help-23

[sterile vault]

I did tan-1 (0.115/4.197)

I got 1.56951673

[worthy hemlock]

Why did you get rid of the negative sign?

[sterile vault]

gonna add the - at the end

cuz ik it's gonna be negative

but I don't think it's right to get

1.56 degrees

is it?

[worthy hemlock]

No

[sterile vault]

yeh that weird

[worthy hemlock]

Because it's negative

[sterile vault]

should be higher number I think

[worthy hemlock]

It's the right value, you need the negative sign

[sterile vault]

oh oki

-1.56951673

so like

in third or 4th

[worthy hemlock]

Look at the coordinates

What quadrant is it in?

[sterile vault]

4th

[worthy hemlock]

So then make the angle in the 4th quadrant

[sterile vault]

oki doki

now 360-1.56951673

[worthy hemlock]

Could be

🤷‍♂️

[sterile vault]

that means yes

alr

[worthy hemlock]

No it doesn't

[sterile vault]

358.4304833

sounds right to me

[safe radish]

@sterile vault Has your question been resolved?

[safe radish]

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[drowsy gust]

i got -4/3 units^2

[drowsy gust]

but in the answer sheet is positive 4/3

[worthy hemlock]

Which one?

[quasi bison]

which one

regardless of which letter you were doing, it's very likely you just screwed up the order for the bounds of integration or the order in which the functions were to be subtracted

[drowsy gust]

for part b

[quasi bison]

did you do int[2,4] (y^2-6y+8) dy?

[drowsy gust]

ye 1 sec

let me take a pic of working out

[quasi bison]

yeah so that's your mistake

no need

[drowsy gust]

nah first i found the anti derivative

[quasi bison]

you should have had -(y^2-6y+8) under the integral

that's your sign error and i think it's the only one

[drowsy gust]

i remember seeing my teacher applying the absolute to the negative area section of the integral

but if there will be future question that would ask for like the net area, or the positive sum of area of an integral

do u get what im sayin?

[safe radish]

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[sterile vault]

When multiplying negative and negative you keep as negative ye?

[sterile vault]

wait no

Ok uhh

my question is

-233.7425+ -202.3823

wouldn't that be Positive 436.1248?

[wide swallow]

no because it's addition

[sterile vault]

could ya remind meh

[wide swallow]

so negative times negative is positive

and negative divided by negative is positive

same with powers so e.g. -1^2 = 1

but addition of negative numbers are negative

for example if i lost 5 cakes and i lose another 5 cakes i dont gain 10 cakes

i lose 10 cakes

[sterile vault]

alr

what I am understanding rn is that

negative x negative= positive

[wide swallow]

so for your problem it'd be so it's negative 436.1248

yes

[sterile vault]

negative - Negative = negative

[wide swallow]

nope

positive

[sterile vault]

hmm

alr

so it is positive 436.1248

[wide swallow]

double negative = positive

no it's negative

because it's addition

[sterile vault]

yee

[wide swallow]

if it was -233.7425- -202.3823 it would be positive

[wide swallow]

but you're adding negative on negative

[sterile vault]

pretend that negative at Ry is positive

[wide swallow]

your answer would be correct even if you didn't add the negative sign because everything squared is positive
but your working would be wrong

[sterile vault]

oh

what I miss?

did I forget to sqrt?

hmm

[wide swallow]

what level of maths is this?

[sterile vault]

yes I did...

I forgot to sqrt lol

this is uhh

Vectors

College

Essential Mathmatics I beilieve

[wide swallow]

I can't help you with the actual question but I can tell you that negative + negative = negative xD

[sterile vault]

oh

uhh

oki oki mb

X_X

it's wrong too lol

eh should I ping for help?

nvm I did it

.close

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[sterile vault]

anyone know what I did wrong?

<@&286206848099549185>

<@&286206848099549185>

[safe radish]

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@sterile vault Has your question been resolved?

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[lean otter]

I don't know how to do this

Can someone please explain?

[hushed mirage]

@lean otter your last answer DB was wrong

[lean otter]

I know

Fixed that

So how do you do this question?

[hushed mirage]

vector times a scalar

[lean otter]

What?

[hushed mirage]

BY = YC*3

[lean otter]

Yes

[hushed mirage]

AB = AX*2

[lean otter]

yes

2AX+3YC?

[hushed mirage]

yeah

no

subtract a

then yes

so AX+3YC

[lean otter]

a+3b ?

[hushed mirage]

yessir

its handy to remember that a vector going from a to b is b - a

[lean otter]

AND YD = b -2a?

?

[hushed mirage]

yes

[lean otter]

DX = -4b + a ?

a - 4b

[hushed mirage]

yep

[lean otter]

thank you

These are easy but I get nervous and then mess up

Suggestions?

@hushed mirage

[hushed mirage]

this is prep for visualizing coordinate systems

here a and b are basis vectors which span the plane

that means any linear combination of them from a starting point can get you to any other point in the plane

[lean otter]

So I can take any route from one point to the other and I'll get same answer?

