help-0

[west geyser]

Someone please help me with this question, it's killing me

I don't even know how to start the solution

it has this solution but I don't understand any of it

[placid zinc]

What is the "join" of two points?

[west geyser]

the line segment, I'd wager

I'm fairly sure that I have to use the Section Formula somehow

but I'm not able to

it just leads me to a very complex solution

and no question would be that complex

god this is literally the first question

[lapis sluice]

,rotate

[oak chasm]

Well, sine gives the y value of a unit circle and cosine gives the x value.

So, x = cos(θ) and y = sin(θ), so y = sin(arccos(x)).

And y = sin(arccos(x)) can be written y = ±sqrt(1 - x²).

[west geyser]

someone please help me I have my maths exam in 6 hours 😭

it will have this topic

[wispy wing]

what about mine?

i need help

[west geyser]

<@&286206848099549185>

[quartz parrot]

evaluate this

I feel like there is a formula im dont know of

[glass lichen]

it's arithmetic

[quartz parrot]

yes

idk

[oak chasm]

@quartz parrot Change it into this:

= (-8 + 1) + (-4 + 1) + (0 + 1) + (4 + 1) + ⋯ + (400 + 1)
= (1 + 1 + ⋯ + 1) + (-8 + -4 + 0 + ⋯ + 400)

[alpine sable]

ig imma use the channel

oh

[oak chasm]

And you can continue simplifying from there.

Like factor out the 4.

And so on.

[rigid kiln]

this is going on forever, should this end up just being 2n+1?

[oak chasm]

@rigid kiln Sorry, this channel is busy.

[rigid kiln]

whoops, sorry if i'm interrupting

yeah, sorry, didn't notice

[alpine sable]

ig ill use another channel then

[rich basin]

(b) please

why that?

[placid zinc]

Why not just compute 1 + z + z² + z³ + z⁴?

[rich basin]

how would i?

[placid zinc]

z² = z•z

So you'd multiply by z itself

[rich basin]

yeah hat is what i did

i made it into exponentual form

and then i change it to (e^(10ipi/4)^n

in which n is for each of the {1,2,3,4}

[placid zinc]

Oh I see, you can't add exponential forms

So, go back to rectangular

[rich basin]

yeah

you mean the polar form?

so what i did was change it to exponential form first

in which i would be able to do the ^n really easily

and then i then change to polar fomr

[placid zinc]

I think you're saying polar when you mean rectangular

If you have it in cosθ + isinθ

[rich basin]

yeah

yeah i changed to rectangular

[placid zinc]

Exponential form is really the polar form

Well, now that you have that, adding them should be easy

[rich basin]

it is pretty difficult

let me write out what i have right now

[placid zinc]

Especially since z² = i

[rich basin]

$1+\cos\left(\frac{5\pi}{2}\right)+i\sin\left(\frac{5\pi}{2}\right)+\cos\left(\frac{5\pi}{4}\right)+i\sin\left(\frac{5\pi}{4}\right)+\cos\left(\frac{15\pi}{4}\right)+i\sin\left(\frac{15\pi}{4}\right)+\cos\left(\frac{10\pi}{1}\right)+i\sin\left(\frac{10\pi}{1}\right)$

[ocean seal]

BorutoEyePower

[rich basin]

wait why is z^2 = i?

this is when i changed into exponential form to easily evaluate or i could of just skipped and use de`moivre theorem

@placid zinc anything i can do, or is this just too much, i could of easily simplify it with an easier step?

[alpine nacelle]

it's a geometric serie

[placid zinc]

So demoivre and exponential are really the same trick

That is, if
z = e^(π/4)
Then you can square both sides to get:
z² = e^(2π/4) = e^(π/2) = i

[karmic quail]

gcf (7, 15)?

[rigid kiln]

7 is prime

[rich basin]

wait how did you reduce z = e^(pi/4)

[alpine nacelle]

z = e^(i*pi/4)

exponential form

[rich basin]

oh yeah

so what is the method you are thinking

so when it is multipled by powered by 2, it would be an imaginary number only

[alpine nacelle]

1 + z + z² + z^3 + z^4 = z + iz + i

that would be a way

[rich basin]

how did you simplify so quickly?

so z^2 would then be i

[alpine nacelle]

so z^4 would be -1

[rich basin]

yeah

[alpine nacelle]

and z^3 = z^2 * z = i*z

[rich basin]

because in a circle, it is on the left hand side of the complex plane

right

okay thanks this helped alot

[alpine nacelle]

or you can just write the sum as (z^5 - 1)/(z-1) = -(z+1)/(z-1) since it's a geometric serie, it works too

at some point you have a bit of calculation you have to do

[rich basin]

is there any way in which i can visualise e^i*degrees

desmos doesn't seem to offer anyway since i = root(-1) is invalid

[alpine nacelle]

e^(i*angle) is of modulus 1

he is on the unit circle

[rich basin]

yeah, so it essentially a circle basically

[alpine nacelle]

then you just have to visualise the angle

[rich basin]

and the n*e^(i *angle) is

and n basically modifies the range

[alpine nacelle]

the distance from the origin

[rich basin]

okay, thanks, this helped alot

[quartz parrot]

so its 400+(8+4+0...+400?)

i dont know what to do with the other braket side

wait

so id get 400+(5050-12)4

so I get 20552

[wispy wing]

can i have some help <@&286206848099549185> ive needed help for 3hrs now

im 15 and im 1 question away from finishing doing 5hrs of maths

i just want to go to sleep, can someone help me on the last question, please?

[oak chasm]

@quartz parrot No, there aren't 400 terms.

[stable dune]

is this right

[oak chasm]

@stable dune Sorry, this channel is busy.

[wispy wing]

@oak chasm can u help me out bud?

[oak chasm]

@wispy wing First, find the coordinates of P. Then, find the angle pointing at P from the origin. Then, the slope of the tangent is perpendicular to that. Now you have the slope and a point on the tangent line, so use point-slope form.

[wispy wing]

theres a graph?

[oak chasm]

Oh, you don't need to find the angle.

You can just find the slope from O to P.

Take the perpendicular slope to get the slope of the tangent line.

Then you have a point and a slope.

[wispy wing]

i hate to be annoying but i dont know how

[oak chasm]

Which part?

[wispy wing]

just the whole thing in general to be honest

[oak chasm]

OK, so x² + y² = 90, right?

[wispy wing]

yes

[oak chasm]

And they give you x = 3.

