help-10

[mellow halo]

First off

The Pythagoras theorem can only be used in certain occasions

Agreed? @plush smelt

[plush smelt]

ye

[mellow halo]

Tell me the formula

[plush smelt]

a2 b2=c2

[mellow halo]

That comes with the theorem

Yup

Wait tf

[plush smelt]

a2+b2

=c3

c2

[mellow halo]

Yes

Look at the triangle

Don’t tell him the answer man

Teach him

[plush smelt]

ong

im looking at the triangle

[mellow halo]

Ye

They’ve only given u the length of 1 side

Or am I wrong?

[plush smelt]

yes

ur right

[mellow halo]

And to use the theorem u need to have 2

[plush smelt]

182.27

[plush smelt]

yeah so pythagorean is impossible to use

[mellow halo]

Do you understand?

[plush smelt]

right?

[mellow halo]

Yes

And that’s why

[plush smelt]

yea

[mellow halo]

H2 = P2 + B2

[plush smelt]

whats that mean

[mellow halo]

You only have P

Hypotenuse = Perpendicular Side X Base

It’s the Pythagoras theorem

[plush smelt]

yea

oh ok

i see

[mellow halo]

Ye

Is that all?

[plush smelt]

well after i realised we dont use pythagorean i was wondering why this person used SOHCAHTOA

[mellow halo]

Similar to the Pythagoras theorem, SOHCAHTOA can also only be used in certain problems

[timid silo]

SOHCAHTOA?

[plush smelt]

or just SIN

myb

[mellow halo]

That is, when you have a right angled triangle, any length of the triangle and an angle other than the 90° one

[timid silo]

What's this

[plush smelt]

sin

[mellow halo]

Sin Cos Tan

[timid silo]

Ohhh

[plush smelt]

oh i see

we have the one length = 183.27

the right angle

and the other angle 86.03

[mellow halo]

Screw the right angle

[mellow halo]

Yup

Remember this: the Pythagoras theorem and SOHCAHTOA can only be used IF you’ve got a right angled triangle (90° triangle)

So basically, ignore the 90°

[plush smelt]

Kk

[mellow halo]

Yup

[plush smelt]

is there a reason

they use Sin?

[mellow halo]

Ye

U got some brushing up to do bro

[plush smelt]

my test is tommorow LMAO

ik

i didnt learn any of this

[mellow halo]

God damn

[plush smelt]

im fucked lowkey

[mellow halo]

Sin = P/H

Fr

Prayers witchu bro

[plush smelt]

thank u

[mellow halo]

Yea so anyways

Np

Lemme show u smth

[plush smelt]

ight

[mellow halo]

This is basically the triangle that they gave u in the Q riff

Right*

[plush smelt]

wait i get that

O/H

P/H

[mellow halo]

Yes

[plush smelt]

that makes sense

its because of what we are given

right

[mellow halo]

Yes

[plush smelt]

we use Sin?

ah ok

[mellow halo]

Sin = Perpendicular / Hypotenuse

U need to find the hypotenuse

[plush smelt]

yea

[mellow halo]

They gave u the perpendicular length thing

And the angle itself

[plush smelt]

yea

[mellow halo]

U can find it

Ez

Which grade u in?

[plush smelt]

9th

going into 10th

[mellow halo]

Damn

I’m in 11

11 started 2 days ago lmao

[plush smelt]

dam

[mellow halo]

Yea ima dip

Peace

[plush smelt]

kk

ty tho

Helped a lot

[mellow halo]

Anytime

[plush smelt]

deadass a real one

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[timid silo]

take log both sides...you'll get to the answrr

[weak yew]

k

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[timid silo]

how would i go about finding the antiderivative for dx/3x

[high lily]

applying constant multiple rule first makes it easier

1/(3x) = 1/3 * 1/x

so you can factor 1/3 out of the integral; focussing only on integrating the 1/x

also pay attention to the note telling you to use abs val where appropriate

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

i see

why would we need an absolute value here

[viscid delta]

Because ln x is not defined for negative x, so the absolute values allow it to work for negative x.

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[iron drum]

<@&286206848099549185>

[timid silo]

use the binomial distribution

[iron drum]

What? Can you explain this process?

[timid silo]

do you know the formula to model the binomial distribution?

[iron drum]

No

[timid silo]

the first step is to find p, the probability of picking a covid case

[iron drum]

Oh and btw it’s THREE people that are randomly surveyed my bad I didn’t include the top part

[timid silo]

?

[timid silo]

how is that relevant to the question

its asking how many people need to be selected

[iron drum]

Oh

Oh ok carry on

[timid silo]

^

[iron drum]

3000/77540

[timid silo]

yup

and whats q then?

[iron drum]

3.86 pc

3.86% chance it’s Covid

[timid silo]

yeah

q=1-p

now plug it all in the equation, and use trial and error to find n

[iron drum]

What is Q mean again

[timid silo]

probability of not selecting someone with covid

x=1 btw

and we want P>99.5

[iron drum]

Woah where did x come from

What the fuck

[timid silo]

.

[iron drum]

Why did x just fucking come

[timid silo]

we want 1 person to have covid in our sample

oh wait

actually, more than 1 lol

so ig use x=0

and then look at the complement probability

[iron drum]

I don’t understand, how you would you write the numbers in the formula for this question

[timid silo]

x=0, P<1-.995, p=3000/77540, q=1-p. Solve for n

[iron drum]

What the fuck? Like maybe can you plug in the numbers into the formula you posted above and write it instead of type?

