help-23

[safe radish]

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[wanton delta]

Hi I’m not sure why I’m getting this probability question wrong could anyone help me?

[quasi bison]

is the marked answer yours or the teacher's?

[wanton delta]

Mine

It’s incorrect

[quasi bison]

you selected that H and T are independent. was this a conscious decision you made?

[wanton delta]

Yea

Is that not true

[quasi bison]

well, let's walk through it.

can you tell me the definition of independent events?

[wanton delta]

Ones that aren’t affected by previous events

[quasi bison]

bad.

there's no notion of "previousness" here to speak of.

[wanton delta]

Yea

[quasi bison]

let's try this again. can you fill in the blank here?

Two events A and B are said to be independent if _________________________________.

i want you to fill in the blank with a statement that involves probabilities.

[wanton delta]

P(A and B) = P(A) x P(B)

[quasi bison]

ok, great.

would rather you didn't use the letter x as a multiplication symbol though.

now, i want you to apply this definition to your pair of events, but before that:

would you like to do it yourself or would you like to be taken through it step by step?

[wanton delta]

taken through it I guess

[quasi bison]

ok

[wanton delta]

I mean it doesn’t work right

So it’s not independent

but I don’t think it’s dependent either

[quasi bison]

why do you think these events aren't dependent?

do you have a definition of "dependent events" on hand similar to the one i had you recite for independent ones?

[wanton delta]

P(A and B) = P(A) * P(B|A)

So it’s dependent and disjoint

[quasi bison]

this isn't a property of a pair of events.
this is simply a statement that is always true (as long as P(B) > 0...)

"dependent" actually means "not independent", nothing more nothing less.

[quasi bison]

yes, those are the correct answer options.

a disjoint pair of events is actually always* dependent.

[wanton delta]

Ohh thank you so much!

[safe radish]

@wanton delta Has your question been resolved?

[safe radish]

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[safe radish]

@twin falcon Has your question been resolved?

[twin falcon]

<@&286206848099549185>

[twin falcon]

<@&286206848099549185>

https://tenor.com/view/this-is-us-rebecca-pearson-somebody-help-me-please-help-me-please-sad-gif-17710090

[clear trout]

what the fuck

[gleaming tundra]

What's your issue?

[twin falcon]

idk how to even start tbh

[gleaming tundra]

https://math.stackexchange.com/questions/2139807/how-is-a-generator-matrix-for-a-7-4-hamming-code-created

[safe radish]

@twin falcon Has your question been resolved?

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[untold topaz]

Learnt log & exponentials

[untold topaz]

Is it correct to say
$$ 22 00 = 4 000 (2)^ (t/9) $$

?

[flat frigate]

kai_funaba

[quasi bison]

aside, not really...

you can say that the population of your town follows $P(t) = 4000 \cdot 2^{t/9}$ yes

[flat frigate]

Ann

[quasi bison]

finding the time at which there are 22,000 people living in it... won't be of that much use

[untold topaz]

Hmm

This line scares me

[quasi bison]

how familiar are you with derivatives

[untold topaz]

We hvn't learnt it yet
Tch' said it's in calc 🥲

[quasi bison]

tf

and you are asked for instantaneous rate of change here??

which is a derivative???

[untold topaz]

@quasi bison If I were to use
h = .001
How would it turn out?

[quasi bison]

well, it'd eventually work out to $22000 \cdot \frac{2^{0.001/9} - 1}{0.001}$

[flat frigate]

Ann

[quasi bison]

but still it is strange that you are forced to do clearly calculus things without calculus

it's cruel

[untold topaz]

How did u come to this tho

[quasi bison]

$\frac{P(t+h) - P(t)}{h} = \frac{4000 \cdot 2^{(t+h)/9} - 4000 \cdot 2^{t/9}}{h} \ = \frac{4000 \cdot 2^{t/9} \cdot 2^{h/9} - 4000 \cdot 2^{t/9}}{h}$

[flat frigate]

Ann

[untold topaz]

Ah right

UGhhhhh

[quasi bison]

$= 4000 \cdot 2^{t/9} \cdot \frac{2^{h/9} - 1}{h}$

[flat frigate]

Ann

[quasi bison]

just some algebra

[untold topaz]

The longer I do it the more concerning it gets

Yep stuck

[quasi bison]

you chose not to do it the way i did it, and landed in the mud.

[untold topaz]

Wait...

I didn't?

I tried to follow u tho

[quasi bison]

you substituted h = 0.001 right away

i didnt do that

i left h as it was

also wait is it one hundredth or one thousandth that you want to use for h

[untold topaz]

0 . 01

[quasi bison]

anyway this is what you should get here

like exactly that

now you can notice that 4000 * 2^(t/9) is just P(t) itself, i.e. the current population

[untold topaz]

Yep got it

[untold topaz]

Oh demmm

What black magic is this

[quasi bison]

black magic?

there's no magic here

[untold topaz]

It's amazing how u can see that u can separate the P(t)

[quasi bison]

i mean i have the formula for P(t) in front of me just as well as you do

[untold topaz]

Do I sub 22k or 4k as 'current'?

