help-42

[golden dragon]

object oriented programming

[hushed badge]

ok

[golden dragon]

computer organization

so far 3

[hushed badge]

ohh

we have 6 courses in each sem

[golden dragon]

thats alot

[hushed badge]

bro let me send you my syllabus

[golden dragon]

so mostly from how you study you watch youtube and practice problem right

[golden dragon]

sure

[hushed badge]

I have also learned computer organization

just took the paper on 1 may

[golden dragon]

the 1010 1000 binary additon thing right?

[hushed badge]

right

micro instructions

registers

[golden dragon]

ok we learn that

[hushed badge]

shift operations circular operations right ??

[golden dragon]

yep

[hushed badge]

memory reference instructions

register reference instructions

right this kind of stuff

[golden dragon]

yessir

i thing some not in my syllabus

but most of them

[hushed badge]

can you send your syllabus

[golden dragon]

computer organisation?

[hushed badge]

yes

[golden dragon]

hold on

this so far

[hushed badge]

you dont have detailed syllabus

[golden dragon]

yep

it just spread out messy

[hushed badge]

ok

let me send you mine

[golden dragon]

gimme 1 min

[hushed badge]

you there ?

[golden dragon]

back

damn ur syllabus is clean

[hushed badge]

lets chat on personal

[golden dragon]

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[calm coral]

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[frosty leaf]

You are playing a game where you start from a very high number. You flip a coin: if it lands heads, you subtract one. If it lands tails, you subtract two. The game ends if you reach either 1 or 0. What is the probability you reach 1?

[dusty crescent]

Ummm... How would you approach this problem?

[frosty leaf]

I drew a graph thingy but idk if it's the right approach

[left sorrel]

you mean states?

[frosty leaf]

like this

[dusty crescent]

Keyword here being "very high number"

starting from 5 might get you biased in one way or another

[frosty leaf]

yeah I was trying to figure out a pattern but I'm not sure how to approach it generally

[dusty crescent]

I mean, the only thing that really matters is the last coin flip isn't it?

if you reach 2

[frosty leaf]

yeah, but if you reach 3 you can get to 1 immediately

[dusty crescent]

I feel like this is a game of even and odd numbers

like, 1 coin flip gets you to an even

another gets you to an odd

the question is are you more likely to end up odd or even

with a large starting number

Or rather

subtracting 1 changes if it's odd or even

and subtracting 2 keeps the oddness (eveness lol)

[frosty leaf]

is it 50/50 then? I feel like it is not for some reason

because 0 seems harder to reach than 1

[dusty crescent]

hmmmm... to get to 0 you need to ensure that you have an even number of 1 rolls

To get to 1 you need to get an odd number of 1 rolls

assuming the initial number is even

but if it's odd it's the other way around

I guess

I think I just confused myself

[frosty leaf]

xd

[dusty crescent]

Hold on, lemme make a program to check this empirically

[quaint sphinx]

it's not

[quaint sphinx]

my approach to this was computing the SM matrix

[frosty leaf]

what's that?

[dusty crescent]

so uhh... on a sample size of 10 I get 5 times 1

and 5 times 0

small sample size ik

but it's funny I got 50/50 lol

did it in excel

[quaint sphinx]

interesting

[dusty crescent]

can't be bothered to do it in python

probably should

and then run a sample size of 100000

or something

[quaint sphinx]

you'd very quickly see the answer with a number as small as 10

[dusty crescent]

no I went with a random large number

545

but I ran the coin flips 10 times

Like so

Ik this isn't mathematical proof but I found it interesting

[quaint sphinx]

call a transition matrix P in canonical form, which can be split into matrices R, S, O, Q

matrix R is the transition matrix for the recurrent classes (in this case, the 2x2 identity for the absorbing states of 0 and 1)
matrix S has the transition probabilities from the transient classes into the recurrent classes (in this case, the 2-by-(n - 1) matrix which has 3 entries of 0.5 and 0s elsewhere)
matrix O is the (n - 1)-by-2 zero matrix
matrix Q has the transition probabilities between the transient classes (in this case, the (n - 1)-by-(n - 1) matrix)

matrix M is the inverse of (I - Q)
the product SM provides the probabilities of starting at a state and being absorbed into state 0 or 1

i computed this product and limited it to infinity to get a result

(there is likely an easier way to do this using high school method, but i recently learned about the SM matrix and wanted to apply it)

[frosty leaf]

interesting

so what's the final answer?

[quaint sphinx]

1/3 to reach 0
2/3 to reach 1

[frosty leaf]

oh nice

that's what I figured too after doing it for small numbers

it looked like it converges to 2/3

very interesting

[quaint sphinx]

yeah it's a quick convergence by geometric series

each term for n > 1 is defined by 1/2 + sum from 2 to n - 1 of (-1/2)^i

[frosty leaf]

yeah I got 1/2, 5/8, 11/16, etc

thanks for the insight

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[dusty crescent]

Just for fun, on a sample size of 1000

[frosty leaf]

nice

[calm coral]

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[chrome barn]

Anyone have any idea on how I could start this problem

[sharp narwhal]

g''(0)/2

[vocal summit]

well, the coefficient for $x^2$ in the maclauran series for g would be $\frac{g''(0)}{2!}$, so just find g''(0) from $e^{f(0)}$ via $f'(0)e^{f(0)}$, and take derivative again.

[chrome barn]

woah

[vocal summit]

huh

[potent lotus]

fish

[chrome barn]

ohhh

okay got it

idk what i was thinking lol

alright thanks

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[lilac shoal]

Why is it B?

