help-27

[sullen osprey]

i had a funeral during this lesson

i learnt this myself so ive learnt it poorly

[drifting sierra]

Let's simplify it 'cause I think you're just having trouble parsing all the parentheses

[sullen osprey]

ok 😎

[drifting sierra]

Drop the forall, we know it's there implicitly, and drop all the (x) parts

So you only write A, B, T, E

[sullen osprey]

ok dropping them mentally

[drifting sierra]

No, write it

[sullen osprey]

oh ok

[drifting sierra]

Write it in LaTeX again, from scratch, after reading the question statement again

[sullen osprey]

(A↔(¬B⋀ (T) ⋁ E)

[drifting sierra]

No space after \

[sullen osprey]

,, (A \iff (\neg B \land (T) \or E

ok there!

[drifting sierra]

\lor not \or

Ok well now you have three opening parentheses and only one closing

[sullen osprey]

,, (A \iff ( \neg B \land (T) \lor E)

[woven radish]

Vaynn

[drifting sierra]

Right, that first parenthesis is useless, that would be for the whole expression after "forall"

[sullen osprey]

,, A \iff ( \neg B \land (T) \lor E)

[woven radish]

Vaynn

[drifting sierra]

Then do you understand that (T) is just the same thing as T?

[sullen osprey]

,, (A \iff ( \neg B \land T \lor E)

[sullen osprey]

that makes sense yeah

[drifting sierra]

You keep undoing your fixes...

[sullen osprey]

oh yeah sorry

,, A \iff ( \neg B \land T \lor E)

[woven radish]

Vaynn

[drifting sierra]

Right, now we can progress

[sullen osprey]

im not used to latex srry

i normally use microsoft word equations

[drifting sierra]

You'll get used to it

So what is $\neg B \land T \lor E$ exactly?

[woven radish]

Nel

[sullen osprey]

Not Blacklisted **and **Temporary Pass **or **Employee

[drifting sierra]

If I give you B = 0, T = 0, and E = 1, what result does that give you?

[sullen osprey]

1?

[drifting sierra]

Why?

[sullen osprey]

Oh do you mean truth table wise?

[sullen osprey]

I was guessing 0 + 0 + 1 = 1 lol

but if you mean truthtables

then how i got to that doesnt really apply

[drifting sierra]

Not really, I mean what steps do you take to calculate it

[sullen osprey]

So you dont mean truth tables?

I calculated 0 + 0 = 0, then I added 1 to it

im confused what you mean

[drifting sierra]

I'm confused too

Why 0 + 0

[sullen osprey]

I'm guessing, because I don't know what I'm supposed to do with B= 0 , T=0 and E=1

So I'm just summing all the values

[drifting sierra]

So if I had given you B=T=E=1 instead you'd have said 3?

[sullen osprey]

i suppose

idk

I guess if B = T and T = E and E = 1

They're all equal to 1

[drifting sierra]

Ok I didn't realize you didn't know what those symbols do

[sullen osprey]

thats maybe another way id look at it

[drifting sierra]

We are doing boolean algebra here

Our only concrete values are 0 and 1

[sullen osprey]

ohh

[drifting sierra]

False and true

[sullen osprey]

thats what i meant by truth table

earlier

[drifting sierra]

Ok but I don't think you have a truth table for that whole expression

[sullen osprey]

good point

[drifting sierra]

If you want to think in terms of truth tables, you can, but you can only use them on basic parts

So for example the result of A AND B is 1 if both A and B are 1, 0 otherwise

[sullen osprey]

yes

[drifting sierra]

That's a description of the AND gate, aka a truth table

Now if you have (NOT A) AND B, you can't use that truth table directly

You first need to do the NOT A part, right?

[sullen osprey]

right

[drifting sierra]

And you do that with another truth table, namely NOT A is 1 if A is 0, otherwise it's 0

Now, what steps do you take to calculate $\neg B \land T \lor E$?

[woven radish]

Nel

[sullen osprey]

NOT -> AND -> OR
i guess

Since if you have to calculate NOT first, and not is connected to the and I suppose it makes sense you do and 2nd

then or last

is that what u mean

[drifting sierra]

That's a reasonable guess but you see how it differs from what you said earlier about the original question statement?

[sullen osprey]

You mean the version I had at the start?

[drifting sierra]

This

[sullen osprey]

oh you mean the actual question

oh

[sullen osprey]

yes

[drifting sierra]

When you see an expression like that, you can be sure that the NOT is applied only to the B, but you can't be sure whether the AND is applied first or the OR is

[sullen osprey]

And is normally first without brackets right?

other than NOT

[drifting sierra]

I said here that it's typical to give AND a higher precedence, meaning it gets applied first

But really you should ask whoever wrote the expression

And in the case you're the one who wrote it, well you need to fix it

[sullen osprey]

So what you're saying is mine is ambigious whether it is and / or first?

Or are you saying I'm making the wrong one be applied first

[drifting sierra]

It's ambiguous, and even if you use the "usual" precedence, it turns out to be wrong

[sullen osprey]

i see

So I need to make it so AND is applied first

then or

is what you're saying maybe..?

[drifting sierra]

No, OR first

[sullen osprey]

why do I want or first?

[drifting sierra]

I don't know why you're still having trouble with which should be first, you read the question statement

[sullen osprey]

let me re-read the question

okok im back

So it checks

Not blacklisted -> Then if you have either pass or employee
Then checks if you meet both criteria?

Is that why or has to come first?

Like it will take the TRUE from the not and the OR
To make the TRUE for the AND to happen

so or has to come first

essentially?

[drifting sierra]

If I understand what you mean, yes

Think of it like a tree

[sullen osprey]

Like

The condition for AND to be true is reliant on the output of NOT and OR

is essentially what I'm saying

And if you evaluate AND first how can you evaluate it properly?

it wont work

[drifting sierra]

    AND
   /   \
 NOT   OR
 /    /  \
B    T    E

[sullen osprey]

yes!

okkk okk

😎 we getting somewhere

[drifting sierra]

So yes, the AND depends on the NOT and the OR

So do you know how to fix your expression now?

