help-0

[topaz scaffold]

t = 24 (in minutes)

[ornate quest]

Okay, so. My book shows set builder notation as:

But goes on to show sets like {2n : n ∈ N} and {n ∈ N : n is even}. Seeing as anything can be on either side, then a set builder expression is really just a series of rules that some item in a set must satisfy. Is there a point to making a distinction between the sides like this?

[ionic jewel]

i mean i think technically the second one would be written {n:n in N, n is even}, but if it gives you that one as an example you can probably do what you want lol

actually i think the first part is kinda whete you "use" the variables, and the second is where you "define" them, but I'm not completely sure formally

if it's clear what the set is, ur probably doing it fine

[ornate quest]

Haaa
So set builder can be used however, its just to make a set definition more clear?

[halcyon granite]

well the n in N can be seen as a rule or an expression but thats not always the case

e.g. {x in R : x > 5} is fine but {x > 5 : x in R} would be a little weird although everyone would get it

same with your thing actually {n in N : 2n} would be weird

because 2n is not a rule you would have to write it as "n is even" or something for it to become a rule

[ornate quest]

Hm... though its a bit clunky when the left side is a unevalulated expression

Because to "get" what is inside the set, you would have to take the elements that satisfy the predicate on the right side and plug it into the left side; going backwards

[halcyon granite]

I dont understand what you mean can you give an example?

[ornate quest]

Well, to get the elements inside the set {2n : n in N}, you would take all natural numbers, and plug them into 2n right? ({1, 2, 3, ...} -> {2, 4, 6, ...}) You can't take "all elements of 2n and evaluate is n in N", because to get the "elements of 2n", you need n

[halcyon granite]

ah I understand yeah but thats just how it is

[ornate quest]

Bleh. So I guess sematically, it would be "take the unknown blob of 2n, and keep only those for which n in N is true"

Which is weird, because that implies that numbers keep how they're generated with them
O' well, this isn't programming, haha

[halcyon granite]

Its just common notation formally/historically I would always use {x in S : P(x)} where S is a set and P a predicate

so I would also prefer the {n in N : n is even}

[warped phoenix]

are you finished w ur q?

[ornate quest]

yep

[alpine sable]

Are you good at math

[ornate quest]

Nope.
Just learning it for fun and Haskell :P
Also, #math-discussion / #chill

[alpine sable]

Are u higher than a grade 8 level

@ornate quest

Cuz i am having a lot of problems

someone pls tell me how i solve 4x = (x+3) + 6-5
or 6-5 + (x+3) = 4x
does x+3 means x3????+? i not rememvber
pls and thank you :>>:>>:

pls and tehank yoeou :>>.<

Someone pls help me

I am needing of help pls

[alpine sable]

👍

[surreal pumice]

What are the two lines I see surrounding a vector? |v|

Does that mean absolute value or magnitude?

or something else?

[ornate quest]

Magnitude irrc

[surreal pumice]

I'm doing some research and there's no clear answer. People are saying that double parallel lines are magnitude

[ionic jewel]

$\norm{v}$ is the normal thing for magnitude, but single bars can also mean it

[ocean seal]

bunny

[ornate quest]

Ooh, we have latex on here. Nice.

[ionic jewel]

if your source doesnt specify absolute value it means magnitude

[surreal pumice]

Ah okay cool thank you guys

[sleek elbow]

wait latex isnt something this discord made?

[ornate quest]

c-can't tell if there is a /s there :P

[sleek elbow]

whats a/s?

[ornate quest]

sarcasm

[sleek elbow]

oh no there isnt

[ornate quest]

I can't tell if you're being sarcastic

[sleek elbow]

nope

[ornate quest]

O-oh
LaTeX is everywhere, haha

[sleek elbow]

wut

where is everywhere

[ornate quest]

Like, every single college paper ever is written in latex

and whitepaper

[sleek elbow]

WHAT

[ornate quest]

And even half the textbooks

[sleek elbow]

wait

so

they format their stuff in latex?

[ornate quest]

Yeah?

[sleek elbow]

lol

[alpine sable]

Can someone help me

Pls

[sleek elbow]

didnt even know

[alpine sable]

It is very easy

But I am autistic

So

[ornate quest]

After reading enough stuff, you can spot latex on sight :P

[sleek elbow]

https://dontasktoask.com @alpine sable

[alpine sable]

Pls hep

[ornate quest]

There isn't even a question on there though.

[sleek elbow]

thats what I thought too

[alpine sable]

This part also

[sleek elbow]

what is the question

@alpine sable

[alpine sable]

All this

Shit

Bad resolution

[sleek elbow]

what is the question, what do you want to find

[ornate quest]

Nono, the thing is, there isn't a question on either of those pictures.

[alpine sable]

Ok so

I have to find the price for this

This is a summer school homework and I don’t understand

@ornate quest

@sleek elbow

[sleek elbow]

Oh so you need to find the measurements of the medal and than find out the price from there?

[alpine sable]

Ye

My mom said she will beat me if I don’t get this right do

So

I am cry soon

[lyric cape]

i

dms real quick

[sleek elbow]

okay so first you gotta find out all the 3 lengths, you do that by dividing the corresponding length of the crest by 20, because the dimensions of the medals are 1/20 of the dimensions of the crest.

The example already there would be 2.4m / 20 = 240cm / 20 = 12cm

After you get all the lengths, you add them up to get the total perimeter of one medal and than multiply that by 20 to get the perimeter of 20 total medals. From there you multiply by 0.25$ and you get your result.

@alpine sable

I have the solution written out if you want to check your calculation after you are done

[sleek elbow]

WHAT

wait you serious?

[alpine sable]

Thanks

Yes lol

[sleek elbow]

thats fucked up

[alpine sable]

I don’t really care what she does as long as she pays my skool

[sleek elbow]

where do you live, like country?

[alpine sable]

Canada

I failed the school year at one of the best schools

So

Rip

Can u send a picture of the calculations

[sleek elbow]

is beating your children not illegal in canada?