[glacial inlet]

if we plug in let’s say 1 for x, we would get -2/1, which is a negative number

E doesn’t correlate with that

so it must be B

[rigid mist]

If you take calculus you’re going to learn to recognize the graph of 1/x really fast (in this case -1/x)

[lilac shoal]

I am going to take further maths next year and that involves calculus I guess

[lilac shoal]

Why does it not correlate? Is it because when x is positive you can't get a negative y number?

[rigid mist]

E is the graph of 1/x

[lilac shoal]

I see

[rigid mist]

It’s Like this, they’re opposites

[lilac shoal]

Oh that makes sense

[lilac shoal]

Thanks

@glacial inlet thanks, I understand your explanation perfectly

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[steady ocean]

okay for this one

I know I want y-y1=f'(x) (x-x1)

and simplify

so to start

f(-1)

3(-1)^2=3 , - 2(-1)=2, 3+2=5+1=6

so (-1,6)

take the derivitive of 3x^2-2x+1

using the power rule

we get 6x-2

sp

y-6= 6x-2 (x--1)

what now?

[rigid mist]

Foil (6x-2)(x+1)

Then simplify

Wait

Oh lol

[steady ocean]

what did I do

[rigid mist]

Plug into the derivative to find the slope of the tangent line at that point and use it for m

[steady ocean]

6x^2+4x-2

[steady ocean]

oh

[rigid mist]

y-y1=f’(-1)(x-x1)

[steady ocean]

so f'(-1) is -8

6x-2

6(-1) -2

y-6=-8(x+1)

and simplfy

y+8x+2=0

right?

@rigid mist all good?

[rigid mist]

Almost

y-6=-8x-8
y=-8x-2

Oh it isn’t slope intercept nvm

Yeah that’s good

[steady ocean]

thanks

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[mossy wigeon]

how do we start with this question

[calm coral]

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[dapper acorn]

here why the position vector of G1 is( b+c+d)/2

[rustic osprey]

Did you mean (b+c+d)/3?

If so, recall that the centroid splits the median in a 2:1 ratio from the vertex

[dapper acorn]

yeah

[rustic osprey]

We have smthn like this

pretend it's more general than an isosceles right triangle

you get the idea

anyway, we know that $$\overrightarrow{OG}=\overrightarrow{OM}+\overrightarrow{MG}$$

[potent lotus]

Civil Service Pigeon

[rustic osprey]

now we want to reframe this in terms of the vertices themselves

any ideas?

[dapper acorn]

no what do this means

[rustic osprey]

basically get the right hand side in terms of $\overrightarrow{OA}$, $\overrightarrow{OB}$, and $\overrightarrow{OC}$

[potent lotus]

Civil Service Pigeon

[rustic osprey]

The vector OM should be an easy start

(note M denotes the midpoint of AB)

[dapper acorn]

yes

[rustic osprey]

so ... ?

@dapper acorn Do yk how to find the position vector OM or not

[dapper acorn]

m-o

[rustic osprey]

(in terms of the position vectors of the vertices)

[dapper acorn]

i can't remember

[rustic osprey]

It's $\frac{1}{2}(\overrightarrow{OA}+\overrightarrow{OB})$

[potent lotus]

Civil Service Pigeon

[rustic osprey]

since it's halfway b/w the pts A and B, you just take the "Average" of their position vectors

does this make sense?

[dapper acorn]

yeah

[rustic osprey]

alright

$$\overrightarrow{OG}=\frac{1}{2}(\overrightarrow{OA}+\overrightarrow{OB})+\overrightarrow{MG}$$

[potent lotus]

Civil Service Pigeon

[rustic osprey]

now we need to deal w/ vector MG

recall what I said earlier abt the 2:1 ratio

how can we rewrite that as a vector in terms of one of the vertices

[dapper acorn]

i think i understood that to find position vector of g here we have to add p.v. of A+B+O/no of vertices

[dapper acorn]

is this what happened here

[rustic osprey]

I mean that's a well known result for the centroid

I was going through the proof of it but you can always look it up

[rustic osprey]

not sure if you're trying to go the centroid of some regular n-something

ig you can always search it up

[dapper acorn]

I started studying vectors now so idk but thanks of help

[jovial eagle]

I have a quick question

[calm coral]

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).

[rustic osprey]

btw do you need anything else

[calm coral]

@dapper acorn Has your question been resolved?

[dapper acorn]

so you were explaining me the proof i didn't understood that but now i also understood the proof you explained and left part from online so thanx for the help

[calm coral]

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[pulsar ice]

Why the bounds for theta in the answer is from -pi/2 to pi/2? Why not from 0 to 2pi?

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[eager forge]

How do i do d

[calm coral]

@eager forge Has your question been resolved?

[eager forge]

<@&286206848099549185>

[potent heath]

Yes?

[eager forge]

how do i do d?

Im not really sure how to start it

[potent heath]

Lemme look at it

So does the approximation you got in part A, does it overestimate or underestimate the volume? So technically you need to give an explanation, if it overestimates and or underestimates the volume of the tank. Hope that helps!

[eager forge]

What about d

[potent heath]

okay i dont think im much of help on this.. im not too sure.. but i would put 5 in (H) and then 6.5 in the lower part.. where (H) but im not to sure sorry! im mostly alg.. sorry!

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[peak mango]

How the heck do I do question 9 b

[dull wagon]

did you do part a)?

[peak mango]

Yes

It's 30.16cm

Sauare

[dull wagon]

sauare?

