dont know where the ellipse is actually positioned

[flint bobcat]

what ya mean?

[severe hawk]

we're not given enough information to determine the shape

[flint bobcat]

😮

can another contrain provide more information?

[severe hawk]

since we're not given the actual ellipse, we cannot know where it's positioned.
OR
since we dont know where the ellipse is positioned, we cannot know the actual ellipse

[flint bobcat]

Like, I thought I had defined the issue with all necessary constraints, but I see it may not be sufficient.

[severe hawk]

to solve (a), you'd also need the minor axis. Since both axes are determined, there's a unique ellipse that would give the 1.2z
to solve (b), you'd need to know the centre of the ellipse, or it's horizontal displacement. Since 1.2z, there would be only a single ellipse that would have that in said position

[flint bobcat]

here it becomes algebra?

[severe hawk]

currently, there are infinetely many ellipses that would give you the conditions of half major axis 9.26 and intersected at 1.2z, just by moving the ellipse and changing the minor axis

a way to solve it (b), would be saying that the center of the ellipse is the corner of the darkgreen rectangle

[flint bobcat]

ok... what if the line 1.2 z passes through the centre of gravity of the quarter ellipse? Or is that too complex or nonsensical a rule?

[severe hawk]

i'm... not sure that's enough

that would make it so it has solution, but i'm not sure if unique

no, it's not unique

[flint bobcat]

wait, gimme a sec to draw from your drawing and explain a better solution

I mean, a better constrain

I intended to make the ellipse defined in such a way that 1.2 is the arithmogeometric mean of b and c

YC = 72.06 based on your approximations.

but I was afraid this method would turn the maths too mind breaking...

I guess this redifines it into not being geometry but arithmetic?

[severe hawk]

what are you calling b and c?

[flint bobcat]

the multipliers by which Y (in the case of C) and X (in the case of B) become the dimensions of the larger figure... their numerical value is close to 1.2 but differs.

[severe hawk]

that still doesnt give a unique solution

[flint bobcat]

no wait, I misspoke

B turns half the width of one rectangle into that plus the distance between it an the perimeter of the outer figure, C turns half the length of one rectangle into that plus the distance between it an the perimeter of the outer figure

[flint bobcat]

are you sure? I think this seems more uniquely solveable now... I will try my hunch and tell you what I find

I think I am solving it, do you wanna see what I come up with?

Here having 15 significant units proves useful.

[severe hawk]

eh

[flint bobcat]

working out arithmogeometric means backwards, retroegineering, takes time

[severe hawk]

with only one of the conditions you said you would STILL not have a unique ellipse

[flint bobcat]

I said? I dont understand : s I think I may have misexpressed myself?

This is what I figure would solve this... I only need to figure the area of the larger figure.

so now the question is geometric again

The ellipse is 6.002976 m wide and 9.27 m long

wait, this is simple math now

what do I do to report this solved?

@severe hawk @quiet zinc thanks for your help you two!

[severe hawk]

you would use .close

[flint bobcat]

.close

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[vague anchor]

S(x)=(x+1)+2(x+1)^2+3(x+1)^3+4(x+1)^4....60(x+1)^60, x not equal to 0
60^2 . s(60)=a(b)^b+b
find a+b

[vague anchor]

<@&286206848099549185>

[lyric obsidian]

!15m

[final saddle]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[terse crypt]

what have u tried?

[tired walrus]

$S(x) = \sum_{k=1}^{60} k(x+1)^k, x \neq 0$

Given that $60^2 \cdot S(60) = a \cdot b^b + b$, find $a+b$

[lyric obsidian]

||geom series||?

[soft zealot]

Ann

[tired walrus]

is this what your question says?

[lyric obsidian]

Thank you Ann

[storm haven]

It doesn't have the same coefficients

[lyric obsidian]

||factor (x+1)||

[vague anchor]

probably
but its in format that i sent

[lyric obsidian]

Mkay

[terse crypt]

but still the k in front?

[lyric obsidian]

There's a workaround

I can explain in #「helpers-lounge」

[vague anchor]

i apologies
i wasnt aware of it but i became aware just after tagging

[lyric obsidian]

You might want to start with this and manipulate what you're given as the sum

[vague anchor]

so what shall i do further

[vague anchor]

i cant understand

my my anyone here

[jagged flare]

isnt this arithmetico geometric series

[vague anchor]

whats that im not taught anything like that

[vague anchor]

sounds good like componendo dividendo

[jagged flare]

https://en.wikipedia.org/wiki/Arithmetico-geometric_sequence

oh nvm

[vague anchor]

i get it

[jagged flare]

well there is the partial sum but its kinda ugly, maybe there are other methods better than it since you havent been taught it yet

[vague anchor]

truly
my teacher has solved this on board
its just he gave us no time to note the solution of this question

[stoic temple]

Like expressing S(x) as a sum of other geometric series with lesser numbers of terms

[vague anchor]

sounds senseless to me
maybe bcoz i dont know that or u dont know something

[gritty solar]

You could differentiate the geometric series

[jagged flare]

alternative: consider S(60)-S(60)/60 and pairing the terms that match the power, notice that a part of it is a geometric series

[vague anchor]

or maybe i would prefer s(60)-61.S(60)

[lyric obsidian]

Sorry, went to do something

[vague anchor]

np

[lyric obsidian]

My point was that you can transform the k in your sum to a 61-k and keep the bounds

[vague anchor]

?