[alpine sable]

If it’s disaplin it ok

And this is technically disaplin

Or however u spell it

[sleek elbow]

"beat up" sounds pretty hard to me

[alpine sable]

Yeah

👀

wtf going on here

[sleek elbow]

yup weird stuff

[alpine sable]

If I don’t get everything right on ma work I get beat

[sleek elbow]

diciplining children apparently legal in canada 👀

[alpine sable]

It is

[sleek elbow]

yeah I believe you

just surprised me

[alpine sable]

Yea

[sleek elbow]

but tbf, I think it was only banned 20 years ago here in germany

[alpine sable]

Oof

[sleek elbow]

okay so try it for yourself first (the exercise) @alpine sable

with the steps I told you

[alpine sable]

Ok I’ll try

[sleek elbow]

and tell me if you get stuck

[alpine sable]

bro, if ur mom hits u gotta tell someone

[sleek elbow]

if its legal and not too much he cant do anything probably

thats like when teachers used to hit their students

probably

[alpine sable]

Ye

i dont believe hitting ur children is legal in canada

[sleek elbow]

I googled

[alpine sable]

Well I got 18 as my number on the question

[sleek elbow]

it is if its dicipline apparently

[alpine sable]

I think it’s wrong

[sleek elbow]

18 for what?

[alpine sable]

The length

[sleek elbow]

length of what

[alpine sable]

And 1167$ for money

[sleek elbow]

can you show your calculation?

like type it out

[alpine sable]

Ok

I’ll send

It’s prob wrong

[alpine sable]

It’s prob wrong

But I tried

Am I right or wrong

[sleek elbow]

where does the 6.4 come from?

[alpine sable]

All the crest added

Is 6.4

[sleek elbow]

oh that works too apparently

let me read the rest...

im not sure what you did, but if 0.32 or 32cm is the perimeter of one of the medals

you need to get the perimeter of 20 off those

from there

[alpine sable]

I rushed since I have 7 mins till my mom checks my work

So

Rip my life

[sleek elbow]

well get it

dw

so the perimeter of one medal is 32cm

whats the perimeter of 20 medals?

[alpine sable]

I don know

[sleek elbow]

20 * 32

[alpine sable]

Oh yeah

[sleek elbow]

which is?

[alpine sable]

640 I geuss

[sleek elbow]

yup

so thats the total length you need

in cm

so now you multiply that by the price

[alpine sable]

Ok

I’ll check

[sleek elbow]

and tell me what you get

[hybrid grove]

hmmm

[alpine sable]

Is it 160

[hybrid grove]

what you guys doing?

[sleek elbow]

I believe so

[sleek elbow]

answering his question

[hybrid grove]

what is it?

[sleek elbow]

look above

this was my calculation btw

80cm / 20 = 4cm
320cm / 20 = 16cm
240cm / 20 = 12cm

So the three lengths of the medal are 4cm, 16cm and 12cm.

The sum of those is 32cm and you need 20 medals, so you need 20 * 32cm total which is 640cm

If each cm costs 0.25$, the cost will be 640 * 0.25$ which is 160$

@alpine sable

[hybrid grove]

oh nice

[south abyss]

Hi.
If the median, unique mode, mean, and range of the five integers 2, a, 5, b, and c are all equal, where 2 <=a <=5 <=b <=c, what is the value of abc?

I don't know how to do it... very confusing.

[hybrid grove]

hmmm that sounds a little too easy

[sleek elbow]

why?

[alpine sable]

What about da crest

[sleek elbow]

what about it?

[hybrid grove]

idk it sounds a little lower than 7th grade stuff

[sleek elbow]

well wdym by that? Do you think the question is too easy or my answer is wrong?

[alpine sable]

It’s cuz Canada is easier than usa

[sleek elbow]

we just needed the crest to get the sides of the medal if I understood correctly

[hybrid grove]

no im saying is that the question looks like a 6th grade question which means he underage for discord

[sleek elbow]

oh idc 😄

[hybrid grove]

kk

[sleek elbow]

still gonna help even if he is

[hybrid grove]

kk

[south abyss]

If the median, unique mode, mean, and range of the five integers 2, a, 5, b, and c are all equal, where 2 <=a <=5 <=b <=c, what is the value of abc?

sorry for repost btw bc i have like 5 minutes to finish this before i hand it to my mom...

[steel palm]

someone help pls

[sleek elbow]

why does everyones mom pressure today, btw I cant help, I have no idea, sorry

[south abyss]

ohh ok

[south abyss]

<@&286206848099549185>

[alpine sable]

I have autism big boy

So I get easier math

And I’m 13

[glass lichen]

what have you tried?

[hybrid grove]

oh sry

[alpine sable]

@steel palm i guess they all have to be 24cm for that to happen

[glass lichen]

you cant have an equilateral triangle which is right...

[steel palm]

in the solutions it shows this but idk how and why it is there

[glass lichen]

yeah

[alpine sable]

i misunderstod sry

[glass lichen]

if you remove the same length, x, from each pipe, they have lengths 24-x,31-x and 38-x

of which form a pythagorean triple

[steel palm]

but the same length has to be cut

[glass lichen]

yes

[steel palm]

from each

[glass lichen]

yeah, you read the question

[paper oar]

210

Just gave the answer cuz u seem in a hurry

I made a = 5, b = 6 and c = 7 because this keeps 5 as the mode (occurs the most in list), 5 as the median (center of list), 5 as mean (2+5+5+6+7 = 25/5 = 5) and 5 as range,7-2 = 5 thus abc = 567 = 210

[steel palm]

last one

<@&286206848099549185>

[glacial hedge]

Generally how do you find if a mtrix is onto? I know it must span R^m but how do u prove that?

Nevermind

hehehe

stooopid

<@&286206848099549185> Is there a linear transformation T such that it is onto and R^1000 --> R^1001

[pallid trout]

Do you remember what function compositions are?

[alpine sable]

isn't this 7 sqrt 2?

[pallid trout]

Now sure what you mean by B, but function composition would solve your first picture

If by B you mean 0, then you'd be wrong 😛

[alpine sable]

he means option 2

[pallid trout]

Yeah, then he'd be wrong

[alpine sable]

is it option 4?

WAIT

no

yeah

for the second one

is it option 4

@-@

tbh I think its option 1

wait no

idk

nvm not option 4

[pallid trout]

For the second one, it starts at -3, up to 3 in (9-x^2) then 3 on the rest. Obviously x-3 is increasing. So we know that 3 -> infinity is increasing.
Due to the square, overalping a negative, you get a parabolla centered on 0. Parabolla has an increasing side (from -3 to 0) and decreasing (from 0 to 3)

[alpine sable]

yeah I think its option 2

@sterile falcon where on khan academy did you see this lesson?

if so, may you provide a link

[dim vector]

i need help with something

Calculate the total surface area, r = 6.9 m and, a = 10 m

[glass lichen]

just plug everything into the formula and simplify

[dim vector]

and how would it be

[jagged raptor]

area of the top and bottom is two circles
area of the side is a rectangle with height a and base the circumference of the circle

[dim vector]

7.33

?