[peak mango]

Squared

[dull wagon]

you shouldn't be getting squared units

[peak mango]

Oh wait

No

Yea I just wrote 30.16cm

I don't know why I said square

[dull wagon]

you are told that this will be the circumference of the circular base

so to find b), they want you to find the radius of a circle with this circumference

[peak mango]

Yea

[dull wagon]

and there's a relation between radius and circumference

[peak mango]

Which is ?

2 pi r

[acoustic temple]

2pir

[peak mango]

And pi r square

[dull wagon]

pi r^2, not relevant to this part

[peak mango]

Ok

So do we do 30.16 division by 2 time pi

And we get4.8

Ohh

Ok

That's it?

[dull wagon]

for the radius, yes

[calm coral]

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[rotund perch]

i did P = 0.5(1-e^-t) + 0.5(1-e^-2t)

i'm not sure how to go from here

does this have something to do with uniform random variables

[glacial oxide]

Hi guys. so i'm going through a 'draw on desmos' phase and I'm not sure what equation to put to get a happy face (like the line)

[final verge]

!occupied

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[glacial oxide]

Hi guys. so i'm going through a 'draw on desmos' phase and I'm not sure what equation to put to get a happy face (like the line)

[urban relic]

very important math question

try a very wide parabola thats a few units down

and restrict it on the sides

something like this maybe

[orchid ridge]

beautiful

[urban relic]

thats so extra

[calm coral]

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[violet solar]

When you merge all trigs 🥹

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[chrome plume]

How to fijd the ingetral of y)

[sharp narwhal]

pfd

[novel token]

!help

[calm coral]

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

[chrome plume]

But what do I do with x^4+4

[sharp narwhal]

sophie germain identity

[chrome plume]

Is this some basic knowledge i should know

Or is there any way without knowing this one

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[clear hill]

hello, I have a simple question. How this equation turns to that (canonic form)
(image 1 is the equation and the second one is the solution)

[verbal finch]

U multiply?

[clear hill]

multiply what exactly?

where the 2 and 3 go

[verbal finch]

What did the LHS become after multiplication?

@clear hill

[clear hill]

what's LHS (im not native sry)

[verbal finch]

Left hand side

Isn't it -x^2 + x + 6?

[clear hill]

wait a sec

[verbal finch]

Mhm

[verbal finch]

U made a mistake

Multiply again

Properly

How did u get 2x

[clear hill]

wait where

I used this

[verbal finch]

How did u get 2x+1?

[clear hill]

idk in english

[verbal finch]

It is not in that form

How could u use that m

[clear hill]

uhh

[verbal finch]

Ok how do you multiply (ax+b)(cx+d)?

[clear hill]

ohhh wrong thing

[verbal finch]

Not this also

[verbal finch]

How do you prove this?

[clear hill]

wait pls

[verbal finch]

Ok

[clear hill]

sry it wasn't me that did this

she doesn't know how she reached that result

we don't know how to start at this point, because the answer is that one

like where is the 2 and the 3?...........

[verbal finch]

Ok so what do you get when u multiply (ax+b) and x?

[clear hill]

what.......?

wdym ax+b

and the x

[verbal finch]

Multiply those both

I'm trying to explain the general thing

[clear hill]

okok

[verbal finch]

So multiply those

What will u get

[calm coral]

@clear hill Has your question been resolved?

[clear hill]

ok forget abt it, thx a lot

.close

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[golden dragon]

how do i solve this

[reef nimbus]

well to solve 1 and 2, you'll have to know 3 pretty much

what do you think the largest possible size could be

[golden dragon]

hmm lemme think

the two possiblities is froom question 1 and 2 right?

when there are no common elements in sets a b or c

[reef nimbus]

right

[golden dragon]

each element of a b and c are unique

[reef nimbus]

yup

[golden dragon]

therefore its a sum?

[reef nimbus]

yup

[golden dragon]

i got 52

so how do i start from first question

[reef nimbus]

that is the first question

[golden dragon]

oh wait what

52 is first question?

[reef nimbus]

yes, it's the largest possible size

[golden dragon]

ah yes yes its the maximum

did i answer the 3 ?

[reef nimbus]

you did first half of 3 (largest possible size occurs when they have no common elements)

[golden dragon]

now uh the minimum hmmm

maximum = not common minimum = common?

vice versa?

so if its common between a b and c does it mean intersection?

[reef nimbus]

well yeah but the problem isn't asking about intersections

[golden dragon]

its asking about min

[reef nimbus]

but you're right that the lowest possible size will be when there are common elements

[golden dragon]

is it possible to find the value

[reef nimbus]

yes

[golden dragon]

we know the value by adding if thats the max

substract?

oh no then

we get 0

[reef nimbus]

think about if one is a subset of the others

[golden dragon]

assume that we try take a set with largest possible size

example in the set there is a b c d e f g ..... so on

like set A

so high chance / probability are there are letter in set B and C is in set A

yes?

[reef nimbus]

I wouldn't think of it in terms of probability

[golden dragon]

ahh my mistake

so means Set B and C is a subset of Set A because all the letters in Set A are also in Set B and C

[reef nimbus]

right

so what's the smallest possible size of A u B u C

[golden dragon]

22?

[reef nimbus]

but A has 30 elements

[golden dragon]

8

[reef nimbus]

how can the union have less than the elements in A?

the union includes everything in A

[golden dragon]

lets see

so what we already know that the largest amount of union is 52

yes

[reef nimbus]

yes, if the sets are all disjoint (no common elements)

[golden dragon]

ok this part im really stuck

[reef nimbus]

you had the right idea already

if A = set of english letters, so |A| = 26. let's say B is the set of consonants and C is the set of vowels

so |B| = 21, |C| = 5

what is the size of the union?

