help-23

[arctic stone]

where does +2 come from

[solid void]

You have $f(-2) = 6$

[flat frigate]

Filquichante

[solid void]

And $f(x) = mx+b$ with $m=-2$ (it's the slope)
So $f(-2) = {-2} \cdot {-2} + b$
The first $-2$ is the slope, and the second is $x$
Then you solve and find $b$

[flat frigate]

Filquichante

[solid void]

@arctic stone

[arctic stone]

I then solve for b

[solid void]

Yep

[arctic stone]

so

-2(-2)+b=-2

4/-2

-2

[solid void]

No, you have f(-2) = 6, not -2

-2*-2 + b = 6

So b = 2

[arctic stone]

Oh I see

I use y1

okay makes sense

[solid void]

YES !

[arctic stone]

thx fliguichante. idk how I forgot it so quickly

.close

[safe radish]

Channel closed

Closed by @arctic stone

Use .reopen if this was a mistake.

[safe radish]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[wispy vortex]

I am confused about coplanar and collinear

my teacher said there needs to be 3 points on a line

[wet pecan]

Coplanar means 'on the same plane'. Collinear means 'on the same line'

[wispy vortex]

for them to be collinear

so for uestion c

she asked

are K and F collinear

but those are only 2 points

so how is it 'yes'

[wet pecan]

Any two points are collinear, because two points define a line

[wispy vortex]

yeah that's what I thought

but my teacher says there needs to be 3??

[wet pecan]

Yeah it's trivially true. Typically for anything interesting you would be talking about 3 or more points

[wispy vortex]

but it is really 2 points on a line needed for them to be considered colinear

[wet pecan]

That's not entirely accurate

[wispy vortex]

ok

[wet pecan]

I guess you can define collinear to be with 3 or more points

[wispy vortex]

why not 2 tho

[wet pecan]

But if you do that then there's no point in asking whether two points are

[wispy vortex]

ok

[wet pecan]

In general, on a fundamental level you shouldn't care about the number of points. Just remember it means 'on the same line'

[wispy vortex]

ok

[wispy vortex]

also for the 2nd part of c

she asks if they are coplanar

and she told us that you need 3 points to make a plane

but K and F are only 2

so how is it "yes, not on plane j"

[wet pecan]

You need 3 points to fix a plane

That's true

It's just that with less than 3 points - in this case two - you have an infinite number of planes that contain those two points. So then yeah you can say they lie in the same plane, but that doesn't give you any information is all.

[wispy vortex]

?

[wet pecan]

The 'not on plane j' just means that if since F is not on plane j, any plane containing F (and K) would not be plane j

[wet pecan]

Which part are you confused about

[wispy vortex]

idk

[wet pecan]

So it makes sense?

[wispy vortex]

no

[wet pecan]

...so what doesn't make sense

[wispy vortex]

idk

when I did geometry on khan academy everything was making sense

but this teacher is giving bad explanations and bad examples for everything

so now it is confusing me

s

so

K and F

are coplanar

if I add another point to the euation

like

if I added

point e

plane

because I need 3 points for them to be coplanar

[wet pecan]

Mhm,

[wispy vortex]

but I only have 2 here

in D it asks

are E, B, and F coplanar

and they are

because those are 3 points

[wet pecan]

Ok so you're right, if you consider K, F, E, they are three points, defining a unique plane, so they are coplanar

[wispy vortex]

but

E B and F

are not coplanar on J

[wet pecan]

Yes

[wispy vortex]

because not all of them are inside the plane of J

[wet pecan]

Exactly

[wet pecan]

Yeah seems like you got it

[wispy vortex]

if we moved E and F into plane J

would they be coplanar on the plane J

[wet pecan]

Yes

[wispy vortex]

but

E B and F

are still coplanar

because they are 3 points

that can make their own plane together?

3 sided plane

[wet pecan]

Technically you can also say like 'point A is collinear.' or 'point A is coplanar'. It just doesn't tell you anything useful. That's why sometimes people might define it to require a certain number of points

Does that make sense

[wispy vortex]

point A is collinear and coplanar by itself?

when I think of collinear I think of 2 points are needed

when I think of coplanar I think of 3 points

[wet pecan]

Yeah. You said 'E, K, F' are coplanar on j if they all lie on j'. This is true. What's also true is that since you need 3 points to define a plane, that E, K, F are always coplanar (if they are also collinear, the plane is not unique).

[wet pecan]

Hence why you might want to define it to require a certain number of points.

[wispy vortex]

so there is only one point really necessary

for that point to be called those

[wet pecan]

Yeah. You normally wouldn't see people saying that though because it's silly. It's just because your teacher is giving these examples that you can think of it this way if it helps

[wispy vortex]

but why is only one point needed for that point to be called collinear or coplanar

I thought co meant more than one

[wet pecan]

These are really basic definitions so you don't need to worry that much about how precisely they are defined. Just remember 'one the same plane/line' and you're good to go. Is one point 'on the same line'? Well sure, it's on any line that passes through it. Are two points? Sure, because you can always draw a line through two points. But with three points, they may or may not all lie on the same line. So that's why usually you only talk about collinearity when there are at least three points

[wispy vortex]

I still don't get why it is 3 points and not just 2

all you need is 2

for a line to be made

you don't need 3

you need 3 for a plane yes

because your making like a triangle

with 3

with 4 points you can make alot more stuff

like a rectangle

but why doyou need 3 for collinear

[wet pecan]

Yeah 4 points define a space

The analogy you're looking for is tetrahedron

[wispy vortex]

can't 3 points define a space

triangle

triangle plane

what if I made a triangle table

the top of the table would be a triangular plane

[wet pecan]

I feel like we're not getting anywhere

[wispy vortex]

you only need 2 for collinear

to make a line

[wet pecan]

And I kind of understand; I have that feeling too sometimes but give it a few days and you'll look back and think why was I ever confused

[wispy vortex]

I don't have days

[wet pecan]

Yea

[wispy vortex]

this assignment is due in hours

[wet pecan]

But you can do the assignment

[wispy vortex]

so you don't need 3 for collinear

only 2

[wet pecan]

Wherever you see 'collinear', replace it with 'on one line'.