[ionic jewel]

,w handwritten style surface area cylinder r = 6.9, h = 10

[dim vector]

I don't know if I'm doing it right

[ionic jewel]

you seem to be off by a factor of 100

[ocean seal]

Results provided by WolframAlpha

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

Requested by 兔子#6562

[dim vector]

oh

[ionic jewel]

aw it didnt work

[dim vector]

732.682

its correct?

;-; ohh

the answer was already

[vagrant rover]

Why does the square root of x^2 = the square root of 25 simplify to plus or minus 5? Specifically why plus or minus 5?

[dim vector]

thanks

[vagrant rover]

Wait but wouldn't ever answer be plus or minus then?

Like any square root

[glass lichen]

that isnt true

sqrt(144)=12

[random flax]

Think about it this way. 5^2=25 we all agree

for (-5)^2 rewrite as (-1)^2*(5^2)=1 * 25 = 25

so you are squaring the negative one separately. And then u get it

[glass lichen]

cause sqrt(144) is 12

[alpine sable]

because its not

-12

xd

[glass lichen]

since sqrt(x) maps from [0,inf) to [0,inf)

[alpine sable]

you are confused with sqrt x^2

[dusk smelt]

@violet ore because by definition square root function of a positive integer is defined as the positive root, not the negative

thats how mathematicians decided to do it, for convenience

[slow hedge]

why is any rows sum in a triangle of consecutive odd numbers is the cube of the row number?

1 * 1 * 1 = 1
2 * 2 * 2 = 3 + 5
3 * 3 * 3 = 7 + 9 + 11
4 * 4 * 4 = 13 + 15 + 17 + 19
and so on

[random flax]

is that the pascal triangle?

[glass lichen]

no

[random flax]

nvm lmao

No clue on arithmetic but lemme try and explain

x^2*x=x^3

@slow hedge

If you notice there the sum of any row can be written as x*x^2

so for row 4 for eg.

4^2=16

and 4*4^2=4^3 as required

now look at the numbers (15 and 17 are one off 16 so they balance out to 2*4^2)

and 13 and 19 are 3 off 16 so they balance out to 4^2*2

so 2* 4^2+2* 4^2=4^2*4=4^3. This can be applied to any row

hope this helps

I don't know how to prove it formally

maybe write it as a set of xs

[slow hedge]

oh ok

I think I got it

[topaz scaffold]

Can you simplify $\frac{a^{100} + b^{100}}{a^{96} + b^{96}}$?

[ocean seal]

RipeOrange

[alpine sable]

🟦🟦🟦🟦🟦🟦🟦🟦🟦
🟦🟦🟨🟨🟨🟨🟨🟦🟦
🟦🟦🟨👁️🟨👁️🟨🟦🟦
🟦👌🟨🟨👄🟨🟨🤚🟦
🟦🟦🟨🟨🟨🟨🟨🟦🟦
🟦🟦🟫🟫🔳🟫🟫🟦🟦
🟦🟦🟦👞🟦👞🟦🟦🟦

[slow hedge]

wow this is cool

[rough compass]

can someone help with this

[warped phoenix]

im just gonna repost ur answers so its way easier for you

• srry for the super late response

how did you get 576*h as the volume?

[topaz scaffold]

24 inches * 24 inches

Oof wrong unit

[warped phoenix]

why 24? 🤔

[topaz scaffold]

Square base of 2 feet

2 feet = 24 inches

[warped phoenix]

ohhhhhhhhhhhhh

how do you know that the width is the same as the length tho? @topaz scaffold

[topaz scaffold]

Cause it says that there is a square base

[warped phoenix]

omg

im so dumb

hahahah right

[sudden tulip]

Hi

Can I ask a question

[warped phoenix]

channel is in use

[topaz scaffold]

Preferably in another channel

[warped phoenix]

ok so

[sudden tulip]

Noted

[warped phoenix]

i get that h(t=0)=0 is saying that when the time is 0

the height is 0

but how do we derive h(t) = 8t from that?

[topaz scaffold]

Yep

[warped phoenix]

is it saying that at 8 minutes, the height is the time?

so like, 8 inches?

[topaz scaffold]

So $\dot h(t)$ is 8

[ocean seal]

RipeOrange

[warped phoenix]

right

[topaz scaffold]

So if we do $\int 8 ,dt$ we get h(t) 8t + c

[warped phoenix]

whats int/dt?

[ocean seal]

RipeOrange

[warped phoenix]

sorry im in alg 2 😅

[topaz scaffold]

Oh

Well you don't need calculus

[warped phoenix]

is this a calc question?

[topaz scaffold]

No it's not

I just default to that 😅

[warped phoenix]

oh ok

nw!

[topaz scaffold]

The height changes at 8 in/min

So every minute the height changes by 8 inches

[warped phoenix]

right

[topaz scaffold]

So we let t be the time passed in minutes

So our height as a function of time would be h(t) = 8t

[warped phoenix]

ohhhhhhhhh

so like after 2 minutes, h(2) = 16

[topaz scaffold]

Yep

[warped phoenix]

ohh i see

[topaz scaffold]

From here we can plug this into our volume equation

[warped phoenix]

so 24 x 24 x 8t

= 4608t

[topaz scaffold]

Yep

Keep in mind this is in cubic inches

[warped phoenix]

oh right

so 384t cubic ft?

cuz u divide by 12 to get ft im p sure

[topaz scaffold]

Well that's where I messed up in my original calculations 😅

[warped phoenix]

ohh haha no worries

[topaz scaffold]

In 3 dimensions, when you scale a side down by an amount s, the entire object gets scaled down by s^3

So you actually need to divide 4608t by 12^3

To convert to feet

You should get $\frac{8t}{3}$

[ocean seal]

RipeOrange

[warped phoenix]

hm

why couldnt we just keep it in cubic feet?

[topaz scaffold]

Yea we could have done everything in feet

I just decided to use inches in the beginning

[warped phoenix]

ohhh

wait

yeah they use normal feet for the square base

i see

8t/3 then yeah thats right

So V = 8t/3, and then we set V = 64? @topaz scaffold

wait but isnt it an issue that 8t/3 is normal feet and 64 is cubic feet?

or does that not matter

@glass lichen ur smart do u mind if i tag u 😅

[glass lichen]

Yes I do mind

[warped phoenix]

sorry 😦

<@&286206848099549185>

anyone?

