help-27

[devout snow]

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

[valid vector]

3 ppl in one thread is diabolical

anyways

[slate ocean]

why are there like 3 people using the same channel what

[dull heron]

i was typing in the other help channel but then it got occupied, my bad

[valid vector]

Let's restrict to the first condition

[restive river]

your good bro

[restive river]

well 300<n<500

n is the 3 digit number were talking about

[valid vector]

Oh my bad

I meant like the portion after that

[restive river]

the number has to be one more than a multiple of 3 +1

which is also 1 less than a multiple of 7

[valid vector]

Where do whole number multiples of 7 start in the 300-500 range

As well as the multiples of 3

[restive river]

multiples of 7 +1

[valid vector]

(The point is to find a "common factor" so we can just apply a simply rule)

[restive river]

300+x=7y

[wooden charm]

7,14,21,28,35,42,49,56,63,70,71,78,85,92,99,106,113,120,127,134,141,148,155,162,169,176,183,190,197

all multiplies of 7 in a range of 200

[restive river]

common factor for multiples of 7 plus 1 and multiples of 3 minus 1

[valid vector]

You did not need to do allat

[restive river]

all of these are divisible

but too much work stone

[wooden charm]

meh

[valid vector]

Let me demonstrate

[wooden charm]

brute forcing it works sometimes

[restive river]

yeah but in a competiton it kills time quickly

theres always a quick solution its hidden in the question

[wooden charm]

I guess you could use explicit arithmetic

[valid vector]

The first starting factor 1 less than multiple of 7 is 302=1+43(7)

[restive river]

no no no plus 1

7n+1>300

7n+1=3y-1

we get 7n=3y-2 where 300<n<500

[wooden charm]

28 multiplies of 7 in 200

approx

66 multiplies of 3 in 200 approx

rounding down

[restive river]

8,15,22

71,141,211

281

351

351 works

[faint kettle]

Now $3y=2+7n$ so we have $$ 3 \vert 7n+2 $$ $$ 3 \vert n+2 $$ in other word
$ n = 3k +1 $ where $k \in\mbb{Z^+}} $

[woven radish]

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

[restive river]

3y=2+7(3k+1)

algebraic approaches is pretty confusing

[faint kettle]

now consider 300 < 21k+ … < 500 instead

[faint kettle]

Here

300 < 21k+8 < 500
292 < 21k < 492

[restive river]

n cant be a fraction

[snow tide]

301 is the first multiple of 7 in that set and 497 is the last
300 + 7, 300 + 14, … 300 + 196
303 is the first multiple of 7 in that set and 498 is the last
304 + 0, 304 + 3, 304 + 6, … 304 + 195
rewrite the 7s
304 + 3, 304 + 10, 304 + 17, 304 + 24, 304 + 31, 304 + 38, 304 + 45, … 304 + 189
every third matches with a multiple of 3
how many multiples of 3 between 189 and 3
67

is that right

liek is 67 correct

[restive river]

multiples of 7 + 1

havent check

[snow tide]

im confident its either 66 or 67

but i could also be rly wrong

[restive river]

7(50)+1 is 351

closest

so ill try

7(45)+1

which is 301

[valid vector]

Oh by the way

[restive river]

yeah?

[valid vector]

As a heads up

!15m

[devout snow]

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.

[restive river]

alright

[snow tide]

@valid vector i was trying to tell that guy that youd be more helpful than me 😭

[restive river]

i can cook on the competition with your help

last year i only threw biscuts everywhere

for fun

i didnt even try that much

this year im locking in

[wooden charm]

This is how far I’ve gotten so far

Brute forcing is fun

[restive river]

its overwhelming though

[restive river]

your the type of person who makes x+5=3 into a multidimensional calculus equation 😭

[wooden charm]

x = -2

e

[restive river]

asy

[wooden charm]

f(x)=x+5

when x=-2, f(x) = 3

[faint kettle]

your question say 3m+1 and 7n-1 but

[faint kettle]

?

[valid vector]

I do not care, pricks are prideful and stupid 😄

[restive river]

"How many numbers between 300 and 500 are one more than a multiple of 3 and one less than a multiple of 7

it can both

i dont know man

you guys know everything i know nothing

[dapper tiger]

let n be such a number
n = 1 mod 3 and n = 6 mod 7
so n = 13 mod 21
n is of the form 21k+13

solving modular systems like this is a result called the chinese remainder theorem

[restive river]

dang😢

maybe im not worth it

i cant do this

[faint kettle]

I’ll following your word then , since $$ 7n+1 = 3y-1 $$
$$ 7n+2 = 3y $$ and since $n,y \in\mabb{Z^{+}} $ we have $$ 3 \vert 7n +2 $$ which mean $$ 3 \vert n+2 $$
We have $n=3k+1$ consider $$ 300 < 7(3k+1)+1 < 500$$ $$ 292 < 21k < 492 $$ then $ k = 14,15,\dots,23 $
And since $ n= 3k+1 $ then $ n = 43,46,\dots,70 $
$$ 7n+1 = 301,323,344 ,\dots , 491 $$

[restive river]

its alright im not going to the competition

[dapper tiger]

no, it's one less than a multiple of 7 C

and one more than 3

[restive river]

im a prick anyways whats the use

[snow tide]

no one called you a prick

[snow tide]

@valid vector was ref to another person here

[restive river]

no no no

its my opponent

im against the best high school in my county

i need your help

like actually

without you guys i might not defeat them

[snow tide]

if you dont place first you dont place first you can do everything in your power to do ur best but like

ive been on both ends of that

ive been against better schools and ive been the best school

theres only so much you can do

[restive river]

we always got 2nd place

1st waited too long

[restive river]

i need help with variables like ABC38=something and word problems

can you give me an example?

[faint kettle]

Sorry if it’s not what you meant but here’s
111 = 100 + 10 + 1
123 = 100 + 20 + 3
Now if we look for ABC is three digits number we have ABC = 100A + 10 B + C

[restive river]

using that how can we solve questions like ABC+CB4=B6D or something like that

[dapper tiger]

by doing what we did earlier

[restive river]

modular arithmetic

right?