Cuz I'm calculating for 22k

My sense of time is kinda messed up at this point

[quasi bison]

your approximate instantaneous rate of change equals $(4000 \cdot 2^{t/9}) \cdot \frac{2^{h/9}-1}{h}$

[flat frigate]

Ann

[quasi bison]

the stuff in parentheses is the current population

in your case that's 22k

[untold topaz]

Thanks so much Ann!!!

.close

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[hasty oasis]

Does anyone know how I would do 2.?

[hasty oasis]

and 3.

[dapper venture]

the gradient of a scalar function of 2 variables is just <df/dx.df/dy>

then plug in x_0 = <2,12>

[hasty oasis]

So the partial derivatives?

[dapper venture]

yes

[hasty oasis]

<df/dx.df/dy> does this mean multiplying them together?

[dapper venture]

no

the gradient is a vector

[hasty oasis]

Ohhh

[hasty oasis]

Thank you!

[dapper venture]

no problem

[hasty oasis]

Do you know about 3.?

I'm not very good at proofs

[dapper venture]

same :/

try writing the partial derivatives as limits and prove they are equal

[safe radish]

@hasty oasis Has your question been resolved?

[gleaming tundra]

^^

[safe radish]

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[wispy flint]

is the answer 0 because x^3 is odd?

[dapper venture]

I think so

[wispy flint]

okay

[raven heart]

You can't just ignore the cos(x^2)

[wispy flint]

thats even so odd * even = odd = 0

[raven heart]

Yeah alright

[dapper venture]

yee

[wispy flint]

alright ty

.close

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[wheat cave]

can someone explain this to me? so in the 4th row I have $x^3$, which is bigger than the cycle, but how exactly does it result in $x^2+0x^1+1$?

[flat frigate]

jigglyproff

[quasi bison]

are you by any chance working with polynomials over F2

[wheat cave]

this is in the extension field of GF(2)

yes

minimal polynomials

[quasi bison]

in that case

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

[wheat cave]

$**x^3 - (x^3+x^2+1) = x^3 + x^3 + x^2 + 1 = x^2 + 1$

[flat frigate]

jigglyproff

[wheat cave]

oh damn

yeah right you just add when subtracting in gf2

okay can you check on my $x^4$\
I just do $x^4=(x^2+1)x=x^3+x$\
then I do modulo, so\
$x^3+x- (x^3+x^2+1) = 2x^3+ x^2 + x + 1$\
which should be $x^2+x+1$ right

[flat frigate]

jigglyproff

[quasi bison]

seems to check out

[wheat cave]

damn, thanks

.close

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[lean otter]

I don't quite get my teacher's solution to this question

What does adding 14 to the series of inequalities imply?

It seems really random

[peak estuary]

now you have 60 numbers between 1 and 59

so two have to be equal

and it cant be two of the same "type" because the a_i are all distinct

[lean otter]

Sorry I still don't understand the logic

Of the steps performed to get to that conclusion I guess

[quasi bison]

i think there might be a better way to format the same logic

[peak estuary]

the steps come a bit out of nowhere

the only motivation is that they work

[quasi bison]

imagine the team keeps a log of the number of games they play

they have a table of 30 columns and each column corresponds to a day, and at the end of each day they write the number of games played so far into that day's column

so they have a strictly increasing sequence of thirty numbers that way

those are the a_i from your prof's work

[lean otter]

alright

I'm with you on that

[quasi bison]

now they write below each day's number

that number plus 14

so they have two rows

together comprising 60 numbers

all in the range from 1 to 59

[lean otter]

Yeah exactly my issue I'd say. Even if I had managed to understand it I don't think I'm going to intuitively think of something similar in an exam environment for example

[peak estuary]

well now you have seen this approach so maybe next time you might find something similar. its ok to not always know all approaches. its ok if you understand some approaches, understand why they work and can then adapt. this will make your toolbox bigger

[lean otter]

Alright I'm with you still

Understandable so far

[quasi bison]

dirichlet principle says there will be two numbers in this table that are the same

[peak estuary]

is pigeonhole also called dirichlet?

never heard that before

[quasi bison]

sure is

[lean otter]

Ohhh that's it

Okay but like

My second confusion is that

How does that conclusion from the pigeonhole principle imply that there is a period of some days in which they play only 14 games?

[lean otter]

Thank you haha. I'm having big Trouble understanding those pigeonhole examples as all of my professor's examples are just completely, seemingly, nonsensical to me at the moment

[peak estuary]

going with Ann's table, the two numbers that are equal cant be in the same row

because in each row are only distinct numbers

so they are in different rows

so a_i = a_j + 14 for some i and j

that means on day i they have played 14 games more than on day j in total

so those 14 games happened on days j+1, j+2, ..., i

which are consecutive

[lean otter]

Sorry I'm going to need a good bit to rationalise all of that one second

[peak estuary]

take your time

pigeonhole arguments are often not easy imo. cause by design they always barely work

so you have to be very precise with putting the puzzle pieces together

[lean otter]

OH I think I get it! We can say that because a_j denotes all the games on the jth day and all the way back to the first day

It just clicked thank you so much your explanation really simplified it

Okay I think I got a rough idea of this now

There is another one which went a bit like,

"Let S denote the set of 6 positive integers below 14. There exists at least two non empty subsets in which the sum of the elements of both subsets are equal"

My teacher'a argument for this eventually made sense, but I didn't find it intuitive at all?