[lyric obsidian]

And then notice that it becomes a geometric series when you add those 2 together

[lyric obsidian]

$S=\sum_{k=1}^{60}k(x+1)^k=\sum_{k=1}^{60}(61-k)(x+1)^{61-k}$

[soft zealot]

;(

[vague anchor]

how

[lyric obsidian]

Because notation

[vague anchor]

im in grade 11 so pls teach at that lvl

[lyric obsidian]

I am, very much

[vague anchor]

oh i see now

[lyric obsidian]

I am ommiting the calculus part of this

[vague anchor]

yep

[lyric obsidian]

You tested a few terms?

[vague anchor]

no just mind

just in mind*

[lyric obsidian]

So now what I want you to do is to add these two series (sorry I have to go eat breakfast!) and hopefully someone else picks up on my idea

[vague anchor]

who? pls Ann

[lyric obsidian]

I'll solve it while I'm eating, see if there's anything good with this method

[final saddle]

@vague anchor Has your question been resolved?

[vague anchor]

nope its pending helper is having breakfast

[lyric obsidian]

You might want to try this out

[tired walrus]

do not rely on me.

[vague anchor]

my my very fast reply

[lyric obsidian]

Considering that I spent a decent amount of time doing the calculations I don't think the way I was presenting was necessarily correct

Or at least, the most efficient manner

We could use calculus but I'm not sure if you've learned that yet

[vague anchor]

its senseless
it might be S(60) - 61.S(60)

i remember my sir had used sum of n terms of GP formula

[lyric obsidian]

Yeah, because its arithmetic geom series

Uggggh I forgot the derivation

This sucks

[vague anchor]

maybe

[vague anchor]

this would be a great time for a mathematician

[lyric obsidian]

?

I feel like I can reasonably say, however, that ||b=61||

[vague anchor]

arent u a mathematician like Ann

[lyric obsidian]

Do not take my word for it

I am not a mathematician

[vague anchor]

r u even a helper indeed

i dont see that tag in ur profile

[lyric obsidian]

Yes

I don't want to get pinged

Anyways, I'm just a high school student but there's a trick here

Something

@vague anchor Are you willing to go through a long process?

I'm gonna do a part of it and leave the rest up to you

[vague anchor]

i dont know
but maybe we can just try

[lyric obsidian]

Mkay

I'm gonna shorthand S(61) to S, just for notational awareness

[vague anchor]

he also had 60 in denominator and sent it to other side to form 60^2

ans is 23

[lyric obsidian]

$S=\sum_{k=1}^{60}k(61)^k$\
$61S=\sum_{k=1}^{60}k(61)^{k+1}$\
$-60S=?$ (hint: this will be telescoping, and then you can find $S$ from here)

[soft zealot]

;(

[lyric obsidian]

Mkay

Well, try this out

[vague anchor]

wasnt i doing same

[lyric obsidian]

Yes

[vague anchor]

i would recomend calling Ann to save time

i got it

its forming GP

.close

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Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[chilly moon]

A cookie jar contains 7 different types of cookies. There are at least 2 cookies of each type. How many ways are there to chooes 3 cookies if the 3 cookies chosen cannot be all the same type?

[chilly moon]

I have this right now:
I find how many cases to choose from for 3 different type of cookies, so I did 7C3 because I believe the amount of cookies is redundant in this case

And then I want to find amount of cases to choose from for 2 different type of cookies, its just 6*7 right? so my answer is 77

question is, is there a way to deduce how many total cookies there are in the jar given this information

[serene panther]

No

It csn be infinite cookies as well your answer would still remain same

[chilly moon]

ok ic

[serene panther]

Atleast 14 cookies are there
That is the only thing you can conclude

[tall breach]

Bro

Isn't it impossible for all the cookies to be the same type

[serene panther]

They aren't though?

[tall breach]

No like

Out of the 3

[mellow vortex]

Yelllow

[tall breach]

Maximum 2 cookies can be same right out of 3

And the third one will be different

Making it satisfy the given condition

[tall breach]

They said all 3 cannot be same

But nothing about two cookies being same

[tall breach]

Wt the