[sullen osprey]

um

truthfully maybe not

but ill have a go

I would imagine I want it like this

So that B and OR will happen first

[drifting sierra]

Yes

[sullen osprey]

So like that ultimately

woooooooo

[drifting sierra]

You don't need parentheses around the NOT B

[sullen osprey]

because it is highest precedence anyway?

[drifting sierra]

That symbol only applies to whatever is directly after

If you want to apply it to multiple things, you use parentheses

[sullen osprey]

So like that i guess

[drifting sierra]

Uh no you messed something up here

[sullen osprey]

oops

oh wait yeah

i accidentally edited the wrong one

and screenshotted that

Like that!

[drifting sierra]

You're missing a parenthesis but that's better

[sullen osprey]

oh at the end

[drifting sierra]

Yes

[sullen osprey]

there we go

[drifting sierra]

Why does it look different

[sullen osprey]

wdym

did i mess up

😨

[sullen osprey]

oh the bracket

word is silly

word is blehhh

word is like that ^

[drifting sierra]

Even there before it looked more similar to the others

[sullen osprey]

could i ask for ur professional advice

on the rules of converting 1st order into CNF

because im slightly confizzled

on how to do that

[drifting sierra]

I'm afraid I need to go but check these out:
https://math.stackexchange.com/questions/214338/how-to-convert-to-conjunctive-normal-form
https://math.stackexchange.com/questions/99068/how-to-write-x-iff-y-in-cnf-form

[sullen osprey]

thank u 🫶

goodluck with what u need to do!

i love how this server has cat emojis

thats so cute

[ionic harness]

(btw I see you opened a new chat for a new question; due to some recent changes, you don't have to anymore! Once you finish with one question, if you have another right after, you can ask someone with a green name to pin your new question, and keep the same channel)

[sullen osprey]

oh

yeah i thought i was supposed to open a new one for every question

[ionic harness]

That's how it used to be, no worries 👍

When I say recent I mean like last week

XD

[sullen osprey]

LOL

ok yeah very recent

[ionic harness]

[sullen osprey]

I'm going to read resources, but if anyone can advice me on turning into CNF then you're a beautiful person

I say read, but im awful at reading I'm gonna watch a yt video on CNF :p

[ionic harness]

I think I saw this in my book let me reopen it

Ah conjunctive normal form is what we call product-of-sums in computer science

Yes b is a great example

Sadly this is going to be nasty

But the process goes like this

1. Turn all implications into 'or's

So see if you can remove the biconditional

From here

[sullen osprey]

Implication -> or
Biconditiona -> or

also?

[ionic harness]

Not quite

So when we say an implication, we say "if p, then q" right?

What do we say if we have a biconditional

[sullen osprey]

if and only if p then q

i think

[ionic harness]

Almost

$p$ if and only if $q$

[woven radish]

Coolempire93

[sullen osprey]

ahh

[ionic harness]

We can split this up into two halves

p if q and p only if q

Do you see how I got there

[sullen osprey]

ohhh

A(x) -> B(x) AND B(x) -> A(x)

like this?

[ionic harness]

Exactly!

Now can you write them with 'or's

[sullen osprey]

ohhh

I see so two step simplification

Biconditional -> Impliciation -> Ors

[ionic harness]

Biconditional -> two implications -> ors yes

[sullen osprey]

thats a bit confusing the 2nd step

[ionic harness]

Do you know how to write $p \implies q$ with or's

[woven radish]

Coolempire93

[sullen osprey]

So you would have this for the first part essentially

[sullen osprey]

not entirely 😅

[ionic harness]

Mm not quite

We'll come back to that though

[sullen osprey]

oh no?

[ionic harness]

Right after I mention how to do this

Which is

$p \implies q$ is equivalent to $\neg p \vee q$

[woven radish]

Coolempire93

[sullen osprey]

ohhhh

if I already have a negate

[ionic harness]

So we can turn any implication into an or

[sullen osprey]

do you think i remove the negate

[ionic harness]

Yes

[sullen osprey]

since the negation of a negation would just be positive yes

[sullen osprey]

ok makes sense 😎

[ionic harness]

So here's how a biconditional looks

$p \iff q$ is the same as $(p \implies q) \wedge (q \implies p)$ is the same as $(\neg p \vee q) \wedge (\neg q \vee p)$

[woven radish]

Coolempire93

[sullen osprey]

So the brackets was bad?

[ionic harness]

(And you can see that the last part is in CNF here already)

[sullen osprey]

its ok now...?

[ionic harness]

Yes but this isn't what's written at the top

[sullen osprey]

wdym at the top

[ionic harness]

Here you have the proposition $\forall x (A(x) \iff \neg B(x) \wedge (T(x) \vee E(x))$, which is really $\forall x (A(x) \iff (\neg B(x) \wedge (T(x) \vee E(x)))$

[woven radish]

Coolempire93

[ionic harness]

So when we do step 1 on this, we get

$$\forall x (A(x) \iff (\neg B(x) \wedge (T(x) \vee E(x)))$$ is $$\forall x \qquad (A(x) \implies (\neg B(x) \wedge (T(x) \vee E(x))) \wedge ((\neg B(x) \wedge (T(x) \vee E(x)) \implies A(x)))$$

[woven radish]

Coolempire93

[sullen osprey]

👀 thats so long and scary

[ionic harness]

Yes, that's why I use variables

I never do it all at once like this

[sullen osprey]

wait so how do u do it

i wanna know the pro method since i gotta do this gang 💔

[ionic harness]

Once we finished here I would've rewritten our original proposition so it was in this form

[sullen osprey]

ohhhh

So all of the left side would be P and all of the right side would be Q

and hten you would've replaced all the real vlaues with p and q

got it 😎

smart method

[ionic harness]

The original proposition: $$\forall x (A(x) \iff \neg B(x) \wedge (T(x) \vee E(x)),$$ which is really $$\forall x (A(x) \iff (\neg B(x) \wedge (T(x) \vee E(x)))$$

I would rewrite it as $$\forall x (A(x) \iff P(x))$$ where $$P(x) = \neg B(x) \wedge (T(x) \vee E(x))$$

Since A is by itself I don't actually need to turn it into a variable

[woven radish]

Coolempire93

[ionic harness]

See if you can follow what I did to get there

[sullen osprey]

understood 🫡

[ionic harness]

Okay

So now see if you can convert the A <-> P

[ionic harness]

Using the steps we established

[sullen osprey]

got it ill give it a go!