The fuck I can’t fucking understand this fucking shit

[timid silo]

the formula is literally here

then we substitute everything in like this:

$0.005>\left(\frac{3000}{77540}\right)^0\left(1-\frac{3000}{77540}\right)^{n}$

[warm shale]

Toby

[timid silo]

then use logs to solve

[iron drum]

Logs?

OH

LOGS

I REMEMBER THAT FUCKING SHIT CUNT YEAH

[timid silo]

lol

[iron drum]

Wait where did the fucking 0.005 come from

[timid silo]

1-99.5%

[iron drum]

Wait this part is completely irrelevant because anything to the power of zero is 1?

[timid silo]

yeah

[iron drum]

So can I just get rid of that shit so I can make it more simple?

[timid silo]

yes

[obtuse pebble]

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[timid silo]

for $-\frac\pi2\leq t \leq \frac\pi2 \ \ -1 \leq \sin t \leq 1 \ \ 0 \leq cos t \leq 1$

[warm shale]

kitchen1112

[obtuse pebble]

@hidden bronze Has your question been resolved?

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[upbeat yacht]

Excuse me, how would you interpret the difference between two functions?

[upbeat yacht]

As in

Do I have to pass "v" to the first function, get its image and then pass the negative image to T?

Is there a way to calculate this T-lambdaIdV into a new function F without the help of the element v?

[warm canopy]

T-λId is the explicit definition of the function, there's no rewriting this into another function

It sends v to T(v)-λv

[warm canopy]

Because Id(v)=v

[upbeat yacht]

ok yeah, I had a hunch that it worked that way, apply linearly those functions, thanks for the confirmation, do you know about Linear Algebra eigenvectors and eigenvalues?

[warm canopy]

I do, pop the question here if you have one, I might not be around but plenty of people here know linear algebra well

[upbeat yacht]

ok thanks

Let me take a few screenshots first

This images are from an explanation (in italian) on how to calculate eigenvalues, what I don't understand is, why is the 0v in the second image not represented as a vector of coordinates and instead as a vector, even though we switched "context".
Also this switching of context (from linear operators to matrices) it's as confusing as it was short(ly explained).

[upbeat yacht]

Here are my couple of questions

[warm canopy]

0_V is just the notation for the zero vector in an arbitrary abstract vector space (not necessarily a vector space consisting of classic vectors like R^n)

For finite dimensional vector spaces, once you fix a basis you can associate the vector space to a vector space of classical vectors

But there's no need to always do that, especially when discussing eigenvectors and eigenvalues which exist in any vector space

[upbeat yacht]

The transformation to matrices in this instance is leveraged to explain a simple way to calculate eigenvalues

What's the connection though?

[warm canopy]

The connection is the eigenvalues and vectors of a linear map are the same as those of a matrix that represents the map

[upbeat yacht]

Yeah it's the zero vector, weird thought that they didn't explicitly say it was supposed to be interpreted as a vector of coordinates since we are in a matrix environment

[upbeat yacht]

Okay nice

[upbeat yacht]

Wouldn't you agree?

[warm canopy]

The language is just kind of interchangeable when working with matrices

The zero vector 0_v in R^n is the same as the vector (0,...,0)

It just saves you having to write the latter every time

[upbeat yacht]

What I'm trying to say here is that in a matrix context there are no vectors v, only its coordinates

so its weird that the zero vector wasn't represented like a vector of coordinates

[warm canopy]

Something like (1,2,3) IS a vector in R^3

[upbeat yacht]

Yeah

And if you have a basis fixated, the coordinates of the vector (1, 2, 3) relating to that basis are unique and represented as a n-tuple, and are exclusively used in a matrix context

Anyway, thank you for putting up with me

Linear Algebra can be a mess sometimes

Have a nice day 🙂

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[hard rivet]

#books-old

[hard rivet]

I need help on mental math and written math

[obtuse pebble]

@hard rivet Has your question been resolved?

[hard rivet]

<@&286206848099549185>

I'm having a hard time to figure it out especially when there is fraction already

[obtuse pebble]

@hard rivet Has your question been resolved?

[tardy epoch]

which problem

[tardy epoch]

9? 17?

[hard rivet]

Mental math 1 to 10
And written math 11 to 20

All of that

I need answers to them

[copper latch]

What have you tried?

[hard rivet]

What?

I just need an answer to the picture i sent to uou

You

[copper latch]

Have you read rules #❓how-to-get-help

While asking questions, make sure to mention all relevant details, including what you have tried and what you're stuck at. Do not expect others to simply solve your questions for you.

[obtuse pebble]

@hard rivet Has your question been resolved?

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[fervent bluff]

Stumped on this one. Let f(x)=5-4x. Solve for f(2-x)

[mild harness]

Cuz wouldn’t it be equally likely maybe I don’t know how to count properly

[blazing sentinel]

You're in someone else's channel

[blazing sentinel]

if f(anything) = 5-4(anything) what do you think you could do with the 2-x

[fervent bluff]

-3x+4, right?