[hushed mirage]

thats exactly right!

following along the parallelogram is helpful, and i think you did that

its easier also to think of the edges of the parallelogram as the basis vectors i and j

[lean otter]

My advice I was asking was more meta: I have the skills but I get nervous. How to not get nervous so I can use my brain to solve the problem and be done?

[hushed mirage]

and say a = i/2 and b = j/4

[hushed mirage]

im not sure tbh. remove all doubt that its correct, and practice and youll gain confidence

[lean otter]

What do you mean remove doubt?

[hushed mirage]

do the logic carefully

if youre smart about it then it will take less steps, leading to less mistakes

[safe radish]

@lean otter Has your question been resolved?

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[spiral patrol]

can someone help pls :))

[worthy hemlock]

Looks like physics

[spiral patrol]

yeaa

Im confused theres no time?

[worthy hemlock]

There is a physics server in #old-network

[spiral patrol]

alright il go there thanks :))

[worthy hemlock]

FYI $$v = \frac{d}{t}$$

[flat frigate]

dldh06

[worthy hemlock]

So you can find time

[spiral patrol]

100m/ 18?

[worthy hemlock]

Sure

[spiral patrol]

i cant chat in the physics server..

[spiral patrol]

i got 5.55 so 6 is my time?

[worthy hemlock]

You don't round

Why should you round?

[spiral patrol]

ow ok

Idk my tcher told me

[worthy hemlock]

You round at the end

[spiral patrol]

okok

So 5.55555556 seconds

im stil not getin the right answer🥲

[weak dirge]

What is this

[weak dirge]

Okay first thing is first you need to solve for time

[inland ivy]

I like how she's wearing hockey skates

[worthy hemlock]

They did

[inland ivy]

[weak dirge]

Do you know the equation to do that

[spiral patrol]

HAHAHA

[weak dirge]

100/18 is not the time?

[spiral patrol]

uhh change in work over change in time?

[worthy hemlock]

Did you read the stuff before?

[weak dirge]

D=t * a^2/2+v*t

[spiral patrol]

ah sht

[weak dirge]

So acceleration is what you got before right

[spiral patrol]

yeps

[weak dirge]

So you need to solve the quadratic for time

[spiral patrol]

oh gosh

can u show me how to get the answer cuz i alredi got wrong in it

I jus wana kno how to get the answer

[unborn snow]

use the formula hegaved you

[worthy hemlock]

Or you can do it yourself and verify

[spiral patrol]

948 W was the right one

okay😔

[unborn snow]

distance=initial_distance+velocitytime +1/2accelerationt^2
or in other words. what KurtDee posted

[spiral patrol]

my distance isnt 100m?

[weak dirge]

It is

[unborn snow]

it is.

[spiral patrol]

So..

Im looking for velocity?

[weak dirge]

No

[unborn snow]

you are solving for time.

[spiral patrol]

i got time 5.555556

dat wasnt right?

[weak dirge]

You have velocity, you already solved for acceleration which some dumb ass told you that’s time

Acceleration = v/t

[spiral patrol]

ooooooohh

dats my acceleration hahaha

aight

[weak dirge]

Yeah

[spiral patrol]

I thought he was a helper

Or she

[weak dirge]

I’m not even a helper

[spiral patrol]

Hahahaha

Aight anyways to find time

[weak dirge]

But you can use that in the equation me and samer posted

[spiral patrol]

Il use da formula u sed

[weak dirge]

Yes

[spiral patrol]

Okoke il try

[unborn snow]

you can ignore initial distance because its 0 in your question

[spiral patrol]

oke

[weak dirge]

I need to go but it should be simple when you solve for time

[spiral patrol]

🥲

[weak dirge]

Acceleration * weight/time I think is the power equation

[spiral patrol]

alright bye thanks 🥲

[weak dirge]

Er

Work = force times distance

[queen parcel]

power = force • velocity

[weak dirge]

So also multiply distance

Or that works yeah

[spiral patrol]

what

im sorry

I got 0.3 seconds i tink its wrong..

[unborn snow]

how did you get 0.3?

[spiral patrol]

T= a/ final vel- initial vel

...gosh

[unborn snow]

what

[spiral patrol]

know what

il sleep

Il ask my classmates tom

Thanks tho 🥲

Sorry for giving u guys headaches

.close

[safe radish]

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[umbral stream]

I never really figured this kind of stuff out and I need assistance.

[safe radish]

@umbral stream Has your question been resolved?

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[arctic stream]

*28. Anton and Bert start at the same time and run towards each other along the same straight road, each at its own constant speed. When they start they are 3 km from each other. How far has Bert run when they meet?

(1) Bert runs 50% faster than Anton.
(2) Anton's speed is 8 km / h and they meet after 9 minutes.*

**Sufficient information for the solution is obtained in: **
A (1) but not in (2)
B in (2) but not in (1)
C in (1) together with (2)
D in (1) and (2) separately
E not through the two statements

[arctic stream]

Can someone explain the solution here?

Should I call the speed of Anton x?