So, 3² + y² = 90. Solve for y.

What do you get?

[wispy wing]

10?

[oak chasm]

Why do you say 10?

[wispy wing]

sorry 9

[oak chasm]

OK, so it's ±9.

3² + y² = 90
9 + y² = 90
y² = 81
sqrt(y²) = sqrt(81)
|y| = 9
y = ±9.

[oak chasm]

@alpine sable Sorry, this channel is busy.

@wispy wing So, what does the problem say about whether y is positive or negative?

[wispy wing]

it doesnt say anything

[oak chasm]

Yes, it does.

[wispy wing]

• ?

[oak chasm]

Why?

[wispy wing]

Because P has a x - coordiante?

[oak chasm]

Sorry?

What is a "x - coordiante"?

[wispy wing]

i dont know

sorry

im trying my best but its 00:38

and im tired

[oak chasm]

It would be better to sleep and then do the problem.

[wispy wing]

its the last problem

ive done about 70 problems in 5hrs, just this to go

[oak chasm]

It's possible to become too tired to do problems or to work your brain so hard that you can't think about anything difficult.

The solution is rest if that's possible, not finishing the last problem right now.

You'll get better results and learn more from it.

[wispy wing]

ive got physics for tommorow though

[oak chasm]

Yes, but like you said, it's only one problem, and it's like a five minute one. You can do it in the morning if you have time.

[wispy wing]

can we just finish it quickly now?

i dont want to go to sleep thinking about it

@oak chasm im sorry if im being annoying

but i really want to get it done

[oak chasm]

No, it's not that, it's that we're supposed to guide you and it's more like I'm doing it for you.

[wispy wing]

its not 5 min for me, ive not even learnt it

[ocean seal]

marc

[oak chasm]

@zinc cosmos https://math.stackexchange.com/a/265700.

[wispy wing]

is there a link i can have on how to learn mine?

[oak chasm]

https://www.bbc.co.uk/bitesize/guides/zqjqsrd/revision/2

[wispy wing]

thanks

ill try my best

[oak chasm]

Point slope is easier than their slope intercept form, though.

[wispy wing]

can u go through with me again?

is x 3?

@oak chasm

mate im trying

[rich basin]

z is unkown, but they want us to expand and simplify

but is there a way to do this a more smarter way than just bruteforcing it?

[warm brook]

@rich basin idk what you mean smarter, I'd just multiply it out

There aren't any tricks in this one in particular

[rich basin]

like how did you expand it?

[wanton folio]

whats up

whats the problem

[warm brook]

Yeah, it's just what you do normally, I don't see any ways to "quickly" do it

Like idk, do the right 2 things first

You mention z is known, what is it dyslexia moment

[wanton folio]

do the last two brackkets first

[glass lichen]

do you know sum and difference of cubes?

[rich basin]

yeah

is it (a-b) (a^2 + ab + b^2)

and (a+b)(a^2-ab + b^2)

[wanton folio]

look

[rich basin]

what about them in here?

[wanton folio]

two key observations

[rich basin]

i don't quite see it

[wanton folio]

1. the multiplication of e^stuff is 1

2. use the definition of the complex cosine

[rich basin]

yeah i got (z+2)(z^2 -2z +4)

which looks like (z+2)^3

but then how do i get this down?

[warm brook]

you sure?

[rich basin]

isn't it?

[bleak ridge]

[warm brook]

(z+2)^2 = z^2 +4z + 4

[rich basin]

yeah?

[warm brook]

z^2 - 2z + 4 \ne z^2 + 4z + 4

[wanton folio]

when you multiply it out you get z^3+8

[rich basin]

yeah

[wanton folio]

what's the problem then?

[rich basin]

i just learned something

about how to fast multiply this

which i never knew

so when you have something that have powers that are negative of each other

it completely removes the powers

[wanton folio]

do you mean a^b*a^(-b)=1?

@warm brook Are you french?

[warm brook]

Why?

[rich basin]

yeah

[wanton folio]

I'm studying french rn 🙂

[warm brook]

very cool

@rich basin I'm still weird about the (z+2)^3 comment and I want to make sure you meant z^3 + 8 or that it was just a mention that it looked kinda like (z+2)^3

as z^3 + 8 is not (z+2)^3, it's z^3 + 2^3

[wanton folio]

actually there's a famous theorem that goes like (a+b)^3=a^3+b^3

[warm brook]

[wanton folio]

but unfortunately it only holds for sedeonic a,b

kirby, veux-tu parler français avec moi?

[warm brook]

eh , pas maintenant

[rich basin]

I actually meant the (a+b)^3

[rich basin]

Like if you take a look at the following equation doesn't this look like (a+b)(a^2 -ab + b^2) ?

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

[warm brook]

what do you mean here

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

[devout summit]

Vieta's formulas would be kinda fast. But it's the same you would get with direct expansion...

[normal chasm]

Can someone help me with question 1?

[glass lichen]

the factorization you had was the complete factorization of z^3+8, since you had scaled up roots of unity

[hushed pasture]

the equation of any circle is: $(x-a)^2+(y-b)^2 = r^2$ where the center is $\Omega = (a,b)$ and the radius is $r$

[ocean seal]

Salah

[normal chasm]

Thank you!!

[wooden crag]

I've looked at explanations of significant figures and it doesn't quite make sense to me
how many significant digits are there in 1.008?

[opal mesa]

4

[wooden crag]

uhhh
4?

2

but isnt 1.080 more precise than 1.08

[opal mesa]

1.008 has 4

1.080 has 4 too

[wooden crag]

1.800 has 2?

[opal mesa]

NO

4

[wooden crag]

4 and 2

[opal mesa]

If there is any N front of the zero then You include it

[wooden crag]

five and five

how do you figure that one out

[opal mesa]

3

[wooden crag]

i wanna say 6 but that sounds wrong

[opal mesa]

3 is the first non zero digit

Sorry one sec

[wooden crag]

i mean, it feels like the entire number conveys information

[opal mesa]

@wooden crag I think there are three basic rules for significant digits

The first one is that any non zero digit is considered a significant digit

Then if there are any zeroes between non zero digits, then you consider them signififsnt too

[ocean seal]

marc

marc

marc

[devout summit]

If $n=pq$ for two primes $p$ and $q$. Then, $\phi(n)=(p-1)(q-1)=70$.

[ocean seal]

Euclid31415

marc

[devout summit]

$\phi(n)=n\prod_{p|n}\left(1-\frac{1}{p}\right)$

[ocean seal]

Euclid31415

[devout summit]

There is no need of that.