[dapper tiger]

yeah or similar ideas, I mean it's just calculations in the end

no matter how you do them

however, if you're going to do a lot of basic arithmetic stuff, modular arithmetic would help a lot

[restive river]

alright lets try ABC5=50x

[faint kettle]

In other words , $$ \overline{a_1a_2\dots a_m} = 10^{m-1} \cdot a_{1} + 10^{m-2} \cdot a_{2} + \dots + a_m $$

[restive river]

i want to find the lowest solution

[woven radish]

C

[dapper tiger]

x is a digit ?

[restive river]

no its a variable 50*x

[dapper tiger]

oh ok so a natural number

[restive river]

yep

[dapper tiger]

so you want ABC5 to be a multiple of 50

which can't happen, bc multiples of 50 end in 0

[restive river]

the smallest 4 digit multiple of 50

but ends in 5

wait thats impossible

[dapper tiger]

yeah I told you

a multiple of 50 is a multiple of 10

so you make things... less interesting

[restive river]

8ABC=50x

[dapper tiger]

A = B = C = 0 give 8000, a multiple of 50

[restive river]

yes

we need the lowest solution

[dapper tiger]

that's the smallest in the 8000

[restive river]

but

[dapper tiger]

you won't have a number in the 8000 smaller that 8000

[restive river]

heres some conditions

we can only use a number once

0 cant repeat

[dapper tiger]

8150

[restive river]

for a b or c

[dapper tiger]

you add 50 until it meets condition

[restive river]

yes

[dapper tiger]

to stay in the multiples of 50

[restive river]

so any variable equation thats equal to a random number times x

makes the variable also a multiple of the random number

[dapper tiger]

try to find the smallest AB42 = 43k

k natural number so AB42 multiple of 43

[restive river]

Smallest 4 digit number of 43

well

1000A+100B+42=43k

A+B=43k-42

[dapper tiger]

wait where did the 1000 and 100 go

[restive river]

we can leave those

[dapper tiger]

ah ?

[restive river]

A and B can remain the same

[dapper tiger]

you're working on numbers here

[restive river]

A is a 4 digit number

[dapper tiger]

no

A is a digit

1000A is the number A000

[restive river]

oh

1000A+100B=43k-42

this makes k greater than 1000

[dapper tiger]

no

43*100 is already greater than 1000 for example

[restive river]

that makes A>4000

?

[dapper tiger]

no, A is a digit

it's between 1 and 9

since it's non zero

[restive river]

a>4

[hollow ice]

In fact k can never be larger than 1000, coz left hand side can be no more than 9900

[restive river]

9900 is the max we can get

the mimimum is 1100

we need the closest number to 1100 that is 43*something - 42

[dapper tiger]

if I have 1000A+100B+42 = 43k, an idea is to look mod 43
so 11A+14B-1 = 0
11A+14B = 1

[hollow ice]

1st step you can take is, since AB42 is divisible by 43, the last digit of k is gonna be 4

[restive river]

we could also just do trial and error

43(20)

nope

[dapper tiger]

trial and error will be kinda long but yeah you can

[restive river]

860

not close

43(25

40+3(20+5)

860+5(43)

215+860

1075

1075-42

1033

1075+43

1118

1118-42

1076

1077

so on

[faint kettle]

That’s gonna be too long

[hollow ice]

Do you know long multiplication? The way you multiply numbers with pen and paper. You can do that and construct the unknown numbers with that

[faint kettle]

try consider AB42 end with 2

and it’s 43k

[hollow ice]

If you feel modular arithmetic is tough, you can always use long multiplication for this

[faint kettle]

so k should end with 4

[restive river]

43(25)+1 makes 1076

43(25)+25

=1100

43(25)-42=1033

43(26)-42=1034

it keeps on going up 1

[dapper tiger]

with modular arithmetic:
AB42 = 1000A+100B+42 = 11A+14B-1 mod 43
so 11A+ = 29B+1 mod 43
A = 30B+4 mod 43
A and B are digits, where B is between 0 and 9
by calculating the ten 30B+4 mod 43, only two cases give A between 0 and 9:
B = 0 gives A = 4, B = 3 gives A = 94 = 8 mod 43
only 4042 and 8342 are multiples of 43 of this form

[hollow ice]

well this is certainly wrong

[restive river]

its 1076

43(27)-42 = 1119

[dapper tiger]

you know you'll have like 200 multiplications to do if you continue this way right?

[restive river]

i know

brute forcing isnt the way to go

i was desperate because i thought i saw a pattern

[dapper tiger]

43*26-42 = 1076

[restive river]

that statement was stupid

[dapper tiger]

like, you add 43 each time

[restive river]

well

43(26)-42

43(25)+43-42

43(25)+1?

thats what i thought

but i might be wrong

[dapper tiger]

you're not wrong

but you're only noticing that 42 = -1 mod 43

you would still benefit from using some arithmetic tho

if you learn the chinese remainder theorem, here there's still the "nice way" of doing it

conveniently, I chose 43 to be prime and the number ends in 42

so the number happens to be 42 mod 100 and 0 mod 43, which gives one possibility mod 4300
so we already know that there are at most two solutions between 0 and 9942

and the theorem also gives the way to compute them

then you find the two solutions 4042 and 4042 + 4300 = 8342

there's also a "pure modulo" way I illustrated earlier

[devout snow]

@restive river Has your question been resolved?

Channel closed

Closed by @sour lagoon

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[versed cipher]

How does one treat this UDL extending on angle member?

[devout snow]

@versed cipher Has your question been resolved?

[devout snow]

@versed cipher Has your question been resolved?

[gray summit]

What is a UDL?

[versed cipher]

Uniformly Distributed Load, the red one

[gray summit]

I didn't understand the question, can you explain

[versed cipher]

Okay

there is a UDL on member BD , and on the left side, it is hovering above an angled member AB,
Should I disregard this excess? or is it part of the equation and would it be transferred to angled member AB?

[gray summit]

Lol, this question is too advanced for my peanut sized brain

[versed cipher]

I feel the same way

T-T

[devout snow]

@versed cipher Has your question been resolved?