So from (1) we have: 1.5x, and we also know their distance from each other.

But we don't know the time, so we can't solve it with information (1) alone.

From (2) we have Anton's speed. We can calculate from here that Anton and Bert have both ran 8km/h * 9/60 = 72/60km at this point, right?

.close

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[tawdry salmon]

shit’s so difficult

[tawdry salmon]

I have radical 18

I simplify that to radical 6 and 3

but website says radical 2 radical 9

which is 3 radical 2

how do I know which to choose when simplifying

the lowest number / factor?

[drowsy moss]

you want a perfect square (if possible)

you can also just fully factor it into primes and see which ones come in pairs

[queen parcel]

@tawdry salmon ^^

The 2 approaches Zybikron mentioned are both helpful 👍

[tawdry salmon]

Ok

[tawdry salmon]

can you explain this approach further

[drowsy moss]

18 factors into 2 * 3 * 3

[tawdry salmon]

But as a rule is factoring to the lowest generally better or does it depends? like if you could get a factor of (random) 9 and 6 or and 2 and 4, would you choose 2 and 4

[drowsy moss]

so you can say $\sqrt{18} = \sqrt{2\cdot3^2} = \sqrt{2}\sqrt{3^2} = 3\sqrt2$

[flat frigate]

Zybikron

[tawdry salmon]

what do you mean by seeing which come in pairs?

[drowsy moss]

if you can see the perfect squares, then sure, do it that way.

18 has two prime factors of 3, so it comes in pairs

[tawdry salmon]

Ok thanks

.close

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[tawdry salmon]

I’ve never seen something more complicated

[safe radish]

@tawdry salmon Has your question been resolved?

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[gloomy thicket]

The Euclidean algorithm for the greatest common divisor (GCD) of two natural numbers
is this: for input x and y , if x and y are equal, that is also their GCD; otherwise,
take the GCD of the smaller one of x and y and the difference between x and y .
Implement this as the function euclid .

[gloomy thicket]

im confused like

gcd x y, if x y are the same then return x

gcd x < y then gcd x x-y

gcd y > x then gcd y x-y

thats literally what that says to do but it doesnt work

[safe radish]

@gloomy thicket Has your question been resolved?

[safe radish]

@gloomy thicket Has your question been resolved?

[dense yacht]

x < y is y > x

[safe radish]

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[echo linden]

How would you factor y-y^4=0

[analog flare]

Well whats the common factor

[rancid wagon]

y(1-y^3)

[echo linden]

It would be y right so would it look like y(1-y^3) yea yea that

But what next

Lol

[analog flare]

Exactly

[rancid wagon]

finished

nothing to do

[echo linden]

Really?

[rancid wagon]

yes

[analog flare]

Do you wanna solve for y?

[echo linden]

This is the original question

Sorry wifi is slow

[analog flare]

So you need to solve for y

[echo linden]

Yea

[analog flare]

Well do you know how to solve
$$(x-4)(x-4) = 0$$

[flat frigate]

Pluton

[rancid wagon]

x^2 - 8x + 16

[echo linden]

Make x-4=0

X=4

[analog flare]

Exactly

[analog flare]

Here you have same thing

2 factors are equal to 0 so 1 of them must be a 0

[echo linden]

Alright so y=0 and then y^3=1 and then cube root?

[analog flare]

Yes

So y is either 0 or 1

[echo linden]

Cube root of 1 is 1 right

Alright so that's it right and then i plug it in to the other equations

Thank you so much

[analog flare]

Yes

[echo linden]

.close

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[plain hedge]

.reopen

[solemn vault]

Hello

[solemn vault]

I got stuck here

Again

So, I got the value of x which is 70°

And I need to get the value of "y"

So, I guess it's 70+30 = 100

[worthy hemlock]

Yeah, that's it

[solemn vault]

Ah

Really

I don't think so

[worthy hemlock]

But it is

[solemn vault]

Cuz my book says its 140°

[worthy hemlock]

Wait, my bad

I flipped something around

[solemn vault]

Ok

[worthy hemlock]

You can use the idea of that point R, it's on a line

[solemn vault]

Yes

Ok

[worthy hemlock]

That blue mark at R is 70

[solemn vault]

R

Hm

[worthy hemlock]

You can find the other angle

[solemn vault]

Hm

[worthy hemlock]

That's in the smaller triangle

[solemn vault]

K

So it would be like

This

70 + R + 30 = 180

??

[worthy hemlock]

Not exactly, give me a sec to draw it

[solemn vault]

Ok

[worthy hemlock]

That orange mark is the angle you can find because it's on a line

[solemn vault]

Ok

Ok

It's

70+that orange line = 180

That orange line = 110°

[worthy hemlock]

Yep

[solemn vault]

And

110° + 30° = 140

@worthy hemlock

[solemn vault]

Can u explain how its 180° I mean there must be a reason

[worthy hemlock]

It's a line, a straight line

[solemn vault]

Ok

The reason is straight line are supplementary

[worthy hemlock]

Yes

Exactly that

[solemn vault]

K

Thx for the help

.close

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[clever oak]

for question 224

[meager chasm]

324 you mean?