(P→Q)∧(Q→P)
Would become (¬P ∨Q) (¬Q ∨ P)
right?

[ionic harness]

Yep

with an AND in the middle

👍

[sullen osprey]

ohh

thank you homie 🫶

I LOVE THE CAT EMOJIS IN THIS SERVER

OMG THEY'RE SO CUTE

[ionic harness]

😆

[ionic harness]

Okay let's see what you have for this now

[sullen osprey]

oop im not quite done yet

[ionic harness]

Take your time, no problem 🙂

[lunar harbor]

[ionic harness]

CSP I might have to leave soon are you familiar with conjunctive normal form/product of sums

[lunar harbor]

ngl I'm just on here waiting for the chicken in the oven to get up to temperature

so um we'll see if I'm even here

[sullen osprey]

I was going to substitute P and Q into the bottom formula

[lunar harbor]

[sullen osprey]

and hopefully that should work right

[ionic harness]

Good but you forgot B, and we won't be substituting yet 👌

[sullen osprey]

oop

whats b again

[ionic harness]

cf.

B(x) = x is on the blacklist

[sullen osprey]

chicken takes so long to cook 😔 it makes me sad

[sullen osprey]

ohh

So it should be like that

[ionic harness]

Good

[ionic harness]

And again an important note about what we ended with is that we are in CNF down there

[sullen osprey]

also between the two (P -> Q) AND (Q->P)
(P -> Q) AND (Q->P)

Do you put an and between those two as well?

[ionic harness]

(well there's a typo at the very end, the AND between not Q and P should be an OR)

[sullen osprey]

So it would actually be

((P -> Q) OR (P ->Q)) AND ((P -> Q) OR (P ->Q))

wait

[ionic harness]

You have all the AND's in the places they need to be 👍 (save for the typo)

[sullen osprey]

or no other way round

[ionic harness]

What am I looking at

[sullen osprey]

i got silly

let me try explain

my brain

One set of P <-> Q becomes that

[ionic harness]

No

That is all of P <-> Q

[sullen osprey]

ohhh

ok yeah i got silly

😼

[ionic harness]

P <-> Q
is
P if and only if Q
is
P if Q and P only if Q
is
Q -> P and P -> Q

[ionic harness]

Great

So now we did step 1 to remove all implications

[sullen osprey]

(¬P ∨Q)∧(¬Q∧P)

so this is the final thing then?

like form

[ionic harness]

From step 1 yes but it will still change

(¬P ∨Q)∧(¬Q∨P)

Again without the typo

[sullen osprey]

ok got it 🫡

[ionic harness]

In step 2, we will simplify the inner expressions

[sullen osprey]

(¬P ∨Q)∧(¬Q∨P)

fixed it

[ionic harness]

In this case not P is just not A

There is nothing to simplify

But

Not Q has a lot

[sullen osprey]

yes

[ionic harness]

So we need to simplify not Q

We have
$$Q(x) = \neg B(x) \wedge (T(x) \vee E(x))$$
so
$$\neg Q(x) = \neg(\neg B(x) \wedge (T(x) \vee E(x))) = ?$$

[woven radish]

Coolempire93

[ionic harness]

I need you to apply negation to Q

So we can simplify

Damn this is actually a ridiculous one to give you without knowing how to do k-maps (interpreting the truth tables into CNF)

[sullen osprey]

my lecturer is evil

you should see the proof he gave me

and keep in mind its my first time ever having to do a proof

[ionic harness]

4 variables is a lot and expanding out across the or's here is going to be a lot as well

I don't think I'll be able to explain the whole thing at once here because I have to go

If you post it in #proofs-and-logic though either someone will come along and explain or I will later tonight

[sullen osprey]

😔 no worries gang ❤️

[sullen osprey]

ill try my best 😤

[ionic harness]

Good luck

Don't forget De Morgan's Law

[sullen osprey]

whats that

he didnt teach me that

😭

😭

[ionic harness]

De Morgan's Law $$\neg (P \wedge Q) = \neg P \vee \neg Q$$ and $$\neg (P \vee Q) = \neg P \wedge \neg Q$$

[woven radish]

Coolempire93

[ionic harness]

But if he didn't teach that then this problem is impossible

[ionic harness]

And once you have everything together, step 3 is to distribute (so turn all the inner parts into CNF using this law: $$A \vee (B \wedge C) = (A \vee B) \wedge (A \vee C)$$ which is the central law of converting to CNF

[woven radish]

Coolempire93

[ionic harness]

This whole process with the previous two steps is so we can get to a form we can apply this central law to

[sullen osprey]

So the top two statements are equal and the bottom two statements are also equal

got it

So anywhere I see that I can turn it into the other one

[ionic harness]

You'll need it even just to do this

[sullen osprey]

And im assuming my goal is to get to ones behind the equal

[ionic harness]

Yes

[sullen osprey]

understood 🫡

[ionic harness]

Step 2 'simplify' is essentially: move all negations so they are beside a proposition

One of your variables that is

No negations over a parentheses

[ionic harness]

Once you have that you can apply the central law to all of the parts

e.g.