She considered the sum of the last 6 integers below 14

So like

9 +... + 14 = 69

[peak estuary]

the question immediately suggests to start counting two things:
number of possible sums
number of subsets

the sums can at most be 69. because thats what you get if you choose the 6 biggest numbers available

[lean otter]

Number of possible sums

[peak estuary]

and on the other hand, each sum is at least what?

[lean otter]

At least 1?

Wait

[peak estuary]

wait I made a mistake. uhm

[lean otter]

Yeah that makes sense

So between 1 and 69?

[peak estuary]

thats too weak. I dont like that

[lean otter]

Too weak?

[peak estuary]

lets count the numbers of subsets for a second

how many can we pick?

[lean otter]

2^6 -1

[peak estuary]

63

[lean otter]

Yep

[peak estuary]

which is smaller

so we cant use "number of subsets > number of sums, so two subsets have same sum"

[lean otter]

My teacher instead

Considered the proper subsets

Exactly because of that

So excluding like, 14

In which the number of (proper) subsets are 62

[peak estuary]

?

wdym excluding 14

[lean otter]

Oh sorry

Excluding 9 I mean

By considering the sum of the highest 5 terms, which ends up being 60

The numbers of subsets becomes 62 (as we have removed one of the subsets)

So pigeonhole becomes applicable

Well that's my teacher's way apparently

[peak estuary]

maybe another perspective (which I havent yet completely formulated in my head but which should lead to the same answer)

if we have eg 9,...,14 as the 6 numbers, then we can go back to what I said earlier about what the sum has to be at least

which here would have to be 9

[lean otter]

Yeah I don't mind solving it in different perspectives either way. Gives me better intuition

[peak estuary]

for the sum to be smaller you need to include a smaller number but that also makes the max possible sum smaller

eg if you include a 1, then the max possible sum is 1+10+11+12+13+14=61

and min is 1

[lean otter]

Okay understandable

[peak estuary]

or if you include 2, the max is 62

and min is 2

[lean otter]

Yep logical

[peak estuary]

if you include 3, max is 63, min is 3

and so on

so we see that its always 60 possible sums

corresponding to the 5 highest numbers

always between [min number, min number + 5 highest numbers]

and now same argument, 60 possible sums, > 60 subsets

and I suck at counting cause thats clearly 61 possible sums

but > 61 subsets so still fine

[lean otter]

Hmmm

Comprehending

Do you mind if I open another help channel in around 30 minutes time? I am currently outside but I want to continue our talk about this

[peak estuary]

yeah thats fine

ping me

[lean otter]

Alright. See you, and thank you

.close

[safe radish]

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[dreamy pike]

titu andreescu and dorin andrica, complex numbers from A to Z, Birkhauser, 2006

[dreamy pike]

Can someone find me the pdf link to this page

of this book

[karmic hedge]

https://thunhan.files.wordpress.com/2008/08/tituadreescu-complexnumbersfromatoz.pdf

[safe radish]

@dreamy pike Has your question been resolved?

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[unkempt marten]

Just trying to double check, but is my process here right?

[unkempt marten]

if im not mistaken my next step would be to cancel out the h's, but id need them on every term in the numerator, which i dont have

[primal granite]

F(x+h) doesn't equal (2x+h)^2

[unkempt marten]

is it 2(x+h)^2?

[primal granite]

Yep

[unkempt marten]

ooh...

alright let me redo it with that in mind

[primal granite]

👍

[unkempt marten]

2(x+h)^2 would be:

2x^2 + 4hx + 2h^2

alright f'(x) = 4x

alright thats good! thank youu!

.close

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[primal granite]

Np :)

[safe radish]

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[cobalt crag]

whats a range in a rational function?

Message #discussion

[final halo]

#discussion message what was the issue with this reply

[safe radish]

@cobalt crag Has your question been resolved?

[safe radish]

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[verbal vault]

the first answer is correct but I have a question regarding the second as Im a little unsure

Sorry for the delay, Im just writing it out for clarity

This doesn’t feel right but it’s the only method I can think of when solving this

nevermind, i worked it out. it looks like it is the right answer haha. just need to have some faith in myself more often

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[cunning shard]

A box contains 6 red, 4 white, and 5 black balls. A person draws 4 balls from the box at random. What is the probability that among the balls drawn there is at least 1 ball of each color?

[cunning shard]

I understand that there are three cases, WBRB, WBRW, WBRR

Total 15C4 ways to choose 4 pairs from the total

not sure how to get the favorable

how many ways to select WBRB?