[clever oak]

yeah

[meager chasm]

also, what's the original question

this seems like the solution set

[clever oak]

its to find the value of csc at those points

[meager chasm]

oh

$\csc = \frac{1}{\sin}$

just use this

[clever oak]

yes

[flat frigate]

Chromium

[clever oak]

which is opposite over hypotenuse

[meager chasm]

wdym 'at those points'

i think you're to find $\theta$?

[flat frigate]

Chromium

[meager chasm]

are $-2\sqrt3$ and $-2$ side lengths

[flat frigate]

Chromium

[clever oak]

yeah

and the hypotenuse is -2 √ 2

but if csc is opposite over hypotenuse i dont see how 2√3 over -2√2 = -2

thats what my answer key says anyway

[meager chasm]

ok first of all, negative side lengths are cringe

[clever oak]

yes

[meager chasm]

i don't even know what you're trying to represent lol

can i have the semantic question?

[clever oak]

its separated in the work booklet

[meager chasm]

please show

[clever oak]

k one sec

[nova creek]

How'd you get -2sqrt(2) for a hypotenuse?

[clever oak]

well a^2 + b^2 = c^2 and i got 8 and had to root 8

[nova creek]

How'd you get 8?

[clever oak]

-2√3^2 + -2^2 = √8

i have to root 8

and then i got 2√2

[nova creek]

(-2 √3)² + (-2)² is not √8

[clever oak]

yeah cause you get 12 + -4 = 8

[nova creek]

(-2)² is not -4

[clever oak]

well is it 12

like the whole thing

[nova creek]

(-2 √3)² is 12

But 12 + (-2)² is not

[clever oak]

lul 16

[nova creek]

Yes, but remember to take the square root

[clever oak]

4

[nova creek]

That's the hypotenuse

[clever oak]

so how is -2√3 over 4 = -2

oops

csc is opposite over hypotenuse

so 4 over -2

= -2

got it thanks

[nova creek]

Csc is hypotenuse over opposite, but yes, that's the right answer

[clever oak]

oh yeah

thanks dio

.close

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[icy pagoda]

I get how they did it, but I don’t understand why they did it this way.

[icy pagoda]

The question said unit vector to the perpendicular of the cross product of a and b plane

What does that mean

<@&286206848099549185>

[worthy hemlock]

It means that vector is perpendicular to a and b

[icy pagoda]

Okay so if the unit vector is perpendicular to two vectors then the

unit vector = cross product of those two vectors / modulus of their cross product

?

[worthy hemlock]

I don't get what you're asking

[icy pagoda]

How do we find unit vector of two vectors?

[worthy hemlock]

The unit vector is the vector divided by its magnitude

[icy pagoda]

Yes

Now how do we find the unit vector that is perpendicular to two vectors?

[worthy hemlock]

You find the cross product to find that perpendicular vector then you take that vector, divide it by the magnitude

[icy pagoda]

Okay

🤝

.close

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[wise cedar]

How do I get the cosine function for this graph

[safe radish]

Please don't occupy multiple help channels.

[wise cedar]

Help

[safe radish]

@wise cedar Has your question been resolved?

[wise cedar]

Hell nah

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[magic plinth]

Which are the two distinct operations required to invert the power 2^3? Are the two inverses you obtain via the application of these operations rational, or real numbers, and why?

[magic plinth]

The inversion of the base (2) is called root extraction and inverting the power (3) is called logarithmization (the root or logarithm). Is what I would answer, but I struggle with understanding wether the operations are rational, or real numbers, and why

[safe radish]

@magic plinth Has your question been resolved?

[magic plinth]

<@&286206848099549185>

[inland ivy]

@magic plinth Can you explain the process of inverting the power?

[magic plinth]

The notion of raising to a power stem from the operation of multiplication. The complexity of it is that it instead involves two numerical entities assuming different operational roles. The base is one and the power itself, is the other. Therefore, the operation of raising the base to a power is not a commutative operation, meaning that the result is not invariant if the base and power would be exchanged. This results in that we have two distinct inverses, one referring to base, and one to the power.

[inland ivy]

So what would be the inverse of per se 4^2

[magic plinth]

would it not be the root of ^2 * 4?

[inland ivy]

$\sqrt{2} * 4$?

[flat frigate]

PapaBread

[magic plinth]

Hmm yes

[inland ivy]

By inverse you you mean like

$\frac{1}{4^2}$

[flat frigate]

PapaBread

[inland ivy]

But written in terms of 2 and 4

[magic plinth]

Im sorry Papabread, I see now that I have to learn what real numbers, irattional and rational numbers are before attempting to help with this question. Thanks for the help nontheless!