[sullen osprey]

wait whats teh central law

[ionic harness]

$$(A \vee (B \wedge C)) \wedge (C \vee D) = (A \vee B) \wedge (A \vee C) \wedge (C \vee D)$$ and now we are in CNF

[woven radish]

Coolempire93

[ionic harness]

central law

[sullen osprey]

ohh

[ionic harness]

Basically the way to turn an or (disjunction) into a conjunction (and)

So now you can turn any dijunction into a conjunctive form

And it's normal once all of the outside operations are conjunctions

[sullen osprey]

this is even harder than i thought it would be 😨

at the start

[ionic harness]

Yeah evil problem

[sullen osprey]

😔 evil

[devout snow]

@sullen osprey Has your question been resolved?

[sullen osprey]

Would this be negated P?

Which would be negated P I'm not sure if the all x thing translates well into CNF or not

Acc im not sure where to go from here 😭

[devout snow]

@sullen osprey Has your question been resolved?

[sullen osprey]

<@&286206848099549185> 🥺 🫶

[sullen osprey]

#proofs-and-logic I've posted this here; since this isn't getting much traction I'll close this and do a different question

.close

[devout snow]

Channel closed

Closed by @sullen osprey

Use .reopen if this was a mistake.

[devout snow]

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[viral rapids]

how to do part b

ik the answer is 112pi/3 for part a

but im getting this for part b which isnt the same number

[fair juniper]

draw the shape

[fair juniper]

notice how there is a concavity

[viral rapids]

ok wb it

[fair juniper]

do you know what your integral is integrating?

[viral rapids]

the top branch volume

i tried doing (top branch)^2-(bottom branch)^2 in the integrand but that also gave a different number

[fair juniper]

graph 2+sqrt(x-1)

[fair juniper]

ok, so when you integrate that function (with the pi r squared stuff) what volume are you making

any ideas?

[viral rapids]

the volume of that piece

[fair juniper]

close

its the whole area from y=0 to the function

[viral rapids]

ohhh ye

so i subtract bottom²

[fair juniper]

there are a couple of ways to get what you want

[fair juniper]

yes, but that will get you the part that has been cut out

[viral rapids]

wait

i see what ur saying

let me try hold on

$\pi\int_0^14^2\dd{x}+\pi\int_1^5\pqty{(2+\sqrt{x-1})^2-(2-\sqrt{x-1})^2}\dd{x}$

[woven radish]

Jash

[viral rapids]

are u suggesting that cuz i already tried that

[fair juniper]

youre so close

[viral rapids]

is it about the region above the top branch

like 2+sqrt(x-1)<y<4

[fair juniper]

what that does is get this

[viral rapids]

OHHHH

[fair juniper]

yeah, you have to break it into two pieces, the top branch and bottom branchj

[viral rapids]

4*4-THAT INTEGRAL

[fair juniper]

yeah, or you could take the whole cylinder and subtract that piece out

[viral rapids]

what cylinder

[fair juniper]

the one you get by revolving y=4

[viral rapids]

oh ye thats what i meant by ths

[fair juniper]

yup

[viral rapids]

omg i got it

tysm

.close

[devout snow]

Channel closed

Closed by @viral rapids

Use .reopen if this was a mistake.

[fair juniper]

decided to make this just because

[viral rapids]

how tf

[fair juniper]

desmos 3d

[viral rapids]

lol thats sick

[devout snow]

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[normal bolt]

Where did i make a mistake in my solution

Anyone here?

The helper

[hollow ice]

I can see where you got t1+t2=2, but what about the other two?

[normal bolt]

By the symmetry of equilateral triangle

[hollow ice]

???

What does the symmetry got to do with this?

all you did is get t1=t2=t3=1/2

thats a degenerate triangle

aka a point

you dont want that

[normal bolt]

okay

What should I do then

[hollow ice]

You have the slope of one side

[normal bolt]

yes

[hollow ice]

find the slopes of other two sides

and then use the slopes and find the similar parametric equations for the other parameter pairs

[normal bolt]

I found the slopes of other two sides in parameter

But when applying m = 2/t+t' for the corresponding sides it is giving me 2=2

[devout snow]

@normal bolt Has your question been resolved?

[devout snow]

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[verbal pasture]

probs stupid simple question I forgot from algebra but what am I supposed to do when the x is in the denominator

[verbal pasture]

is it divide or multiply

[young spade]

multiply by $\frac x{\tan 19^\circ}$ on both sides.

[young spade]

It will cancel out neatly and you will get your "solution" for x.

There are other ways but this is the most direct one.

[verbal pasture]

thank you Dexter

[young spade]

np

[verbal pasture]

.close

[devout snow]

Channel closed

Closed by @verbal pasture

Use .reopen if this was a mistake.

[devout snow]

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Send your question here to claim the channel.

Remember:
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[hollow heart]

Number 9

[hollow heart]

,rccw

[hollow heart]

Where did i go wrong?

[devout snow]

@hollow heart Has your question been resolved?

[devout snow]

@hollow heart Has your question been resolved?

[toxic flower]

ig there's a typo in qsn

for the given answer, there must be a 's' in numerator

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[hidden rain]

anybody able to offer some help

[winter patrol]

what have you tried?

[hidden rain]

since i've got a, b, and the angle in between them i can use the law of cosines right

[winter patrol]

Yes

[hidden rain]

to find c

since i have c now i can use law of sines to find B and A?

[winter patrol]

Yes, well use it to find A

and you can get B from angle sum

(since a isn't the longest side, you wont have to worry about ambiguous case when determining A)

[hidden rain]

going to send a few more rq

so...
A = 222
B = 200

for resultant force i would take (222)^2 + 200^2

[winter patrol]

don't forget to sqrt

[hidden rain]

got it

for angle B would i take tan(200/222)

then for sin(a) i would take sin(222/299)

the 299 being what i got for the resultant force

and tan(B) would it be tan-1(200/22)

[winter patrol]

you're mixing up the trig function with their inverses

[hidden rain]

so for angle B it would be the inverse?