[wispy vortex]

ok

so here is a line

no collinear yet

now there is point A

so collinear but nobody really cares about it because it sounds goofy

now

there is collinear

more collinear

[wet pecan]

Does that make sense now? You can ask, are three points collinear? That means, 'are three points on one line'? And you look at the given three points. Can you draw a line through them? Maybe, maybe not. If you can, then they are on one line. So they would be collinear.

[wet pecan]

Yup

That's all correct

[wispy vortex]

still collinear

so here

they are collinear

because you can draw a line through them

but they are collinear on plane J

because F is outsid

outside

[wet pecan]

Yeah

So you've got it

[wispy vortex]

but in C

the pt 2

asks

if

K and F

are coplanar

and coplanar needs 3 points

on the same plane

[wet pecan]

If it's really bugging you you should tell your teacher the question is wrong

[wispy vortex]

this teacher has retired

[wet pecan]

Because they've defined it in one way and asked you a question that can't be answered with that definition

[wispy vortex]

k

[wispy vortex]

so what is your definition of coplanar

[wet pecan]

'contained in one plane'. that's it

[wispy vortex]

so any amount of points?

one

2

3

4

5

6

[wet pecan]

sure

[wispy vortex]

can be coplanar

ok

[wet pecan]

why not

[wispy vortex]

but

[wet pecan]

Just that anything less than three is silly

[wispy vortex]

K and F

[wet pecan]

But you can still say it

[wispy vortex]

are not contained

in plane J

[wet pecan]

That's right.

[wispy vortex]

so why did my teacher still say "Yes"

[wet pecan]

But there exists another plane that contains them

[wispy vortex]

imaginary plane?

cause I don't see any other planes on that image

do I just make up a plane

[wet pecan]

Yes.

[wet pecan]

Yes.

[wispy vortex]

ok I made a plane B

out of nowhere

it magically appeared

because I wanted it to

[wet pecan]

shruggie

sure

[wispy vortex]

hm "are they coplanar?"

now they are because I just want them to be

plane C

poof

I can just do that?

[wet pecan]

yeah

[wispy vortex]

two random points floating in space

[wet pecan]

yeah

[wispy vortex]

and I can say they are coplanar

[wet pecan]

yeah

[wispy vortex]

because I can just make up a plane

ok

cool

[wet pecan]

But notice you cannot do that if the four points are e.g. vertices of a triangular based pyramid (tetrahedron)

[wispy vortex]

and E B an F are coplanar because I can just put a box around them and say they are in a plane

[wispy vortex]

hm

[wet pecan]

Yes

[wet pecan]

(because you can't put a 3D thing inside a 2D thing)

[wispy vortex]

oh

ok

[wet pecan]

just like you can't put a triangle in a line

[wispy vortex]

line is one-d

[wet pecan]

(but you can put it in a plane)

[wispy vortex]

triangle plane is 2d

[wet pecan]

Yeah. so you can't put 2D inside 1D

[wet pecan]

Pretty easy right

[wispy vortex]

ye

thank you alot

[wet pecan]

lol

np

[wispy vortex]

.close

[safe radish]

Channel closed

Closed by @wispy vortex

Use .reopen if this was a mistake.

[safe radish]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[arctic stone]

,rotate

[arctic stone]

kinda feel like I am missing a step

[gusty trench]

it looks right so far

i agree with everything up to 2x (3ln(10) - 5ln(3)) = 5ln(10) - 6ln(3)

[arctic stone]

idk how to close

[rough storm]

how would you get x by itself? :3

if you have ax=b

what would you do to solve to x

[arctic stone]

looking at mathaway that's the final answer

[rough storm]

do you see how you can get that from this?

[arctic stone]

idk am I suppose to flip it on the divisor

then change the + - signs

?

[rough storm]

well first answer this

how would you solve for x

[arctic stone]

idk what you mean

[rough storm]

if you have 2x=5 how would you solve for x

[arctic stone]

5/2

[rough storm]

what did you do

to solve

[arctic stone]

divide

[rough storm]

yes

[arctic stone]

okay I see so divide the left side to shift to the right

[rough storm]

mhm

[arctic stone]

,rotate

[arctic stone]

it works!

[safe radish]

@arctic stone Has your question been resolved?

Channel closed

Closed by @arctic stone

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[thick stag]

when it comes to existential generalization and similar topics im having a hard time trying to figure out what value to assign to the variable a.

for example this problem, i don't understand the logic behind assigning 3-x/2 to a without looking at the solution

[thick stag]

please ping me

[safe radish]

@thick stag Has your question been resolved?