[wild gale]

M@ocean seal#cô0796

W uu Z a loo muumuus I no

[merry coral]

what are the odds of getting a clump of n ones:

in a x * y grid if the odds of a one is 50% (o)

first off I assume I need the formula for the amount of corners, sides and middle squares

x must be greater than or equal to 2 and same with y

so corners = 4

sides = 2((x-2) + (y-2))

middle squares = (x-2)*(y-2)

now I'm lost

[alpine sable]

\begin{equation*} \left{ \begin{aligned} &x+2y-3z=4 \ & 3x -y +5z=2\ &4x+y+(a^2-14)z=a+2 \end{aligned} \right. \end{equation*}

[ocean seal]

Luis Henrique Schwab

[alpine sable]

Determine the value of a so the linear system has no solution

Can someone give me a hand?

[ionic jewel]

is this a linear algebra course?

[zealous adder]

<@&286206848099549185>

How do I find the equation for this

[alpine sable]

yes

[ionic jewel]

in that case to have no solution it needs to row reduce into something that ends in
0 0 0 | C, where C is not 0

which means 0x+0y+0z is not equal to 0

which is no solutions

[alpine sable]

thanks, got it. I used gaussian elimination but was conceptually confused

[ionic jewel]

that was the intended solution

[alpine sable]

got to 0x+0y+(a^2-16)z=a-4 but didnt know how to get from there

[rocky dock]

for the first one i got

x int: 3/4
y int: -3/5

but the stupid graph wants whole numbers

what should i put as the points, i need a straight line?

<@&286206848099549185>

[final cradle]

H

[soft zodiac]

how can i get the vector perpendicular to another vector?

[pearl marlin]

can you draw both the lines?

[rocky dock]

it just lets u pick the points

only whole numbers

@pearl marlin

wont even let me round

[pearl marlin]

hmm you alredy solved this

[rocky dock]

it was wrong

big fat red x

[pearl marlin]

try to draw this in desmos

[pearl marlin]

these are parallel lines

[topaz scaffold]

8t/3 is in cubic feet if you work out all the units

[topaz scaffold]

Correct

[warped phoenix]

ohhhh dividing it by 12^3 made it into cubic feet?

i see

[topaz scaffold]

That just converts from cubic inches to cubic feet

[warped phoenix]

ohhhhhhhhhhhh

rightttttttt

i see now

ty!!!!!

[topaz scaffold]

No problem

[rocky dock]

@pearl marlin I did

They’re the same. Did you see a whole number on desmos?

[left echo]

PART 1

y= 3log3(x-4)

• has a vertical stretch by 3
• vertical asymptote is x=4

PART 2

• after the transformations are applied the new equation will be: y= 3log3(x-3)+2
• the domain of the new equation is: x⋲〈3, + ∞〉, you can find the domain from the equation of a log function by finding where the equation is defined. The domain of a logarithmic function are all the values for x in which x-3 of the logarithm is positive.

x-3>0 - move the constant to the right hand sign and change the sign ↓

x>3 - find the intersection ↓

x > 3

x⋲R

x⋲〈3, + ∞〉

can somebody lmk if my answer to this question is right pls

[pastel bobcat]

bro we cant give answers to ur homework, test and exams question

sorry

@left echo

[left echo]

dude i have the answer just someone lmk if its right

and its not a test or exam

[pastel bobcat]

u sure abt that?

[topaz scaffold]

Could be a hw question

[left echo]

so ??

its literally just an assignment

[pastel bobcat]

wait lemme open rules

if its a hw question then its allowed

lemme see the question

[left echo]

i already sent it up there

[high thistle]

k so can anyone solve this quesiton

and explain logic also?

[topaz scaffold]

Channel occupied rn

[topaz scaffold]

Please find another channel

Oh well

[high thistle]

ok

i didnt knpwo

[left echo]

sorry

LMFAO

[topaz scaffold]

So you already done part 1 of question 3?

[left echo]

i did the whole question

i just want to make sure it’s right

[topaz scaffold]

Ohhh I see

[left echo]

yea

[topaz scaffold]

It looks good

[left echo]

alright thanks

[opal wyvern]

Does the function solve the hypotheses of the Mean Value Theorem on the given interval?
f(x)=e^-4x, [0,1]

Yes, f is continuous and differentiable on R, so it is continuous on [0,1] and differentiable on (0,1)

If it satisfies the hypotheses, find all numbers c that satisfy the conclusion of the Mean Value Theorem. (Enter your answers as a comma-separated list)

f'(x)=-4e^-4x

f'(c)=(f(b)-f(a))/(b-a)

f'(c)=e^-4-1

So e^(-4)-1 = -4e^-4x

Help me solve for x

[topaz scaffold]

Use natural log

[formal sun]

Modular exponentiation 3^65 mod 7 , (1,4 5,10,20,25) ,3^1) mod 7 = 3 , (3^2)^2 mod 7 = 4 m (3^5)^1 mod 7 = 5 , (3^5)^2 mod 7 = 6, (3^10)^2 mod 7 = 2, (3^5)^5 mod 7 =3 ,3 *4 *5 *6 *2 * 3 this isn't making sense to me the remainder is suppose to be 5 but the way I did it which I though was the right way ended up giving me the wrong answer, any help is greatly appreciated . thank you

[opal wyvern]

@topaz scaffold I literally used a website to find the apparently singular value of c and webassign said it was wrong

[topaz scaffold]

So you got $e^{-4}-1 = -4e^{-4x}$?

[ocean seal]

RipeOrange

[opal wyvern]

Correct

[topaz scaffold]

,w solve e^{-4}-1 = -4e^{-4x}

[ocean seal]

Results provided by WolframAlpha

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

Requested by RipeOrange#4907

[topaz scaffold]

It seems to be complex

[opal wyvern]

That seems a little complicated

[topaz scaffold]

I think we got a negative input to the natural log

[opal wyvern]

Can you help me solve -4e^(-4x)=e^(-4)-1?