[winter patrol]

trigfunction(angle) gives the ratio
inversetrigfunction(ratio) gives the angle

[hidden rain]

so to find angle B it would be tan-1(200/222)

[winter patrol]

yes

[hidden rain]

and then for tan(B) it would be tan(200/222)

[winter patrol]

no

B isn't 200/222

[hidden rain]

oop

i see

it would be what i got for angle B

[winter patrol]

tan(B) would just be the ratio you used / applied the inverse to to get B

[hidden rain]

hm

[winter patrol]

don't overthink

[devout snow]

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[devout snow]

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[viral kernel]

What are aleatory variables used for in probability?

[stone stump]

the wikipedia article is a good start

[viral kernel]

calculate the number of times heads comes up out of 10 tosses

he asked me to do it with random variables but I don't know if I did it right

X(>=8)={0,1,2}

[stone stump]

what is the 2 doing there

[viral kernel]

sorry

[stone stump]

ok the edited version makes even less sense

[viral kernel]

I need to see the probability of getting more than 8 heads out of 10 tosses. I have coded the set as X={0,1,2} where 0=8 heads, 1=9 heads and 2=10 heads.

[stone stump]

ok why not {8,9,10} ?

or at least 0 = 0 tails = 10 heads?

[viral kernel]

why I started from the numbers at the beginning

[stone stump]

it ultimately doesnt matter much but it will probably lead to errors

[viral kernel]

P(X>=8)=P(0)+P(1)+P(2)=(10!/8!2!2^10)+(10!/9!2^10)+(10!/10!2^10)

=7/128

I don't understand why I used the aleatory variable here, isn't this the same way of using counts and favorable sequences?

to find the probability I mean

[stone stump]

then treat it as a way to have nice notation for what you want

for now

P(X>=8) encapsulates the entire result in very short notation

[vernal grail]

i think we can imagine a random variable as :

$\Omega$ all possible paths of the ball

[woven radish]

alee

[stone stump]

there are situation were it wont be as simple without a random variable

[vernal grail]

X cylinder number

X = 5 is the column 5

and X >= 5 are all column 5 , 6 ecc

[woven radish]

alee

alee

[vernal grail]

does this makes sense ?

[viral kernel]

yes

[vernal grail]

this is X>=1

with n = 2 and p = 1/2

X = number of heads in 2 coin tosses

every column is {X=k} and the height is P(X = k)

and the colored colums are 2 so P(1 <= X <= 2) = P(X=1)+ P(X=2)

[devout snow]

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[viral kernel]

thanks

[devout snow]

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[quartz bloom]

Q1 (b)

[quartz bloom]

I wrote it as

xn = (-1)^n / (1+1/n)

I know limit of denominator converges to 1 and numerator is divergent

But from this information, i don't know how to reach the exact answer

[hollow jolt]

-1^n flips between 1 and -1

[placid plank]

Try 2 subsequence

[quartz bloom]

Whats that

[placid plank]

x_phi(n) with phi Strictly increasing, but ig if you're asking it means you shouldn't do it like that..

[quartz bloom]

Some other way ..?

[placid plank]

ig saying it "flips" is enough then, usual demonstration is supposing it converges, then all subsequence have to converge to this limit

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[quartz bloom]

.close

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[ionic trout]

<@&286206848099549185>

[devout snow]

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[raven kayak]

Hello, could someone check if this proof looks good please?

\begin{Lemma}
Suppose $A$ and $B$ are sets. Then, $A \subseteq B$ if and only if $A \cap B = A$.
\end{Lemma}

\begin{Theorem}
Suppose $C$ is a set. Then, there exists a unique $A \in \powerset{C}$ such that, 
for every $B \in \powerset{C}$, $A \cap B = B$.
\end{Theorem}

\begin{proof}
Let $A \in \powerset{C}$.
Either $A = C$ or not.
Suppose $A \neq C$.
If $B = C$, then there is a $x \in B$ that is not in $A$.
So, $A \cap B \neq B$.
Now suppose $A = C$.
For any $B \in \powerset{C}$, we have that $B \subseteq A$.
By lemma 1, this means that $A \cap B = B$.
Therefore, $A = C$ is the element of $\powerset{C}$ such that for every $B \in \powerset{C}$, $A \cap B = B$
\end{proof}

[woven radish]

Mor Bras

[raven kayak]

\begin{proof}
Let $A \in \powerset{C}$.
Either $A = C$ or not.
Suppose $A \neq C$.
If $B = C$, then there is a $x \in B$ that is not in $A$.
So, $A \cap B \neq B$.
Now suppose $A = C$.
For any $B \in \powerset{C}$, we have that $B \subseteq A$.
By lemma 1, this means that $A \cap B = B$.
Therefore, $A = C$ is the element of $\powerset{C}$ such that for every $B \in \powerset{C}$, $A \cap B = B$
\end{proof}

[woven radish]

Mor Bras

[native gyro]

Yea the end is wrong

Yea now its good

[devout snow]

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[devout snow]

@raven kayak Has your question been resolved?

[raven kayak]

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[devout snow]

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[deft flower]

how do we solve this, with n = 2013 and if it could be generalized to n it would be better

[mystic scarab]

King's property/rule, I believe

Inspired from the fact that π/3 + π/6 = π/2

[deft flower]

i tried it but got nowhere

[mystic scarab]

Where did you get stuck?

[deft flower]

uh wait gimme a minute imma send a pic

[sand dove]

What is tan(pi/2 - t) for you?

[deft flower]

-tan(t)

[sand dove]

No that's not it

[deft flower]

-tan(-t) ?

tan(t) ?

[sand dove]

Let's not guess and try to actually find it

What's sin(pi/2 - t)

[deft flower]

cos(t)

oh then it is 1/tan(t)

[sand dove]

What's cos(pi/2 - t)

[sand dove]

Yes

[deft flower]

but then how do i proceed

[sand dove]

?

[deft flower]

it seems like it got more complicated

what's the next step

[sand dove]

Try it again, it's not gonna be complicated

Write the integral correctly

[deft flower]

alr imma be back in a bit !

is this correct?