I had -4 and i realized I just did it wrong

[mild harness]

5-4(2-x)

[fervent bluff]

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[spring mountain]

X(5 - X) = 7 (X - 9) [PQ FORMULA]

[spring mountain]

The correct answer is
X1 = 7
X2 = -9

I somehow got - 8 and 6

[slow sonnet]

7*-9 is 63

@spring mountain

[spring mountain]

Oh my god

UGHHHHHH

appreciate it 😅😅

[slow sonnet]

I just did 3+3 = 9 ten minutes ago while I was solving an integral

happens

[spring mountain]

yeah kinda awkward but thank you! ❤️

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[hexed tundra]

hi!

[hexed tundra]

I'm looking for help modeling blackjack draws for a program I'm building

currently, the solution i have is to generate all unique dealer hands, then multiply them by some factor that's a function of the contents of the dealer hand

then add this sum up

the function is a factorial of hand length

minus cards that are 16 or less

[obtuse pebble]

@hexed tundra Has your question been resolved?

[hexed tundra]

The challenge is that aces have dual values

I need to model a stand on 17 value rule in this way

[obtuse pebble]

@hexed tundra Has your question been resolved?

[hexed tundra]

<@&286206848099549185>

[solar remnant]

Idk about black jack rules / draws
Maybe rewrite it with maths for us

[hexed tundra]

Hi

Cards are ranked from 2 to 10 in value

Face cards have value 10

There is a special case, the ace, which has value 1 or 11

@solar remnant

Still here?

The dealer draws until their hand has a value of 17 or higher

[hexed tundra]

I need to determine the number of permutations that can be validly generated for a single combination algorithmically.

[solar remnant]

there's no bijection so no permutations possible if differents cards can have same value

Your program is supposed to know how many ways there's to obtain a hand value of 17 ?

@hexed tundra

[hexed tundra]

any hand value between 17 and 21, inclusive

[solar remnant]

Do you started your program on github ?

[hexed tundra]

i have part of it

it's in Haskell

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[hexed tundra]

sorry

@solar remnant looking for other approaches

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[oak lichen]

quick stuppid question

[oak lichen]

im having a brainshit

for derivatives

6x^2/3

6*2/3=4 x

what is one less power for 2/3

like for example for 3x^2

one less power would be one

im having a brainshit

[bold bane]

2/3 - 1

[oak lichen]

so -1/3

?

[bold bane]

Yes.

[oak lichen]

oh and u make it denominator

thanks

ims tupid

[fierce lagoon]

Relatable

[oak lichen]

.close

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[compact gate]

@drifting nova Use the quadratic formula.

[drifting nova]

I did quadratic formula myself didnt get one of the answer choices

[compact gate]

what did you get?

can you post a picture

[drifting nova]

Sorry its sideways

In the discriminant the denominator of b just dissapears

In the answer choices

[compact gate]

so a = 1
b = -p/5
c= n

[drifting nova]

Yeah

[compact gate]

let me think about it.

[compact gate]

So this is the correct first step

Next you would want to combine p^2/25-4n into one fraction

then pull out the denominator outside the square root. Then simplify

I'll leave the details to you.

[drifting nova]

How?

4n is in the numerator

[compact gate]

what do you get as a combined fraction?

We can use the following fact:

$\sqrt{\frac{a}{b}}=\frac{1}{\sqrt{b}}\cdot\sqrt{a}$

[warm shale]

Goose on a Moose

[drifting nova]

Okay so 1/5*radicalp^2-4n

[compact gate]

yee

[drifting nova]

But then b would be divided by 2 at the front and 10 on the back

[compact gate]

$\frac{p/5\pm 1/5\sqrt{p^2-100n}}{2}$ Do you have this?

[warm shale]

Goose on a Moose

[drifting nova]

Where did we get 100?

[compact gate]

So we have √(p^2/25+4n)

To combine the fraction, we need to get p^2/25 and 4n to have the same denominator. Easiest way to do this is to multiply 4n by 25/25

hence the 100n

[drifting nova]

Ohhhhh

[drifting nova]

We cant do this without doing that first?

[compact gate]

yeah.

I have to attend other things right now. If you need further assistance, maybe someone else can jump in.

cheers

[drifting nova]

Thanks

Please someone hep

[drifting nova]

I got this far

[obtuse pebble]

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[drifting nova]

.close

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[oak lichen]

how do u solve sin^-1(2x) using unit circle

[pallid flame]

solve???

[oak lichen]

evaluate*

cos(sin^-1(2x))

[warm canopy]

you should draw right triangle with some labels relevant to sin^-1(2x)

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[gilded needle]

any thoughts?

[ashen wagon]

4.65

[gilded needle]

any other possibilities?

[ashen wagon]

i dont think so

[gilded needle]

consider the possible arrangements of the house, school and library

like what order they could occur in as you go from left to right

[ashen wagon]

if the school were on the right

the library could be on the left right

[gilded needle]

yes

[ashen wagon]

it could be 1.3 too then

no it coudln't

0.65

[gilded needle]

yes

he would have to go from the house to the school and back (8 miles) and from the house to the library and back (2x, where x is the distance between the house and the library), and those have to add up to 9.3

[ashen wagon]

yeah

could u write an abosolute value equation for this

[gilded needle]

i think you would still have different cases to consider

[ashen wagon]

how so?

[gilded needle]

well for example you'd want to analyze what happens if the library is between the school and the house

absolute values don't really play a role there

[ashen wagon]

oh okay

[gilded needle]

i mean maybe you can express it in terms of absolute values in a clever way that i'm not seeing

[ashen wagon]

so for problems like this you'd really just have to use logic?