[quasi bison]

well, how many ways are there to draw 2 black, 1 red and 1 white?

[cunning shard]

6C2* 4C2 *5C2?

there is no blue btw

[quasi bison]

why 4C2? and why 5C2?

[cunning shard]

we want 2 black from 5, so 5C2

similarly 2 white from 5 white, so 5 C2

[quasi bison]

no, we want only 1 white...

we're considering the case where we draw 2 black and 1 each of the rest

how many ways to select WBRB?

also i misidentified which of your choose functions went to which color

so i should have asked you why 6C2

rather than just 6

(or 6C1 if you're a formalist/bureaucrat)

[cunning shard]

in WBRB is each of these choices are independent?

[quasi bison]

they're independent per color

the choices of white are independent of the choices for red and for black

[cunning shard]

I was wondering why it’s not $645*5$

[flat frigate]

dotdoc.

[cunning shard]

choosing 4 so can’t I choose one by one?

As you said the correct way is 5c2* 6*4

Thank you

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[flat perch]

Find the number that this series converges on:
9/17 + 3/17 + 1/17 + 1/51 + ...'

[flat perch]

Is it ummm 27/34?

Because first term * common ratio =

[halcyon fog]

i need help

[flat perch]

9/7)(1- 1/3)

[dapper venture]

#❓how-to-get-help

[flat perch]

which is 27/34?

right?

[flat perch]

did u forget me lol xD

awww man

yeah I know

[dapper venture]

this is correct btw

[flat perch]

tysm

.close

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[covert cosmos]

proof that 2*(n)! / (2n)! = 1/(2^(n-1) * (2n-1)!!)

[flat frigate]

toby____

$\frac{2(n!)}{(2n)!} = \frac{1}{2^{n-1}(2n-1)!}$

[safe radish]

@covert cosmos Has your question been resolved?

[merry pelican]

Is that factorial applied twice or is it double-factorial

[covert cosmos]

$\frac{2(n!)}{(2n)!} = \frac{1}{2^{n-1}(2n-1)!!}$

[flat frigate]

nusuntrares

[covert cosmos]

what do you mean?

(2n-1)!! means only odd numbers

[merry pelican]

Do you define $n!! := (n!)!$ or $n!! := n(n - 2)(n - 4)\hdots$

[flat frigate]

mlkkei

[merry pelican]

<@&268886789983436800>

[covert cosmos]

second

[white umbra]

do you know what induction is?

[covert cosmos]

ok

[covert cosmos]

use induction

just proof it

[white umbra]

Are you asking me to prove it, or are you asking for help proving it yourself?

[covert cosmos]

i ask you to do it

[white umbra]

We don't do that here.

!nosols

[safe radish]

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.

[covert cosmos]

bruh

[covert cosmos]

I calculated the expression but I am not sure that it is correct, so please prove it

[white umbra]

Again, we don't do that here.

There's no giving out answers in this server

[safe radish]

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[covert cosmos]

i've got memeed

[reef talon]

what's the question here

[white umbra]

Proving this

[covert cosmos]

I have reached this tie and I just want to see if it is correct

it was ok until this keyboard warrior @white umbra came and started to close my channel

[reef talon]

is this the question? prove or disprove the correctness of the equation?

[covert cosmos]

prove

[reef talon]

who tasked you with this question

oh wait I just saw the !!

[covert cosmos]

no one just practicing for college admission

[reef talon]

is this the inital question?
or that you have a solution and you ask someone here to verify it

[covert cosmos]

this is the initial question

[reef talon]

is that limit of the n'th derivative of this function (at x=0)?

[covert cosmos]

2*(n)! / (2n)!

yes

[reef talon]

looks gross

[covert cosmos]

sry

[reef talon]

nah dw

[covert cosmos]

2*(n)! / (2n)!

[safe radish]

@covert cosmos Has your question been resolved?

[safe radish]

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[little marlin]

i'm confused

[safe radish]

@little marlin Has your question been resolved?

[bitter galleon]

let me figure this out

so you input variables for the present ages of both girls

so in this case: J+5:S+5 = 3:5

and J+10:S+10 = 2:3

now

J+10:S+10 is also equal to =20:30

and so J=10

ad S=20

so honestly

i have no idea how thats wrong

since 15:25 is also 3:5

[little marlin]

lemme show u the solution

[bitter galleon]

alright

[bitter galleon]

so you were right

[little marlin]

i don't understand how if use song's age to calculate x i get a negative number...?

[bitter galleon]

i honestly dont get this work

[little marlin]

😭

[bitter galleon]

you got the answer right

[little marlin]

i don't get why x negative tho..?