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[sharp condor]

Please help

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[deep oak]

I need to find the derivative using the chain theorem for this problem. I’m working on it but got stuck

[deep oak]

Is it right for me to say that this equation can be re-written as v^3(1+v^2)^1/3

And if so does this problem have 3 layers (like one inner and two outter shells for the chain theorem)

[pine shell]

Is that square root or cube root

[wild copper]

Use multiplication formula for derivation first

[pine shell]

If you use the uv formula it'll be h'(v)= (v^3)((1+v^2)^1/2)' + ((1+v^2)^1/2)(v^3)'

[hushed mirage]

f'(x) = d/dx x^(1/2) = 1/(2sqrt(x))

[hushed mirage]

no. it has 1 product and 1 composition

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[lean otter]

I got (x+1)(x-1)(x+2)(x-2)

I don't know if there is another step where I do something with the positive and negative values.

.close

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[thin bridge]

that's the simplest full factorisation but it was unecessary to get that

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[lean otter]

wdym?

[thin bridge]

it wasn't necessary to fully factorise the quartic polynomial to answer the question being asked

[lean otter]

I got x-1 at one point

[thin bridge]

in fact it wasn't necessary to actually do any factorisation at all

[lean otter]

how so?

please explain.

how else am I supposed to get the answer

[thin bridge]

an application of factor theorem is sufficient

i.e. plug in x=2 to see if (x-2) is a factor,
plug in x=1 to see if (x-1) is a factor

[lean otter]

but those just equal 0

x-2

2-2

right?

[thin bridge]

plug x=2 into the polynomial to see if (x-2) is a factor

i.e. is
x^4 - 5x^2 + 4 equal to 0 when x=2

and if it is, then by factor theorem (x-2) will be a factor

[lean otter]

ok

so is 1

so i was right

c is the answer

cuz they both equal 0

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

can someone confirm this?

[queen parcel]

Looks good to me

[lean otter]

shen from league?

@queen parcel

[queen parcel]

Nope - just a nickname 🙂

[lean otter]

.close

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[slim pawn]

I’m having trouble understanding what example 5 is asking

[nova creek]

Evaluate the given limit

[slim pawn]

but what do i sub in for f(x)

[nova creek]

f(x) is the graph

[slim pawn]

exactly so how do i sub in a piece wise function

because its made of more then one equation

[nova creek]

Do you remember how to evaluate the limit of f(x)/g(x)?

Where f and g are any functions

[slim pawn]

i belive i do

[nova creek]

It's the limit of f / limit of g, right?

[slim pawn]

ya

[nova creek]

So the limit of (f(x))²/(x²-1) as x->2 becomes?

[slim pawn]

lim of f(x)^2/ lim of x^2-1 as x become 2?

[nova creek]

Yes

[slim pawn]

thank you ill try it and see how it goes

is the answer 3?

[nova creek]

Ye

[slim pawn]

thank you!

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[royal needle]

this is given information from the question

[royal needle]

the question want to me to show this

i know u can factor this quadratic equation but idk how do you continue later

[elfin yew]

Use trig identities

[royal needle]

uhhhh

[copper rock]

Hmm

Mm Mitchell

[meager chasm]

tan x = sin x/cos x

then cross multiply

[royal needle]

im given that information

i dont need to solve that

do i

[elfin yew]

No?

[royal needle]

idk how to prove 3cos^2(x) + cos(x) - 2 is equal to 0

[elfin yew]

Oh

Just do subtitution

u = cosx

[royal needle]

i got the quadratic

like (3u-2)(u+1) in your u = cos x

then i suppose u+1 = tanx(2sinx) according to the info i got (if i use it like that)

then i just stopped

[elfin yew]

??

Solve for both u's

[royal needle]

i dont need to solve

i need to show its 0

the question just said "Show that 3cos^2(x) + cos(x) - 2 = 0"

[meager chasm]

$\frac{\sin x}{\cos x} = \frac{1 + \cos x}{2\sin x}$

[flat frigate]

Chromium

[royal needle]

ye ik that but am i suppose to complicate it lol

[meager chasm]

2 sin²x = cos x + cos²x

afterwards shouldn’t be hard

[royal needle]

k i deal with this later

how the fu

im solving the quadratic

not the tan x = blah blah thing

uhh im so braindead rn

omg got it

.close

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[vague void]

I need help trying to figure out how to use wolframalpha,

[vague void]

,w solve (a+b-c+2d-e=0,a+c+d-e=0,a+b+2c+d-2e=0,4a+3b+c+6d-5e=0,-2a-3b-5c-2d+5e=0,[a,b,c,d,e])

[flat frigate]

Results provided by WolframAlpha

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

Requested by Chad#8741

[vague void]

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[quasi bison]

do you know how to plot points on a coordinate grid?

or are you saying your teacher did not teach you how to do that

...

wym "both"

"i know how to plot points on a coordinate grid, but my teacher didn't teach me"?

https://www.google.com/search?channel=trow5&client=firefox-b-d&q=introduction+to+cartesian+coordinate+system

suggest this as a starting point

[blissful gate]

Khan academy is well liked for this, maybe try that

[quasi bison]

that is also an option

[blissful gate]

Don't mess with Wikipedia, it can be tough to read

[quasi bison]

i posted a google search link :p

[blissful gate]

Yeah I just mention that because the first result for me clicking your link is wiki, and I couldn't understand math wiki pages until like year 3 of uni lol

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[minor bobcat]

when its

asking about thr headingf

does it mean the angle i circled

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[placid flower]

Can u talk here @Enoo58#9243

[placid flower]

@coarse cape

[coarse cape]

yes

about?