[gilded needle]

like either way the total distance traveled is |2x| + |2y| where x and y are the coordinates of the school and the library, assuming the house is at 0 and is between the house and the library

so you'd have to solve |2x| + |2y| = 9.3

where if y is the house, you know |2y| is 8

so it becomes |2x| = 1.3

that's the case where they're on opposite sides

[ashen wagon]

|2x+2y|=9.3 is this the same thing?

[gilded needle]

if they're on the same side of the house and the house is closer then it would be |2x| + |2(y-x)| = 9.3

so it's still casework

[ashen wagon]

okay that makes sense

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[glad violet]

Did I do this wrong?

I double checked it through mathway and it gave me this answer instead..

[lethal bone]

it is 12

[gilded needle]

you factored t^3 + 8 wrong

the second factor is not right

[glad violet]

also

i just realized

in mathway

i put

=

on accident

yeah this

works

also @keen garden did i

?

I thought

a^3 + b^3 = (a + b)(a^2 - ab + b^2)

[gilded needle]

yes this is right

[glad violet]

isnt that what i did

[gilded needle]

oh wait sorry yeah i misread

[glad violet]

all g ty

[gilded needle]

12 is correct

[glad violet]

ty 🙂

.close

[obtuse pebble]

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[willow ravine]

Complex Variables:

Start with the first one please. z^2

I don’t remember the formula or name of the theorem I am supposed to be used if someone could remind me.

[obtuse pebble]

@willow ravine Has your question been resolved?

[slim cove]

just parameterize it

easiest way

[lethal bone]

what would a parameter look like in C?

z,i -> r and theta/

?

[slim cove]

make a new help channel if you have a question (and aren't the original poster), I don't really wanna spoil it for poisonedfood lol

[lethal bone]

oki

[willow ravine]

Yeah, Eric loves making all 4 of my brain cells go into overdrive.

A cruel friend he is.

[slim cove]

Lol

[slim cove]

looks good!

[willow ravine]

One moment

Work problems

[lethal bone]

incredible

[slim cove]

complex numbers are quite the beauty :)

[lethal bone]

like mustard

[obtuse pebble]

@willow ravine Has your question been resolved?

[willow ravine]

Does this look correct?

[slim cove]

seems right to me

[willow ravine]

Sorry, I have a new person (an old lady) that I am training periodically. She is supposed to be well versed with a computer just to even get this job and she tried to download Chrome by just typing “chrome” in the search bar of Microsoft edge and hitting enter.

She just stared at her screen and asked me why the download wasn’t starting.

Anyways, back to my miserable problems.

[slim cove]

Rip

[willow ravine]

Checks out?

[slim cove]

where did the i go in 64i?

wait nvm

why do you have pi/2 instead of 3pi/2

should be cos 3pi/2 + i sin 3pi/2, no?

[willow ravine]

Shoot!

-64/3

[slim cove]

otherwise looks good

[willow ravine]

Alright.

I think this rings a lot of bells and makes sense that the other two I can do on my own.

Thank you @slim cove

.close

[obtuse pebble]

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[slim cove]

np

[obtuse pebble]

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[smoky panther]

is this correct? i f

[smoky panther]

how do i solve this?

[nocturne minnow]

Do you know what each of the terms means?

[smoky panther]

yea i just need to know which numbers it is talking about

and in which order

[nocturne minnow]

You mean, like for example, 20, 31, 17, what number relates to IQR, median, and range?

Because I'm assuming it's going to be in order, meaning the first value is IQR, the second is median, and the third is range

[smoky panther]

but dont i need more than 3 numbers?

[nocturne minnow]

No?

[smoky panther]

yea

[nocturne minnow]

What does vertical line with the blue dot represent?

[smoky panther]

median

?

[nocturne minnow]

Yes

So what's the median?

[smoky panther]

17

oh so

the only option there is d

[nocturne minnow]

Yes, that's it

[smoky panther]

thankyou

also how do i solve this?

[nocturne minnow]

Don't know, sorry

[smoky panther]

okay one more question i got

.close

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[uneven palm]

think of 3g(1) as 3 * g(1)

is that your doubt?

[nova violet]

I’m just not sure I’ve been stuck on this for a while now

[uneven palm]

hmm well.. what part of it are you confused on?

[nova violet]

isn’t g(3) 18?

idk it was much easier to do one step functions

[uneven palm]

oh you're talking about b

no that's incorrect, I think you missed a negative

[nova violet]

what about a?

yeah you’re right it’s -18

[uneven palm]

what's -15 + 2?

[nova violet]

13

negative

[uneven palm]

yeahh not positive

[nova violet]

yeah

idk how can I break problem c up

[uneven palm]

first, you should solve g(1)

[nova violet]

hm

solving for g(1)

@uneven palm

would it just be -2?

[uneven palm]

yup

[nova violet]

right

[uneven palm]

now the question is asking for 3 * g(1) - 4

but you already solved for g(1), so you can plug that in for it

[nova violet]

-10?

idk now I need to write it out

[uneven palm]

yep well done

[nova violet]

can you help me with the rest?

[uneven palm]

yea sure

[nova violet]

for problem d

hm

this one looks a bit more

[uneven palm]

yeahh

so basically like we did for C, you want to simplify the functions first

what's f(3)?