[topaz scaffold]

Well we would divide both sides by -4

$-\frac{e^{-4}-1}{4} = e^{-4x}$

[opal wyvern]

Got that

[ocean seal]

RipeOrange

[topaz scaffold]

Then take the natural log on both sides

You'll get $ln(-\frac{e^{-4}-1}{4}) = -4x$

[ocean seal]

RipeOrange

[opal wyvern]

got that

wait what happened to ln(e)

[topaz scaffold]

Here we see the problem

[opal wyvern]

is that just 1

[topaz scaffold]

ln and e cancel

Since they are inverse functions

[opal wyvern]

got it

[topaz scaffold]

$ln\big{(}-\frac{e^{-4}-1}{4}\big{)}$

Do you see the problem with this part?

[opal wyvern]

tf is big

[topaz scaffold]

Oof that's not the right thing

[ocean seal]

RipeOrange

[topaz scaffold]

Here

[opal wyvern]

i have 0 idea how to solve from there

[topaz scaffold]

Well $\frac{e^{-4}-1}{4}$ is positive

[ocean seal]

RipeOrange

[topaz scaffold]

So the entire thing inside the ln() is negative

And you probably know that ln() isn't defined in the negatives

At least for the real numbers

[opal wyvern]

yea i definitely did know that for sure

[topaz scaffold]

So you'll have to go into the complex world to solve this

And I don't think this is what's supposed to happen here

So something must be wrong before this step

[opal wyvern]

Unfortunately I think everything is correct

[topaz scaffold]

I'm not sure what the mean value theorem is

[opal wyvern]

https://www.webassign.net/scalcet7/mean-value-theorem.gif

[topaz scaffold]

Ok

Our a and b here would be 0 and 1 correct?

[opal wyvern]

Yes

[topaz scaffold]

$-4e^{-4c} = \frac{e^{-4*1} - e^{-4 *0}}{1-0}$

[opal wyvern]

whered you get the 5

[wanton turtle]

-4e^-4c

Not e^-5c

[ocean seal]

RipeOrange

[topaz scaffold]

Ah sorry typo

[opal wyvern]

pretty sure thats what i got too

[topaz scaffold]

Hmmm yea

,w solve -4e^{-4c} = \frac{e^{-4*1} - e^{-4 *0}}{1-0}

[opal wyvern]

Christ

There is no way he meant to make the problem that hard

[wanton turtle]

Do you have an answer key?

[topaz scaffold]

Something isn't right here

[wanton turtle]

I'm getting c as (1/4) . ln(4/1-e^-4)

[opal wyvern]

@crude dockp[e

oops

[topaz scaffold]

How

[opal wyvern]

@topaz scaffold why did you do it as c-4e^-4c ?

Why are you subtracting from c

[wanton turtle]

I'll send the steps

[ocean seal]

Results provided by WolframAlpha

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

Requested by RipeOrange#4907

[topaz scaffold]

Ehhh

[topaz scaffold]

Cause I typed in "solve for c" 😅

No wonder it was using the Lambert W function

[topaz scaffold]

Wouldn't ln(4/1-e^-4) still be complex tho?

[wanton turtle]

No

The denominator is positive

[topaz scaffold]

,w evaluate ln(4/(1-e^-4))

[ocean seal]

Results provided by WolframAlpha

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

Requested by RipeOrange#4907

[topaz scaffold]

Well nice

[opal wyvern]

this is madness

[wanton turtle]

@opal wyvern is that the answer?

[opal wyvern]

im scared to try cause i got limited attempts

[wanton turtle]

It's basic logarithms if you know LMVT

[opal wyvern]

What steps did you take

idk lmvt

[wanton turtle]

It's the mean value theorem

What you did so far

I just continued from where you stopped

[wanton turtle]

^ from here

Then my image

[opal wyvern]

So how did we continue from e^(-4x)=e^(-4)-1

oh im looking

[wanton turtle]

Check the image I sent

[wanton turtle]

^

[topaz scaffold]

Something isn't right tho

[opal wyvern]

Is from step 1 to step 2 accurate?

[topaz scaffold]

e^(-4)-1 is negative

[opal wyvern]

Not sure if that exponent should be negative

[wanton turtle]

yes, and the LHS is negative too

[topaz scaffold]

Oh yes there

[wanton turtle]

it is, the derivative of e^-4x is -4e^-4x

[wanton turtle]

took ln on both sides

[topaz scaffold]

Where did I mess up then? 🤔

[wanton turtle]

or you could just invert the signs

[opal wyvern]

you sure lol

[topaz scaffold]

Wolfram Alpha has failed me

[wanton turtle]

simplify this and you get 1 - e^-4

[opal wyvern]

But like

You sure

[wanton turtle]

yes

[opal wyvern]

@topaz scaffold Can you confirm

[topaz scaffold]

Yea

[opal wyvern]

I love how it "simplified" it into something way more complex

[wanton turtle]

[topaz scaffold]

Turns out Wolfram Alpha showed the complex answers and didn't show the real ones

Thanks Wolfram

[opal wyvern]

Why do you have ln4-4c from 4e^(-4c)

[topaz scaffold]

Now I feel stupid

[topaz scaffold]

Took the natural log on both sides

[wanton turtle]

ln(4e^(-4c)) = ln 4 + ln e^-4c = ln 4 - 4c

[opal wyvern]

thanks

So there is only one value of c?

[wanton turtle]

seems to be the case unless i'm missing something

[opal wyvern]

Alright Im about to click submit

lets see

Wow

It worked

Life is suddenly halfway decent

[topaz scaffold]

Aha

[winter linden]

Hey guys i have a small question here... if i have x + y = 14 and talking y to the other side.. is it legal to write in after 14? Cuz like for standard froms the variables are always before the numbers soo yeah?

[topaz scaffold]

Yea either way works

Cause addition is commutative

[thorn kindle]

x + y = 14 => x = 14 - y

is that what you mean?

[opal wyvern]

that is accurate math

it doesn't matter where the y is

it could be x = -y+14

[thorn kindle]

that is true over every field

[topaz scaffold]

Wdym over every field?

[opal wyvern]

in addition and subtration, as a numerator, as a denominator, as an exponent

probably elsewhere

[thorn kindle]

every field of numbers where addition is defined as it normally is

[opal wyvern]

-y * 14 = 14 * -y

[winter linden]

Ohhh ok ok thank you

[topaz scaffold]

You saying like complex numbers and irrational numbers?

[opal wyvern]

How can you possibly solve this

Guess?

Its just a graph no function

ok yea i just guessed lol

it was right

[thorn kindle]

take for example the irrational numbers (10 + sqrt2) and (-sqrt(2))

[topaz scaffold]

I was seeing if he meant that or just types of numbers 😅

[topaz scaffold]

Those aren't equal (?)