[bitter galleon]

how do you get a negative

show your work plzzzzz

[little marlin]

if I do 3x + 5 = 2x, I get x = 5
but
if i use song's age i get 5x+5=3x
x = -2.5 ??

am i supoussed to assume that x can't be negative or am i mising smth

[bitter galleon]

how does 5x+5=3x

the way they use the variables for the question confuses me

[little marlin]

ratio of J:S five years later is 3:5, so 5x is his age after 5 years
in 5 more years (10 years) their age ratio is 2:3
so 5x + 5 = 3x

[little marlin]

that's the way i have been taught 😭

[bitter galleon]

where did the +5 come from

if your trying to find the ratio of their ages in 10 years

[little marlin]

the 5x is his age 5 years from the present, add 5 to that we get 10 years from the present

[bitter galleon]

so what are we supposed to "equal" that to?

[little marlin]

= 3x (as the ratio after 10 years is 2:3)

[bitter galleon]

im stupid

hold on

but that would not be 3x

try to solve it this way

3x:5x
2x+5:3x+5

5x=3x=5

so it would be 2.5

not -2.5

it would not be 5x+5

it would be 3x+5

there is NO other way

[little marlin]

ooooh okayy

i seee

[bitter galleon]

you get it?

Im honestly confused why they teach you age questions like this

[little marlin]

yepp tyyyy

[little marlin]

😭

[bitter galleon]

honestly, life hack here: Just make 2 variables

J+5:S+5 = 3:5
and J+10:S+10 = 2:5

and just multiply the ratio on the right by 10

or 5 here

oop @little marlin close the channel please

its .close

[little marlin]

ookay tyy!!!

.close

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[hallow moss]

Their claim is that the event of 1st empty 2nd chocolate contains symmetric elements to the event of 1st chocolate 2nd empty, such that their probabilities are the same

I don’t get where’s the bijection and symmetry here

By bijection, are we saying that the elements in 1st=E 2nd=C are the same as 1st=C 2nd=E?

.close

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[charred flame]

Not sure how to get an exact answer from this

[dapper oak]

There you go!

Follow the instructions (like rationalization)

[charred flame]

ah

so wait

when they ask for sin(135), I just give the sin of the reference angle?

[dapper oak]

no

there are a set of rules

in different quadrants

we get different answers

[charred flame]

man

okay I must have forgotten that specific detail

[dapper oak]

dont worry

revise it and you are fine!

[charred flame]

any idea what it is called so I can look for it in my book?

[dapper oak]

those are basic trigonometric rules

i will give you a list!

wait

Allied angles

[charred flame]

what the heck

[dapper oak]

yes

[charred flame]

okay

not in my book but it's in my notes lol

[dapper oak]

that is also a valid method to remember

[charred flame]

so the reference angle is in q2

Oh you know what I have a different physical book than the online book I am using for this class

[charred flame]

so is the 1/sqrt2 just something you know to equal that?

[dapper oak]

yea

here:

[charred flame]

and is there a way to get the answer if you don't know all of the conversions?

man trig is wild

[dapper oak]

some common functions to keep in mind!

also if you are not sure

you can derive it

[charred flame]

wow I haven't even seen this

so the answer you gave was incorrect

the 1 in the numerator is also a 2

so now I am even more confused

[dapper oak]

Yes that's called rationalization

[charred flame]

oh yeah that adds up now nvm

well I sure hope the exam this week allows notes because trig has a lot of stuff to just remember

ty for the help

[dapper oak]

Np!

[charred flame]

.close

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[delicate pumice]

Did i set up my limits correctly

[safe radish]

@delicate pumice Has your question been resolved?

[delicate pumice]

<@&286206848099549185>

Is this right

[steep vessel]

can u write in chat the equation?

[delicate pumice]

Huh

1/((z^2)+(x^2)+(y^2))^(1/3)

Idk how to use latex

[gleaming tundra]

Have you been introduced to different coordinate systems?

Things like cylindrical and spherical coordinates?

[delicate pumice]

Yes

[delicate pumice]

I alr did it just need to know if im right

[gleaming tundra]

Oh oops didn't look at the latter half of the page for some reason

Oh my god

Why the decimals

Your work looks accurate, but I'm having trouble following your train of thought

[safe radish]

@delicate pumice Has your question been resolved?

[delicate pumice]

.close

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[exotic yoke]

please help me i am so lost on my homework

i am alsmot done for the night

😭😭😭

[dapper venture]

take partial derivative of f in respect of x then t

[exotic yoke]

i was doing that, but i’m so lost like for the work

[dapper venture]

where are you lost

[exotic yoke]

my brain fried from homework 😭😭😭 sorry

[lilac patio]

do you know how to find a partial derivative?@exotic yoke

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@exotic yoke Has your question been resolved?

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[grim plover]

Okay so I tried applying this formula here

[grim plover]

But theres a problem

since there can only be 8 cubes

Therefore either there will be 1 less red, blue or green cube

depending on which case I consider I get the answers 560, 420, 1260

the correct answer is 1260

[hardy lion]

Lemme check gimme a sec

[grim plover]

how do I know in an exam which case is the correct?

(1260 case is the case when 1 green cube is left out)

[lean otter]

i believe you have to first choose the 8 cubes

then with that, permute it

[grim plover]

i think thats what i have done

[lean otter]

so like

[grim plover]

but the problem is how do I know which cube should I discard to get the correct answer?