[mystic scarab]

Forget for a while this exercise

How would you simplify $\frac{1}{1 + u^{-1}}$?

[woven radish]

Alberto Z.

[sand dove]

Technically yes but you could find an easier way to write 1/(1+(tan(t))^-n) as Alberto is suggesting

[deft flower]

i don't know 😭

multiplying top and bottom by u ?

what do u exactly mean by simplify

[sand dove]

That's exactly it

[deft flower]

oh right i see it now

it would've been so much easier

[sand dove]

So one term is 1/(1+u)

The second is u/(1+u)

Their sum is 1

[deft flower]

and it gives exactly 1

yup now i understand thanks so much @sand dove @mystic scarab

❤️‍🩹

[mystic scarab]

Exactly, that's the neatest approach, I'd say

[deft flower]

well have a good day both of you and thanks !

.close

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[viral lynx]

How is it not indeterminate? o+ in the denominator tends to infinity

and o+ in the log gives -infinity, no?

So - infinity/infinity

Which is indeterminate form right?

[scarlet sequoia]

Are you sure?

[viral lynx]

I mean I know im wrong I just dont see why

Hiiiii chartbit ❤️ (Or in this case chartBITE)

[scarlet sequoia]

x tends to 0^+, so these will be small positive numbers, close to zero

not infinity

[lyric hornet]

1/x tends to infinity as x->+0, but not x itself

[viral lynx]

Ohhhhhhhhhh

I see my mistake

Ok thanks guys!

❤️

.close

[lyric hornet]

<@&268886789983436800> again

[devout snow]

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[upper schooner]

.(They're just trying to get me to say hiiiiiii to you )

[lyric hornet]

how you doing catbit

[upper schooner]

.Also hiiiiiiii

[upper schooner]

.Not too bad, I can't complain, hope you're good too

[lyric hornet]

.Doing very good myself, hope you took care of some of that stuff you said you would and that you hadsome very restful and much needed regenerative sleep

[devout snow]

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[sudden flower]

1+1 ؟

[trim pewter]

_ _

[acoustic leaf]

Don't use the help channels for troll questions.

.close

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[willow helm]

[devout snow]

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[robust bobcat]

for 30 how is he able lto get the second y derivative in the form it is written. I have found y'' = -2x(x+y)/y^2 but I cannot see a way to make x+y = y^-3

[robust bobcat]

one sec I was diffing wrong

yet I still cannot see the simplification. You have -2(x+y^2 y'*y')/y^2 which doesn't seem to lead anywhere

nvm

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[supple lily]

Solve this interval please

[supple knot]

,rotate

[scarlet sequoia]

It's of the form dx/(a+bx), so you can easily solve it, it will be something with ln

[supple lily]

I tried solving it with substitution but I think the answer wasn’t right.

I just want someone else to solve it for reference sake just in case I am doing some silly mistake

[scarlet sequoia]

What was your substitution?

[supple lily]

Taking the denominator as t. dt= -k2tan theta dx

[scarlet sequoia]

So

[woven radish]

Modus

[supple lily]

Yeah

[scarlet sequoia]

It becomes

[woven radish]

Modus

[supple lily]

Yeah then it just becomes natural log of t

[scarlet sequoia]

Ye

[supple lily]

And then I apply the limits. I’ve done it correctly right ?

[scarlet sequoia]

Yes, basically

[woven radish]

Modus

[supple lily]

For context this was to calculate the expression for capacitance in a given problem so I ended up solving it in term of tan theta but they did it in terms of d and a wherein tan theta = d/a so I’ll have to do the entire thing again prolly.

But atleast I integrated it right

[scarlet sequoia]

Put tan(theta) = d/a (it doesn't depend on x, so it won't change anything)

[supple lily]

It becomes somewhat complicated with the full expression

But I’ll have to see. Thanks for clearing it up though.

.close

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[last apex]

Why can't I use the geometric mean theorem here to find length of segment PA?

[last apex]

It seems plausible to use PA = sqrt(13)

[void fox]

you can use similar triangles here

[last apex]

I know I can, but I'm asking why I can't use the geometric mean theorem here

It's just one thing I noticed

[drifting sierra]

Because you don't have RP and PO?

Can you state the theorem you're referring to?

[last apex]

Oh frick, yeah

I'm dumb

I forgot 13 isn't the product

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[sullen osprey]

Does my answer to this question ^ seem correct?

[devout snow]

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✅ Original question: #help-27 message

[bright phoenix]

\begin{tikzpicture}[every node/.style={circle, draw, minimum size=8mm}]
% Staff nodes (spaced vertically)
\node (Logan) at (0,5) {Logan};
\node (Jamie) at (0,3.5) {Jamie};
\node (Rowan) at (0,2) {Rowan};
\node (Taylor) at (0,0.5) {Taylor};
\node (Rudy) at (0,-1) {Rudy};

% Task nodes (spaced horizontally and vertically)
\node (Planning) at (6,5) {Planning};
\node (Publicity) at (6,3.5) {Publicity};
\node (Sales) at (6,2) {Sales};
\node (Marketing) at (6,0.5) {Marketing};
\node (Development) at (6,-1) {Development};
\node (Industry) at (6,-2.5) {Industry};

% Edges representing skills
% Logan: planning, sales, marketing, industry
\draw (Logan) -- (Planning);
\draw (Logan) -- (Sales);
\draw (Logan) -- (Marketing);
\draw (Logan) -- (Industry);

% Jamie: planning, development
\draw (Jamie) -- (Planning);
\draw (Jamie) -- (Development);

% Rowan: publicity, sales, industry
\draw (Rowan) -- (Publicity);
\draw (Rowan) -- (Sales);
\draw (Rowan) -- (Industry);

% Taylor: planning, sales, industry
\draw (Taylor) -- (Planning);
\draw (Taylor) -- (Sales);
\draw (Taylor) -- (Industry);

% Rudy: planning, publicity, sales, industry
\draw (Rudy) -- (Planning);
\draw (Rudy) -- (Publicity);
\draw (Rudy) -- (Sales);
\draw (Rudy) -- (Industry);

\end{tikzpicture}

[woven radish]

Guido

[bright phoenix]

there we go thats a better graph

soo is it bipartite

?