[thorn kindle]

add them

[topaz scaffold]

Oh

I would assume that also works

That's beyond my realm tho

[topaz scaffold]

10 (?)

[thorn kindle]

Yes

That is a rational number!

[topaz scaffold]

Am I getting quizzed at this point?

I'm lost

[hasty solstice]

Can someone help me solve this?

I'm out of ideas

[topaz scaffold]

You can easily find the percentage of smokers with cancer to all the smokers

And similarly with the non smokers

By doing #of smokers with cancer/total number of smokers

Same thing with non smokers

Then compare those percentages

[hasty solstice]

I don't get the 5%LOS thing

Thanks for the efforts though

[topaz scaffold]

It's greater than 5% yea

Maybe compare to the total

That wouldn't make sense tho...

Idk

[leaden jetty]

How do i calculate this ?

---

there are 3 slots
The 1 slot can contain any of (0,1,2,3,4)
The 2nd slot can contain any of (0, 1,2,3,4,5,6)
The 3rd slot can contain any of (0,1,2,3,4,5,6)
What is the no of combinations possible if repeatation is allowed ?

[ionic jewel]

5*7*7

number of options for the first times number of options for the second times number of options for the third...

[leaden jetty]

ok

[rugged maple]

<@&286206848099549185> <@&286206848099549185> <@&286206848099549185> can y’all do my exam for me i got $500

[topaz scaffold]

Nope

[rugged maple]

big brrr

[topaz scaffold]

Must resist temptation

[alpine sable]

anyone know linear algebra here?

[topaz scaffold]

The tiniest bit

[alpine sable]

I need a proof looked over 😭

to make sure im not crazy

[topaz scaffold]

I can't do proofs well 😅

Sent it anyways, someone else probably knows

[alpine sable]

question was If $u,v$ are vectors in $\mathbb{R}^3$ and $||u||=1$, $||v||=2$ then $u\cdot{v}\neq{10}$

[ocean seal]

jswatj

[alpine sable]

I know this statement is true

but my proof is wild

[midnight sundial]

do you know a particular inequality that bounds |u*v| ?

[alpine sable]

not at all

[midnight sundial]

hmm

it is immediate from Cauchy-Schwarz inequality

[alpine sable]

Yeah

i said

since |u|=1 then one of its variables has to be -1 or 1

and thus all the rest 0

so i chose one

then did the dot product

but in the second one, theres several cases like one variable can be -2,-1,0,1,2

[midnight sundial]

the vector u = (1/sqrt(2), 1/sqrt(2), 0) also has |u|=1

[alpine sable]

but in every single case its less than 10

[alpine sable]

😮

shit

[midnight sundial]

assuming they mean the normal euclidean distance

[alpine sable]

hmmm

well

[midnight sundial]

have you learned Cauchy-Schwarz or AM-GM?

[alpine sable]

i learned AM-GM

i can use that?!

wait

cauchy-schwarz is like the triangle inequality

wait maybe i have learned that

hold up

no i haven't

😦

[midnight sundial]

since the dot product is \sum u_i v_i, you could use the inequality ab <= 1/2(a^2+b^2) on each term

it won't give you as good a bound as Cauchy-Schwarz but you just have to show |u.v| < 10

actually what definition of dot product do you have

is it \sum u_i v_i or |u||v|sin(theta)

[alpine sable]

the first one

[vale wigeon]

cos(θ)

not sin(θ)

[alpine sable]

^

[midnight sundial]

oh yeah cosine not sine

[alpine sable]

hmm

im thinking

[fervent grail]

a²-a+1/a²+a+1=3/2 , find a+1/a

[fervent grail]

Pls help

[midnight sundial]

use different channel

like #help-3

[alpine sable]

top is (a-1)^2/a^2+a+1

anyway

[fervent grail]

Can u tell the ans ?

Is the ans 2 or -3/2 or any other ?

@alpine sable

[midnight sundial]

#❓how-to-get-help

[alpine sable]

u can literally symbolab that

[fervent grail]

?

[vale wigeon]

nilesh, we do not give answers here

also you're missing parentheses

did you mean $a^2 - a + \frac{1}{a^2} + a + 1 = \frac{3}{2}$?

[ocean seal]

Ann

[alpine sable]

oh

that would be easier

[fervent grail]

Nope
I was saying a²-a+1 / a²+a+1 = 3/2

[jagged imp]

thats what ann wrote.

[teal epoch]

$\frac{a^2-a+1}{a^2+a+1}=\frac{3}{2}$?

[ocean seal]

Nonexisty

[vale wigeon]

if you wanted to write $\frac{a^2 - a + 1}{a^2 + a + 1}$, then you should have written \verb|(a^2 - a + 1)/(a^2 + a + 1)| so that it gets read correctly.

[ocean seal]

Ann

[alpine sable]

you'll just get $a=-\frac{3\sqrt{5}\pm 5}{2}$

[ocean seal]

jswatj

[vale wigeon]

in any case! this problem is literally no harder than a quadratic equation.

you can reduce it to one.

[fervent grail]

Yes

[vale wigeon]

just by doing algebra to it.

[alpine sable]

^

quadratic formula

[fervent grail]

What will be the value of a+1/a , I am getting -3/2 is it right ?

[alpine sable]

no

[vale wigeon]

do you mean $\frac{a+1}{a}$ or $a + \frac{1}{a}$?

[ocean seal]

Ann

[alpine sable]

lmaooooo

[fervent grail]

Second one

[alpine sable]

wtf

[fervent grail]

a + 1/a

Or a^2+1/a

[vale wigeon]

NILESH

HOW MANY TIMES DO WE HAVE TO TELL YOU

when you write fractions in plaintext

[alpine sable]

its -7

a=-7 then

[vale wigeon]

you need to use parentheses

[alpine sable]

just go

[vale wigeon]

i wouldn't be so sure of that

[alpine sable]

?

actually in both cases

its -7

[vale wigeon]

i think you may have messed up

[alpine sable]

?

[vale wigeon]

the equation reduces to a^2 + 5a + 1 = 0

-7 is not a solution of that by any means

[fervent grail]

:=|

[vale wigeon]

yes, imagine

if you want people to understand you

you have to use parentheses

[teal epoch]

You need to or else people may misunderstood you

[fervent grail]

Is it right ?