[golden dragon]

52

[reef nimbus]

there are 52 distinct letters of the alphabet?

[golden dragon]

there are only 26

[reef nimbus]

right

[golden dragon]

so you want to narrow it down

[reef nimbus]

idk what you mean by narrow it down

[golden dragon]

mistake agian

[reef nimbus]

A u B u C is the total of the distinct elements of A, B, and C

[golden dragon]

got it

its like number of toys intead type of toys

[reef nimbus]

how about another simple example

let's say you have 3 classes A, B, and C

A = {Tom, Dick, Harry}. B = {Tom, Dick}. C = {Tom}. what is A u B u C

[golden dragon]

tom dick harry

[reef nimbus]

right. so |A U B U C| = 3 = |A|

if A (the larger set) contains B and C, then |A U B U C| = the size of the larger set

[golden dragon]

so in A ur saying theres no duplicate right?

[reef nimbus]

sets never have duplicates

[golden dragon]

example like tom tom tom and dick harry

[reef nimbus]

exactly, since there is only 1 Tom

[golden dragon]

woah my brain opening

so if i would give another example

lets say set A theres every letter which is 26 alphabet

b have 7 and c have 3

ok wait

em

there are letter in b and c that comes from A

in set A represent a -z

so my mistake before was taking set A subtracting my own subset

[reef nimbus]

yea

[golden dragon]

its like terminating ur own children

not include with it

hmm

is it 26 then?

[reef nimbus]

yeah

[golden dragon]

gosh bro

dang it

[reef nimbus]

but back to the stated problem, what would be the smallest possible union

[golden dragon]

30?

[reef nimbus]

yup

[golden dragon]

maaaaaaan

waaaa

dang didnt know it was easy but make it so complicate

[reef nimbus]

just takes practice

[golden dragon]

so to summarize the biggest set is the min of the union

which include every children in there " no duplicates"

[reef nimbus]

yes, if the biggest set contains the others

[golden dragon]

or for instance merge

we blend the c and b into A

right?

[reef nimbus]

union is kind of like merging, yeah

[golden dragon]

why i didnt think of that earlier

sheeesh

did we solve the 3 yet

or havent

[reef nimbus]

yeah

the first condition is when all sets are disjoint (no common elements)

the second condition is when A contains B and C

[golden dragon]

so its more like writing a hypothesis?

[reef nimbus]

not sure what you mean. I think so yea

[golden dragon]

i have a hard time listen to lecturer tbh

luckily you explained for me clearly

[tight lichen]

hi, sorry to interrupt but i think that q.2 is 8

im not really getting it

[golden dragon]

💀

how

[tight lichen]

because c is 8

[reef nimbus]

they are asking about the union of A, B, and C

since A has 30 elements, the union must have at least 30 elements as well

[tight lichen]

they are asking probability of A or probability of B or probability of C

not and

[golden dragon]

uh this is sets topic

not probability

[reef nimbus]

there is no mention of probability

[tight lichen]

ok im dumb, i get it now

thanks

[golden dragon]

@reef nimbus can you verify this question if i did correct

[reef nimbus]

not too sure on these problems, but I think the first one is wrong

should be 65, since you take |N| + |S| + |M| and subtract those who are in at least 2 classes, so 75 - 10 = 65

[golden dragon]

ww

oww

[reef nimbus]

but I'm not 100% sure. haven't done these in a long time

[golden dragon]

ur in cs course?

[reef nimbus]

then part 2 should be 90 - 65 = 25... maybe

i'm not in school anymore

[golden dragon]

dang

[reef nimbus]

you could ask in #discrete-math or #probability-statistics , they might have better help

[tight lichen]

i was thinking 1 is 55

[golden dragon]

@tight lichen 💀

bro how long you been in this chat

[tight lichen]

i just came in

[golden dragon]

ahh

[tight lichen]

i walk around see if i can learn anything

[golden dragon]

did you learn anything

[tight lichen]

for q1, o think the outer of your venn is correct but the overlapping bits are wrong

the overlapping bits should be a total of 10

[golden dragon]

hmm

try check the discussion

search laughingcoder

[tight lichen]

and the union is correct

[golden dragon]

he helps me yesterday

[tight lichen]

so my final answers is Q1. 55, Q2. 35, Q3. 45

[golden dragon]

how did you prove q1 tho

[tight lichen]

30+10+5+10

[golden dragon]

10 is the flower shape right?

[tight lichen]

the 4 overlapping circles if thats what you mean

[golden dragon]

yes

the one i wrote 10 - 8

yes?

[tight lichen]

this is my venn, see if it makes any sense to you

[golden dragon]

got it

[tight lichen]

the question never direcly specifies you need to know exact values for the students doing 2 things, you just need to know how many

in total

[golden dragon]

thats i never know the value inside does petal right?

it could be random

[tight lichen]

yes

but you know it adds up to 10

[golden dragon]

but the question give 10 in at least two of three courses

which is 2 <= 3

[tight lichen]

i correct my venn the =10 should be =8

my bad

[golden dragon]

8 but break it down and distributed between the 3 petal

yes?

and added up to 10

[tight lichen]

im not quite sure what you are saying

what grade are you in

[golden dragon]

freshman

in computer science

[tight lichen]

ah

[golden dragon]

yeh

hbu

[tight lichen]

GCSE Y10

[reef nimbus]

how did you find the number for each class 30, 10, 5

[golden dragon]

i substract the flower in the middle

[tight lichen]

the given number - 10

[golden dragon]

and add the sides

[golden dragon]

yes

[reef nimbus]

but there's a small part that should not be subtracted

the middle intersection we know is 2

call the others a, b, and c

so the students taking N alone would be 40 - 2 - a - b

[golden dragon]

so the 3 sides we know added to 10? right?