[empty gyro]

The choice is rather odd. You could have proven the same thing with choosing x=a

Heck, you could choose a=2

oh I'm dumb

a<1, not 1<a

[thick stag]

yeah

im just having a hard time understanding the logic

[empty gyro]

You need an $a$ that satisfies $2-x<a<1$. So any line equation that fits in this region will work

[flat frigate]

SWR

[empty gyro]

So you want an equation of the form $a=mx+b$ where $a(1)=1$ and $-1<m<0$

[flat frigate]

SWR

[empty gyro]

So let's try, as an example, $m=-\frac{1}{2}$, since it's right in the middle.

[flat frigate]

SWR

[thick stag]

sorry im back

[empty gyro]

We solve for $b$ in $1=-\frac{1}{2}(1)+b$ and get $b=\frac{3}{2}$

[flat frigate]

SWR

[empty gyro]

$m=-\frac{1}{2}$ is just an arbitrary choice. Any $m$ so long as it is between -1 and 0 will work.

[flat frigate]

SWR

[thick stag]

right

that makes sense

i also think that

if i isolate

$2-x<a<1$

[flat frigate]

3m41r

[thick stag]

a here

i can just choose the midpoint no

[empty gyro]

Looks like it, yea

[thick stag]

someting like this

ty

.close

[safe radish]

Channel closed

Closed by @thick stag

Use .reopen if this was a mistake.

[safe radish]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[snow zealot]

Does anyone know how to graph this?: lim as x approaches -∞ f(x) = 0
lim as x approaches -4 from the left f(x) = 3
lim as x approaches -4 from the right f(x) = 1
f(-4) = 1
lim as x approaches 1 from the left f(x) = -1
lim as x approaches 1 from the right f(x) = -1
f(1)=5
lim as x approaches 3 from the left f(x) = 2
lim as x approaches 3 from the right f(x) = 3
f(3)=2
lim as x approaches ∞ f(x)=∞

[snow zealot]

putting the points f(3)=2, f(1)=5, f(-4) = 1 is all i know how to do on a graph.

.close :/

[safe radish]

Channel closed

Closed by @snow zealot

Use .reopen if this was a mistake.

[safe radish]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

Channel closed

Channel closed due to the original message being deleted.
If you did not intend to do this, please open a new help channel,
as this action is irreversible, and this channel may abruptly lock.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[austere hinge]

Hi, does anyone have any resources on finite groups of lie type (and the conditions under which they are simple)? I've found some, but I really don't understand what a group of lie type is (and the resources I've found for that aren't helping).

[safe radish]

@austere hinge Has your question been resolved?

[safe radish]

@austere hinge Has your question been resolved?

[visual kraken]

#book-recommendations

[safe radish]

@austere hinge Has your question been resolved?

Channel closed

Closed by @austere hinge

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

Please don't occupy multiple help channels.

Channel closed

Closed by @limber heath

Use .reopen if this was a mistake.

[safe radish]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[chilly seal]

Hello.

[chilly seal]

I was wondering about a probability issue.

In 2 roles of a six-sided dice, there are 36 possibilities.
If we want to predict the sum of the roll, 7 has the highest chance of being correct. If we want to predict it 20 times in a row however, 7 would possibly only appear 3 times. Is it possible to predict the sum of the roll MORE times? Thanks : D

[gusty trench]

is there something i'm misunderstanding about the problem setup here? if you rolled a dice 20 times in a row, the smallest each roll could be is 1 so the smallest sum you could possibly get is 20. how is it possible that 7 would appear?

[chilly seal]

not the total sum of 20 rolls

in a throw of 2 die

[mortal sandal]

if i understand what you're asking the answer is no, aside from getting lucky

[chilly seal]

i see

i thought there was some way to increase the chances

[mortal sandal]

well if you guess 2 every time, you're definitely gonna be less successful

[chilly seal]

it's just frustrating that it's not possible.
(there's a game i play with this system)

[hard crest]

the events are independent - they don't influence each other - so the best you can do is just guess 7 every time

[mortal sandal]

but on no particular guess does it make sense to guess anything but 7

[chilly seal]

yeah. that would lose me more coin

[chilly seal]

I figured. Thanks !

[hard crest]

you have a 1/6^20 chance of predicting it 20 times in a row which

,calc 6^20

[flat frigate]

Result:

3.656158440063e+15

[chilly seal]

it doesn't. I was just wondering if there's some kind of manipulation I don't know

yeah

so 7 would probably come 3 times

[mortal sandal]

on average yeah 3.333...

[chilly seal]

i mean, if there was a way to predict it more times

[mortal sandal]

if you want to predict the number of times a sum shows up then you're probably best off going with 2

less variance there

[chilly seal]

yeah

last time i played the game

7 came twice

but the way the game works

I lose more coin than i gain

if 7 comes only 2 times

[mortal sandal]

oof

[chilly seal]

Thanks for your time. : D

[mortal sandal]

np

[chilly seal]

Have a nice day

[mortal sandal]

you too

[chilly seal]

.close

[safe radish]

Channel closed

Closed by @chilly seal

Use .reopen if this was a mistake.