Channel closed

Closed by @versed cipher

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[open spear]

I am struggling with putting up the sums using the summation symbol

[pseudo basin]

you need to figure out the rule by which the terms in each summation are obtained

some of these look pretty counterintuitive...

[open spear]

i got the first 3, but i dont know how to continue...

[pseudo basin]

d, f and i think h may be quadratic sequences

given that the differences between adjacent terms seem to be constant

kind of painful to figure out exactly what those are tho

e is quite clearly the sum of a GP

[open spear]

i have no idea what d could be

[open spear]

yh i got that, but i just dont know how to express it into the summation symbol

[pseudo basin]

well, how can you write the n'th term of the sequence {2, 4, 8, 16, ..., 1024}?

[open spear]

not sure

its just doubling it

[pseudo basin]

you said you get that it's a GP

do you know how to write down the n'th term of a GP

[open spear]

what is GP?

[pseudo basin]

geometric progression

[open spear]

ah

[pseudo basin]

why did you say you get it when you didnt even know what i meant by GP until now

[open spear]

sorry, i understood it wrong

i dot

i dont usually do maths in english

so some of the terms are new to me

[pseudo basin]

what's your native language

[open spear]

german

but its fine

[pseudo basin]

ok i dont speak that anyway

[open spear]

its fine haha

but how can i learn to build the sum of these things?

like i tried searching online, cant find anything. I dont know if there are set rules to follow

[pseudo basin]

you should not expect yourself to be able to come up with a formula for an out-of-nowhere sequence of numbers

bc this is not really a thing that can be done at all

[open spear]

well, thats kind of what i need to do ... XD

[pseudo basin]

but for 2, 4, 8, 16, ..., 1024 you can recognize these numbers as powers of 2

2^1, 2^2, ..., 2^10

[open spear]

yes

i got those

i just cant figure out d and e

[pseudo basin]

i am talking about e

[open spear]

ok

[pseudo basin]

do you get this

[open spear]

yes, it makes sense

and i cant seem to find the pattern for d?

[pseudo basin]

i think it's going to be something like $a_k = Ak^2 + Bk + C$ with some constants $A$, $B$ and $C$.

[woven radish]

Ann

[open spear]

wtf

how do u know?

[pseudo basin]

honestly it is kind of the best you can come up with if the first 3 terms don't follow any specific pattern and are all you're given

this exercise is dumb

[open spear]

haha

and this is the first exercise we get with this new topic, smh

weird ahh Teacher

just asked ChatGPT, he got this

but it only works for the first 2

i think im just going to skip it

[devout snow]

@open spear Has your question been resolved?

Channel closed

Closed by @open spear

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[bleak arrow]

what does it mean when a variable is a placeholder

[stone stump]

context?

[bleak arrow]

let's say the electric field due to a line charge = lambda/(2pie_0x), is "x" a placeholder here?

[stone stump]

well in the sense that it occupies some space and you later will plug in some other thing for it, sure

[bleak arrow]

.close

[devout snow]

Channel closed

Closed by @bleak arrow

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[lofty verge]

$\text{Heavy or tall husbands}=470\\frac{2}{3}\text{tall husbands are also heavy}\\frac{1}{2}\text{heavy husbands are also tall}$

[woven radish]

UCYT5040

[lofty verge]

how do i solve this?

[jaunty mantle]

If you just type words outside of the $ signs it will just be normal text

Like this $x = 3$

[woven radish]

frosst

[lofty verge]

ik but im just doing latex idk why

lol

[jaunty mantle]

Draw a diagram id say

[lofty verge]

like venn diagram?

let me try this

[jaunty mantle]

I honestly would say just start writing out the numbers

If two thirds of the husbands who are taller than their wives are also heavier

How many people is this

[lofty verge]

not sure

but i also just realized that means 1/3 of tall husbands are not heavier

but still not sure where to start here

[jaunty mantle]

But how many people is that

Like a numerical value

Not a percentage

[lofty verge]

i dont think its 470 because thats # of husbands that are tall, heavy, or both

[jaunty mantle]

<@&268886789983436800>

[jaunty mantle]

What is 470

Be specific

[lofty verge]

husbands whose wives arent heavy and tall

?

[jaunty mantle]

That many?

How many husbands are there?

[lofty verge]

530 husbands

total

[jaunty mantle]

Ok and you’re saying 470 of them are what

I don’t think 470 is the right number

And let’s not use the opposite statements, let’s focus on what the question states for now

[lofty verge]

okay

[jaunty mantle]

2/3 of 530 husbands are what

[lofty verge]

i dont think they are anything

because not all 530 husbans are taller than their wives

[jaunty mantle]

Ok

Let’s suppose there are x husbands that are taller than their wives then

Now what would we say

[lofty verge]

2x/3=husbands both tall and heavy

[jaunty mantle]

Ok what about the other way round

[lofty verge]

like (3*(husbands both tall and heavy))/2=x

or did you mean with the other condition?

[jaunty mantle]

Yeah the other condition

[lofty verge]

ahh ok then y/2=husbands both tall and heavy=2x/3

where y=heavy husbands and x=tall husbands

[jaunty mantle]

Hmm

[lofty verge]

only 6% of test takers got this one right

so it must be pretty hard

oh and also the correct answer is 400 if that helps

[jaunty mantle]

This is very confusing indeed

So I would think about it like this

Red is both heavier and taller

Green is heavier but not taller

Blue is taller but not heavier

And outside is not taller and not heavier

There are 530 people in this diagram

[lofty verge]

outside is 60?

[jaunty mantle]

Yes

That’s good

Let’s deal with the first statement

[jaunty mantle]

Look at this

So taller means red + blue

Also heavier ok

So red / (red + blue) = 2/3

Do the same for green

[lofty verge]

red/ (red + green) = 1/2

[jaunty mantle]

And we also have green + red + blue = 530 - 60

Ok let’s now call x = green, y = red, z = blue

So we have the 3 equations…?

[lofty verge]

3y=2(y+z)
2y=(y+x)

x+y+z=470

[jaunty mantle]

Do you think you can solve this

[lofty verge]

i think so, let me try

y=188

z=94

x=188

that cant be right

y should be 400 right?