[nova violet]

11?

[uneven palm]

yep

[nova violet]

then you’d multiply that by 2?

22

[uneven palm]

yep

and finally, add it to g(2)

[nova violet]

one second just want to write this

-8?

then add them

14 final answer

[uneven palm]

yupp

[nova violet]

for the last one it’s gotta be 3

but I need to show my work lmao

[uneven palm]

yeahh

[nova violet]

hm

what equation can I set up to show

or can I just show plugging in numbers

[uneven palm]

well you're basically just plugging in -18 for g(x)

which makes an equation to solve

[nova violet]

thanks

[obtuse pebble]

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[obtuse pebble]

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[pine storm]

Can anyone take a look at this question and give me some suggestions about how I can improve my answer? I'm not sure whether my proof in the red rectangle is correct. Any help would be appreciated.

[pine storm]

for ease of reading:

here are all the relevant definitions for this problem, I think?

[obtuse pebble]

@pine storm Has your question been resolved?

[pine storm]

For future reference as well, can anyone tell me how should I structure my questions better? Thanks.

[timid silo]

You probably structured it quite well, it's just a lot of reading and pretty sure no one really wants to do that.

[pine storm]

Unfortunately I wasn't able to make it shorter because of how the problem is presented 😦 but I'll keep in mind to keep my questions concise next time around. Thanks!

[timid silo]

Nah its good you gave a proper background needed. I myself haven't really heard of that stuff before, but if I wanted to try to help you its good you provided the definitions. I feel like there's just a low chance someone in the help channels have learned this and remembers it still.

Higher chance of getting an answer by posting in the topic channels, say #proofs-and-logic

[pine storm]

Good point! It seems like help channels are mainly for pre-uni math problems, I'll try to ask around in other channels. Thanks a lot 😄

[obtuse pebble]

@pine storm Has your question been resolved?

[obtuse pebble]

@pine storm Has your question been resolved?

[pine storm]

!close

.close

[obtuse pebble]

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[wide orchid]

Is this correct? Plz tell me if there are any errors.

Till the general solution.

[timid silo]

,w int (sec(x)/cosec(x))^4

[warm shale]

Results provided by WolframAlpha

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

Requested by Toby#2275

[wide orchid]

hmmm, mine looks wrong

[wide orchid]

can you check my work and tell me where did i go wrong?

[warm canopy]

your work is fine!

[timid silo]

sec^2=1+tan^2

[wide orchid]

alright. thanks for checking it guys :)

my C comes out to be 2/3 - pi/4 , it can't be solved further right? i just replace C with 2/3 - pi/4?

@warm canopy @timid silo

[timid silo]

yeah

thats right

[warm canopy]

how are you getting 2/3-pi/4?

[wide orchid]

by replacing x by pi/4 in the equation

[warm canopy]

right but then what

you need to choose c such that $F(\frac{\pi}{4}) = \frac{\pi}{4}$

[warm shale]

ΣAC

[warm canopy]

bruh idek what was wrong there with \left( \right)

[wide orchid]

ohhhh , right
then it will be

C=2/3 right?

[warm canopy]

i wrote that verbatim im sure lmao

yeah 2/3

bingo

[wide orchid]

:D , thank you so much

got it

.close

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[teal fjord]

if i had to sketch a velocity time graph for this problem how would i find all the specific measurements: A ball is dropped from a height of 5m, it bounces to half the height of the previous bounce each time. You may assume the velocity reverse direction instantly when it hits the ground

sorry bout this i'm new to physics

[obtuse pebble]

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[teal fjord]

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[solid bear]

Is this ok?

I got a bit confused at the end

[gilded needle]

how did you go from $\frac{-1 + 2x^2}{x+2}$ to $\frac{-1 + 2x}{2}$?

[warm shale]

Bungo

[gilded needle]

assuming i'm following correctly

[solid bear]

I simplified the x in the numerator and denominator

I think

that's what I tried at least @gilded needle

[gilded needle]

by what operation? the two fractions are not equal in general, as you can see by plugging in say x=1

[solid bear]

ok

so I change that

[cedar lichen]

(x + a)/(x + b) ≠ (1 + a)/(1 + b). You can't just divide some terms by x without dividing the others as well

[solid bear]

oh ok

[fickle turret]

Also you missed the - before the x^2

[solid bear]

where

[gilded needle]

-(x^3 + 2x^2) is -x^3 - 2x^2, not -x^3 + 2x^2

going from the first line to the second on the right side

[solid bear]

ohhh so the sign changes

it doesn't affect too much though

this is the new result

[fickle turret]

It "only" makes the difference between - and + infinity

[solid bear]

yeah you're right

[solid bear]

is it right now?

[fickle turret]

Can you show your whole arguementation?

[solid bear]

yes

[fickle turret]

1. Is it -2x^2 or 2x^2 now? Doesn't get clear in your notes

2. Bottom there is missing x goes to infinity

[solid bear]

-2x^2

[fickle turret]

3. Leaving the -1 and the +2 is not valid even though it does not matter

[fickle turret]

Then leave the right side away

[solid bear]

i thought you have to choose the one with the highest value only

[fickle turret]

Yes, you're right about that. You can just write numerator degree is bigger than denominator degree. But leaving a term out without justification is odd

But besides that it's good and correct 👍

[solid bear]

yayyy noted it thank you

[obtuse pebble]

@solid bear Has your question been resolved?