[hardy lion]

280+420+560=1260

Checks out to me

[grim plover]

where da 280 come from?

[hardy lion]

Can you show work

[grim plover]

its pretty rough and my phones dead rn so i will type it

[hardy lion]

Discard red, blue, green respectively

Use the formula and thats what you get

[grim plover]

discarding green is giving me 1260. let me redo the calculations

[lean otter]

,w (9 choose 8) * 8!/(2!*3!*4!)

[flat frigate]

Results provided by WolframAlpha

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

Requested by lexqal#0

[hardy lion]

Oh thats a good way to do it

[grim plover]

why u do 9C8?

[lean otter]

im learning, thought of this from our help channel with the ternary string question 😎 👍

[grim plover]

Thats like the combinations

[solemn vault]

Yes

First you choose 8 cubes out of 9

Then arrange it

[lean otter]

you are choosing 8 cubes from 9, then you are ordering those 8 cubes

[grim plover]

ohhh

that makes sense

@hardy lion it seems it was a calculation error (using my method)

tysm!!

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[west steeple]

I think the result is 1260

[safe radish]

@west steeple Has your question been resolved?

[solemn vault]

It is

1260

[safe radish]

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[whole forge]

Having trouble constructing the sequence I have in mind over here properly

How would i use a recursive definition to create such a sequence?

[potent osprey]

Why do you want a recursive definition

[whole forge]

It doesnt have to be recursive

i dont see any other way to construct the sequence

but if there is, do let me know

[potent osprey]

$a_{3k + 1} = 1, a_{3k+2} = -(k + 2), a_{3k + 3} = (k + 2)$

With $k \geq 0$

[flat frigate]

_daili

_daili

[whole forge]

ah. Edit: also realised my proof above in the picture itself is wrong, it should be 3k+1 not 1+2k.

okay thank you very much

.close

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[hasty oasis]

How do I do these?

[hasty oasis]

Do I just do normal epsilon delta and sub in (x,y,z) for x?

[lean otter]

polynomials are continuous, so direct-substitution works

for (a)

but yes

you just plug it in and then use the epsilon delta argument

[hasty oasis]

How would I write the epsilon delta arguement for 3 variables?

[lean otter]

https://www.youtube.com/watch?v=PEgjh5YAN5k

[hasty oasis]

Thank you!

[lean otter]

no worries

[hasty oasis]

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[lament nacelle]

Hi, I don’t understand how step 1 to step 2 is made?

[thin bridge]

factorisation

[lament nacelle]

I thought it would become log(2x)^4x + log(2x-1)^3

is that not how it works?

[delicate bobcat]

Instead of doing that

They saw the 4x and the + 3 both had a

Log(2x-1) multiplied to them

They pulled out log(2x-1) as a common factor from both

Leaving behind (4x + 3)

It's not the only way to solve you could have done it the other way using the logarithm properties

[lament nacelle]

Ohhhh I see now

Yeah, I understand, thanks!

[delicate bobcat]

Yw!

Good luck

[lament nacelle]

.close

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[misty shoal]

I need help with this double integral please someone help

[safe radish]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[grim socket]

I don't see the interval on second integral

the first is from 1 to 2

the second is from ? to 1

[misty shoal]

2

[grim socket]

if you need a hint try to set parametet t = xy then dt = y

[misty shoal]

Here

Yeah

I’m stuck I’m not getting the right answer

[stoic dune]

Your work looks solid

[misty shoal]

So then

How is it still wrong

[grim socket]

Here is my solution

Your answer must be right

[safe radish]

@misty shoal Has your question been resolved?

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[radiant gale]

cant figure out what im doing wrong here
the dimensions are definitely right (the dimension of any pn polynomial space is n+1)
with that fact the three polynomials should constitute a basis for p2 (there are 3 polynomials, each with a degree of 2 or less, satisfying the basis thm.)
and then the three polynomials that they gave are said to span the subspace H, meaning that if they are independent (and i checked they are) they should serve as a basis for H

[drowsy moss]

The dimension of P2 is 3, not necessarily H

it would only constitute a basis if they're linearly independent.

[hot thistle]

can you show your work for linear independence? because there must be an error somewhere

this set is not linearly indep

[radiant gale]

hold on getting the picture of my work now

Ignore the stuff that's partially offscreen I kinda shoved this work in between other stuff

[drowsy moss]

your second matrix is incorrect

2/7(15) -2 is not 26/7

similarly
3/7(15) - 7 is not -19/7

[radiant gale]

yeah youre right

fixed my arithmetic and now ive got a free variable

so no independence

would it be too much to ask for some help on finding a working basis then

bc im not actually sure where to start lmao

[hot thistle]

row reduce your matrix to rref

the pivot columns in the rref matrix correspond to basis vectors in the original matrix

[radiant gale]

oh okay

yeah wait this does sound familiar

ty

i think i can solve it now

hold on

i already know my pivot columns

i dont need to row reduce further i already know my pivots are R1 and R2

[hot thistle]

then you have your basis vectors

[radiant gale]

but they arent correct

assuming im not misinputting the vectors somehow

[hot thistle]

show us your answer

[drowsy moss]

why is that not correct?