[ocean seal]

marc

[devout summit]

That's what I did

[ocean seal]

marc

[devout summit]

No. I assumed n=pq first and then wrote phi(n)

Oh no I didn't notice

Then, if $n=p^aq^b$ then, $\phi(n)=(p-1)p^{a-1}(q-1)q^{b-1}$

[ocean seal]

Euclid31415

[devout summit]

I think you just have to consider different cases and check...

[alpine sable]

hey could someone look at this and let me know if it’s right

[stiff pawn]

it is

[alpine sable]

thanks

[silver current]

help?

[fierce basalt]

what have u tried, and what do u need help on?

i suppose a hint could be use logs

[fierce fern]

what is the two possible roots of √36?

[silver current]

I tried dividing 40 by 200

so 5.7^x =0.2

[dim oasis]

Most would say it’s $\pm6$ and I guess that works

The step where absolute value comes into play is often skipped

[ocean seal]

lexitorius

[fierce fern]

then is -6 counted as a root of 36

[dim oasis]

It shouldn’t really be but usually yeah

[fierce fern]

ok thanks

[fierce basalt]

yup and then u use log

do you know logarithms?

[fierce basalt]

if you do ${log{5.7}{.2}}$ in a calculator it should work

[ocean seal]

Swag369

[fierce basalt]

idk latex properly lol my b

[silver current]

what is latex

[fierce basalt]

its the fancy math typing

you knoow how the tex bot had the +/- 6

thats written with the $$

[glass lichen]

$\log_{5.7}(.2)$

[ocean seal]

Mosh

[glass lichen]

also it said to use log not log_(5.7)

[fierce fern]

value of √25/36?

[rich basin]

Oh right i got it thanks

[placid zinc]

What can't you plug in?

[fierce bluff]

ah thats it?

so 1 and square root +-2

[glass lichen]

Can you plug in -100000?

[placid zinc]

You're right, you can't plug in 1,√2,-√2

You can't plug in even more than that, though

[fierce bluff]

what else?

I think its just asking for domain

[placid zinc]

Number under a square root can't be negative.
x + 1 ≥ 0

[fierce bluff]

ah oh yeah

could u show how I would write using interval notation and union symbol

[placid zinc]

Those are your discontinuities. Everything else is continuous

[-1,1) U (1,√2) U (√2, inf)

[fierce bluff]

yeah ok thats what i was thinking for union

interval notation is the line right?

[stiff pawn]

should be closed on -1

[fierce bluff]

yeah square on -1 for me as well

ok thanks

[alpine sable]

based on vector addition, what is a+b?

we know a+b is line segment AC

now if we have that OC starts at the midpoint then we need to scale this (a+b) by something that would cut it in half

thus we get 1/2 * (a+b)

did i make a feasible problem based on the figure on the first picture?

(i only used the left figure the right figure is only optional)

[knotty badge]

Where is that on a ti-84

Where is Z Alpha

[rich basin]

oh right that make sense, thanks

[knotty badge]

Guys please I need help

[wary stream]

If this is a legit exam, we cannot help you with exams

[alpine sable]

lmao mans asked for help on an exam

<@&268886789983436800>

[sly mantle]

@knotty badge is this a timed graded test?

[knotty badge]

It’s not a “legit” exam

[sly mantle]

"legit"?

[knotty badge]

It is a test indeed

But it’s not graded as a test

It’s practice and it’s still graded

And it’s stressing me out

I just need to know where z alpha button is on ti-84

[wary stream]

So what's the ruling on helping out this person?

[knotty badge]

I really don’t need help on the question I just need to know where that button is

That’s all I need guys

[wary stream]

I'm not even sure if that button does exist on a calculator. If that's probability and stat, and you need z, there's a z table that you can look up

[tired drum]

How would you do this, I get X>-7 but for some reason I don’t get how it’s also X<5/2

[wary stream]

What's the full question?

[knotty badge]

Hey if you had round to fourth decimal and your answer is 0.4469 would it now be 0.45

[bleak ridge]

No

It would be if it was 0.4496

[knotty badge]

So 0.447 would be correct right

[wary stream]

No

Count how many decimals you have

[wary stream]

How many does this value have?

[tired drum]

You just need to find what x is

[knotty badge]

0.4469=0.447

[wary stream]

It says 4 decimal places, you start counting after the decimal point

[bleak ridge]

Yes

0.4470 is the same as 0.447

[wary stream]

I don't think so, because it looks like those values are what makes the denominator undefined

[frozen echo]

Can anyone tell me how to calculate solstice equinox?

I'm trying to understand the formula I have to use.

[knotty badge]

Is a number unusual if it’s less than 0.05 or greater

[tired drum]

That’s what your mean’t to do with inequalities…

[devout summit]

Bring the 1/(x+7) to LHS and you get\\$\frac{2}{2x-5}-\frac{1}{x+7}=\frac{2(x+7)-(2x-5)}{(2x-5)(x+7)}<0\\\implies \frac{19}{(2x-5)(x+7)}<0\implies \frac{1}{(2x-5)(x+7)}<0\\\implies (2x-5)(x+7)<0$

[ocean seal]

Euclid31415

[oblique stirrup]

how do i approach this question?

If cos(10)=sin(θ)and 0<θ<90, then
θ =  ? degrees

[devout summit]

What is the complementary of 10 degrees?

[alpine sable]

$x^2 -2(y)^2 = 1$
solve pls

[ocean seal]

Aldrich

[alpine sable]

ans x=3 y=2

[austere urchin]

you can use a trig identity with the complement angle, yeah

[alpine sable]

cos10=sin(90-10)

theta=80

[devout summit]

Its even in the name. "complementary sine" or "cos"

[alpine sable]

help a mf out

[tired drum]

Ohh thanks

[shy thicket]

pft idk

[oblique stirrup]

sin80?

[devout summit]

Yes! sin(80 degrees)=cos(10 degrees)

[oblique stirrup]

it says my answer is wrong, it wants it in degrees so i did sin(80) on the calc and got 0.98

oh thhe key said 0.8

ok

[alpine sable]

thats nice but where do we get an integer solution from?

[devout summit]

0.8??

[oblique stirrup]

80 degrees

mb

[alpine sable]

Ye guys study a lot

I am busy in PUBG

What a mess....