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[timid silo]

how many elements are in S_3?

[noble kindle]

6

[timid silo]

so what possible subgroup sizes are there?

[noble kindle]

Divisors of 6, so 1,2,3,6

[timid silo]

yup, can the subgroup of order 6 be cyclic?

[noble kindle]

No, since S3 can’t be generated by any one of its element

Like 3 cycles can’t be generated by 2 cycles, I’m guessing

[timid silo]

yup

also, its not abelian

and cyclic groups need to be abelian

[noble kindle]

Oh true

[timid silo]

finally, do you know your small groups?

[noble kindle]

What are small groups

[timid silo]

like do you know what the only group of order 3 is?

[noble kindle]

Might know it by another term

[timid silo]

(or alternatively, groups of prime order are cyclic)

[noble kindle]

im not sure

[timid silo]

there is a unique group of order 3, namely Z/3Z, the unique cyclic group of order 3

similarly, for 2 and 1

[noble kindle]

ok

sorry, im still not understanding how that is related

[timid silo]

the subgroups of orders 1,2,3 of S_3 are all automatically cyclic because of their orders

[noble kindle]

right, because 2 and 3 are prime.

[timid silo]

yup

[noble kindle]

and subgroup of order 1 is just {e}

So it is at most 3 now?

[timid silo]

yeah

its exactly 3

[noble kindle]

ok, but what about normal subgroups?

since its asking for number of subgroups that are both normal and cyclic

[timid silo]

cyclic subgroups are automatically abelian

abelian subgroups are automatically normal

[noble kindle]

hmm, ok then the answer is 3

ill think about it more

thanks

[timid silo]

yeah np

[noble kindle]

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[hallow hatch]

Ms. Q gave George the following problem: "License plates in Aopslandia consist of six upper-case letters. For example, two possible Aopslandian license plates are $ABCDEF$ and $AAAOPS.$ No two license plates are the same. How many possible Aopslandian license plates are there which contain at least four $A$'s?"

George got the answer $\binom{6}{4} \cdot 26^2,$ but Ms. Q told him this was the wrong answer!

(a) How did George arrive at his answer?
(b) Why is George's answer wrong? Should the correct answer be smaller or larger than George's answer (and why)?
(c) Write a solution to Ms. Q's problem, explaining in complete sentences what the correct answer to the problem should be and why.

[warm shale]

banan|crab

[hallow hatch]

i figured out how george got his answer

but i dont know what he did wrong

i cant figure it out

ill then try to attempt part c myself

anyone?

does george overcounnt AAAAAA $\dbinom{6}{4}$ times?

[warm shale]

banan|crab

[tired skiff]

aaah sorry i wish i knew about permutations ><

[hallow hatch]

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[timid silo]

try multiplying $\begin{pmatrix}1&a\0&1\end{pmatrix}\begin{pmatrix}1&b\0&1\end{pmatrix}$

[warm shale]

Toby

[slim cove]

lol toby that's what I said yesterday

[noble kindle]

1 b+a

0 1

[slim cove]

not quite

check your b+1 thing

[noble kindle]

oops

[slim cove]

yup

so which group does it look like it behaves like

out of the answer choices

[noble kindle]

Z under addition

[slim cove]

yep

what is the isomorphism from Z to G?

[timid silo]

🙃

[noble kindle]

via a ---> [1. a]
0. 1

[slim cove]

yup

and then the things you need to make sure you know how to prove are

1. that's an isomorphism, and
2. Z under addition is not isomorphic to any of the other choices (hint: cyclic)

[noble kindle]

okok i think I can do those

ty

.close

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[river plank]

Hi this is question from Oxford. I am stuck at the first part. I don't have any idea how I could show that M2 and M3 lie in the same orbit

[warm canopy]

try showing they're row equivalent

[obtuse pebble]

@river plank Has your question been resolved?

[river plank]

Done

[river plank]

Now, how do i show

[warm canopy]

congrats!

[river plank]

that

M1 and M2 are in different orbits

also found out A btw

[river plank]

by constructing a specific counterexample?

[river plank]

I think that since both have determinant zero so are not invertible hence the only way their orbits can be equal is if they are row equivalent but clearly they are not. Is this good enough?

[warm canopy]

im not sure im convinced they can only be in the same orbit if they are row equivalent?

(i dont know one way or the other, if thats true then yeah thats a fine argument)

[river plank]

ok, Thanks!

[river plank]

I guess since every invertible matrix can be decomposed into elementary matrices. then for them to lie in the same orbit we must have that that there exists A2 for every A1 A1M2=A2M3 and conversely as well

but this can only be achieved if M3=A3M2 for some intvertible matrix A3

but every invertible matrix can be decomposed into elementary matrices

so it follows that m3 an m2 must be row equivalent

[river plank]

Is this good now?

[warm canopy]

every invertible matrix can be decomposed into elementary matrices

this was the fact i was missing!

looks good to me

[river plank]

Thanks!

[river plank]

Any hint for the last one?

So far what i

i've gotten is that

both M1 and M2 can only be stabilized by the elemntary matrices corresponding to scaling the third row by an arbitrary constant

but don't know why this should be the case

[warm canopy]

Eh I think there's more in the stabiliser, for example you could add the bottom row to another

[river plank]

yeah you're right

i just solved it for arbitray matrix and got free 3rd column as free parameters

and the rest as they should be

[river plank]

I have another question do I continue this session or close it and open another one?