[safe radish]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[grave epoch]

i got this answer right but i guessed it, when i solved it, i got $x^2 - 2x +1$ can i get some help pls? <@&286206848099549185>

[flat frigate]

Bahnies

[safe radish]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[thin narwhal]

show your work

[grave epoch]

ok gimme a sec

here

[thin narwhal]

where did you get (x-1)^2

[grave epoch]

it's in the turning point formula

[thin narwhal]

ah vertex formula

[thin narwhal]

can you show the whole question here

[grave epoch]

alr

[thin narwhal]

thats the same image

well ok

whats the vertex of the parabola

[grave epoch]

1, -4

[thin narwhal]

ok

plug that into the formula you know

[grave epoch]

$(x-1)^2 - 4$ mb i forgo to subtract the four but that still doesn't give me my answer

[flat frigate]

Bahnies

[thin narwhal]

the thing you learned is

$y = a(x-h)^2 + k$

[flat frigate]

insult

[thin narwhal]

right

[grave epoch]

well idk what a is equal to

[thin narwhal]

where (h,k) is the turning point or vertex

[thin narwhal]

well you dont just go around assuming what it is

so my suggestion is find one point which the parabola passes through

that isnt the vertex/turning point

[thin narwhal]

.

based on the graph

[grave epoch]

the y intercept it 0,-2

[thin narwhal]

thats good

so you know that your equation should have (0,-2) as a solution

correct

meaning when x = 0, y = -2

so when you substitute it in

[grave epoch]

ohh ok

[thin narwhal]

$-2 = a(0-1)^2-4$

now solve for a

ill double check wait

[flat frigate]

insult

[thin narwhal]

wrong sign sorry

@grave epoch

now solve for a

[grave epoch]

is a equal to 2?

[thin narwhal]

yes

now you have the equation in vertex/turning point form

[grave epoch]

ok..

[thin narwhal]

$y = 2(x-1)^2 - 4$

[flat frigate]

insult

[thin narwhal]

now all you have to do is turn it into general/standard form

[grave epoch]

ohh i understand know

thx

.close

[safe radish]

Channel closed

Closed by @grave epoch

Use .reopen if this was a mistake.

[safe radish]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[devout maple]

Context: Finding a global minimum of an n-dimensional non-convex function using gradient descent.
From what I gather, gradient descent is not guaranteed to find a global minimum.
Perhaps I'm being a bit naive/stupid here:
Could we not find a global minimum by finding a local minimum and seeing if there's more than one intersection on a derivative of the minimum and restarting the gradient descent there, iteratively?

[devout maple]

While I really like math, I'm still very early in my math adventures to be gentle

by derivative I mean:
in 2D: a line
in 3D: a plane
...

please ping / ping reply when replying

wait this might blow up in complexity, wouldn't it.

[wet pecan]

I assume you mean by 'intersection on a derivative' that you want to try to solve the equation to see if the local minimum is non strict

In theory you can do this but the point of gradient descent is to avoid solving anything

[devout maple]

What's non-strict?
again, pretty new

[wet pecan]

Oh just non unique

[devout maple]

ah

Ya so we're trying to find the global minimum of an n-variable function

I'm messing around with machine learning hence I'm researching this

[wet pecan]

Mhm

[devout maple]

Maybe I'm misunderstanding the purpose of gradient descent

[wet pecan]

In general I think if your function is non convex it's not a great situation. Have you heard of duality?

[devout maple]

nope

[wet pecan]

You can look into that. The objective function of what's called a 'dual problem' is always concave iirc so you there are good properties there

[devout maple]

Thanks, I'll research a bit more and come back if I have more questions. Cheers

.close

[safe radish]

Channel closed

Closed by @devout maple

Use .reopen if this was a mistake.

[safe radish]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[mental echo]

Hello! Is there a certain method to easily find the primitive of a function, like an algorithm to follow? For example how do I find the primitive function of f in this case?

[haughty abyss]

helo

[thorn meteor]

nah u just can do
f(x) = 1 + (x-2)/x^2 + 1

and after that u can find the derivative with derivative of the quotient

i know that much, if there's another way to solve it, i didn't achieve it yet

[mental echo]

But how does it help me find the primitive?

Let F:R->R be the primitive, then F(x) = x +1/2ln(x^2+1) + another term

?

The primitive should look something like that, right?

Do let me know if I am wrong

[haughty sigil]

the algorithm is pretty much, what derives to this function

ie inverse of the derivation rules

for example: f’ = 1 what is f

well its generally
f’ = (x + C)’ = x’ + c’ = 1

where C is any constant

so just split the function into 3, one per each addition and then perform and inverse of either a log(x)’ or (f/g)’

[safe radish]

@mental echo Has your question been resolved?

Channel closed

Closed by @mental echo

Use .reopen if this was a mistake.

[mental echo]

Aight, I'll give it a few more tries

Thank you!

.close

[safe radish]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[uneven knoll]

$\int\frac{\cos x-\sin x}{\cos x+\sin x}\dd x=\int\frac{\cos^2x-\sin^2x}{1+2\sin x\cos x}\dd x=\int\frac{\cos(2x)}{1+\sin(2x)}\dd x$

[flat frigate]

pirateking0723

[uneven knoll]

now let $u=1+\sin(2x)$ then $\frac{\dd u}{2}=\cos(2x)\dd x\implies\int\frac{\cos(2x)}{1+\sin(2x)}\dd x=\frac{1}{2}\int\frac{1}{u}\dd u=\frac{1}{2}ln|u|+C=\frac{1}{2}ln|1+\sin(2x)|+C$

[vague slate]

Ok so

[flat frigate]

pirateking0723

[uneven knoll]

is this correct

[vague slate]

Yes

[uneven knoll]

how to convert into the RHS of this

[vague slate]

You can write 1+sin2x as (sinx+cosx)^2

[uneven knoll]

ah yes i am so stupid bro

i just used it some steps ago

[vague slate]

Yeah

[uneven knoll]

things are clear now

[vague slate]

Great

[uneven knoll]

tysm

.close

[safe radish]

Channel closed

Closed by @uneven knoll

Use .reopen if this was a mistake.