[oblique stirrup]

dead game in NA, i bought the game and tried it for like 5 minutes and refunded because the movement is so slow

feels like the map has 0 gravity

[alpine sable]

I see

Things are opposite in my country

[oblique stirrup]

yeah i know its popular outside

[alpine sable]

Valorant ?

That was going HOT

It's all like CS : GO

[devout summit]

Pell equation

[oblique stirrup]

eyes going bad x.x

[alpine sable]

Same here.

[fierce bluff]

did anyone get sqrt(a)/b

can someone explain why theres a - on the numerator?

[oblique stirrup]

it means the result is negative unless the result is negative and therefore positive, i think

[rich basin]

how do we do (d) ?

[rich basin]

<@&286206848099549185>

[oblique stirrup]

to find a coterminal angle u just add/minus 360 right

[devout summit]

Yes

[oblique stirrup]

hell im inputing wrong answers .-.

actually blind

[rich basin]

anyone help me do (b)

[alpine sable]

use the formula sqrt(x²+y²)

[small helm]

well, you know that BD and DC are the same length as the other vectors since it is a square

and since it's a square BD and DC forms a 90 degree angle with the other lines that are connected to it

[rich basin]

yeah

[small helm]

so

[rich basin]

i got it

how would you find the intersectiion

would that jsut be the mid point for (e)

[small helm]

you basically just find the equations for BD and DC

then solve for x,y

[rich basin]

Okay i got it

[small helm]

since x,y in the two equations should be the same

[rich basin]

how do we do (d) for this?

i mean (c)
\

[small helm]

if you're trying to find the two point

and correct me if I'm wrong about this

but if you're just trying to find the two points that divide DE into 3 equal parts

then you can just use the equation to find the distance between two points

except instead of knowing the second point, you have to use the slope and the distance

[rich basin]

what equation would that be?

[rich basin]

I can't really see images dammit

[small helm]

ok so

what you're trying to find

[rich basin]

so i did what you told me to

to find the vectors between hem

[small helm]

is a point that is on the line DE

[rich basin]

whichc is {4 , -4}

and then i divided it by 3

wait what is the right answer

[small helm]

uh

lemme check

[rich basin]

how would we do (d) then?

like i don't know how can i reprsent it using d and e?

[small helm]

ok so

you have the equation of DE

in the form of y=mx+b

then you have then equation to solve for the distance

[rich basin]

yeah

what about (e)?

[small helm]

not sure

[ripe tendon]

does the order of these symbols matter?

like could I do L | D' and it would be the same thing?

[small helm]

no

[ripe tendon]

no to the top or bottom

sentence

[small helm]

the order does matter

that says the probability of D' given that L is true right?

[ripe tendon]

im seeing it for the first time right now im not quite sure what it means to be honest

I'd assume you are right

[surreal meadow]

P(A n B) = P(A | B)P(B) is kind of circular here lol

[small helm]

I think what it's supposed to mean is just P(AnB) = P(a)P(b)

because if we're just talking about intersection, then it doesn't have to be conditional

[rich basin]

this is the solution for (e)

but i can't put in for some reasons, the screenshot?

okay now it works

but why is it that the they change the vectors to become (b-a)

and not (a+b)

[plush moon]

what happend to this : f(x) = 4x+1

to be this : g(x) = ( 4x+1 ) + 3

i need some help

[graceful cave]

Isn't that g(x) = f(x)+3?

[plush moon]

no i mean the move

[hardy gulch]

so what does it mean to +3

what move is it
and by how many units

[plush moon]

Offset 3 units to the right

Offset 3 units to the top

Horizontal stenosis with a factor of 3

Vertical dilation with a modulus of 3

[plush moon]

one of this

[hardy gulch]

yeah

what would +3 look like

suppose i had a bunch of points on the graph

i +3 to all of them

what happens

[plush moon]

ssoooo whats the answer from here :) ?

@hardy gulch

[rich basin]

(A)

[clever folio]

Maybe it's a symmetry type thing?

Like the triangle formed by OP and the x axis seems similar to the one formed by OR and the x axis

Which makes me feel like it would be -3+4i or something.

But that's just a vague guess.

[rich basin]

@clever folio you got it somehow, but i don't know how multiplying it by i makes it like that?

isn't it rotating by 90 degrees anti-clockwise

when multiplied by i

[clever folio]

I didn't multiply by i

Are you asking because you wanna do it that way?

[rich basin]

what did you do?

just want your thinking

[clever folio]

I looked at the geometry being shown

Take the perpendicular to the x-axis and P and look at the triangle formed. It is similar to the one you'd get if you take the perpendicular to the x-axis thru R

[vale wigeon]

yes, multiplication by i corresponds to a 90° ccw rotation about the origin.

[clever folio]

Using complex numbers will probably be a lot more elegant and easy lol

Would what I did be equiv to multiplying by i^2?

Nah wait nvm lol

[vale wigeon]

i mean this question mentions complex numbers directly

[rich basin]

wait if you multiply by i^2]

it wouldn't be -3+4i

[clever folio]

Yeah I said nvm because I wa being dumb with that

Lol

[rich basin]

I will be back, I have to go for something really quick. Be back

[carmine lion]

how to solve such systems quickly

yes

[rich basin]

okay i am back @clever folio

[clever folio]

I think you are just mixing something up when you mult by i is all.

It seems like it works when you do that.

[rich basin]

like aren't they perpendicular once i saw it?

[clever folio]

What do you mean?

They're perpendicular yah.

[rich basin]

so can you go with your method?

@clever folio what did you exactly do to find that solution?

[clever folio]

Make two triangles by drawing vertical lines from P and R to the x axis

Notice that these two triangles are basically the same but one is rotated and translated

[rich basin]

yeah

[clever folio]

So the lengths correspond in such a way that to get from the origin to R you would go left by 3 (so -3) and up by 4

It's a more complicated solution than just multiplying by i

[rich basin]

wait so how is the triangle rotated?

is it triangle rotated by the origin?

or is it from the placement?

[clever folio]

It's rotated and translated

[rich basin]

Rotated from what reference?

[clever folio]

Rotate the OP triangle I mentioned clockwise and shift it left

Seeing as this is not the solution the exercise intends I doubt it's worth worrying a bunch about lol.

[rich basin]

what i'm confused is what axis are you using to rotate them?

[clever folio]

I'm not sure how to better explain rotating a triangle clockwise. :/

You see the op triangle I'm talking about right?

[rich basin]

yeah

but then like where are you rotating it around?

and by how many degrees?