Review your definition.

[sullen osprey]

Yes its bipartite, because it has no interset connections

None of the people connect

[sullen osprey]

And none of the skills connect to eachother either

[bright phoenix]

Good shit

[sullen osprey]

So is what I posted earlier true

Like this is correct for both

[bright phoenix]

Yes

[sullen osprey]

ok cool 😎

[devout snow]

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[bitter garnet]

Counterexample? What's that? How do I do that?
Prove or give a counterexample: If 𝑈 is a nonempty subset of 𝐑2 such that
𝑈 is closed under addition and under taking additive inverses (meaning
−𝑢 ∈ 𝑈 whenever 𝑢 ∈ 𝑈), then 𝑈 is a subspace of 𝐑2
.

[wicked turtle]

if it were R1 instead of R2, could you find a counterexample?

[bitter garnet]

Uh

[lyric hornet]

As for what a counterexample is, it is a case where the opposite holds. If we're trying to prove the statement true in general but we find one case where such a U is NOT a subspace of R2, then the statement fails in general and thus not generally true.

[bitter garnet]

Okay, thank you

i think I am just gonna go to sleep

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[bitter garnet]

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[vital saffron]

lowk need help with the answers to all 25

this is my current key

[devout snow]

Please post images (such as PNGs or JPGs) of the question rather than other filetypes such as PDFs which have to be downloaded. Non-image downloads can potentially contain viruses or other security risks.

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[vital saffron]

wair

imma put photos

[lunar harbor]

<@&268886789983436800>

<@&268886789983436800>

idk if y'all kick me for spamming

[vital saffron]

it’s not.

it’s a mock on amc8

[copper harbor]

either way it seems to be pirated content

[lunar harbor]

alright I'm out

[delicate dust]

lol at .gg/exams watermark

[lunar harbor]

[copper harbor]

well either its a leak, or its a pirated copy of someone else's mock they wrote

[delicate dust]

.close

.close

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[supple knot]

<@&268886789983436800> spam

[devout snow]

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[restive river]

help

[restive river]

i was looking for questions on my prep for selective school and i need explanation

<@&286206848099549185>

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[devout snow]

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[idle citrus]

back again, if my whole number is negative 3, when i put it over one for slope do i keep the one positive?

[idle citrus]

so if the slope is -3/1 would i go down three then one to the right

or make the 1 negative and go to the left

i think i know the answer i just wanna double check

[urban harbor]

whether you go 3 down, 1 right OR 3 up, 1 left you end up making the same line

[idle citrus]

yes but im asking specifically if the numerator is negative

[urban harbor]

$\frac{-3}{1}=\frac{3}{-1}=-\frac{3}{1}$

these are the same number

[idle citrus]

like im asking if its neg three as a whole number and im putting it over one does the one stay positive

bcs -3/-1 isnt the same as -3/1

[woven radish]

Soosh

[idle citrus]

for slope u keep the one positive right? im just tryna make sure im moving the right way

[urban harbor]

you can move the negative sign around, but don't introduce a whole new one if thats what your question is, that would be incorrect

[idle citrus]

okay thanks!

[pseudo basin]

are you asking how to write down the answer to a "find the slope" question? or about the graphical procedure of doing it

[idle citrus]

for example my equation is y = -3x -1
so when im doing the slope to graph im asking do i do -3/1 (keep one pos) or make the entire fraction neg (-3/-1)

im thinking to keep 1 pos but i keep thinking too literally when im actually solving it and i wanna double check

like would i go down 3 to the right 1

and not down three to the left 1

[pseudo basin]

-3 means (-3)/1 with one negative only
(-3)/(-1) would have the negatives cancel out and become positive 3, which is the wrong slope

[idle citrus]

i thought so ty

[pseudo basin]

right 1, down 3 is your best bet

right 1, up/down whatever

[drifting sierra]

The slope is the amount you go toward the positive y over the amount you go toward the positive x

Positive y is usually "up", positive x is usually "right"

If you go away from the positive direction, you're going negatively in that direction

[idle citrus]

thank you!

.close

[devout snow]

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[devout snow]

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[cobalt crystal]

how do i determine this type of point

[cobalt crystal]

i know f'(x) >= its growing

if f''(x) > 0 its local minimum

[polar chasm]

do u know what f(x) is or do u only have the graph?

[cobalt crystal]

nah i just used it as an example

[stone stump]

do you mean "how do I determine whether a point is a saddle point?"

[cobalt crystal]

i just need to know how you know if it is a sattelpunkt

[cobalt crystal]

yes

[round raptor]

-# is that german

[stone stump]

yes

well the first derivative has a minimum at that point

which you can see in your image

and how do you find minimums?

[cobalt crystal]

i know 2nd derivative has to be > 0 for it to be a minimum

basically we look for
f'(X) = 0

and put in the same x for f''(x)

if its >0 it is a minimum

[stone stump]

its a minimum of f, yes

[cobalt crystal]

if its < 0 it is a maximum

[stone stump]

how do you check whether x is a minimum of f' ?

[cobalt crystal]

basically same process i guess

[stone stump]

yes

[cobalt crystal]

but why would that matter here

are you suggesting for saddle point we go 1 derivation deeper?

[stone stump]

look at the picture

saddle point of f = minimum of f'

[cobalt crystal]

yeah its just a random pic i found, i just need to know how to get the saddle point in general

[stone stump]

the derivative does not cross the x-axis

because that would mean that the slope switches from positive to negative or vice-versa

[devout snow]

@cobalt crystal Has your question been resolved?