[vague slate]

Wlc

[uneven knoll]

have a great day/night

[vague slate]

Same to u

[safe radish]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[hardy meadow]

In the Venn diagram, shade the region (A∩B)U(C'∩ A)

[vague slate]

What is the problem...?

[hardy meadow]

I wrote it

[vague slate]

I mean where are you stuck at

[hardy meadow]

I'm not really sure how

A∩B would be A and B which is the intersection between A and B?

But I'm not too sure what U means

[vague slate]

It means union

Like A union B = Region of A and Region of B united

[hardy meadow]

So all of A and all of B?

[vague slate]

Yes

[hardy meadow]

So then C' is C not so that's everything that's not C?

[vague slate]

Yes

[hardy meadow]

Ohh

[vague slate]

Hmmm yes

[hardy meadow]

Let me try

I got it

thanks

.close

[safe radish]

Channel closed

Closed by @hardy meadow

Use .reopen if this was a mistake.

[safe radish]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[magic tusk]

how would we graph it

[glacial sonnet]

do you know what f(x) = |x| looks like

[magic tusk]

uh

like a V shape

[glacial sonnet]

yes

then each of those functions is a trasformation of |x|

[magic tusk]

ok yeah makes sense

so |x-4| --> V starts at x = 4

|x-4| --> V starts at x = 4 and y = 1 --> (4, 1)

[glacial sonnet]

that is the idea yes

[magic tusk]

thank you

.close

[safe radish]

Channel closed

Closed by @magic tusk

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[magic tusk]

I just closed this channel but i need help

[magic tusk]

😭

how would I do (7a)

so first i'd need to sketch x^2-9 as it usually is sketched

so a parabola

and then i'd need to flip it in a way if that makes sense

[lean otter]

out of curiosity, is this the Haese AA HL textbook?

[lean otter]

what happens to f(x) when we compose it with |x|

[magic tusk]

cambridge

also sorry i figgured it out

[lean otter]

you're an IB candidate?

[lean otter]

nice!

[magic tusk]

never heard of what that is

how do i do this one

I can solve for x by letting y = 0

right?

$0 = 3^x - 3$

[flat frigate]

pixel

[magic tusk]

ping when reply pls

[lean otter]

yh

[magic tusk]

how would i solve for y

[lean otter]

let x = 0

[magic tusk]

y=3^0 - 3

y = -2?

that doesnt seem right tho

how would i draw it

[lean otter]

y = |-2|

[magic tusk]

hmmmmmmm

y = 2

interesting

how would i draw

tho

[safe radish]

@magic tusk Has your question been resolved?

Channel closed

Closed by @magic tusk

Use .reopen if this was a mistake.

[safe radish]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[lean otter]

divide by 2 then

what was your method ?

[vale oriole]

how do u think this solves it?

[lean otter]

arent all wins/loss alike ?

[vale oriole]

this method implies that after the two wins you are picking either 3 losses in a row then 3 wins, or 3 wins in a row then 3 losses

right lets focus on 6C3 for the moment

we have 3 losses and 3 wins to place

[topaz tree]

WWWLLL
WWLWLL
WWLLWL
WWLLLW
WLWWLL
WLWLWL
WLWLLW
WLLWWL
WLLWLW
WLLLWW
LWWWLL
LWWLWL
LWWLLW
LWLWWL
LWLWLW
LWLLWW
LLWWWL
LLWWLW
LLWLWW
LLLWWW

[vale oriole]

lets say from the 6 we pick 3 wins and place them like this WWW---

then from the remaining 3 we can place them as so which is 3C3

[topaz tree]

i guess you're right there are 20 ways to arrange 3 wins and 3 losses

[vale oriole]

but you could start with 6 and choose 3 losses instead like this ---LLL

then from the remaining 3 wins we can place them in the open spots

but notice how those 2 situations are the exact same

so we have to divide by 2 to account for the double up in situations

@solemn flume

[topaz tree]

yo there are 20 ways

i wrote them

[topaz tree]

no those two are completely different scenarios

[vale oriole]

they are the same scenario

they both end up as WWWLLL

[topaz tree]

yes but they don't double up cuz we only considered starting with WWW

we don't consider LLL

[vale oriole]

6C3 means to have 6 spots and choose 3 of them

[topaz tree]

yes

and there are 3 wins to fill 6 spots and 3 losses to fill remaining 3

[vale oriole]

I know its 20 now, im missing a 2C1

but still

my point stands

[topaz tree]

what is your point?

[vale oriole]

they are the same scenarioi

u could start with 3 losses and place them at the end

or start with 3 wins and place them at the start

and they end up as WWWLLL

[topaz tree]

yes? and then?

[vale oriole]

the reason im wrong is because i omitted a 2C1

[topaz tree]

why 2C1?

[vale oriole]

the correct answer would be 6C3*2C1 / 2

2 options win or loss

choose 1

so it would be choose 3 out of the 6 slots and then choose either a win or loss

[topaz tree]

ah yes, but her approach is correct as well

[vale oriole]

yes it was correct

[half rune]

yes

It often happens that you get stuck or confused with tasks, exercises or math problems because the answer they give you is wrong (be it a book, teacher, etc.), it is not necessarily that you have done it completely wrong.

[safe radish]

Channel closed

Closed by @solemn flume

Use .reopen if this was a mistake.

[safe radish]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[burnt grail]

how do i solve c (i)?

[burnt grail]

i cant find any value to sub in

[split fulcrum]

dr du moment

[burnt grail]

ikr

[split fulcrum]

is this y11

[burnt grail]

no

yr10

[rough storm]

do you know what a triple zero mean?