[reef nimbus]

but a + b is not 8

because a + b + c = 8

[tight lichen]

yes

[golden dragon]

but the purpose of question does not ask us to find the intersection right?

[reef nimbus]

the answer for exactly one course should be 53

for at least one course should be 63

and for zero courses should be 90 - 63 = 27

those are my answers

[golden dragon]

i need use the same formula right?

[reef nimbus]

call the other intersections (besides the middle one = 2) as a, b, and c

then the number of students taking only N = 40 - 2 - a - b

similarly, those only taking S = 20 - 2 - b - c

and M = 15 - 2 - a - c

add all these up gives you 75 - 6 - 2a - 2b - 2c = 69 - 2(a + b + c) = 69 - 2 * 8 = 53

[golden dragon]

damn

[reef nimbus]

I think

oops, then total taking at least one course would be 63, you had that right

[golden dragon]

cuz the middle is overlapping

so i need remove n i m i s

2 times

but for question 2

why do i need to substract total with previous answer but not will the total of student who enrolled in courses

which is 75

[reef nimbus]

some of those 75 will be "double counted"; i.e., in more than one course

sorry I gotta go

[golden dragon]

ohh its okay

thanks for the help!

[reef nimbus]

I recommend asking in #discrete-math to confirm

[calm coral]

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[normal wasp]

The distance between A and B cities is 24km. to meet each other 2 cars came out so they meet each other at the same time. After they meet the car that came out of A made it to B in 16 minutes and the second one made it to A in 4 minutes. Need to find the speed of each car

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@normal wasp Has your question been resolved?

[calm coral]

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[remote mural]

need help for (v)

its so difficult

i wanted to use induction

for r = 1 we have

$\sum_{n=0}^1 s(1,n) = s(1,0) + s(1,1) = 0 + 1 = 1 = 1!$

[potent lotus]

Derivative

[remote mural]

assume it holds for case k, thus we have:

$\sum_{n=0}^k s(k,n) = k!$

[potent lotus]

Derivative

[remote mural]

now we need to check case r = k+1

$\sum_{n=0}^{k+1} s(k+1,n) = \dots$

[potent lotus]

Derivative

[remote mural]

i have trouble with this i dont know what to do

[calm coral]

@remote mural Has your question been resolved?

[remote mural]

<@&286206848099549185>

nobody knows how to do it?

<@&286206848099549185>

[remote gorge]

what?

[remote mural]

i listed my problem above

[remote mural]

problem (v)

i showed my process but i am stuck at the last inductive step

[remote gorge]

k

[remote mural]

ye can u help?

[remote gorge]

ill take a look

it looks really hard sorry

im only 13💀

[remote mural]

no problem

i sent the problem to math stack exchange

its getting downvoted

[remote gorge]

k

[calm coral]

@remote mural Has your question been resolved?

[remote mural]

well here is the problem again

need help for (v)

<@&286206848099549185>

[calm coral]

@remote mural Has your question been resolved?

[calm coral]

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[glass lagoon]

I want to prove that dimKerA+dimKerB ≥ dimkerAB

I know that KerB is included in KerAB but not more

I dont know how to continue

<@&286206848099549185>

[calm coral]

@glass lagoon Has your question been resolved?

[glass lagoon]

.close

[calm coral]

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[calm coral]

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[finite lily]

I have to make a exponential formule for K with this information how do i do that?

[vital raven]

substitute the points given into the formula y = bg^x

you have 2 points so that gives you 2 equations. 2 equations, 2 unknowns (b and g).

[calm coral]

@finite lily Has your question been resolved?

[finite lily]

i dont get it

[finite lily]

do you mind writing it out?

[vital raven]

The point (1,2) satisfies y = bg^x, so 2 = bg^1

do the same with the point Q(3, 4) and that will give u another equation

[finite lily]

and how do you get the answers of b and g

[vital raven]

so the 2 equations are: 2 = bg^1 and 4 = bg^3

you can divide the second equation by the first to get g^2 = 4/2 = 2

[calm coral]

@finite lily Has your question been resolved?

[calm coral]

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[remote mural]

Hey, I need to show that if $\mathbb{A} = {A_i : i \in \mathbb{N} }$ such that each $A_i$ is countable and infinite, then $\cup \mathbb{A}$ is countable and infinite.

[potent lotus]

ecoproducts

[remote mural]

Can anyone please give me pointers for this problem?

I started things off by saying that any countable and infinite set will have the same cardinality as N. Hence |N| = |A| is a given.

[clear delta]

look up the proof that rational numbers are countable

think about how it works, you should be able to adapt it to this problem

[remote mural]

Plus, we know that each set $A_i$ is countable and infinite. We can then say that $\mathbb{A}$ is a countable and infinite set of countable and infinite sets. Although sets' elements are usually unordered, we can index them as long as they are countable, hence the i-index of sets $A_i$. Using the same reasoning, we can order the elements of $A_i$ and return the j-th element of the i-th set $A_i$, with $i, j \in \mathbb{N}$. This ordering is some bijection $f_i : \mathbb{N} \rightarrow A_i$, which must exist as $A_i$ is countable. We can say that because $|A_i| = |\mathbb{N}|, f_i(j)$ functions exist.