[jaunty mantle]

Doesn’t line 1 say y = z

[jaunty mantle]

From here

No it doesn’t

I’m dumb

Hmm

[lofty verge]

these equations look perfect too

idk whats wrong

[jaunty mantle]

Ok suppose 400 was y

Then 400/(400+ blue) = 2/3

[lofty verge]

blue=200?

impossible tho cuz x+y+z=470

and blue+y is already greater than that

[jaunty mantle]

There is some number z of people who are taller than their wives, of the z people, 400 are also heavier, and at a proportion of 2/3

Yeah

So then there are 600 husbands

400 is not correct

Unless they mean 1060 people are all husbands

We can try that

[jaunty mantle]

Put this as = 1000

And solve it again

[lofty verge]

oh yep thats what they mean

1060 couples = 1060 men & 1060 women

[jaunty mantle]

Oh that wording threw me off

[lofty verge]

same

i got the right answer this time

[jaunty mantle]

I suppose 1060 married couples means 1060 pairs of people

[lofty verge]

y=4000 and z=200 and x=400

thanks for your help

.close

[devout snow]

Channel closed

Closed by @lofty verge

Use .reopen if this was a mistake.

[jaunty mantle]

Yeah this first instinct was right

A diagram helps a lot

@lofty verge

[lofty verge]

i agree

i will make sure to do that in the future

[devout snow]

Available help channel!

Send your question here to claim the channel.

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

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

[jaunty mantle]

Especially with these prooortion stuff

[jaunty mantle]

It’s so hard to read and process and hold all that info in your head

Diagrams make it easy to see what’s going on

[lofty verge]

yeah

sorry i closed the channel so early lol

[jaunty mantle]

All good

👍

.close

[devout snow]

Channel closed

Closed by @jaunty mantle

Use .reopen if this was a mistake.

[winter torrent]

@jaunty mantle has your question been resolved?

[pseudo basin]

.coose

[devout snow]

Available help channel!

Send your question here to claim the channel.

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

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

[ruby talon]

I had a problem in another help channel but it closed after I fell asleep. Here's the problem restated:

Hello all. I have a problem that I have been stuck on and want some guidance on.

Say you have a Mafia game with x town and y mafia. The total number of players is x+y, the combined total of town and mafia. The town win if all mafia are eradicated, and the town lose if there are ever more mafia than town. Assuming the town kills one random player and the mafia kills one random town every day/night, what is the probability that the town wins?

I have made the following recursive formula: f(x,y) = (x/(x+y))f(x-2,y)+(y/(x+y))f(x-1,y-1) where f(z,0) = 1, f(z,z+1) = 0, x and y are positive integers, and x+y is an odd integer (trust me, this matters). It is easy to find a solution for f(x,1) with double factorials, but it seems that going any higher is harder. I would appreciate any help.

To clarify: I'm sure the recursive formula is right, but I want a general solution to the problem.

[jaunty mantle]

You problem seems to not be well defined

What is a player/how many players are there

[ruby talon]

a player is either a town or mafia

and it can be any positive (in this case, odd) number

essentially its just the combined total of all town and all mafia

i suppose i shouldve also said that mafia wins when there are more mafia than town and town wins when there are no more mafia

[devout snow]

@ruby talon Has your question been resolved?

[ruby talon]

i edited the original post so thats its more clear

[cinder nova]

$f(x, y) = \frac{x}{x+y}f(x-2, y) + \frac{y}{x+y}f(x-1, y-1)$\\
$f(x, y) = $\
$\frac{x}{x+y}\qty(
\frac{x-2}{x+y-2}f(x-4, y) +
\frac{y}{x+y-2}f(x-3, y-1)
)$\
$ + \frac{y}{x+y}\qty(
\frac{x-1}{x+y-2}f(x-3, y-1)

• \frac{y-1}{x+y-2}f(x-2, y-2)
  )$

[woven radish]

Sepdron

[devout snow]

@ruby talon Has your question been resolved?

[ruby talon]

oh hi bot

um yeah that is what happens when you apply the formula twice

[cinder nova]

yeah, just expanding it to try to make a pattern clearer

if we only expand the f(x-1, y-1) repeatedly, we get\
$f(x, y) = \sum_{n=0}^{y-1} \frac{(y)_n}{g(x+y, n+1)}(x-n)f(x-n-2, y-n)$\
where\
$(a)_b = a(a-1)(a-2)\dots(a-b-1)$\
$g(a, b) = a(a-2)(a-4)\dots(a-2b+2)$

[woven radish]

Sepdron

[cinder nova]

@ruby talon do we have x>y here?
if not, what happens if it's like f(-1, 1)?

[devout snow]

@ruby talon Has your question been resolved?

[ruby talon]

yeah we assume so

for initial conditions at least

obv "if mafia wins" you can get f(2,3) or f(4,5) (or generally f(z,z+1) but thats the only case that those values should be expected to be put into f

and in those cases its always 0

[cinder nova]

alr, I think it might be useful to make x=y+a then

[cinder nova]

also this is wrong, I misread something

this should be right I think\
$f(x, y) = \frac{y!}{(x+y)!!} + \sum_{n=0}^{y} \frac{(y)_n}{g(x+y, n+1)}(x-n)f(x-n-2, y-n)$\

[woven radish]

Sepdron

[cinder nova]

$f(y+a, y) = \frac{y!}{(2y+a)!!} + \sum_{n=0}^{y} \frac{(y)n}{g(2y+a, n+1)}(y+a-n)f(y+a-n-2, y-n)$\
$f_2(a, y) = \frac{y!}{(2y+a)!!} + \sum{n=0}^{y} \frac{(y)_n}{g(2y+a, n+1)}(y+a-n)f(a-2, y-n)$\

[woven radish]

Sepdron

[ruby talon]

interesting its less recursive now, per se

[devout snow]

@ruby talon Has your question been resolved?