[burnt grail]

triple roots

[rough storm]

yes

[burnt grail]

or 3 solutions

[rough storm]

no

it means it has a root with multiplicity 3

notice the usage of 'a'

[burnt grail]

alr

[rough storm]

like how you would say (x-2)^2 has a double zero at x=2

[burnt grail]

it has a single root of muplicity 2?

[rough storm]

yes

same thing

so it means P(x) has a root of multiplicity 3

[burnt grail]

but how

i need to prove it

[rough storm]

whats the form of a polynomial with a root at x=a with multiplicity 3

[burnt grail]

uh

im not sure

[rough storm]

do you agree (x-2)^2 has a double zero at x=2

a repeated root

[burnt grail]

shouildn;'t it only have one

cus that equals 0

[burnt grail]

oh alr

then yes

[rough storm]

so

what would be a triple root? what would be the form of a factor that is a triple root

at x=a

[burnt grail]

x^4-x^3-x^2-x?

[rough storm]

don't expand it or do anything

if (x-1)^2 has a double(2) root at x=1

what would be a polynomial with a triple root at x=a

[burnt grail]

(x-1)^3

[rough storm]

this would be a triple root at x=1

but we don't know what the triple root is

so you can make it a variable

1->a

(x-a)^3

[burnt grail]

ohhh

[rough storm]

but you know P(x) is also a quadratic

and what can you multiply a cubic by to get a quadratic

[burnt grail]

x

[rough storm]

not quite

if you multiply just by x, you are going to assume it has a 0 at x=0

so what would be something more general to multiply by

if we have a linear factor that has a 0 at b

[burnt grail]

ax+b?

[rough storm]

what would it look like

[rough storm]

thats good!

but we can assume a=1 in THIS case, because the leading coefficient of P(x) is 1

but not for (x-a)^3

we can make it (x+a)^3 just to be easier

so

putting it all together

what would we get

[burnt grail]

so would i expand in this case

[rough storm]

you would expand (x+a)^3 (x+b) yes

and then match coefficients

[burnt grail]

wait wouldnt i have to fully expand (x+a)^3?

[rough storm]

yes

[burnt grail]

oh

[rough storm]

you can use pascals to do this

[burnt grail]

idk cubic for this

i can only expand (x+a)^2

[rough storm]

well you can expand (x+a)^2

then multiply (x+a)^2 out by x+a

[burnt grail]

oh yea

[rough storm]

then multiply that by (x+b)

[burnt grail]

i swear i did something wrong

i got such a messy reuslt

[rough storm]

well

you can combine all the terms with an x^3,x^2,x and constants together

so you'll have x^4+(constants n stuff)x^3+(even more stuff)x^2+(stuff)x+(other stuff)

[burnt grail]

bruh thats so painful to do

[rough storm]

$x^{4}+x^{3}b+3ax^{3}+3ax^{2}b+3x^{2}a^{2}+3xa^{2}b+xa^{3}+a^{3}b=$

did you get something like this?

[burnt grail]

hold on

yes

but cant we combine the last two expressions?

[flat frigate]

🫎 Chmoosey 🫎

[rough storm]

$x^{4}+(3a+b)x^{3}+(3ab+3a^{2})x^{2}+(3a^{2}b+a^{3})x+a^{3}b$

[flat frigate]

🫎 Chmoosey 🫎

[rough storm]

?

[burnt grail]

yep

[burnt grail]

cus they both have a^3

[rough storm]

we're trying to get it to match the polynomial P(x)

because P(x) has a triple zero, and a linear factor

[burnt grail]

oh yea

fair enough

[rough storm]

so what does this tell you about 3a+b

and

3ab+3a^2

you can use any two of these equations honestly

[burnt grail]

3a+b = -3

[rough storm]

mhm

[burnt grail]

soooo

[rough storm]

actually

lets use 3a+b and a^3b

[burnt grail]

alr

so 3a + b = -3

[rough storm]

so 3a+b=-3 and a^3b=-6

[burnt grail]

and a^3b = -6

[rough storm]

mhm

[burnt grail]

yep

[rough storm]

you can solve for either a or b in first equation

then sub it in for a or b in the second equation :)

hmm

[burnt grail]

im losing brain cells

by substituting b

i got

a^3(1+a)=-2

[rough storm]

gra, it might be easier just to use one of the other equations

[burnt grail]

a=1

it's one of the solutions

roots*

a=1

wait no

it's not

[rough storm]

i would use 3a+b=-3 and 3ab+3a^2=-15

more mangeable

all you have to do is solve a quadratic

ah wait

[burnt grail]

(2a+5)

(a-1)

wait so a could equal 1

:o

and -5/2

[rough storm]

wait

note if a=1

(x+1)^3

[rough storm]

yes

and you can use the other equations to solve for b if you know a=1

[burnt grail]

so b = -6

[rough storm]

so putting it together

what does P(x)=

[burnt grail]

it should give the original question

wtvr p(x) equals to there

im not bothered writing it all out

[rough storm]

we had

(x+a)^3(x+b)

and we matched coefficients so this should =P(x)

[burnt grail]

so how should i start the proof

[rough storm]

so if $P(x)=x^{4}-3x^{3}-15x^{2}-17x-6$

[burnt grail]

cus everything i've done in on scrap paper

[flat frigate]

🫎 Chmoosey 🫎

[rough storm]

P(x) also equals

[rough storm]

I would just say something along the lines, if P(x) has a triple root, it must also have a linear factor, as it is quartic polynomial

[rough storm]

and since the leading coefficient is 1, P(x) can be written like

let us see if there exists a and b such that P(x) is equal to this, then you go through your calculations

[burnt grail]

we learn calculus in a few months

so yea

but polynomials suck

especially graphs

[rough storm]

do you see how the process we went through proves theres a triple zero though?