[clever folio]

So rotate that triangle clockwise until P is on the x-axis lol

[rich basin]

okay

[clever folio]

Now shift that triangle left until its hypotenuse lines up with OR

The vertical and horizontal legs give you the coords of R

[rich basin]

so let me assume that the triangle is OPR?

[clever folio]

No

Different triangle

Make a perpendicular to the x-axis thru P

[rich basin]

with the triangle or with the entire box?

[clever folio]

Call the pt on the x-axis B or something. This triangle would be OPB or whatever

[rich basin]

if triangle, what triangle would the points be?

[clever folio]

Just told you the pts

[rich basin]

okay

so the triangle is OP with the x axis that is perpendicular to the line P

[clever folio]

No that is not what I said.

[clever folio]

See this

[rich basin]

is it a right angled triangle?

[clever folio]

Yes it is a right triangle

[rich basin]

yeah that is what i was thinking about

[clever folio]

Ah okay

[rich basin]

so looks like this right

[clever folio]

Yep basically

[rich basin]

okay

[clever folio]

Rotate it clockwise and shift it left til it lines up with OR

[rich basin]

yeah

[clever folio]

You know the vertical and horizontal legs of the new rotated shifted triangle

They tell you the coords of R

[rich basin]

like this right?

the little slash up the top denotes the rotation

[clever folio]

Yeah looks right

[rich basin]

oh right, that is pretty good. Thanks

what would exactly be the complex nubmer way of looking into this?

[clever folio]

Multiplying by i

It'll be way simpler than this lol

i(4+3i)=4i+3i^2 =4i-3=-3+4i

It gets the same answer in much easier way.

[rich basin]

so how you identify that they are peerpendicular

is it through using the geometric theorem, that bisecting diagonals within a square are perpendicular to each other

forming a right angle triangle

[clever folio]

You mean the two numbers -4+3i and 3+4i?

[rich basin]

yeah

[clever folio]

I'm not exactly sure I know there are formulas and stuff for it but I haven't dealt with complex numbers in a while

[rich basin]

okay thanks, this helped alot

[clever folio]

I think they are orthogonal (perpendicular) when the complex dot product is zero. So that would be (a+bi)f(c+di)=0 where f stands for taking the conjugate.

Don't quote me on that though lol.

[devout summit]

Well, real part of that product=0 implies perpendicular

[cedar prism]

Hello guys I need help can anyone help me ?

[alpine sable]

send the q

[cedar prism]

I think I solved a but i didn't understand what do they mean by the minimum values

[alpine sable]

Let OA = a and OB = b.
Find vector of OD and then ratio of OF:FD and EF:FA

coord bash wud be a fun way to do this

[vale wigeon]

have you ever encountered the min() function before?

[cedar prism]

No I don't remember such a thing

[vale wigeon]

min(x,y) is equal to whichever of x and y is smaller

for example min(420, 69) = 69

this problem tells you that in fuzzy logic val(P & Q) = min(val P, val Q)

[cedar prism]

Oh ok tbh honest I wasn't expecting it to be that simple ... thank you so much ... I kinda feel stupid now :)

[cedar prism]

So I got a equal what should I do there's no maximum

Or minimum

[vale wigeon]

what do you think

what could min(x,x) possibly be, if not x

[rich basin]

can you please show me that, i tried using z = 3-2i and iz = 3i+2 and when i multiply them together, i get Re(iz^2) = 12

[devout summit]

I meant for non-zero complex numbers $z_1$, $z_2$ they are \perpendicular iff $Re(z_1\overline{z_2})=0$

[ocean seal]

Euclid31415

[rich basin]

aren't all real numbers complex?

[devout summit]

Yes. They are a subset of complex numbers

[rich basin]

isn't this a non-zero compelx number?

[devout summit]

I didn't understand

[rich basin]

like if you take a look at 3 -2i there are information about the complex number already, so it wouldn't be zero

[devout summit]

Yeah it totally applies here

[devout summit]

See that line over the top of z2 here? That is taking the conjugate

[rich basin]

Okay, aren't conjugates just refelctions?

[devout summit]

Yes they are reflections about real axis

[rich basin]

so if you have a position vector for them, and draw a line to them. They would just form a rombus

[devout summit]

$\overline{3-2i}=3+2i$

[ocean seal]

Euclid31415

[devout summit]

We just flip the sign of imaginary part

[rich basin]

but then perpendicular is more visualised through geomtry

[devout summit]

This fact holds

[devout summit]

That's just an opinion

[rich basin]

I mean, in order to start with this you need to first prove it with something you know

which is geometry

[devout summit]

It starts with Euler's formula and why multiplication by complex numbers corresponds with rotations and scaling

[rich basin]

yeah

oh right i get it

i could start visualising it in euler's formula

okay thanks

[balmy zinc]

Hi i dont know how to do these questions

[alpine sable]

The lesson you're training on is transformations of functions, right?

[balmy zinc]

yes

[alpine sable]

Okay. Let's see c)i). The way it's displayed can be a bit confusing, but when you multiply both sides by 2 you can start to understand what you're supposed to do.

y = 2h(x). Which means that for each input of x, you'll get twice the output of the original function.

So in the original function, if x = 0, y =1. In this newer one, if x = 0, y = 2(1)

[alpine sable]

If you apply this to all points, you'll end up with a curve that's stretched vertically by a scale factor of 2.

[real crescent]

hi can someone help

[balmy zinc]

@alpine sable so do i multiply y-coordinates by 2?

[alpine sable]

Yea

I have this cheat sheet that my teacher actually gave me in case I'm not available to explain in detail the rest

It seems you're just doing the stretches now so don't worry about the rest.

[alpine sable]

But if I'll explain the horizontal stretch, when here in c) ii) it tells you that when you enter an X into the function, you'll half that X before you input it.

Originally X = 2 gave you Y = 1, but here since you half the X before adding it in, you'll need the X to be 4 to give you the same Y = 1.

[balmy zinc]

what would i be dividing the x-coordinates by?

[alpine sable]

Actually you'd be multiplying them by 2.

Since you need twice the X to give you the same Y.

It's like someone weakened the X and you're giving it reinforcements.

But don't worry it's usually only the f(x) changes that are counter-intuitive.

[balmy zinc]

ohh i see

thanks i understand it more now

[tacit kernel]

The sum of two numbers is 13. Twice the larger number is 8 more than 4 times the smaller. What are the numbers?