[ruby talon]

i wonder if, similar to how you found a pattern in the summations for one f, that you do the same thing for the second and end up with a double summation

[ruby talon]

oh wait no bc there would be a... new summation every recursion

[ruby talon]

i suppose i couldve pinged helpers but like whats the point lol

[kind tiger]

you could maybe try doing smth with generating functions

ok thats seems pretty useless

honestly i have a bad feeling there just isnt going to be a closed form

i thought looking at a simpler problem (y=1) might help and even that seems like there wont be a nice solution

$f(2n, 1) = 1 - \frac{\sqrt{\pi}}{2} \cdot \frac{\Gamma(n+1)}{\Gamma(n + \frac{3}{2})}$

[woven radish]

Acman

[kind tiger]

$f(2n+1,1) = 1 - \frac{2}{\sqrt{\pi}} \cdot \frac{\Gamma(n + \frac{3}{2})}{\Gamma(n+2)}$

[woven radish]

Acman

[kind tiger]

so you get $(1 - f(2n+1,1))(1 - f(2n,1)) = \frac{1}{n+1}$

[woven radish]

Acman

[kind tiger]

idk if that means anything, i dont see this generalising in any way tho

i suppose using these results its possible to probe a bit deeper at f(n, 2)

but thats getting messy real quick

unless ive made an error somewhere

$f(2n, 2) = \frac{1}{2(n+1)} \left[n - \frac{\sqrt{\pi}}{2} \sum_{i=1}^n \frac{\Gamma(i)}{\Gamma(i+\frac{1}{2})}\right]$

[woven radish]

Acman

[kind tiger]

$f(2n+1, 2) = \frac{2}{2n+3} \left[n - \frac{2}{\sqrt{\pi}} \sum_{i=1}^n \frac{\Gamma(i + \frac{1}{2})}{\Gamma(i)}\right]$

[woven radish]

Acman

[kind tiger]

oh and wolfram is being very kind and saying those sums can be simplified

$f(2n,2) = \frac{1}{2} - \frac{\sqrt{\pi}}{4} \cdot \frac{2n+1}{n+1} \frac{\Gamma(n+1)}{\Gamma(n+\frac{3}{2}}$

[woven radish]

Acman

[kind tiger]

$f(2n+1,2) = \frac{2n}{2n+3}\left[1 - \frac{4}{3\sqrt{\pi}} \cdot \frac{\Gamma(n + \frac{3}{2})}{\Gamma(n + 1)}\right]$

[woven radish]

Acman

[kind tiger]

i was hoping theyd both end up in the form $q(n) \left[1 - C \frac{\Gamma(...)}{\Gamma(...)}\right]$

[woven radish]

Acman

[kind tiger]

for some rational function q and constant C but alas

though perhaps we can posit that $f(2n, m) = p_m(n) - q_m(n) \frac{\Gamma(n+1)}{\Gamma(n+\frac{3}{2})}$ for some rational functions $p_m, q_m$

[woven radish]

Acman

[kind tiger]

and a similar result for f(2n+1, m)

i hope that a) i havent made any (serious) errors and b) any of that is of use to you haha

if you want me to explain how i got any of these feel free to ping

[devout snow]

@ruby talon Has your question been resolved?

[devout snow]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[true vapor]

Can I calculate a) using the impulse formula? I tried it doesn’t give me the correct answer so idk

[cyan palm]

how to use this bot?

[devout snow]

@true vapor Has your question been resolved?

[fossil locust]

https://en.wikibooks.org/wiki/LaTeX/Mathematics and also this help channel does not belong to you

[fossil locust]

no, you need to use conservation of momentum of a)

[devout snow]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[devout snow]

Available help channel!

Send your question here to claim the channel.

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

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

[swift knoll]

A tournament is a (directed) graph with exactly one edge connecting a pair of vertices (with the edge directed in one of its two possible directions). A king in a tournament is a vertex $v$ that can reach all other vertices within two steps. That is, for all vertex $u$ other than $v$, we have either $v\to u$ or $v\to w\to u$ for some $w$. I'm trying to prove the claim that every tournament has a king. \

I'm just looking for feedback for my proof. I'm a bit uncertain about the conclusion. \

Proof. Let $v_1$ be an arbitrary vertex. If $v_1$ is a king, we are done. If it isn't, this means $v_1$ can not reach some vertex $v_2$ within two steps. This means the out-degree (i.e. number of outgoing edges) $d^+(v_2)$ is greater than $d^+(v_1)$. This is because we have $v_2\to v_1$ (recall it's a tournament) and if $v_1\to w$ for some $w$ then we have $v_2\to w$ (otherwise we'll have $v_1\to w\to v_2$, contradicting that $v_1$ is not a king). Continuing this way, we can find $d^+(v_1)<d^+(v_2)<\cdots$ and terminate at some vertex $v_k$ since there are only finite number of vertices.

[woven radish]

psie

[swift knoll]

Now, is v_k a king? Can I just argue by contradiction and claim v_k has to be a king?

[devout snow]

@swift knoll Has your question been resolved?

[devout snow]

@swift knoll Has your question been resolved?

Channel closed

Closed by @swift knoll

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[odd elk]

hi

[odd elk]

y = ln x^e^x

how do I do this?

since:
u = e^x
u' = e^x

please correct me. this is my solution:

y' = (1/x^e^x) (x^e^x)
y' = x^e^x / x^e^x
y' = 1

[winter patrol]

y' = (1/x^e^x) (x^e^x)
how are you getting that

[devout snow]

@odd elk Has your question been resolved?

[odd elk]

based on the given formula from my prof

sorry for the late response

[winter patrol]

what formula

[odd elk]

y = 1/u × u'

oh hey its you! how are you man?

[winter patrol]

ok, but your u here is
x^e^x

[odd elk]

yeah

[winter patrol]

not e^x

[odd elk]

oh typo sorry

i'm really sorry. it's a typo

...

[celest zealot]

so what is it supposed to be?

[odd elk]

oh wait the first one was correct

[winter patrol]

what first one

[odd elk]

the given before I edited it...