[vale wigeon]

no

[alpine sable]

that is very wrong

[vale wigeon]

^

that is very wrong and you need to relearn algebra

[teal epoch]

How did you do the third step it felt so wrong-

[alpine sable]

also this is wrong too

[fervent grail]

🥲

[alpine sable]

i don't think it reduces

[fervent grail]

Than how to do this ?

[alpine sable]

only the top is factorable

bottom has no real solutions

[vale wigeon]

jswatj i'm talking about the whole equation

[alpine sable]

ohhhh

sorry

[vale wigeon]

(a^2 - a + 1)/(a^2 + a + 1) = 3/2
multiply both sides by 2(a^2 + a + 1) to get 2(a^2 - a + 1) = 3(a^2 + a + 1)

[alpine sable]

Yeah yeah

thats right

[vale wigeon]

do algebraic cleanup to get to a^2 + 5a + 1 = 0

[alpine sable]

wait my answer is still right tho

it is -7

[fervent grail]

Is it right ?

[teal epoch]

,rotate

[alpine sable]

cause he's asking what is $a+\frac{1}{a}$

[ocean seal]

jswatj

[vale wigeon]

now it's okay nilesh

@alpine sable it's -5 tho

[fervent grail]

Okay thanks :)

[alpine sable]

oh

[vale wigeon]

continuing from what i had

a^2 + 1 = -5a
(a^2 + 1)/a = -5
a + 1/a = -5

[alpine sable]

ripppp

nvm

[alpine sable]

I am completely stuck in this problem
$0\leq\theta\leq 2\pi, 0\leq z \leq 3, 0\leq r \leq\sqrt{81-z^2}$

[ocean seal]

Fung

[alpine sable]

Not sure where I am going wrong

[restive finch]

Hello I have a strange question, but I wonder why it is not possible to calculate the logarithm of a negative base. If I have a negative base and

an odd exponent, then the result should be negative so for example:
$(-2)^{3}$

[ocean seal]

Florian

[alpine sable]

@restive finch Channel is currently occupied

[restive finch]

Okay, what shoud i do?

[alpine sable]

Find a question channel that is currently not under use

[restive finch]

Okay, sorry

I am new here

[alpine sable]

All good 😄

[restive finch]

Ist just fucking basic stuff

[alpine sable]

Make sure to check out #rules and #info

[restive finch]

Thanks

[alpine sable]

Going to repost my question above the spam

Not sure why this problem is being marked incorrectly by my hw system, but here are the coords I got: $0\leq\theta\leq 2\pi, 0\leq z \leq 3, 0\leq r \leq\sqrt{81-z^2}$

[ocean seal]

Fung

[alpine sable]

<@&286206848099549185>

Any guidance on where I might have gone wrong would be super helpful 😄

Thank you in advance

[final viper]

how can we start opening this

to get tan values

🤔

[topaz scaffold]

You know these identities?

[final viper]

yes

and i expanded as well

but still

¯_(ツ)_/¯

like after expanding if you take lcm

then lmao

[topaz scaffold]

Hmm

[final viper]

or wait

rhs is 0

what if i take the lcm then sent denominator to rhs

sus

numerator = 0

wait lemme try

[topaz scaffold]

Oki

[sand pecan]

Would it be possible to reverse engineer a modified geometric sum function to get the components it is made up of?
Like for example the typical sequence function is
a * (1-r^n)/(1-r) and the components for that would be a + ar + ar^2 + ... + ar^n
But what if I had a function which is
a * (1-r^(n/2))/(1-r)? if it's possible how would I go about attempting to reverse that and find the components? what are some strategies I can use to help with this

[vale wigeon]

no idea what you're talking about

if all you know is the value of $\frac{a(1 - r^{n/2})}{1 - r}$ and nothing else, then you cannot extract from it the values of $a$, $r$ or $n$ on their own

[ocean seal]

Ann

[sand pecan]

What I mean is you know how the sequence for a * (1-r^n)/(1-r) a + ar + ar^2 + ... + ar^n right,
So what I'm trying to do is find what a + b + c + d + g + ... x = a * (1-r^(n/2))/(1-r)

[vale wigeon]

...

a * (1-r^(n/2))/(1-r) + 0 + 0 + 0 + ... + 0

[gray isle]

maybe they want long division?

[sand pecan]

That is not what I am looking for

[vale wigeon]

you're not making yourself clear at all

[sand pecan]

Sorry I do not know the proper vocabulary for this stuff
Hmm, so first let's say I have the geometric sum function I'll call that s(n) and that is a * (1-r^n)/(1-r)
So we know that s(n) = a + ar + ar^2 + ... + ar^n

But in this hypothetical let's say I had no clue s(n) was equal to a + ar^n + ar^2 + ... + ar^n
I only know that it's equal to some sequence that happens n times
How can I go about determining what that repeating sequence is without knowing it beforehand?

[vale wigeon]

you can't

there are many sequences of n terms which add up to a given pre-determined value

[sand pecan]

I think I'm confused on what you mean by pre-determined value

Because n can be any positive integer, I feel that I need a specific generic sequence

[vale wigeon]

...oh wait

i think i just understood what you were trying to say

you have an unknown sequence $a_1, a_2, \dots$ and a sequence of sums $s_n$ defined as $$s_n = a_1 + a_2 + \dots + a_n$$ for every natural $n$, and you want to recover $a_n$ from $s_n$?

[ocean seal]

Ann

[sand pecan]

Yeah that's it

[vale wigeon]

ah then that's easy

$a_n = s_n - s_{n-1}$

[ocean seal]

Ann

[sand pecan]

Hmm well I'm not trying to recover the actual value of 'a' at 'n'

I guess I'm trying to recover the entire sequence so I don't need to use the function s(n) anymore
Instead if I wanted to get s(3) I'd be able to do a + ar + ar^2 + ar^3 because I figured out that the sequence is just ar^(ascending natural number)
But I didn't know that at first, I had to figure it out myself

[vale wigeon]

hghghrhgrhgrh

you asked for the sequence

and yet, when i gave you the formula to determine each term of the sequence,

suddenly it's not what you're looking for

[jagged imp]

Are you sure what Ann gave you isn't what you're looking for? this sounds exactly like what she gave you. if you have a_n, you have recovered the entire sequence.