What does the 8 more than 4 times the smaller mean?? 😖

[alpine sable]

The smaller multiplied by 4

Then add 8 to that

[tacit kernel]

oooh

so 2x = 4y + 8???

[alpine sable]

Yep

[tacit kernel]

Alright thanks

[alpine sable]

nn

[thick ferry]

can someone help answer q 74 and 76

[vague coral]

f(g(x)), you replace the x in f(x) with g(x)

[balmy zinc]

@alpine sable wait for questions like y= h(x/(random number)), would it mean you always multiply x-coordinates by 2 since you need twice the x before halving the fraction to get 1?

[alpine sable]

You multiply the x by that number

In this case it was x/2 so you multiply it by two.

If it was h(2x) you divide by 2

Really just do the opposite of whatever he does to the x

[alpine sable]

But I think you got the general idea right.

[thick ferry]

ok thank youuu

[static kindle]

what do the square brackets mean?

[vale wigeon]

nothing

[fierce linden]

Is my algebra right so far?

<@&286206848099549185>

[spring aspen]

yesss

[fierce linden]

tyty

[young warren]

how is my answer not correct, shouldnt the product rule be used here>

[devout summit]

That last "u" should have been "c"

[hollow pelican]

hi guys how do i differentiate this?

[graceful yoke]

You bring -2t behind t and multiply it by t to which gives you (-2t^2)e and then you subtract 1 from the power which gives you a power of -2t-1.All together the answer would be s'=(-2t^2)e^-2t-1

[young warren]

thank you 🙂

[hollow pelican]

ohh i can apply properties of log for e also?

[tawny fable]

Hey there,
I need to find the value of sup(A), inf(A), min(A), and max(A). Or - prove that it doesn't exist.

My see that there is not sup(A) and max(A) but I don't know how to write it down and prove it.
I showed that 2 is inf(A) and min(A), but I couldnt continue from there.
I am sitting on this one for hours... any clue?

[devout summit]

No! This is incorrect!

[forest jacinth]

Using the product rule, $\frac{ds}{dt} = \frac{d}{dt}(t)e^{-2t} + t\frac{d}{dt}(e^{-2t})$

[ocean seal]

JarrGarde

[forest jacinth]

This should be alright, although I am a bit tired

[static kindle]

how do you find the bearing?

[hollow pelican]

i See

[graceful yoke]

what did I do wrong when differentiating I used the power rule (nX^n-1)and simplified the expression.

[devout summit]

Yes and then we get $\frac{ds}{dt}=e^{-2t}-2te^{-2t}$

[ocean seal]

Euclid31415

[devout summit]

That power rule is for differentiating x^n with respect to x… and not for things like n^x

[graceful yoke]

Oh why is the order switched around in this case?

[devout summit]

You treated e^{-2t} as power rule…

And moreover it has an extra “t” multiplied on the outside!!

[snow pawn]

hello

If you have a limit function, where f(0) is not defined

the assignment states that f(0) = 1 because of this

and then they ask me find f'(0) and f''(0). not sure how to begin this assignment..

[devout summit]

Would be helpful if you posted the question...

[alpine sable]

Why is the marked thing there? Like how did we get that its equal to the right hand or left hand side?

[rustic junco]

How do you find x?

[chilly panther]

You can work out how many degrees total there are in a 5 angled shape by dividing it into triangles and using the fact that the sum of the angles in a triangle is 180 degrees. That should start you off at least

[rustic junco]

thanks

[fresh fulcrum]

I had a question on a number theory exam.

P=110322353973061799

P contains a five character HEX subliminal message.

Using n=10007 as the base extract the five decimal coefficients of the polynomial generated by n that P represents.

Not expecting to get a full answer but maybe some pointers to how do I go about solving this.

[fallow cradle]

Hi, this is a pretty dumb question but I can't seem to figure out where I went wrong?

I am given this equation and Im trying to find the roots

while not knowing what y is

So I found this as a root

but the actual answer is

What gives?

And delta (the stuff under the root idk what it's in english) would be 4y+8 which the correction confirms

Nvm im rlly dumb

[bold panther]

helppp how is this wrong? if tanx = 2t/(1-t^2) then isnt sinx the value on the numerator @-@?

[karmic falcon]

hi

i have an exercise

given the are of circular sector= 16pi find the perimeter of this sector

[graceful yoke]

16 pi?

wait no 8 pi + 8

[digital needle]

hi i need help

with this ques

[old sluice]

Tan is undefined when cos is 0. Therefore, tanx = sinx/cosx

So yes, sinx is the value of the numerator

[pliant zephyr]

n=25

[digital needle]

ik

just figured it out

[graceful yoke]

How do you algebraically prove that sqrt(2) is irrational?

[bold panther]

then y my answer wrong T_T

[karmic falcon]

how did you do it

[old sluice]

What does it say when you switch the two?

I'm very sure that sinx is the numerator

[graceful yoke]

You can find the radius of the segment by square rooting the 16 and taking the pi away and then you do pi times 2r which is 8 pi which is the curve part of the segments length then you just add 2r to 8 pi for the 2 radius length lines attached to each end of the curve and go into the middle of the circle

Oh wait I think I forgot about the angle part with fetur/360

[bold panther]

it says that its wrong :(

[old sluice]

🤨

[graceful yoke]

Yea idk then 😂

[karmic falcon]

are you sure that the radius is found that way. I think the formula is S=(n*r^2*pi)/360

where n is the degree of the angle

[graceful yoke]

Yea your right I forgot about the angle part at the begging sorry 😅

[old sluice]

Sounds like a problem with this thing you're using. I don't know why does it error like that

[karmic falcon]

no problem

[old sluice]

Does it count brackets?

[karmic falcon]

i'm asking on #geometry-and-trigonometry any help is appreciated

[bold panther]

no :(

im just gonna skip that problem and do a different one

[orchid python]

how can I solve this?

anybdoy??

[spring kiln]

use rules for manipulating det

let me get a good reference for you

https://www.vedantu.com/maths/properties-of-determinants

There we go

Let me know where you think we should start. Hint: transposing the original matrix means its det is still = 3

We can also swap rows and columns without changing the value of the det

except in sign !

3 -> -3

etc

@orchid python

[chrome ice]

does sin(a+b/2) have a formula of its own?