:3

[burnt grail]

well we broke it down

into (x+a)^3 (x+b)

since a = 1

i mean

since it has a coeff of 1

and then we expanded it

then by using simultaneous equations and comparing to the original quesiton

we are able to find the values of a and b

such that it equals p(x)

[rough storm]

so we can write P(x)=(original polynomial) and as the factorized version

which has a triple zero

[burnt grail]

yes

[rough storm]

so P(x) must have a triple zero

:)

[burnt grail]

wait what are the zeros for the expanded equatrion

the factorised one

a=1, b=-6 and a = -5/2?

[rough storm]

well only one of them will make sense

a=-5/2 doesn't make sense for 3a+b=-3

so a=1

[safe radish]

@burnt grail Has your question been resolved?

Channel closed

Closed by @burnt grail

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[mossy jackal]

Let ABCD be a square with side 16 cm and M, N the midpoints of sides AB, BC. If P belongs to CD, PC over PD is equal to 1 over 3 and NQ is the bisector of the angle MNap, Q belongs to MP demonstrate
That Amdf over Apcbm is smaller than 4×MQ over 5×QP

[mossy jackal]

kinda dont know what to do

[safe radish]

@mossy jackal Has your question been resolved?

[safe radish]

@mossy jackal Has your question been resolved?

Channel closed

Closed by @mossy jackal

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[lone arch]

So:

[lean otter]

ok

[safe radish]

@lone arch Has your question been resolved?

[safe radish]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[safe radish]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[dapper grail]

I feel like im doing this completely wrong

this is differential equations where I have to prove that the first equation is a solution to the second one

[rapid flare]

Would probably be easier to calculate the second derivative of the first equation and substitute into the second

[dapper grail]

but wouldnt the second derivative of the first equation be 0??

[hoary wind]

Product rule

[dapper grail]

😭 i dont see a y variable though

unless y' is d/dx 💀 but i dont think so

[lament gyro]

can i get a answer

[hoary wind]

!occupied

[safe radish]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[hoary wind]

y' and dy/dx mean the same thing

In either case the derivative of (4-x)e^-x is not 0

[dapper grail]

but it's with respect to y not x right that's my issue right now 😭

do d/dx and dy/dx mean the same thing is that it?

[hoary wind]

dy/dx means the derivative of y with respect to x

d/dx (something) means differentiate that something with respect to x

[dapper grail]

me when im stupid

ok ill resolve rq

@hoary wind

i think i got it?

2e^-x = 2e^-x

therefore solution :D

[hoary wind]

Looks reasonable to me I generally refrain from equating both sides and ending up with 1=1 or whatever when doing proofs as you run the risk of irreversible operations

I kept things split up and it worked quite nicely as you just end up with y being in the derivative

Which is exactly what you got given at the beginning so it's just QED job done

(forgive the handwriting, I did not put enough effort into the aesthetic of my midnight scrawling)

[dapper grail]

thank u 🙏 i might ask another quesiton later if im confused

.close

[safe radish]

Channel closed

Closed by @dapper grail

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[mellow veldt]

quick question is this identity correct ? $\cos(\theta) + \cos(2\theta) + \cos(3\theta) = cos(2\theta) + 2\sin(2\theta)\sin(\theta)+1$.

[flat frigate]

Coolgamer2000(xbox)

[eternal carbon]

no

[river field]

Does not appear to be the case

[light shoal]

plug in theta = 0, it's not true even in that case

[eternal carbon]

^ graphing not necessary

[mellow veldt]

thank you, lol i had a brain fart

[river field]

Graphing very helpful to check things like these

[eternal carbon]

not.. really? in cases where it's clear equality is false

all i said is that it wasn't necessary

[river field]

I didn't even have to look at the expression, just typed it over and saw the graphs don't overlap

best way to check

[eternal carbon]

there's no "best"

[eternal carbon]

but in this case that's not even true because of how absurdly slow it is

it's like saying you need to graph to check if x is equal to x^2

[river field]

took less than 10 seconds

[eternal carbon]

takes less than half a second to plug in x = 0 bro

your point is non-sensical

[river field]

😛

[safe radish]

@mellow veldt Has your question been resolved?

[safe radish]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[tepid crypt]

I was looking for an S-curve function to use in my robotics project. I am using stepper motors and wanted to add acceleration to it. My small brained self was only about to remember the sine function. So I had a look at it and at first I wasn't really getting anywhere. But then, by accident I squared the sine and it's perfect. But I don't really understand what I did. What does it mean if I square the sine.

[tepid crypt]

Here's the acceleration profile I came up with by taking the sin^2 of colum A.

What are the implications. When I overlay both functions I can kinda make sense. But I feel like there's a deeper meaning to it. It's telling me something that I don't understand. Some sort of relationship.

Ok, my last message sounds like wizard talk

[quiet plume]

I mean they are similar. You want only the S-behaviour, which happens over half a period of sin/cos.
It just turns out that $\sin^2(x) = \frac{1-\cos(2x)}{2}$, so you've "isolated" the S-behaviour by shifting it up from the cosine wave.

[flat frigate]

Azyrashacorki

[quiet plume]

What I mean is the S shape is the same as the one you had on the sine wave to begin with, just translated so that it starts at 0 (and up as well so as to not go into negatives)

[tepid crypt]

Can you elaborate the last sentence that starts with "It just turns out that..."? What do you mean by "shifting it up fromt he consine wave"?