[devout snow]

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[night rune]

is there a way to eyeball the standard deviation?

[devout snow]

@night rune Has your question been resolved?

[devout snow]

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[cold nebula]

I am so confused i dont even know where to start

[cold nebula]

I would assume this tends to 0 so 0<x[n]< smht i need to figure out that tends to 0

[polar chasm]

|sin| can be bounded by 1

and u can probably also simply write 3^i instead of (i+1) 3^i

[cold nebula]

uhhh

ic

but wouldnt the sins add up

to form an n

oh i see now

thanks

[polar chasm]

Every single one of them can be bounded by 1, so it would be 1 + 2 + 2^2 + ... + 2^(n-1)

[polar chasm]

also, it's probably more appropriate to bound the absolute value of this expression (it could tend to sth negative)

[cold nebula]

i gues

[keen hawk]

there are tricks to find the sum of the denominator, the sum of n * 3^n and so on

but you don't even need to do that

[keen hawk]

oh right, this was mentioned before

[polar chasm]

u dont even need geometric probably

clearly the denominator sum is > 3^n and the numerator sum is < 2^n

and that means that its < (2/3)^n -> 0

[cold nebula]

andd how abt sum from 1 to n of cos k / n*n

[polar chasm]

$\frac{\sum_{k=1}^{n}\cos\left(k\right)}{n\cdot n}$

[woven radish]

MathIsAlwaysRight

[polar chasm]

this?

[cold nebula]

ye

[polar chasm]

again, try bounding it

[cold nebula]

it gives 1/n

which is 0

[polar chasm]

yeah

bounding cos by 1 results in n in the numerator and n / (n * n) = 1/n -> 0

[cold nebula]

so this bounding thing can be used wiht cos aswell alright

[polar chasm]

Yeah, both sin and cos have values between -1 and 1

so in absolute value, <= 1

[cold nebula]

👍

[devout snow]

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[potent compass]

Question o please

Could you check the process bc I got it wrong and idk why

[heavy current]

4(x - 1) = 4x - 4 :p

you wrote 4x - 1 instead

[potent compass]

Ty

I’m I doing something wrong

It keeps giving me 6x = 6x

[heavy current]

nothing here looks wrong so far

just keep on simplifying

using a ||logarithm|| might be a good idea c:

[potent compass]

Idk what Tht is

[heavy current]

ah

[heavy current]

okay so, nothing here is wrong, but this probably isn’t helpful if you can’t use logs

my new suggestion is to rewrite 3^2/2^6 as 3^2/(2^3)^2 on the right hand side

[potent compass]

Why is toesy here

Nvm

I’m seeing things

[uncut crow]

[void fox]

[devout snow]

@potent compass Has your question been resolved?

[devout snow]

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[astral marsh]

Can anyone help regarding 8-13? I'm just not 100% sure about my answers

[astral marsh]

,help

[woven radish]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[astral marsh]

,list

[woven radish]

My commands!

Use ,ls to obtain a briefer listing, and use ,help <cmd>to view detailed help for a particular command, or ,help to view general help.

If you still have questions, talk to our friendly support team here.

Meta

View or set meta-information about me.
​ ​ ​ ​ ping: Check heartbeat and API latency.
​ ​ ​ ​ help: Bot and command usage information.
​ ​ ​ ​ list: Lists all my commands!
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Page 4/4

[devout snow]

@astral marsh Has your question been resolved?

[worldly lagoon]

I think x should be 2, right?

[potent compass]

where ole

Pls

And is tht luka

[worldly lagoon]

Yes that is luka

Because i thought(im not sure) that 2^3x=2^6
3x=6
x=2
And 3^x=3^2
x=2

I don’t know any explanation but this seems logical to me

[devout snow]

@astral marsh Has your question been resolved?

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[devout snow]

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[meager delta]

Hii

[meager delta]

I need help

Integrals give me a headache

[supple knot]

You should split the 4x-10 more intelligently. Use (x^2+2x+2)' = 2x+2 so that 4x-10 = 2(2x-5)=2(2x+2-7)

Then you'll have -14/(x^2+2x+2) and use completing the square

[meager delta]

Uhm

Can u do it on paper pls

[supple knot]

No

I gave you more than enough information to do a couple more steps

[meager delta]

But I don't understand

...

And why -7

[supple knot]

Read the previous eqn

[meager delta]

I give up

[supple knot]

!done

[devout snow]

If you are done with this channel, please mark your problem as solved by typing .close

[meager delta]

Best help

[supple knot]

Did you have a question

[supple knot]

Didn't seem like it

[meager delta]

Man I don't understand

Please

[supple knot]

.

[meager delta]

I read it

[supple knot]

-5 = 2-7

[meager delta]

Oh

Gay

[meager delta]

And how does that even help me

[supple knot]

Keep reading

[supple knot]

The beginning part

[supple knot]

Then read this part

[meager delta]

I will go hang myself

!close

.close

[devout snow]

Channel closed

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[meager delta]

Bruh

[devout snow]

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[meager delta]

Help

[willow helm]

!xy

[devout snow]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[meager delta]

Dude the

...

[willow helm]

[willow helm]

Will we get the original problem or no

[devout snow]

@meager delta Has your question been resolved?

[willow helm]

😭

[gritty terrace]

is the question $\int x^2 \sqrt{a x^3 + x} dx$?

[woven radish]

Pseudo (Cat theory #1 Fan)

[meager delta]

yeah

[willow helm]

I thought that was a 4

[meager delta]

its a 4

[gritty terrace]

...

[willow helm]

[gritty terrace]

so, $\int x^2 \sqrt{4x^3 + x} dx$?

[woven radish]

Pseudo (Cat theory #1 Fan)

[meager delta]

yes

[willow helm]

$a \neq 4$, unless said $a = 4$

[woven radish]

USS-Enterprise

[gritty terrace]

what have you tried so far

[meager delta]

this

[meager delta]

what is that language

[gritty terrace]

i don't think this will work very well unfortunately