[potent lotus]

ecoproducts

[remote mural]

Now, I let $g : \mathbb{N} \times \mathbb{N} \rightarrow \cup \mathbb{A}$, given by $<i, j> \in \mathbb{N} \times \mathbb{N}, g(i, j) = f_i(j)$.

[potent lotus]

ecoproducts

[remote mural]

I'm stuck right here. How do I show that g is well defined?

[remote mural]

Sure. Will do.

Does the beginning of my reasoning align with that proof?

[clear delta]

seems to yeah

if you can assume N^2 is countable that will be helpful i think

but... i'm not sure if you can assume that given that it's basically what you're trying to prove

[remote mural]

I actually continued the proof after this step. Let me just try to prove that g is well defined, and I'll get back to you.

[calm coral]

@remote mural Has your question been resolved?

[remote mural]

I'm stuck, my bad. How do I show that every output $f_i(j) \in \cup \mathbb{A}$? To finish proving that $g$ is well defined.

[potent lotus]

ecoproducts

[remote mural]

Would appreciate some help!

[clear delta]

probably with the definition of union

but are you just constructing a mapping from N x N to UA?

keep in mind the sets might not be disjoint

[remote mural]

Yeah.

Well, a mapping, which I prove is a function.

Then, I would use the diagonal argument (?) to prove that there exists a bijection from $\mathbb{N} \times \mathbb{N}$ to $\mathbb{N}$, i.e. $|\mathbb{N} \times \mathbb{N}| = |\mathbb{N}|$.

[potent lotus]

ecoproducts

[clear delta]

if that function isn't an injection or a surjection i don't know how helpful it'll be for you

[remote mural]

Then, using g, we'd be able to conclude that $codom(g) = \cup \mathbb{A}$ is countable and infinite.

But I guess that for that to work, g must be a bijection?

[potent lotus]

ecoproducts

[clear delta]

yeah

you either need a bijection

or a surjection and an injection

the second one is often easier

[remote mural]

Is this not how I should take things?

Setting up a mapping, etc.

Are there simpler ways of solving the problem?

[remote mural]

Using rational numbers, or something?

[clear delta]

what i would do here is show that 1) it's infinite, and 2) it's at most countable

[remote mural]

No g needed?

[clear delta]

like bijection? no

the Ai could be something really stupid like:
A0 = { 1, 2, 3, 4, ...}
A1 = {0, 2, 3, 4, ...}
A2 = {0, 1, 3, 4, ...}
A3 = {0, 1, 2, 4, ...}

[remote mural]

Yes, I know. Thing is I tried setting up g to come up with some sort of proof... But if we don't need that, don't know where to start things off...

[clear delta]

bijections are often really annoying to make

[remote mural]

This is what my professor suggested us.

So yeah.

[remote mural]

Considering that, do I need to go back to finding a bijection? Or nah?

[clear delta]

no

[remote mural]

What am I supposed to do with the g that I just defined then?

[clear delta]

uh i think they want you to use it to show that the set is at most infinite: that you can define a surjection from N x N onto UA

[remote mural]

If there exists a surjection from N x N to UA, that implies that UA is infinite?

[ancient thistle]

the question is why g wouldn't be well-defined

what a weird thing to ask

[clear delta]

no, it implies that UA is at most countably infinite

[remote mural]

I'm not sure what that means.

[clear delta]

well imagine UA was {0}

then you could still have a surjection of N x N -> UA

[remote mural]

Sure.

But UA is finite.

[clear delta]

but you know that it's not as big as R, since then there wouldn't be a surjection N x N -> UA

ie UA is countable

[remote mural]

Oof, this is very confusing. Haven't seen "at most countably infinite" in class.

[ancient thistle]

it just means that [ \abs {\Union\AAA} \le \abs\N ]

[calm coral]

@remote mural Has your question been resolved?

[remote mural]

.close

[calm coral]

Channel closed

Closed by @shell hamlet

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[calm coral]

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[remote mural]

On April 30th, 15:23 GMT time, Kendrick Lamar drops his diss track, "euphoria", directed to Drake. (x^2 + 64) hours later, Drake drops a response to Kendrick Lamar, titled "Family Matters".

Sometime later, Kendrick Lamar drops his diss track, "meet the grahams", in response to "Family Matters".

The time between Kendrick Lamar's "euphoria" and "meet the grahams" is 3.52430554167 days.

Find)
a) The amount of hours it took for Drake to drop "Family Matters" after "euphoria"
b) How much faster Kendrick Lamar was (represented by a percentage) in dropping his diss track compared to Drake
c) The time "meet the grahams" was dropped
d) The time "Family Matters" was dropped

[acoustic temple]

alright bro 😭

Let's break it down step by step:

a) To find the amount of hours it took for Drake to drop "Family Matters" after "euphoria", we need to solve for x in the equation (x^2 + 64) = 3.52430554167 days.

Converting days to hours, 3.52430554167 days = 3.52430554167 * 24 = 84.5833334 hours.

So, (x^2 + 64) = 84.5833334

Subtracting 64 from both sides, we get x^2 = 20.5833334

Taking the square root of both sides, we get x ≈ 4.537391 hours.

So, it took approximately 4.537391 hours for Drake to drop "Family Matters" after "euphoria".

b) To find how much faster Kendrick Lamar was in dropping his diss track compared to Drake, we need to compare the time it took for Kendrick Lamar to drop "meet the grahams" with the time it took for Drake to drop "Family Matters".

Kendrick took 3.52430554167 days, which is 3.52430554167 * 24 = 84.5833334 hours.

The difference in time is 84.5833334 - 4.537391 = 80.0459424 hours.