[winter patrol]

what given

retype exactly waht you mean

[odd elk]

since:
u = e^x
u' = e^x

[winter patrol]

is incorrect

[odd elk]

u represents the exponent of the problem

how?*

[winter patrol]

you're starting with
$$y = \ln(\red{x^{e^x}})$$
in your first layer of chain rule, $u = \red{x^{e^x}}$

[woven radish]

ℝαμOmeganato5

[odd elk]

okay then what next

[winter patrol]

first apply that

[odd elk]

to what formula

[devout snow]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[winter patrol]

that should be y'

.reopen

[devout snow]

✅

[odd elk]

.reopen

[winter patrol]

first apply that

[odd elk]

how 😞

[winter patrol]

exactly as the formula says

[odd elk]

oh to the formula from earlier

[winter patrol]

don't do anything else more

worry about the rest later

[odd elk]

okay wait

[winter patrol]

leave u' as u' for now in this first step

[odd elk]

still have the outcome of 1

if that's the case

[winter patrol]

no

try to follow my instructions exactly

[odd elk]

got it sir

[winter patrol]

use what i stated for u
apply
y' = 1/u × u'
and leave u' as u'

do nothing else

[odd elk]

oh then 1/x^e^x

[winter patrol]

no

[odd elk]

oh...

can I have a hint..

[winter patrol]

read what i've said carefully

don't overthink

[odd elk]

can I use logarithm properties instead? is it possible?

[winter patrol]

no

don't overthink

this is an exercise of simple replacement

you already have everything needed for this step

[odd elk]

you mean i'm cancelling out u' for now, right?

[winter patrol]

no

[odd elk]

then wdym by leaving u' as u'

[winter patrol]

u' is u'
don't do anything with it

just leave it there exaclty as it is untouched as u'

do not attempt to differenitate that in this step

[odd elk]

then this is it

[winter patrol]

no

leave u' as u' does NOT mean erase u'

its still there

[odd elk]

oh then (1/x^e^x)(u')

[winter patrol]

yes

[odd elk]

alright

[winter patrol]

that's the first application of chain rule

$$y' = \frac{1}{x^{e^x}} \cdot u'$$
$$y' = \frac{1}{x^{e^x}} \cdot \blue{(x^{e^x})'}$$

[woven radish]

ℝαμOmeganato5

[winter patrol]

now you apply chain rule again for the blue part

[odd elk]

then it is x^e^x for the blue part

[winter patrol]

no

how are you getting that

[odd elk]

from my mind

(not being sarcastic)

[winter patrol]

don't do this in your head

[odd elk]

noted

[winter patrol]

this derivative is actually quite tedious

and requires a few applications of chain rule

[odd elk]

I hate chain rule

wait

isn't chain rule formula is

a^u ln a u'?

[winter patrol]

if you had a^u, yes
but your base here is x, which makes things more complicated

[odd elk]

oh yeah, nvm nvm

you know what, I quiti

i'm sleepy i'll just do this tomorrow

i'm starting to see the actual pattern, but i'll continue this instead since I can't fight the tiredness

thanks man! @winter patrol

.close

[devout snow]

Channel closed

Closed by @odd elk

Use .reopen if this was a mistake.

[devout snow]

Available help channel!

Send your question here to claim the channel.

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

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

[shy vault]

I need to solve 16a, all I've done so far is written a composite function, that is equivalent to f^3(x)

but i don't see how that helps, since there are 3 unknowns (or well the x isn't a problem really, but a and b being in the same equation is)

[sand dove]

So you've computed f^3(x) given f(x)?

What did you find?

[shy vault]

uh a(a(ax +b)+b)+b

[sand dove]

And when you develop?

[shy vault]

gimme a sec

[shy vault]

a^3x + a^2b + ab + b

[olive snow]

Seems good

[shy vault]

I used a calculator for the 64 part

but a has to be 4?

[olive snow]

Yes

[shy vault]

and b is 1?

okay yay

yippiee tyyy

.close

[devout snow]

Channel closed

Closed by @shy vault

Use .reopen if this was a mistake.

[devout snow]

Available help channel!

Send your question here to claim the channel.

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

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

[lusty tapir]

integral of 1/(x^4- 2x^2 - 3) dx

[lusty tapir]

original question was

integral of 1/([x^2-4] root{x+1})

i took x+1 = u^2

got x^2-4 as the denominator

[lusty tapir]

i have this in form of u

if i take x^2 here at t so 2xdx = dt dx=dt/2roott

so this becomes 1/(2{root t}[t^2-3t-3])

now t = smth else ^2 i tried and it just a circle

[lusty tapir]

got back to here

[wind mason]

I don't think you need to

You can factorize x^4-2x^2-3

Hint: ||Let u=x^2||, but not in the integral

[wind mason]

Then proceed with partial fractions

[lusty tapir]

mere sub?

[wind mason]

?

[lusty tapir]

mere substitution?

[wind mason]

What I want you to do is

[wind mason]

You don't need to sub in the integral

[lusty tapir]

just used to simplify and factorize or find A B C in partial fractions

[wind mason]

Never heard of it

Anyways

What did you get for the factorization?

[lusty tapir]

(x^2+1)(x^2-3)

i got it from here

1 more question i had

1/(x[x^5+1])

[wind mason]

Ah

You can just take out the x^(5)

Let me show

[sacred salmon]

Someone help with 2 4 and 5 please

[wind mason]

$\int\frac{1}{x(x^5+1)}dx=\int\frac{1}{x^6(1+x^{-5})}dx$

[woven radish]

;(

[wind mason]

!occupied

[devout snow]

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

[wind mason]

@lusty tapir This is a really nice technique known as the "Caveman Technique"

[lusty tapir]

oh

so name take x^-5 = t

bring x^6 to numerator

we get -5log |1+t| + c

yayyyy

[lusty tapir]

first time hearing about it

my teacher didnt teach with name of techniques or any technoques he wants us to just use our own logic and figure it out

[wind mason]

Lol

[lusty tapir]

can you tell me names of other such techniques?

[lusty tapir]

i know there DI for uv integation and kings smth smth but idk how to do them

[devout snow]

@lusty tapir Has your question been resolved?

[wind mason]

I can't really name them off the top of my head

[devout snow]

@lusty tapir Has your question been resolved?

[devout snow]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[velvet wren]

Where did i mess up

<@&286206848099549185>

[devout snow]

@velvet wren Has your question been resolved?

[devout snow]

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.