[placid flower]

That set

[coarse cape]

oh of course

what is your question?

[placid flower]

$\mathbb{Q}:={\frac{a}{b} \mid a,b\in \mathbb{Z}, b\not=0}$

[flat frigate]

JUGisMUG

[placid flower]

What is that | a, b

[coarse cape]

ah right

this is standard notation

you can read it like this

{ stuff | condition that stuff needs to fulfill}

so in our case its the fraction

where a and b are intergers and b is not zero

[placid flower]

That $\in$

[flat frigate]

JUGisMUG

[coarse cape]

$x \in \mathbb{Z}$

[flat frigate]

Enoo58

[coarse cape]

this means x is an integer

[placid flower]

If that Z was N

That would mean x is natural

[coarse cape]

yes

and writing $a,b \in \mathbb{Z}$ is just out of laziness but everybody understands it

[flat frigate]

Enoo58

[placid flower]

But what's it without laziness

[coarse cape]

$a\in \mathbb{Z}$ and $b\in \mathbb{Z}$

[flat frigate]

Enoo58

[placid flower]

Oh

Yes

[coarse cape]

sorry i dont understand thisn otation

are you asking if $\mathbb{Q}=\mathbb{Z}$

[flat frigate]

Enoo58

[placid flower]

Q is an extension of Z

[coarse cape]

yes it is but there is more to it

you would need to know about equivalence relations and classes

[placid flower]

?

[quasi bison]

what is the question at hand

"what is Q in general?"

is it that?

[placid flower]

Nah, how to write the set of Q

[quasi bison]

oh, you're practicing set-builder notation?

the whole equivalence relations and classes shit is only if you want to know how to construct Q formally from the integers. which seems kind of orthogonal to your goal rn

[coarse cape]

yeah i might have gone to far

[placid flower]

[safe radish]

@placid flower Has your question been resolved?

[placid flower]

Can I do $\mathbb{Q} = {\frac{x}{y}: x, y \in \mathbb{Z}, \mathbb{Z}}$

[flat frigate]

JUGisMUG

[coarse cape]

i have never seen it but im not sure if its absolutely wrong

i guess technically its not wrong but the second Z is redundant because it doesnt add any new information

[placid flower]

it does

[coarse cape]

but what is wrong is that y is allowed to be zero

this cannot be

[placid flower]

Oh crap

Can I do $\mathbb{Q} = {\frac{x}{y}: x \in \mathbb{Z} y \in \mathbb{Z} y \not\in 0, \mathbb{Z}}$

Oh no forgot 0

[coarse cape]

first of all you dont need to put a \mid behind every expression

one is enough

[placid flower]

How would that be?

[coarse cape]

{ stuff | condition that stuff needs to fulfill}

beacuse its written like that

and if you put more in its not clear which condition it is supposed to fulfill

[flat frigate]

JUGisMUG

[coarse cape]

and the last part where you write $y\not\in 0,\mathbb{Z}$

[flat frigate]

Enoo58

[coarse cape]

is wrong

if you want to tell y cant be zero with set notation you should write $y\not\in {0}$

[flat frigate]

Enoo58

[placid flower]

Can I do $\mathbb{Q} = {\frac{x}{y} \mid x \in \mathbb{Z} y \in \mathbb{Z} y \not\in {0}, \mathbb{Z}}$

[flat frigate]

JUGisMUG

[coarse cape]

why the last Z tho?

[placid flower]

Becuase rational numbers are integers too

[coarse cape]

and put commas between logical statements

so

[placid flower]

Which fraction didnt include

[coarse cape]

$\mathbb{Q} = {\frac{x}{y} \mid x \in \mathbb{Z} ,y \in \mathbb{Z}, y \not\in {0}, \mathbb{Z}}$

[flat frigate]

Enoo58

[placid flower]

If I put a comma between them, it will become har to differentiate between conditions and elements

[coarse cape]

yeah you have your integers already for y = 1

[placid flower]

But I represent those in fractions

[coarse cape]

you dont need the last Z

[coarse cape]

yes its standard notation to say $\frac{x}{1}=x$

[flat frigate]

Enoo58

[placid flower]

But I represent that as fractions and integers can't be fractions

[coarse cape]

yes they can if the denominator is 1

they are one and the same

[placid flower]

Damn

But is that a fine way of writing it?

[coarse cape]

which one?

and for writing what?

[placid flower]

This $\frac{x}{1}=x$

[flat frigate]

JUGisMUG

[coarse cape]

yes because addition subtraction and multiplication work exactly as in the integers

for example $\frac{x}{1}+\frac{y}{1}=\frac{x+y}{1}$

[flat frigate]

Enoo58

[placid flower]

Damn so $\mathbb{Z} = {\frac{x}{1}, \frac{x+x}{1}, ...}$

[coarse cape]

yes essentially

altough this notation is wrong

you would just say we identify the elements of Z with x/1

[flat frigate]

JUGisMUG

[placid flower]

Now?