Now, we find the percentage difference: (80.0459424 / 84.5833334) * 100 ≈ 94.665%.

So, Kendrick Lamar was approximately 94.665% faster in dropping his diss track compared to Drake.

c) The time "meet the grahams" was dropped is April 30th, 15:23 GMT + 3.52430554167 days.

d) The time "Family Matters" was dropped is April 30th, 15:23 GMT + 4.537391 hours.

[remote mural]

MB BRO

uhhhh

na

very wrong

[acoustic temple]

WHAT

[remote mural]

very very wrong.

To say the least

the answers would be like

[deep perch]

copypasta? 💀

[acoustic temple]

Chatgpt

dropped that banger

[remote mural]

a would be atleast above 3 days
b would be atleast above 200%
c would be atleast above 30 minutes and below a day
d would atleast be above 3 days

actually when did kendrick drop meet the grahams

ye52 minutes but still idk how to solve the question the intended way

[acoustic temple]

hmm

i suggest

listening to some Taylor swift

[remote mural]

sir

sir hell no

[mint scarab]

i never trust it

[exotic falcon]

!nogpt

[calm coral]

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).

[calm coral]

@remote mural Has your question been resolved?

[umbral gorge]

wrong

[calm coral]

@remote mural Has your question been resolved?

[calm coral]

@remote mural Has your question been resolved?

[calm coral]

@remote mural Has your question been resolved?

[calm coral]

@remote mural Has your question been resolved?

[mint scarab]

poor miky

someone help

[remote mural]

yes bro plz

[remote gorge]

isnt this 0 instead?

[calm coral]

@remote mural Has your question been resolved?

[languid isle]

what the hell is this 💀

this question is lacking information, you need to drop a diss track against your professor

[remote mural]

i MIGJT have to

Whats lacking tho

i probably still have my work written down on a paper

so i can show where i got to

[languid isle]

question doesn't say the time between family matters and euphoria so it's not fair (it doesn't specify x is the time between meet the grahams and family matters)

so unless you go and search the time for that you can't do it, in which case you'd just search up the time for everything and do 0 math

[remote mural]

oh I thought abt it like this

u can write the time between family matters and euphoria by doing

(time euphoria released + (x^2 + 64)) - (time euphoria released)

[calm coral]

@remote mural Has your question been resolved?

[regal bolt]

what probably makes this super easy is to just draw it as a timeline

yeah and by drawing the timeline i realize that we have absolutely not enough info?

[regal bolt]

and this person is 100% correct

you can express all the answers in terms of x though if you want

e.g. the answer to a) is "it took drake x^2 + 64 hours because that's what the fucking question states"

and the answer to d) is "april 30th plus x^2 + 64 hours, or converting 64 hours to 2 days + 16 hours, may 3, at 7:23 GMT, plus x^2"

without knowing x you cannot simplify that further

[remote mural]

dont we know the time of meet the grahams too?

[remote mural]

yeah, i got stuck tho

i tried to do like

time between euphoria and meet the grahams - time between meet the grahams and family matters = time between euphoria and family matters

[regal bolt]

yea april 30 + 3.524... days

but we have absolutely no info about fam matters

we can only phrase it in terms of the unknown x

[remote mural]

damnn

alr

tytysm

i thought it was solvable

.close

[calm coral]

Channel closed

Closed by @shrewd vault

Use .reopen if this was a mistake.

[subtle oriole]

Can someone explain how I'd get the answer to this?

The correct answer is "3" but I don't see how they got that...

[jovial narwhal]

check the y-coordinate of the graph when x=-2

[subtle oriole]

That was too easy (face palming myself)

[grim patio]

miky are you here

[remote mural]

yes

[remote mural]

.reopen

hi

[calm coral]

✅

[calm coral]

@remote mural Has your question been resolved?

[dense falcon]

!help

[calm coral]

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

[dense falcon]

,list

[potent lotus]

My commands!

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If you still have questions, talk to our friendly support team here.

LaTeX Rendering

Render LaTeX code and configure rendering options.
​ ​ ​ ​ ​ ​ ​ ​ ​ ​ tex: Render LaTeX code.
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Guild administration
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[tough grove]

miky can close the channel, your question I think is incomplete so can we help you into it

@remote mural

[languid isle]

💀

[remote mural]

YOYOYOYOYOY

BRO SORRRY

[remote mural]

i thofijt dis guy had an answer

i was gonna close after like a day if he didn't respond

my bad

.close

[calm coral]

Channel closed

Closed by @shrewd vault

Use .reopen if this was a mistake.

[tough grove]

It’s ok

[tough grove]

Yeah, it seems that way

[hollow totem]

what the heck

.reopen

[calm coral]

✅

[hollow totem]

.close

[calm coral]

Channel closed

Closed by @hollow totem

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[frosty crane]

Hi, I need help with this combinatorics exercise pls.
Given the grid:
(a) How many ways can one go from square A to square B always moving one square to the right or one square up?
up?
b) How many ways if you have to go through the X square?

[calm coral]

@frosty crane Has your question been resolved?

[calm coral]

@frosty crane Has your question been resolved?

[frosty crane]

<@&286206848099549185>

[tough birch]

Ok

[frosty crane]

a?

[prime mauve]

In a) I think there are 330 possible ways.

[prime mauve]

in b) 90

[calm coral]

@frosty crane Has your question been resolved?

[calm coral]

Channel closed

Closed due to timeout

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[calm coral]

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>.
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• Be polite and have a nice day!

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[stiff dragon]

how do i find x :')

[eternal shard]

try substitution

[potent lotus]

𝔸dωn𝓲²s

[eternal shard]

find your solutions in terms of u and then resubstitute

[stiff dragon]

oh..