[quiet plume]

Well the S you want looks like this :

It's the same on a cosine wave, since sine waves are just shifted sine waves.

[tepid crypt]

Right, got that.

[quiet plume]

But as per the identity I gave you, sin^2(x) is equivalent to taking that exact segment on the cosine wave, reflecting it about the x axis, and shifting it up by 1/2.

So they're linked in that, it's the exact same pattern. The squaring didn't really distort anything, it just changed the position of the segment so that it lies above the x axis

[tepid crypt]

Uhm. Makes sense.

So first I have cos(x). Then I shift it on the x-axis. Now I have 1-cos(x). Now the values are between 0 and 2. (1-cos(x))/2. And multiplying the x with 2 is just a side affect that allows me to use values from 0 to 90 for theta instead of 0 to 180?

Did I get that right?

[quiet plume]

Yeah exactly

It squishes the pattern inside the interval 0,pi/2

[tepid crypt]

Could you say you double the frequency?

[quiet plume]

Yep

[tepid crypt]

Uhm. What exactly does 1 - cos(x) do? What does subtracting cos from 1 produce? It seems to shift the graph both in x and y direction.

[quiet plume]

See it as multiplying cos by -1, which reflects about the x axis, and then shifting up by 1

[tepid crypt]

Oh..... It's the same as -cos(x)+1

[quiet plume]

Yess

[tepid crypt]

Awesome!

.close

[safe radish]

Channel closed

Closed by @tepid crypt

Use .reopen if this was a mistake.

[safe radish]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[tight marten]

Is it right to think disjoint events cannot be independant, while independant events can either be or not be disjoint?

[buoyant shadow]

that can't be true

if independent events can be disjoint, it means some disjoint events are independent

[raven ether]

Why are new questions being answered but two people have been waiting over 30 min with no responses

[light shoal]

it's possible if one of the events has zero probability
otherwise you are correct

[light shoal]

it's not a queue, people answer whatever they feel like answering

[tight marten]

i went to 2 different sources and this is what i gathered "Independent events are unrelated events. The outcome of one event does not impact the outcome of the other event. Independent events can, and do often, occur together." and

[light shoal]

independence means $P(A \cap B) = P(A)P(B)$, disjoint means $P(A \cup B) = P(A) + P(B)$. You always have $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. If you assume $A$ and $B$ are both independent and disjoint then the latter formula becomes $P(A) + P(B) = P(A) + P(B) - P(A)P(B)$, which simplifies to $P(A)P(B) = 0$. Which is true if and only if $P(A) = 0$ or $P(B) = 0$.

[flat frigate]

Bungo

[buoyant shadow]

if events are independent, and it's not a 0 probability case, they are not disjoint

[flat frigate]

Boi

[outer hull]

this implies, but is not implied by, P(A \cup B) = P(A) + P(B)

[safe radish]

@tight marten Has your question been resolved?

Channel closed

Closed by @tight marten

Use .reopen if this was a mistake.

[safe radish]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[agile wraith]

why does the answer above satisfy this?

anyone with analysis knowledge please help explain

[glacial sonnet]

S-{m} = [m-1, m)

m is still an upper bound of that

and m - epsilon is not an upper bound for every epsilon > 0

since S-{m} intersect (m-epsilon, m) = (m-epsilon, m) is nonempty

so m is the least upperbound

[safe radish]

@agile wraith Has your question been resolved?

[agile wraith]

Ok wait in simpler terms I am confused because when you subtract m from S what essentially are you doing?

[gentle lintel]

Taking it out of the set

[cerulean crane]

Well, m was our maximum, and we had everything between m-1 and m in S, so we had things arbitrarily close to m but nothing greater

So if we take out m, we have things arbitrarily close to m but now we don't have m itself, and we have nothing greater than m, so m is our supremum

[agile wraith]

ahhh ok so basically the set becomes

[m-1,m)?

[cerulean crane]

Yes, in fact that's the definition of that set

[agile wraith]

ok ok

that makes a lot of sense

[safe radish]

@agile wraith Has your question been resolved?

[safe radish]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[vital root]

Relearning a few things for math finals but I’ve completely forgotten how to do this, I did p(a) times p(b) to get p (a and b) but I get .88 and on the cheat sheet it says it should be .14, am I missing a step?

[stiff phoenix]

I think itts best ot visualize tthis in a venn diagram first

mb my t key is bugging

[safe radish]

@vital root Has your question been resolved?

[stiff phoenix]

ok so i think the interpretation of the question is important here

the question said that given all people who have tiktok, 65% of people have insta

and 22% have a tiktok

so the answer (of the percentage of people having BOTH instagram and tiktok) is...?

and question b requires bayes' theorem, is that correct?

@vital root

[safe radish]

@vital root Has your question been resolved?

Channel closed

Closed by @vital root

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[gentle edge]

Can someone help me for vales of t pls

[gentle edge]

.rotate

[desert pasture]

,rotate

[gentle edge]

Thx

[potent seal]

Do you know the quadratic formula?

[gentle edge]

Yes

But if there a way to do it without using formula

[potent seal]

Hmm there is a way

Just by seeing the equation, can you tell what the sum and product of roots are?

[gentle edge]

Yes

[potent seal]

Ok cool so
If the roots are a,b
Then we know a+b and ab

[gentle edge]

But these will be imaginary

Wait no

Never mind

[gentle edge]

Ye