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[sacred cloak]

hello

[sacred cloak]

requesting help with this problem

we are given the function f(x) = ax^3 + ax^2 + x

determine how the amount of critical points

is dependent on the value of a

and the character of those critical points

I've derived the function and completed the first part of the question

don't know how to get the character of the points

[astral lodge]

f'(x)=3ax^2+2ax+1

so $x=\frac{-2a\pm\sqrt{4a^2-4(3a)(1)}}{6a}$

[woven radish]

vengeance

[sacred cloak]

I used the quadratic formula yes

[astral lodge]

Then just use determinnat

if determinant>0, 2 real sol

if determinant=0, 1 real sol

if determinant<0, no real sol

[sacred cloak]

ah yes

what do I do after that

because

[astral lodge]

Just analyze when discrim>=0

[sacred cloak]

f''(x) = 6ax + 2a

[astral lodge]

And say those are the a-values for which critical points exist

Yes

[sacred cloak]

when a = 3

determinant = 0

f''(3) = 18x + 6

don't know how to interpret that

[gray summit]

You're supposed to get -11 + 3√3 for the first part. And in the [2-1], it's actually [2-18]. Therefore, the final answer would be 5 + 3√3 as intended

[devout snow]

@sacred cloak Has your question been resolved?

[sacred cloak]

.close

[devout snow]

Channel closed

Closed by @sacred cloak

Use .reopen if this was a mistake.

[devout snow]

Available help channel!

Send your question here to claim the channel.

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

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

[trim quarry]

Can anybody help me with this problems ive been trying for hours and i dont know what else to input

[abstract zodiac]

Share the problems

Don’t ask people to commit before they’ve seen the problem

dontasktoask.com

[trim quarry]

Give me a freaking second damn

The first photo is the supposed example

[pseudo basin]

how did you conclude C=2? and what's that 12B/12 doing on the other side of it?

also that's a weird way to write pi

[trim quarry]

Thats the formula for finding the period

Can anyone in this discord actually help me

[pseudo basin]

ok first off please hold off on the entitlement.

2pi/B equals the period, yes.

but the way you wrote it, in the place where the period is supposed to go, you put 12B/12 -- which is just B.

i'm asking you what's up with that.

[trim quarry]

To get B on one side

[pseudo basin]

no, hold on

ok let me ask it again this way

[pseudo basin]

what is the period of your function as read from your table of values?

[trim quarry]

12

[pseudo basin]

are you sure?

[trim quarry]

No

[pseudo basin]

the entire interval covered by the table is [0, 12] which does have length 12, but does that constitute exactly one complete period?

i, for one, don't think so -- you've got those 2's in the y-row which are not 12 units apart on the x-axis.

do you see what i am talking about?

@trim quarry

ok imma need to disappear for a few

[trim quarry]

A little

Im still working through it

Try period 8 which is pi/4

I redid this is what I got so far

[pseudo basin]

ok the inside of the sine is now correct but the amplitude and midline arent

your formula has the midline as 0; surely this isnt the case?

[trim quarry]

I got it from the graph

[pseudo basin]

so you're certain the midline is 0?

[trim quarry]

Not in the first solution I sent but I change it in the latest photo to 4

The amplitude I counted from one end to the other and got 8

[pseudo basin]

i do not see midline=4 in this formula. i see no D term at all, which as-written would mean it's zero. is it cut off?

[pseudo basin]

wdym by "one end to the other"?

[trim quarry]

The dots

Thats how they said to find it in the example

[pseudo basin]

i think maybe you are confusing yourself, or not articulating yourself well about what you did, or some mix of the two.

[trim quarry]

I can draw a line if its easier

[pseudo basin]

but the highest and lowest y-values are 6 and 2 respectively.

the difference between them is 4.

half that gives the amplitude.

[trim quarry]

So 6-2/2 is how you get the amplitude

Ok

[pseudo basin]

brackets!

you can get the amplitude by many different ways.

you can either take (max - min)/2.

or, if you already know the midline, you can find it as |max - mid| or |min - mid|, no halving.

[trim quarry]

Ill use this formula from now on

And you said my period is correct?

[pseudo basin]

your period and phase shift are, yes.

also, if you're using desmos, you can always write your formula into it to have that graphed alongside your points.

so you can easily see if your graph matches.

[trim quarry]

is the midline just the exact middle of the graph

Thats how I got 4

[pseudo basin]

this 4?

[trim quarry]

This is what I got right now

What do we use the midline for again

[pseudo basin]

...

ok you are dropping the numbers around like rotten eggs

[trim quarry]

What do you want me to do im not good at math im trying

[pseudo basin]

i want you to stop trying to hold a million things in your head and mess them all up, first and foremost

[trim quarry]

Ok, right now 2 is the amplitude pi/4 is the period and so now I just need to figure out D and C to finish the rest

[pseudo basin]

the general formula for a sinusoid is: $$y = A \sin\paren{B(x-C)} + D$$ where:
\begin{itemize}
\item $B$ is the stretch factor, which indirectly controls the period -- in your case, $B = \frac{\pi}{4}$ is correct
\item $C$ is the phase shift, which for you is $4$ --- and $4$ was correct, and you had no reason to change it now
\item $A$ is the amplitude, which in your case is $2$
\item $D$ is the midline, which in your case is $4$
\end{itemize}

[woven radish]

Ann

[pseudo basin]

i hate to say this, but

task failed successfully

what is that =4 doing? why isn't it a +4?

i can understand the dropped y=, but the plus sign should not have transmogrified into an equals.

[trim quarry]

Yea

Just a brain fart there

But yea 2sin(pi/4(x-4))+4

@pseudo basin

[pseudo basin]

what

yes now it's OK

[trim quarry]

Let me try it

Thank you

[devout snow]

@trim quarry Has your question been resolved?

[devout snow]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[jade jackal]

pls someone help me idk where to start

[grand warren]

Applies the logarithm in the equation

[jade jackal]

presh

.close

[devout snow]

Channel closed

Closed by @jade jackal

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[storm sphinx]

Slope fields

[storm sphinx]

I'm not sure how to get the answer

dy/dx=0 at y=-1

However, when you differentiate all of the equations, they all only have x

[lunar harbor]

This is saying that when y = -1, dy/dx = 0

so you can solve for the value of x such that this happens in each case

and check if dy/dx is indeed 0

but this is too much work

Just recall that you can "trace" the solution in a slope field

since we have that the graph of the particular solution goes through (0,0), the graph of the particular solution looks like this

it should be a bit "smoother", but you get the idea

[storm sphinx]

i see, but if we use the graphical approach, then how do we know it is y=e^(-x)-1 instead of y=e^(-x^2)-1?