[coarse cape]

what is x?

i think best would be

$\mathbb{Z} = {...,\frac{-2}{1},\frac{-1}{1},\frac{0}{1},\frac{1}{1}, \frac{2}{1}, ...}$

[flat frigate]

Enoo58

[placid flower]

Damn

[coarse cape]

one rule when using variables like x,y and so in is: it should always be clear what they are

[placid flower]

And what if $\mathbb{Z} = {..., \frac{x}{1} \mid x \in {\mathbb{..., -3, -2, -1, 0, \mathbb{N}}}$

[flat frigate]

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

[placid flower]

[coarse cape]

$\mathbb{Z} = {\frac{x}{1} \mid x \in {\mathbb{..., -3, -2, -1, 0, \mathbb{N}}}$

[flat frigate]

Enoo58
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[coarse cape]

ah i see

$\mathbb{Z} = {..., \frac{x}{1} \mid x \in {{..., -3, -2, -1, 0, \mathbb{N}}}$

$\mathbb{Z} = {\frac{x}{1} \mid x \in {..., -3, -2, -1, 0, \mathbb{N}}$

[flat frigate]

Enoo58
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[coarse cape]

no this wont work beacuse x can be N and this doesnt make sense

[placid flower]

Damn

[coarse cape]

$\mathbb{Z} = {\frac{\pm x}{1} \mid x \in \mathbb{N}}$

[flat frigate]

Enoo58

[coarse cape]

im not sure but maybe this would work

[placid flower]

Damn i see

[coarse cape]

but i would stay away from stuff like

$\mathbb{Z} = {\frac{x}{1} \mid x \in \mathbb{Z}}$

[flat frigate]

Enoo58

[placid flower]

Damnn cool

[coarse cape]

if you want to know why you can look at russels paradox

[placid flower]

Anyways $\mathbb{Q}={\mathbb{Z}, \frac{1}{1}, \frac{2}{2}, ...}$

This seems wrong but idk hoe to write this without set-builder notation

[flat frigate]

JUGisMUG

[quasi bison]

it is wrong.

Z isn't an element of Q

[placid flower]

My book says it is

[quasi bison]

does your book specifically say Z is an ELEMENT of Q?

i don't believe that.

are you sure you are not confusing element with subset?

[placid flower]

It says "the set Q is the extension of the set Z"

[quasi bison]

verbatim?

"the" extension?

[placid flower]

Subset is an element aswell

[quasi bison]

no

[coarse cape]

no

this is wrong

[quasi bison]

subset does not mean the same thing as element

[placid flower]

Wait what

[quasi bison]

{1,2} is a subset of {1,2,3} but {1,2} is not an element of {1,2,3}

[placid flower]

But as it is in the set, isn't it still an element

[coarse cape]

there is a big difference between $x$ and ${x}$

[flat frigate]

Enoo58

[quasi bison]

the problem is that you are trying to conflate two different notions of "in"

this would be like saying that, because all members of a certain fitness club also happen to be members of a boating club, the fitness club must also somehow appear as a member in its own right in the boating club.

which is nonsense as is hopefully clear to you

[placid flower]

Oh I see

[quasi bison]

anyway,

you might perhaps (but probably don't) want to write $\bQ = \bZ \cup \text{something}$,

though this ``something'' takes more effort to properly describe than it's worth.

[flat frigate]

Ann

[placid flower]

But then how can I write Q without set builder notation?

[quasi bison]

are you forced to do that by a ridiculously worded problem?

if not, i wouldn't bother trying.

[placid flower]

Not really

I mean it is possible to just end it with ...

But what would be the starting and stuff

[coarse cape]

yes you can

https://proofwiki.org/wiki/Rational_Numbers_are_Countably_Infinite

[broken yew]

$$\bQ = \left{\frac ab : a,b\in\bZ, b\neq 0\right}$$

[coarse cape]

in proof 1 there is a list notation

[flat frigate]

Shuri2060

[broken yew]

Just in case, since I didn't see a correct definition above.

[quasi bison]

they said without set builder notation.

[broken yew]

yh and idk what is going on with the above written set builder notations which look wrong or off to me (where are the set brackets for all of them)

[coarse cape]

if you copy paste discord eats them away

[coarse cape]

this is essentially that list

[placid flower]

Wow

Complex

[coarse cape]

yeah this is a genius way in getting every rational number

[placid flower]

Thanks everyone

[safe radish]

@placid flower Has your question been resolved?