[spring kiln]

https://socratic.org/questions/how-do-you-prove-sin-a-b-sin-a-b-sin-2-a-sin-2-b

[subtle seal]

Good day! I'm trying to understand polynomial factorisation in the complex set then coming back to R set. say for example we have $P(x)=x^4+x^3+x^2+x+1$ what do we do in order to transform it into $z^2-\alpha$ to solve for the complex roots. for P(x) our prof multiplied P(x) by (x-1) to get $z^5-1$ and then find its roots. How do we figure out that?

[ocean seal]

nas

[visual lava]

any helpers

nvm brb

[placid zinc]

@subtle seal
One way might be a geometric series:
1 + x + ... + xⁿ = (x^(n+1) - 1) / (x - 1)

In general it's a common factoring trick you'll get used to:
(x - 1)(1 + x + ... xⁿ) = x^(n+1) - 1

[subtle seal]

excellent thank you! that makes sence

I suppose this is solely for polynomials with 1 as the coefficient for all its terms correct?

[placid zinc]

Yeah there's no similar trick

Well, if they're all the same coefficient you could factor it

[subtle seal]

also after I find the complex roots for x^(n+1)-1 how do I rewrite p(x) as a factor of those roots? do I need to divide p(x) over (x-1)? for the final result

[placid zinc]

The roots of x⁵ - 1 are
1
e^(πi/5)
e^(2πi/5)
e^(3πi/5)
e^(4πi/5)

@subtle seal
If you take any of those to the power 5, you get 1

[alpine sable]

Does someone know why the truth table of logical implication is like it is?

[subtle seal]

that makes sense

[alpine sable]

why is it b ?

[placid zinc]

Or, sorry, double all of those exponents

[subtle seal]

so $x^5-1=(x-1)(x-z_{1})(x-\overbar{z_{1}})(x-z_{2})(x-\overbar{z_{2}})$ Now how do we get p(x)?

[placid zinc]

I messed it up haha

[ocean seal]

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

[subtle seal]

the other z1 is the conjugate

don't know how to add the bar in latex sorry xD

do we divide it over (x-1) to get back to p(x)?

[placid zinc]

z1, -z1, z2, -z2 are all of the roots we've found

So you have the factored form of P(x) right there

[subtle seal]

what about the (x-1) that we initially multiplied by to get x^5-1?

[placid zinc]

I'm confused. Are you starting with P(x), or trying to find P(x)?

[subtle seal]

Starting with it and trying to find its factorised form

[placid zinc]

z1, -z1, z2, -z2 are all of the roots we've found
So you have the factored form of P(x) right there

Note that multiplying by x - 1 put it into a form that gives the roots easily

[subtle seal]

I see the root of x-1 is irrelevant

perfect! thank you very much. you cleared up a lot of confusion

[placid zinc]

Np! Feel free to ask if you need anything else

[haughty forge]

Does anyone know what does x #=3 means?

Or like x + y #> 5 for example?

[vernal thunder]

If the x was negative the graph would be decreasing

Now if you were to plug any value to x you should get a y value that lies on graph a

[azure jungle]

Need help with derivations

[alpine sable]

Don't know anything about derivations

[oak basin]

May someone help me understand a question I have about a math problem it's kinda confusing

It's: If RS= 23-2x , ST = 9x-5 and RT = 39 Find RS

[alpine sable]

Sure

[oak basin]

I have edited it in the message :D

I can send a pic if that helps at all :3 but idk if It should adds to 39 because it's on a line yk and the line total is 39 so yeah

[orchid python]

how can I solve it @spring kiln

[tight locust]

homogenous systems of linear equations are equations which lack a constant term, hence the zero vector is always a solution.

the trivial solution is the zero vector.

hence you want to find the values of a such that the only solution to this system is the zero vector.

@orchid python

[karmic sandal]

If x = - 1 that would be positive right?

[alpine sable]

what would be positive?

[karmic sandal]

X+2 part

(+) (-) (this part can also be positive right? )

@alpine sable

[alpine sable]

I don't get your question to be honest

[warm brook]

6(-1-3)(-1+2)=(+)(-)(+) so

[karmic sandal]

But that one is decreasing graph right?

Which is contrary to what I sent above

[warm brook]

no because -1 is not in (-infty,-2)

[karmic sandal]

Loool, I am a dick for sure

Thanks @warm brook

[low dove]

I need to describe a class of sequences that satisfy this condition on the image. Like is it convergent or not, is it fundamental or not and so on. I guess that all x(n) , n > n(epsilon) don't belong to epsilon neighborhood of point a. So can I say that this sequence is unbounded?

[robust phoenix]

is it possible to multiply those ?

2 matrices

[low dove]

I think no because the number of columns of the first matrix does not equal the number of rows of the second matrix but I am not sure

[robust phoenix]

Yeah it aint possible

What's the area of this ?

[small bear]

Do you know Pick's theorem?

[robust phoenix]

yeah

I have an answer but i'm not sure

I think it

[small bear]

Use it, another way would be to construct a square, and find the area of that square, and remove the three other triangles, let me make an image

What did you get

[robust phoenix]

I did all of that and got 14

I just want to double check

[small bear]

That looks good

[robust phoenix]

you sure it's 14 ?

[small bear]

Yes, substituting i=12 and b=6 into the formula gives 14

[sharp fjord]

should that marked square be there or not, because I have notes saying it shouldnt and im confused

[low dove]

I think it should be there

[wind hawk]

yeah the square seems fine, but im not sure where the (x+2)(x-1) came from

[shut temple]

How can i solve this w/o l'hospital

[vital inlet]

13 pls

[elfin talon]

d

[warm brook]

Here’s a (big) hint:
[\frac{f(ax)}{g(bx)}= \frac{ax}{bx}\cdot \frac{f(ax)/(ax)}{g(bx)/(bx)}]

[ocean seal]

kirby

[warm brook]

substitute f(ax) and g(bx) for sin(ax) and sin(bx) respectively where a=7pi and b=2pi

[elfin talon]

X-sqrt(5) = 0
Therefore x = sqrt(5)

Also:

x + sqrt(3) = 0
Therefore x = - sqrt(3)
S.S = {-sqrt(3),sqrt(5)}
So d is the right answer

[shut temple]

ah thank you so much

[warm brook]

@shut temple yes, it’s pretty clever, but for things like this, i’d just recommend putting it into a form of a limit that you already know the answer to (if possible) then proceed

[alpine sable]

for this division statement i keep getting -x is that right?

[zealous ruin]

Is that long division?