[lunar harbor]

$e^{-x^2}-1$ is symmetric abt $x=0$

[woven radish]

Civil Service Pigeon

[lunar harbor]

(note that this is the same thing as saying that the function is even)

[storm sphinx]

i see, thank you!

.close

[devout snow]

Channel closed

Closed by @storm sphinx

Use .reopen if this was a mistake.

[devout snow]

Available help channel!

Send your question here to claim the channel.

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

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

[lucid rock]

Prove that in worst case if you apply pruning in an Voracious algorithm for solving for example TSP The complexity in worst case with Big-O notation will still be n!. How could we prove this

[lucid rock]

I want to understand it deeply better

[devout snow]

@lucid rock Has your question been resolved?

Channel closed

Closed by @lucid rock

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[rotund linden]

Plz I need help on trigonometry

[devout snow]

Please don't occupy multiple help channels.

[rotund linden]

Plz can u help me with trigonometry

[devout snow]

@rotund linden Has your question been resolved?

Channel closed

Closed by @rotund linden

Use .reopen if this was a mistake.

[devout snow]

Available help channel!

Send your question here to claim the channel.

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

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

[copper sigil]

<@&286206848099549185>

this is my first time asking questions on this discrod server, is this how I do it?

[devout snow]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[lavish nimbus]

You usually wait for 15mins before pinging helpers.

[copper sigil]

oh okay, my bad, first time doing this

[lavish nimbus]

Apply the formula of volume of a cone, and substitute the values. You should find a relation between the big and small one.

(I'm uncertain on this one, but I guess it's worth a shot)

[copper sigil]

so the volume of the icecream is 9 pi

but I still don't know the radius of the icecream

[lavish nimbus]

How did you get it?

[copper sigil]

isn't the volume of the cone 18 pi?

[lavish nimbus]

My bad, you're right. I thought you meant the big one is 9 pi.

[copper sigil]

and if it says that the volume of the ice cream is half the volume of the cone, than doesn;t that means the volume of the ice cream is 9 pi?

[lavish nimbus]

You should find a ratio of $$\frac{r_s}{r_b}=\frac{h_s}{h_b}$$

[woven radish]

Good

[lavish nimbus]

You know $r_b$ and $h_b$, now isolate $h_s$.

[woven radish]

Good

[devout snow]

@copper sigil Has your question been resolved?

[chilly quest]

Hey sir! I am just curious about that question . Is the answer 4.76?

It would be awesome if u could help me check 🙏

[copper sigil]

now i'm even more confused

😭

[chilly quest]

I solved it this way I am no sure if this works

[devout snow]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[copper sigil]

wait, I think you solve this using similiar triangles

[chilly quest]

What’s that

[devout snow]

Available help channel!

Send your question here to claim the channel.

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

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

[copper sigil]

wait, give me five minutes to solve it, I send a picture to you after

[chilly quest]

Oki thanks

[astral lodge]

ezpz similar triangle

you should also use constant of proportionality

[chilly quest]

What’s that

[astral lodge]

so if the ratio of volumes is 1/2, side ratio would be 1/cbrt(2)

[copper sigil]

it's like triangle abc is similiar to triangle edc

[chilly quest]

Oh fire

[astral lodge]

very good method to use

so the depth would just be 6/cbrt(2)

[chilly quest]

Ohhh

[chilly quest]

Is that the real name for similar triangles ?

[astral lodge]

it is close to it

[chilly quest]

Oh

[astral lodge]

like

[chilly quest]

Thanks

[astral lodge]

you welcome

[copper sigil]

I got 7.35 cm

[chilly quest]

Hi pro , do u think u can calculate that answer

[chilly quest]

We got different answer we dun know who correct

[copper sigil]

sorry for the messy handwriting and working out, but that's how I got the answer

[astral lodge]

oh ok

,w 6/cbrt(2)

[woven radish]

Results provided by WolframAlpha

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

Requested by vengeance0548

[chilly quest]

Thanks

[astral lodge]

yeah same as you

[astral lodge]

use my method

certified method fr

[copper sigil]

oh okay, thxs

[devout snow]

@copper sigil Has your question been resolved?

[devout snow]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[vocal mural]

The recursion I found is P1:P3 was:

$a_{n+1} = a_n + (n+1)$

[woven radish]

eugene

[vocal mural]

Should I be using Partial Fraction Decomposition'?

What am I trying to find here in Problem 3?

Expanding the generating function?

[low plover]

from where you all get these questions?

[vocal mural]

on my paper

[low plover]

do you get paper as softcopy?

[vocal mural]

I have both but my work is written on physical paper

[low plover]

ok

[vocal mural]

Im going to use Partial frac decomp and then find the simplified gen function

but i am not sure why i am using my previous recursion

[low plover]

this question is of which grade?

[vocal mural]

its combinatorics

[low plover]

grade?

[vocal mural]

undergrad

[low plover]

?

[vocal mural]

wdym grade

[low plover]

class

like 11th 12th or college

[vocal mural]

college

undergraduate

[low plover]

ok

[vocal mural]

nvm i dont think i am using PFD

[wicked rover]

$a_n=a_{n-1}+n$ for $n\ge1$

[woven radish]

ロケットジャンプ

[wicked rover]

use this to compute the generating function
$$f(x)=\sum_0a_nx^n=a_0+\sum_1a_nx^n$$

[woven radish]

ロケットジャンプ

[vocal mural]

Ohh okay

[wicked rover]

btw those sums are 0 to infty etc

[astral lodge]

what kinda shorthand notation

golly

[devout snow]

@vocal mural Has your question been resolved?

[wicked rover]

if the index range is clear then we can sometimes even leave it out entirely 🙂

[astral lodge]

that just feels weird though 😭

[wicked rover]

$\sum a_n$ enjoy