[safe radish]

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[viral beacon]

hey

[viral beacon]

I don't understand something

We have this differential equation whose characteristic polynomial and roots are:

I don't see why the homogeneous solution can be rewritten like this:

and why the particular solution is obvious and can be written like this:

[pulsar condor]

$Z(r)=r^4-2k^2r^2+k^4=(r^2-k^2)^2=[(r-k)(r+k)]^2$

[flat frigate]

Mosh

[pulsar condor]

so the roots are +-k, and to make them independent you need xe^-k and xe^k

[viral beacon]

I know the roots are +-k that's what's written on the picture. This does not answer my question, how do we find this form of the expression

this expression

[pulsar condor]

the homogenous solution is $Ae^{kt}+Bte^{kt}+Ce^{-kt}+Dte^{-kt}$, then just put the exponential definitions in for sinh and cosh

[flat frigate]

Mosh

[safe radish]

@viral beacon Has your question been resolved?

[viral beacon]

and why the particular solution is obvious and can be written like this: ?

[pulsar condor]

the forcing function is not of a form from the homogenous solution

so your Ansatz just be C

[safe radish]

@viral beacon Has your question been resolved?

[safe radish]

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[junior wagon]

Not sure what to do with this

[plucky elk]

what class is this for

[junior wagon]

Probability and statistics II

second year stat course

[plucky elk]

i guess strong law of large numbers is what you're looking for

[junior wagon]

my professors did not cover this

They pulled a "you should've done that in first year" on us 😑

I'll look into it, thanks

[plucky elk]

damn you took a year of prob and stats and didn't learn LLN? you got scammed

ask for a refund

https://en.wikipedia.org/wiki/Law_of_large_numbers

[junior wagon]

.close

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[nova creek]

Start by graphing it

[timber atlas]

Ok

[crude prism]

find the point(s) where x is zero

[timber atlas]

2 and -2

[crude prism]

yes!

can we use both of them?

as our answer?

[timber atlas]

I think so

[crude prism]

look at the quation

equation*

[timber atlas]

We cant use the 2

[crude prism]

yes!

and why? 👀

[timber atlas]

Because it is greater than or equal to 0

[crude prism]

u got it! :D

[ancient escarp]

bro what

[crude prism]

we could use it if it was equal to zero btw

[ancient escarp]

that's not

what

[nova creek]

I'm sorry

What

[crude prism]

the equation says the answer has to be <=

[ancient escarp]

that's not how you do this

[crude prism]

??

[nova creek]

It never says x has to be greater than or equal to 0

[ancient escarp]

you're finding where the equation is less than or equal to zero

[crude prism]

he said we can't use '2' because its greater

[ancient escarp]

???

[crude prism]

not that we use it because its greater

[ancient escarp]

"use"??

[crude prism]

wording was confusing

[ancient escarp]

everything that just happened is confusing

[nova creek]

If you plug in 2, (2)² - 4 ≤ 0 --> 4-4 ≤ 0 --> 0 ≤ 0

That works

We should have a range of answers

[crude prism]

no but the equation says <=0

so the answer needs to be <= zero

[ancient escarp]

believe it or not, 0 is in fact equal to zero

[crude prism]

2 isn't

[ancient escarp]

2 isn't the...

[crude prism]

i'm talking about 2, not zero

[ancient escarp]

yknow what

forget it

[ancient escarp]

start back here

[nova creek]

We don't care about whether x is ≤ 0, we only care if x²-4 ≤ 0

[crude prism]

ohhh

my b

i misunderstood the question

@timber atlas u alive, mate?

[timber atlas]

Ye sorry I was just with the teacher

So what should I do?

[ancient escarp]

this

[timber atlas]

Ok

[nova creek]

Can you see which x values make the function negative, aka, less than 0?

[timber atlas]

Ye

-2

[nova creek]

It should be a range of values

[timber atlas]

-1

[nova creek]

Not just -2 and -1. -1.5 works, π-2 works

Any value between what and what works?

[timber atlas]

Between and -2 and 0.1

[nova creek]

Why 0.1? Look at the graph. Doesn't x=1 give a negative value for the function?

[timber atlas]

Yes

[nova creek]

So clearly 1 works

[timber atlas]

Yep

[nova creek]

Between what two values is the function negative?

[timber atlas]

-2 and 0

[nova creek]

Why?

[timber atlas]

I’m a little confused

[nova creek]

Look at where the function is below the x axis. It's still below the x axis for some numbers past 0, so clearly 0 isn't the upper bound

[timber atlas]

Yh true

So between zero and x infinity?

[nova creek]

Look at the graph again. The function at x = 3, for instance, is above the x axis

So clearly it's not all the way to infinity

[timber atlas]

I’m confused

[nova creek]

Give me a second and I'll draw a picture

[timber atlas]

Sure

[nova creek]

Gimp is updating, so it might take a second

The red area is the bit under the x axis, correct?

That bit is less than or equal to 0

[timber atlas]

Ok

[nova creek]

Do you agree with me on that?

[timber atlas]

Yes

[nova creek]

Can you see that bit I showed in the previous image is between these two lines that I've drawn?

[timber atlas]

Yep

[nova creek]

So, between what two x values is the function negative? Between what two x values is that red bit situated? Between what two x values is contained between those red lines?

[timber atlas]

Between -2 and 2