[hot thistle]

that should be correct

[radiant gale]

that's what i wanna know!

[hot thistle]

where does it say it’s wrong

[radiant gale]

maybe i did something wrong in another part of the problem?

[hot thistle]

show us all your answers

[drowsy moss]

all of your answers in the original problem were incorrect.

[hot thistle]

in the future if it says “at least one,” make sure to share that with us

[radiant gale]

my b

[hot thistle]

but yes the other two answers are incorrect

no problem

[radiant gale]

adjusted the answer for a bc i have the actual dimension now

and b was easy enough bc it's just true or false

but why was b wrong?

[hot thistle]

recall the definition of a basis

[radiant gale]

set of vectors that you can make linear combinations of to get some vector space

[hot thistle]

no

[drowsy moss]

you're missing a key part of the basis definition.

[hot thistle]

a basis of a vector space is:
a linearly independent set that spans the space

key term being linearly independent

also note how the answer to (a) was 2, meaning the number of basis vectors is 2

[radiant gale]

oh shit i read the basis thm. more closely and it says that too lmao

[hot thistle]

but this set has 3

[radiant gale]

idk why but i thought the basis thm just skipped that req

sorry sorry

ty guys i was gonna be tearing my hair out before long

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[rain quarry]

can someone please point out where i am going wrong here/

[drowsy moss]

you subracted wrong, which why your answer is negative,
but you got 256/3 why did you double it?

[rain quarry]

im not sure that it was bc i subtracted wrong

maybe it is because i didn't add y=4x instead?

but that would equal to 512/3 and that was wrong too

[drowsy moss]

area between curves is the integral of top function - bottom function. your top function is 4x.

[rain quarry]

omgg ur right

i literally told myself not to mix them up and still did 😭

[drowsy moss]

i'm still confused why you're wring 512/3?

[rain quarry]

like why i submitted that as an answer?

bc 512/3 = 8^3/3

also i didn't enter -256/3 as an answer bc it didn't make sense to me

should i be adding

bc if i subtract i get a negative answer

[drowsy moss]

now you're multiplying by an extra x

[rain quarry]

typo

so like this

which makes more sense

as to why my answer was negative

ty for the help 🙏🏼

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[kindred panther]

Could some confirm for me if I picked correctly

[kindred panther]

.

[proper crypt]

yes that is correct

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[kindred panther]

.close

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[scenic pier]

Is there a way to represent this sum of matrix as a polynomial on a single matrix?

[scenic pier]

For the moment, I’ve thought about quaternions but they don’t have exactly the same format

I tried multiplying them and they are part of a group so I believe it’s impossible

[safe radish]

@scenic pier Has your question been resolved?

[scenic pier]

<@&286206848099549185>

[safe radish]

@scenic pier Has your question been resolved?

[safe radish]

@scenic pier Has your question been resolved?

[safe radish]

@scenic pier Has your question been resolved?

[covert yoke]

@scenic pier those matrices with b, c, and d almost look like if you square them you might get one or more of the others.

Let's call the four matrices I, B, C, D. It looks like B^2 = C^2 = -I, and D^2 = I

You mentioned that they were part of a group, if that's the case the group will have generators, but only a cyclic group will have a single generator

So if you can characterize this group and show that it is not cyclic, then it will have multiple generators, and therefore cannot be characterized in a single polynomial equation.

Hope this helps!

[safe radish]

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[solar cradle]

When you divide 18 by 3 where does the 6 go at the top or bottom

[ivory gulch]

Numerator

18 cancels to 6 and 3 cancels to 1

[solar cradle]

Ok thanks

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[solar cradle]

How do you simplify this when there’s no division bar or anything

[quasi bison]

do you know your exponent laws?

specifically $a^m \cdot a^n = a^{m+n}$?

[solar cradle]

I think so

[flat frigate]

Ann

[quasi bison]

do you know this one

[solar cradle]

Umm

I think Mayby

[quasi bison]

yes or no

there's no maybe, either you know this or you don't.

was gonna tell you to apply this law now (twice)

[solar cradle]

What do you mean twice

[quasi bison]

two times

or only once if you're ok applying it to more than two powers being multiplied.

[solar cradle]

Hmm you sound like your name

[quasi bison]

i beg your pardon?