I am done with this one

I need help with this.

[warm canopy]

Jheeze I can't help much with this

Maybe utilise lagranges theorem to limit down your options

The first one is not so bad, the others im not so sure on

[compact shadow]

Check exp(G) for each G, where exp(G) is the maximal value of order of elements in G

For C_5 and D_10 maximal n is clearly 5 and 10 respectively (if D_10 means the dihedral group with 20 elements , Wikipedia version )

For the rest two clearly 6 since

6! is minimal n! who is divided by 36

[gilded needle]

for D_10 it depends on which convention is used for the dihedral group of the n-gon, some authors call this D_n and others call it D_2n

assuming it means the dihedral group of the 5-gon, note that this is a group of certain kinds of permutations of 5 elements

[compact shadow]

Then 5

Depends on which version he uses, if dihedral group of 10 elements then 5

[obtuse pebble]

@river plank Has your question been resolved?

[river plank]

yeah i mean the symmetry group of 5-gon

[river plank]

I am done with the first two now trying to figure out the 3rd one

[compact shadow]

And told you this

[river plank]

So for C2C2C2

exp(G)=2?

[compact shadow]

36 doesn’t divide 5!=120, so minimal has to be 6

[river plank]

since every element has order 2 in C2C2C2

[compact shadow]

No that page is turned over

exp(G) was for C_5 and D_10

[river plank]

Oh got it

[compact shadow]

Since they have elements of order 5 so minimal has to be 5

[compact shadow]

Just this for the rest two

[gilded needle]

for C2xC2xC2, a hint: note that in that group there are three distinct elements of order 2 that commute, so you need that property in some subgroup of S_n as well

[river plank]

Not sure how this also applies to C2C2C2 i think you are just referrrinh to the 4th one

[gilded needle]

think about what condition is needed for two transpositions in S_n to commute

[river plank]

that they be disjoint

[gilded needle]

right

so that tells you right away that n has to be at least how much?

[river plank]

n>=6

[gilded needle]

yep

now can you show that n=6 works?

[river plank]

yeah let me try

[river plank]

<(1 2) (34) (5 6) > this should work

[compact shadow]

Just order

S_n n<=5 doesn’t have subgroup of order 36 was my whole point since 36 doesn’t divide 120

And they are clearly subgroups of S_6 so minimal is 6

[river plank]

but c2c2c2 is of order 8

so doesn't that just give us n>=4

[compact shadow]

Oh

Okay fourth one is solved

Let me rethink about third one

Okay this:

Suppose that image/ of (1,0,0),(0,1,0),(0,0,1) are (a,b),(c,d),(e,f) respectively in S_n

Then those (a,b),(c,d),(e,f) mustn’t overlap otherwise

You will have an element of order 3

Say b=d, you have (a,b) and (b,c) then (a,b)(b,c)=(a,b,c) is of order 3

Contradiction since elements of C_2^3 have order power of 2

The only way you can make them not overlap is n>=6

n<=5 it has to overlap

[river plank]

Yeah, got it

Thanks!

[compact shadow]

Np

[river plank]

but confused about

the

4th one

[compact shadow]

36 doesn’t divide 120

[river plank]

i get this

but

how do i construct a subgroup of S_6 isomorphic to

S_3crossS_3

[compact shadow]

{g from S_6: g({1,2,3})={1,2,3}, g({4,5,6})={4,5,6}}

[river plank]

So sym{1,2,3} union sym{4,5,6}

[compact shadow]

Cartesian product

Not union

[river plank]

lift to S_6

lifted*

[compact shadow]

(123)(45),(45),(13)(46) those

Just their Cartesian product

[compact shadow]

Easy to prove this is a subgroup

Having h, k from S({1,2,3}) and S({4,5,6}), define g from S({1,..,6}): g(i)=h(i) for i =1,2,3, g(i)=k(i) when i=4,5,6, mapping (h,k) to g

[distant fossil]

https://gyazo.com/0a6fd0b448acb89a6da86ad555855dec

HOW ME DO MATH

[timid silo]

f(x) - f(2) = f(x) - 1

in this case

[distant fossil]

what

Where did the x-2 go

[distant fossil]

Explain pls

[timid silo]

the x -2 is still there

just saying that the numerator

is f(x) - 1

plugging 2 into the f(x) you get 1

[distant fossil]

yea

[timid silo]

have you done any work

[distant fossil]

i got 1-sqrt(x-1) all over sqrt(x-1) and all again over x-2

idk how to simplify it

here let me sketch out what i have

https://gyazo.com/c7774d16b9ac4a73fff32dea2e28fe38

cause i turned the 1 into a Sqrt x-1 over sqrt x-1

[obtuse pebble]

@river plank Has your question been resolved?

[void pelican]

maybe ask in #groups-rings-fields @river plank

[obtuse pebble]

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[obtuse pebble]

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[inland lava]

You’re provided a list of arrays , 1-26, they’re assigned to the alphabet as each. E.g A is 1 and B is 2.

Your mission is to re-range the arrays (inverted) Z-A but the numerics are negative.

Re-read the new array return the sum of the new array (hint: answer is postive)

If the negative returns a valid equality test == or >= or <=. E.g a>=b then add it to the sum as a positive value (a=ax2).