[eternal shard]

ye..

[stiff dragon]

is there another way?

bc theteacher didnt tell us to do it that way

[eternal shard]

what did they tell instead

[stiff dragon]

well they didnt rlly say anything but

we werent rlly doing substitution

[eternal shard]

well then what is your point

[stiff dragon]

if theres another way to do it

[eternal shard]

it's just a tool

[spice viper]

i mean id assume its something with logs or ln

[eternal shard]

to get a better overview of the given equation

[spice viper]

but the substitution is easier

[stiff dragon]

we didnt learn that yet

[eternal shard]

log wont probably work here anyway

well another way woul maybe be factoring

but that seems more difficult

[spice viper]

i mean ur not gonna get marked down for substituting

[eternal shard]

substitution is not mystical

[potent lotus]

𝔸dωn𝓲²s

[eternal shard]

which looks familiar and so on

you can basically do the same without substitution too

but it looks ehh

[stiff dragon]

ok

[eternal shard]

!nosols

[calm coral]

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.

[stiff dragon]

.

u know i have the answer key right..

lol

plus answers help me learn better

but either way ty

[eternal shard]

that didnt make sense but sure

[stiff dragon]

.close

[calm coral]

Channel closed

Closed by @stiff dragon

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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!

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[sand dove]

!help

[calm coral]

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

[sand dove]

<@&286206848099549185> i just had a quick question involving matrix multiplication

[eternal shard]

just ask it

[sand dove]

So when im doing these rows

i feel like im missing a step

cuz i know that the final matrix needs to be a 2 by 2

but when im getting these numbers itll be a 4 by 4, so what am i supposed to do? do i subtract the numbers

[eternal shard]

how come 4 by 4?

[haughty gust]

should be $0 \times 11 + \frac{1}{3}\times 18$

[potent lotus]

°Jason Parker°

[eternal shard]

the dimensions are 2 by 2 so it will be 2 by 2

[haughty gust]

Remember that when you have a row and column, you "multiply" them by taking each entry and multiplying, then adding up all the results

[sand dove]

ohh so i add them

[calm coral]

@sand dove Has your question been resolved?

[calm coral]

Channel closed

Closed due to timeout

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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>.
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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.

[rich tangle]

so

ion know how to do dat

find the y intercept and then a second point on the line?

so y intercept is 0, 9

[eternal shard]

y = mx + b

b is your y intercept

for m take two points

,,m = \frac{y_1 -y_2}{x_1 -x_2}

[potent lotus]

𝔸dωn𝓲²s

[rich tangle]

so

which 2 points do i take

[eternal shard]

Any random

[rich tangle]

asnywhere on the line?

[eternal shard]

Yes

[rich tangle]

aight

[eternal shard]

Yea

(-6,6) and (-12,3)

Now try apply

[rich tangle]

alright

y = (-1/6) x + 9

?

[eternal shard]

It cannot be negative

It's increasing not decreasing

[rich tangle]

oh

so would i choose 2 diff points then on the posuitive side

[eternal shard]

,,m = \frac{6-3}{-6-(-12)}

[potent lotus]

𝔸dωn𝓲²s

[rich tangle]

o

[eternal shard]

evaluate and done

[rich tangle]

alr

so m = 1/2

?

alr tysm

[eternal shard]

thank you too

[rich tangle]

y = (1/2)x + 9

preciate

it

.close

[calm coral]

Channel closed

Closed by @rich tangle

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[calm coral]

Available help channel!

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• 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.

[pliant sedge]

is there some way i'm missing to easily find all pairs of integer solutions, w and n, for an equation of the form w = (α - βn + β)/n for some given alpha and beta? (context: β is the width of a border around n boxes that are w pixels wide, on a screen that is α pixels wide, such that they fit exactly to the edges of the screen, with a border either side of the first and last box)

[pliant sedge]

currently looking at the equation w = (1001 - n)/n (i.e α = 1000, β = 1), but for some reason i find myself doing this kind of thing a lot, be it how wide buttons on a screen can be to have equal spacing, or whatever, i'm always trying to find pairs of integer solutions for simmilar equations to this, so a generalized method would be nice

[ancient thistle]

rearrange the equation into
[ nw = \alpha - \beta n + \beta \Implies n(w + \beta) = \alpha + \beta ]

[ancient thistle]

then the only solutions correspond to factorisations of α + β

[pliant sedge]

im not quite sure i understand what you mean, so say i have n(w+1) = 1001, then what?

[ancient thistle]

,w factorise 1001

[potent lotus]

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Requested by tearfox

[pliant sedge]

OH, prime factorisations?

[ancient thistle]

so you can have
1001 = 1*1001, 7*143, 11*91, 13*77

[autumn warren]

Number theory ftw.

[pliant sedge]

right, so i could have 11 boxes, 90px wide, for example, is what you're saying?

[ancient thistle]

so the possible solutions are
n = 1, w = 1000,
n = 1001, w = 0
n = 7, w = 142,
n = 143, n = 6
n = 11, w = 90
n = 91, w = 10
n = 13, w = 76
n = 77, w = 12

[pliant sedge]

okay, nice, i apparently fucked up the original equation, bcs if there are 11 boxes, there are gunna be 12 empty spaces between/around them, so that's 11*90 + 12 which is 1002px wide, not 1000px wide... but i think if i fix the original equation i should be good from there xDD

thank you!

.close

[calm coral]

Channel closed

Closed by @pliant sedge

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[calm coral]

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.

[trail jolt]

hi