[solar cradle]

Like smart

Usually Ann or teachers names

I have a teacher named Ann

[quasi bison]

this is news to me lmao

[solar cradle]

Yeah Ann. Suzanne . Carol

Edward

Smart names

So I plus them together

[lilac patio]

old names

[solar cradle]

But you agree yeah artic

[lilac patio]

back to math

[solar cradle]

Okay so I plus them

[quasi bison]

well, yes, you add the exponents together

remember that lone 8 is 8^1

[solar cradle]

Ahh okay

So 8^-5plus 8=8^-4 then plus 8^-4and it’s 8^-8

[quasi bison]

bad notation

[solar cradle]

Sorry

[quasi bison]

but yes 8^(-5) * 8^1 * 8^(-4) = 8^(-5 + 1 + (-4)) = 8^(-8)

[solar cradle]

How do I do it@like you did

[quasi bison]

what a coincidence that im planning to go into teaching job-wise

[solar cradle]

My teacher told me if I have negative exponent to put it under 1 so 1/8^-8

Oh really

My teacher sometimes says stuff like why did I chose this

So your old aswell

Not in a bad way though

Okay so is it 1/8^-8?

Or that’s wrong yeah

[lilac patio]

1/8^8

flipping it to the denom makes the sign change

[solar cradle]

Oh yeah! I forgot

Changed my pic aswell

Okay is that completed

[lilac patio]

yeah

[solar cradle]

Okay thanks!!!!

Thanks guyyys!

[solar cradle]

Thanks Anna!!!

See ya guys later

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[pulsar night]

Can someone check this for me

[lilac patio]

whats x

[pulsar night]

the first column im assuming

[lilac patio]

why would it equal 1

[pulsar night]

i just took the only value in the first column

[lilac patio]

thats not how it works

that 1 means 1x

in the 2nd column, 1y

third, 1z

[pulsar night]

oh

wait so then 1x = 7?

[lilac patio]

-7

[pulsar night]

oh yea my b i meant -7

ok i see

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[solar cradle]

I know it’s completely wrong but I thought those brackets are to multiply each other?

[lilac patio]

send original problem

you should have used exponent rules instead of applying the exponents

since the base is the same

[broken yew]

The last is wrong.

nvm theyre all wrong

the power of x is wrong.

[dense prism]

Actually only -27 is wrong

The net power of x is correct

2+4-3=3

[broken yew]

????

[solar cradle]

Umm

[lilac patio]

but still

[broken yew]

bruh

thats clearly not what they meant

[dense prism]

I mean sure but yeah somehow it worked out for them

[solar cradle]

I just thought the brackets meant to multiply each other

[dense prism]

Coincidence

[broken yew]

[dense prism]

Nothing wrong in that thought my mate

[broken yew]

correct but your working appears as if you did not do that

[solar cradle]

The top equation is

[broken yew]

and just assumed the power does not distribute properly

[solar cradle]

Soo

Where did I mess up

[broken yew]

also the massive spacing between each bracket made me think it wasnt

[broken yew]

see this

[solar cradle]

Sorry

[lilac patio]

you messed up by using exponent rules first

[dense prism]

Did you intentionally write only one x is each bracket

[lilac patio]

needlessly complicating it

[dense prism]

So that the net power is 3

[solar cradle]

Yea

[dense prism]

Lmao

[solar cradle]

Like -3 timez x?

[broken yew]

You should show all steps in working and not skip maybe

I still am not confident they have understood correctly

[dense prism]

Yeah show your full work

[broken yew]

at this stage

[dense prism]

Yeah

[solar cradle]

Who has understood?

Me or the people that made the question

[dense prism]

Wait answer me this

What is (5x)² equal to

[solar cradle]

Itsss

5 times 5

Because it’s a number

[dense prism]

And?

[solar cradle]

So it’s 25

[dense prism]

$(5x)²$

[flat frigate]

radiation1

[broken yew]

what happened to the x.

[dense prism]

You're just ignoring the x it seems

[lilac patio]

combine them first! dont multiply it out like that

[solar cradle]

Ummm

[dense prism]

Yeah that's easier but I just wanna check whether they have understood it or no

[solar cradle]

But I thought the 3 is related to the x

[broken yew]

[lilac patio]

$A^B\cdot A^C=A^{B+C}$

[dense prism]

They can take -3x as the common base

[flat frigate]

Arctic

[solar cradle]

Like if it was just x and it was power to 2

[broken yew]

please leave that

[solar cradle]

Then it becounes 2x

[broken yew]

for later

[lilac patio]

they already did a question like this

[broken yew]

First please can we make sure they can expand this expression

[lilac patio]

they arent understanding that just since it has a variable you can still apply the same rules to it

[solar cradle]

Uhhh

[broken yew]

,,(5x)^2=???

[solar cradle]

Well I thought it’s 25x but I think it’s supposed to be 25x^2???

[lilac patio]

$5^2 \cdot x^2$

[broken yew]

yes the latter. why do you think its 25x

arctic please.

[flat frigate]

Arctic

[solar cradle]

Because x is a variable and 25 is 25 multiplied by the variable

So

[broken yew]

ok first of all

[solar cradle]

I’m multiplying the x

[broken yew]

variables are just numbers.

[solar cradle]

By multiplying the number

[broken yew]

They are no different from numbers.

Ok?

we could let x be 3 for example

[solar cradle]

Wait soo x

Ok

[broken yew]

and suddenly i was asking you this:

,,(5\times 3)^2=???

[broken yew]

And the answer is not 25 times 3