Like if you recover a_n=ar^n from s_n=a * (1-r^n)/(1-r), then you can calculate s_n=a_1+a_2+a_3+...a_n=ar^1+ar^2+ar^3+....ar^n

[boreal frost]

bit of real easy maths but i haven't taken a maths class or done maths in ages... plus im garbage at googling, i can't find a calculator or formula to do this for me

lets say i want to increase 100 by 5%, 50 times... it would be 105, 110.15, 115.7625, etc. etc.

but i dont wanna just increase the answer by 5%, 50 times cus that takes forever

what's an easier way to do this?

[jagged imp]

,calc 100*1.05^50

[ocean seal]

Result:

1146.7399785754

[boreal frost]

so x*(percentage increase)^(however many times)

[jagged imp]

yes. The idea is that a 5% increase consists of multiplying by 1.05, so doing that 50 times consists of multiplying by 1.05 50 times thus multiplying by 1.05^50

[boreal frost]

❤️

[sand pecan]

I think I get what you mean now sneaky, I am trying to rearrange the equation to become just ar^n but get stuck because there's 2 exponents being added together I'm not sure what to do about

[jagged imp]

so first thing is that your formula is wrong here and it should be a * (1-r^(n+1))/(1-r)

but it will be the same process

sounds like you have something like $a_n=\frac{-ar^{n+1}+ar^n}{1-r}$ but with the indexes moved down 1 because you used that $1-r^n$ formula instead of $1-r^{n+1}$. change it to what you should have with the right formula, then factorise

[ocean seal]

Sneaky

[jagged imp]

specifically, factor as much as you can from the numerator and a 1-r will cancel leaving you with ar^n

if you need more guidance after correcting your formula or w/e post your work that far and I'll try to explain further

[sand pecan]

I agree that it's n+1 instead of n but do you know why half the images on google have 1-r^n

[jagged imp]

cant speak for the entirety of google but this one for instance is missing a term

should be an ar^n

[sand pecan]

oh I see

[jagged imp]

they're treating a as the first term

and we're treating it as the zeroth term

basically

well not basically its literally just that

[sand pecan]

That factors nicely

[jagged imp]

does that mean you're done?

[sand pecan]

nope

[jagged imp]

ok, show me where you're stuck then.

you cantake a picture or just write it in plaintext

[sand pecan]

Right so I'm trying to do it with the function s(n) = a * (1-r^(n/2 + 1))/(1-r)
and I get:
(a - ar^(n/2 + 1) - a + ar^(n/2))/(1-r)
(ar^(n/2)-ar^(n/2+1))/(1-r)
ar^(n/2)(1-r)/(1-r)
= ar^(n/2)
I went wrong somewhere

[jagged imp]

the error is in the first line. should be (a - ar^(n/2 + 1) - a + ar^((n-1)/2+1))/(1-r)

Which is ugly, should be $\frac{a-ar^{\frac{n}{2}+1}-a+ar^{\bold{\frac{n-1}{2}+1}}}{1-r}$

[ocean seal]

Sneaky

[jagged imp]

you didnt replace n with n-1 properly

[sand pecan]

ohh I gotcha I guess I got confused with the fraction

[jagged imp]

yeah i was confused why it was wrong when i first saw it too tbh

[sand pecan]

so for this one is it possible to do anything with those exponents other than factor out the +1?

[jagged imp]

Sure, you can do a very similar factorisation to the last one

if you want you can then factor 1-r on the denominator as (1-sqrt(r))(1+sqrt(r)) and one of those cancels after factoring the numerator

this is what it ends up being if you do all that

if you'd consider that simpler than what you've got go for it

i don't think you can do any "simpler" than that though

[jagged imp]

(i.e. factor out ar^(n+1)/2 from the numerator)

[night berry]

So there's an algorithm I need ideas for.

The problem: I'll try to not be too vague: Say there are 5 iPhone7's. 10 iPhone8's. They are under the category of iPhones, where we have 15 iPhones total. Similarly, we have other types of phones in different numbers which are further in the category of just phones. We may have laptops, USBs etc too. Kinda like sun in a solar system in a galaxy.

Now, say someone buys an iPhone7, so a specific iPhone7 is automatically assigned to the person. Now, if I want to PHYSICALLY access the phone from the inventory of electronic goods, what should be my approach/algorithm to do so?

If you have any mathematical algorithmic ideas pls ping

[random flax]

I have never learnt mathematical algorithms but I want to shoot a blind dart until one of the geniuses in this DC comes along

[random flax]

I give up. I was just going to say, select phone select iphone select iphone 7

[shrewd otter]

Mathematical algorithm?

This seems like a data structure.

Namely, something akin to a file system.

[night berry]

Hmm I'm pretty confused too

[quaint pond]

Can someone familiar with probability help me understand this?

(if the channel isn't taken already)

[vale sapphire]

I think I can do this

What in particular about this do you have trouble understanding?

[quaint pond]

the second point

[vale sapphire]

Right

give me a sec

[quaint pond]

and also saying A (_ S means that an event A has happened within S or what

(_ is the weird symbol

[vale sapphire]

$\subseteq$ ?

[ocean seal]

Syst3ms

[quaint pond]

yeah

^^^^^^^^^

[vale sapphire]

Ok, so this has to do with how we formalize probabilities

The objects in probabilities are really the same as basic basic set theory

Just with different names

In this case S is a set, but in probabilities it's a universe

if A ⊆ S, A is called an event

fancy probability word for subset

[quaint pond]

So A is a subset of S?

[vale sapphire]

Yup

As a side note, if you talk about it from an event standpoint

A ⊆ B means that event A implies event B

Let's do two coin throws

There are 4 possible outcomes : {(H,H),(H,T),(T,H),(T,T)}

This is your universe

The set of all possible outcomes

[quaint pond]

so how come A implies B?

like if we know that in a set A one is head

[vale sapphire]

take {(H,H)} ⊆ {(H,H),(H,T)}

all we're saying is that if we got two heads, in particular our first throw was a head

[quaint pond]

ohhh

yeah

[vale sapphire]

the event "get two heads" implies "the first throw was a head"

But it also implies "the second throw was a head"

[quaint pond]

yeah

right

[vale sapphire]

We say two events are disjoint if they can't both happen at once

In set theory terms, this just means their intersection is ∅

[quaint pond]

isn't that an impossible event

[vale sapphire]

Yup

For example, {(H,H),(H,T)} ∩ {(T,T),(T,H)} = ∅

You can't both have "the first throw is heads" and "the first throw is tails"

[quaint pond]

why not?

sorry if thats a stupid question

[vale sapphire]

the first throw only