[alpine sable]

ye

im trying to solve for the oblique asymptote in a rational function

[zealous ruin]

https://www.emathhelp.net/calculators/algebra-1/polynomial-long-division-calculator/?numer=x^3-1&denom=1-x^2

Here's a step-by-step solution

[alpine sable]

im getting the same as this

but i dont understand the bottom part

[zealous ruin]

https://www.mathsisfun.com/algebra/polynomials-division-long.html

Retrace the example problem here, it might help you understand the principles of long division better

[alpine sable]

ohh makes sense tysm helps a lot

also does horizontal asymptote on a slant jus mean oblique asymptote?

[zealous ruin]

English is not my first language, no idea. Let's ask <@&286206848099549185>

[alpine sable]

oh ok tysm for the help I appreciate it

How do i derive this

I got but i'm not sure

Didn't simplify yet

[crimson blade]

make it into an easier to use number. to start, you could change the sqrt^3(x) to x^1/3

then take the inverse so you can move it up to the numerator

then just use the power rule to differentiate it normally

[alpine sable]

Yeah i did that's what i got but everyone in the class has a different answer

[crimson blade]

what was their answer?

[crimson blade]

oh yeah

because the differentiation is linear, you could take out the 8 first

to go from d/dx[8/sqrt^3(x)] to 8 • d/dx[1/sqrt^3(x)]

[alpine sable]

Hmm

[graceful yoke]

Oh so derive does mean the same thing as find the derivative of.

[alpine sable]

Yes it does

[crimson blade]

yeah that is the way

8 • d/dx[1/sqrt^3(x)] then becomes 8 • d/dx[x^(-1/3)]

[graceful yoke]

Sorry I was unsure of the word so I didn't help

[alpine sable]

It's okay

[alpine sable]

You mean -2/3?

[crimson blade]

then 8(-1/3x^(-4/3))

why would it be -2/3?

[alpine sable]

1/3-1 is -2/3

[crimson blade]

its -1/3-1

so its -4/3

because when you move the term from the denominator to the numerator, the power is inversed

[alpine sable]

OH

Right

[crimson blade]

you get it?

[alpine sable]

Yep

Thanks

[crimson blade]

np

[ionic inlet]

I don't really understand how to do this. I'd know how to do it if the 5^(sec(-5x)+1) wasn't there or if it was only that but I don't know how to do it with both of them

[clever folio]

Does u=sec(-5x)+1 not work?

I have not done any calc in ages so I could be wrong tho lol

[devout summit]

Yeah it should work

[clever folio]

That makes sense lol

[clever folio]

Try u-sub

[bleak ridge]

You sub

[alpine sable]

what is X in this case ?

[spring kiln]

Did you finish the last problem with det?

[shut temple]

What the fuck happens here

+-

[bleak ridge]

Sus

[zealous ruin]

I guess it's a shorthand for "do it for both +inf and -inf", since the limits differ

For +inf it's +inf, for -inf it's 3/4

[bleak ridge]

Well wouldnt the limit not exist if it didnt approach the same value

[zealous ruin]

Wait, the other way around

[bleak ridge]

I would imagine it would have to approach the same value cause its one limit

[shut temple]

It is 3/4 for both

No

I am dum

[zealous ruin]

It grows linearly when approaching -inf

[shut temple]

Oh tru

Sorry

[wispy olive]

Can someone explain why the answer is 3¹ + 3² + 3³ + 3⁴?

[zealous ruin]

3^n is the number of n-letter words

And n can only go up to 4

_ means that any letter can occupy this space
1-letter words have a form of _, 2-letter words _ _, and so on

For each _ underscore you can choose one of three characters

[wispy olive]

Why the 3^n series?

Why are we multiplying A, B and C?

[zealous ruin]

Let's create all one-letter words in this language:
A, B, C

Let's create all two-letter words in this language:
AA, AB, AC
BA, BB, BC
CA, CB, CC

[wispy olive]

Yes.

Why do we multiply the values?

[zealous ruin]

When creating 2-letter words, you fill each gap in _ _ with an arbitrary letter, A B or C. You can choose the 1st letter in one of 3 ways, same goes for the 2nd letter, they're independent

[wispy olive]

I see.

I know that.

[zealous ruin]

Can you elaborate on your problem a little more?

I assume that all operations in ~code~ are O(1) - then the function itself is O(1)

[wispy olive]

I want to know why they are raising A,B,C(3) to the power of something?

Channel busy, #computing-software #numerical-analysis .

[zealous ruin]

Power of 3? You mean 3^whatever or whatever^3?

[wispy olive]

Yeah.

[zealous ruin]

Which one?

[wispy olive]

1st.

[broken pine]

https://cdn.discordapp.com/attachments/269573202018041856/901568637947768892/Screen_Shot_2021-10-23_at_1.31.24_PM.png

anyone know? please explain!

[zealous ruin]

Alright, let's create a language with a maximum of 4 letters, and the letters are {A,B,C,D,E,F} this time. How many 1-letter words can you create?

[wispy olive]

Channel busy and which AoPs book is that?

[broken pine]

uhhh geometry

[wispy olive]

Hmmm.

[broken pine]

chapter 7

[wispy olive]

That fucking chapter made me quit that book.

[broken pine]

can ya help?

[zealous ruin]

Ancient Greeks sure had it rough :P

[wispy olive]

There should be.

[broken pine]

alright so thats a no for help right?

[wispy olive]

Another channel.

Channel busy.

[broken pine]

alright ty

[zealous ruin]

That's right. You have only one space for a letter, so you have 6 choices

[wispy olive]

xD.

[zealous ruin]

What if you had 2 letter words?

[wispy olive]

I see I see.

Let me see.

[zealous ruin]

Hint: 2-letter words in this language are just a concatenation of two 1-letter words

[wispy olive]

No wait.

Right?

[zealous ruin]

How did you arrive at that conclusion?

[wispy olive]

6 × 6.

Since 2 lettered.

[zealous ruin]

That's right

[wispy olive]

Ahaaaa.

[zealous ruin]

You choose a single letter arbitrarily (6 choices)

And then you choose another letter, arso arbitrarily, another 6 choices

[wispy olive]

Wow man you are awesome!

Are you a professional teacher or something?

[zealous ruin]

Not yet!

[wispy olive]

And sorry for being dumb and typos, it is 2:00 AM and I am sleepy lol.

[zealous ruin]

No worries

[wispy olive]

Well let us leave this channel cause within like 10 minutes 2 people wanted this place.

[zealous ruin]

disappears into a fine mist