If the new sum of array is greater than 9e9 or 100 then print it as "You’re a heartbreaker" if not print "ahh"

Test cases -> 427```
can anybody help me solve this

[obtuse pebble]

@inland lava Has your question been resolved?

[obtuse pebble]

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[near mulch]

i tried doing this myslef and i wasnt getting the right answers cause no way in hell the f(x) is -11.99600 im so lost

[obtuse pebble]

@near mulch Has your question been resolved?

[forest sinew]

should just be calculation, which part are you confused about exactly?

[near mulch]

the subbing in part

[forest sinew]

just use a calculator

do you understand what f(x) means?

like given some x

[near mulch]

yea

when i sub it in tho

i get a number no where near 4

so i think im doing something wrong before thay w the function

[obtuse pebble]

@near mulch Has your question been resolved?

[devout sable]

It’s 0/0 form

Factorise

[obtuse pebble]

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[wide orchid]

i have a doubt in integrals.

integration of 1/x dx is log x + c
so, integration of -1/x dx should be -log x + c right?

my book says other wise, it says -1/x dx = -log x + log c

[wide orchid]

i am confused

[haughty coyote]

Log is a bijection so it doesn't matter

It still amounts to "+ any constant"

[bleak maple]

even if it wasnt bijective it would still be fine

[haughty coyote]

The choice of c differs but that's it

[bleak maple]

log(c) has a single well defined value

also you should probably specify x > 0 here

[haughty coyote]

There would be constants for which it wouldn't work if not surjective

[wide orchid]

alright, so if i just do + C instead of +log C, the answer would still be correct

[bleak maple]

yeah

[wide orchid]

okay. but when i solve with log c the answer comes out to be c = e
when i solve with + c , c comes out to be 1

still both are right?

[bleak maple]

those c's are diff so yeah

[haughty coyote]

But there's a constant in both cases. It amounts to the same formula. It's just that log c is much rarer to use

[bleak maple]

oh right it needs to have an image of R since c is arbitrary

[wide orchid]

okay, C = 1 looks better , so ill use that :D

thanks

[timid silo]

it's mainly used to simplify the expression in solving

[wide orchid]

.close

[obtuse pebble]

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[hasty flint]

Can someone point out what I did incorrectly cause this is driving me insane.

I'm trying to integrate 1/(1-x^2). I turn it into 1/((1-x)(1+x)). Then I have A/(1-x) + B/(1+x) where A + Ax + B - Bx when I cross multiply
A+B = 1 and A-B = 0. That means A = B = 1/2. Then I end up with 1/2(ln|1-x|+ln|1+x|) + c as my answer

[hasty flint]

but that's incorrect

[timid silo]

what's the given answer

[hasty flint]

but if I factor out a -1 at the beginning, then I get the right answer

it's just 1/2(ln(1+x)-ln(x-1)) + c

[hexed blaze]

Maybe they want you to

Simplify moẻ

More

ln(a)-ln(b) can be ln(a/b) too

[hasty flint]

no the answer is literally the same thing as mine

[hexed blaze]

Also remember, it's the abs value

[hasty flint]

except one sign is flipped

and I can't figure out why

[hexed blaze]

Lemme check

Tho I'd like to show you a faster method

∫ 1/(1-x²) dx right

[hasty flint]

ye

[hexed blaze]

∫ 1/(1-x)(1+x) dx

½∫ [(1+x)+(1-x)]/(1-x)(1+x) dx

½ln|(1+x)|+½ln|(1-x)| +C

It's + not -

[hasty flint]

yeah but if you factor out a -1 in the beginning

like

-∫ 1/(x²+1)

[hexed blaze]

Why you factorize out -1

It's the different problem

[hasty flint]

wdym

[hexed blaze]

Because the original one there's no -

[timid silo]

this is wrong factorisation

[hasty flint]

oh sorry

I meant -∫ 1/(x²-1)

[hexed blaze]

If you want it that way

[hasty flint]

cause this is the answer

[bleak maple]

are you confused because you have |x-1|?

|x-1| = |1-x|

[hasty flint]

yeah I know

[hexed blaze]

Yeah i think he confused that

[hasty flint]

wait

½ln|(1+x)|+½ln|(1-x)| +C

[hexed blaze]

• ∫ 1/(x²-1) dx
  = -½ ∫ [(x+1)-(x-1)]/[(x-1)(x+1) dx
  = -½∫ 1/(x-1) - 1/(x+1) dx
  = -½ ln|x-1|+½ln|x+1|+C

[timid silo]

they are convertible

rationalisation

[hasty flint]

yeah so I got that too

but since |x-1| = |1-x|

isn't that just ½ ln|x-1|+½ln|x+1|+C except a sign is flipped

I'm so confused

cause ½ln|(1+x)|+½ln|(1-x)| +C is not the same as -½ ln|x-1|+½ln|x+1|+C unless I'm tripping

<@&286206848099549185>

[timid silo]

Wait

https://www.integral-calculator.com/

[hasty flint]

yeah I'm using that site and they just factor out the -1, so I can't tell what is incorrect

[obtuse pebble]

@hasty flint Has your question been resolved?

[hasty flint]

Ig this is a better way to put it. If I did, where I screw up

[timid silo]

There is a direct formula for it

[bleak maple]

\int 1/(1-x) isnt ln|1-x|
other than forgetting a constant, you also need a negative sign