help-33

[boreal zenith]

ohh yeah I used the If then one

[jagged relic]

That's "implies"

[boreal zenith]

so id have to use a different one than that?

[jagged relic]

You should not use it, it's not a simplification

[boreal zenith]

is there an easy way to tel which one is right to use? or just a guess

[jagged relic]

You only need "and", "or", and of course "not"

[boreal zenith]

okay let me try again ill send it in chat

this look better or still off?

[jagged relic]

You don't understand the goal

The goal is to have an expression where negation symbols only appear directly before predicates

Parentheses and quantifiers are not predicates

[boreal zenith]

how would I even do that if the there exists and the for all are different?

[jagged relic]

Well you don't seem to understand how negating works with quantifiers

Here's an example statement: there does not exists a rational p such that p^2 = 2

Written symbolically: $\neg \exists p \in \mathbb{Q}, (p^2 = 2)$

[elfin berry]

Nel

[jagged relic]

An equivalent statement would be: for all rationals p, p^2 = 2 is false

Symbolically: $\forall p \in \mathbb{Q}, \neg (p^2 = 2)$

[elfin berry]

Nel

[jagged relic]

Do you understand how these are equivalent?

[boreal zenith]

ohh okay wait so with the part in parentheses I was getting to the rigth thing but with the quantifiers I was wrong

[jagged relic]

Not really

Here I used a negation before a parenthesis but in reality in my example the parenthesized expression is a single predicate

[jagged relic]

Let $R(p): p^2 = 2$ be the predicate, then my example is $\forall p \in \mathbb{Q}, \neg R(p)$

[elfin berry]

Nel

[boreal zenith]

okay

so does mine have the quantifiers down? the predicate just needs work?

[jagged relic]

Here? There's still a negation in front of the "there exists" quantifier

[boreal zenith]

is it not supposed to be? i thought to keep it equivilant when you switch the quantifier you have to leave a negation there

[jagged relic]

I'm not sure what you mean

[boreal zenith]

wait would the there exists be a for all still?

[jagged relic]

Look at my example again

I went from $\neg \exists p \in \mathbb{Q}, R(p)$ to $\forall p \in \mathbb{Q}, \neg R(p)$

[elfin berry]

Nel

[boreal zenith]

yeah take away the negation and switch the quantifier

then apply the negation to the predicate?

[jagged relic]

Apply the negation to whatever is next

[boreal zenith]

like this?

[jagged relic]

,, \neg \exists x \forall y (\neg O(x) \lor E(y))

[elfin berry]

Nel

[jagged relic]

(that's the question)

[boreal zenith]

okay i put the new negation now and distribute it to the parentheses

[jagged relic]

No the negation goes to the next part, in this case the next quantifier

[boreal zenith]

and then i also use demorgans law

[jagged relic]

Once you've fixed that part, yes

[boreal zenith]

okay let me try

THIS

[jagged relic]

You haven't fixed that

[boreal zenith]

not for all y?

so it would be there exists y?

[jagged relic]

Yes

[boreal zenith]

and is the other part right?

[jagged relic]

Yeah that's just De Morgan's

[boreal zenith]

so basically what you do is you take the not at the beginning and distribute it to the rest of the equation

Right?

[jagged relic]

Wdym distribute

[boreal zenith]

im thinking like distributive property its similar

like you multiply the sign like you multiply a negative sign in -(1+1) to -1-1

[jagged relic]

If you're thinking the first negation goes to both the "forall y" and the parentheses, then no

"not for all" and "there exists" are not the same thing

[boreal zenith]

but it switches it?

theyre interchangable when they come in contact wiht a not?

[jagged relic]

It switches and passes the negation on

[boreal zenith]

Okay

[jagged relic]

,, \neg \forall x, P(x) \iff \exists x, \neg P(x)

[boreal zenith]

and it just keeps passing the negation to the end of the equation or the end of a single predicate?

[elfin berry]

Nel

[jagged relic]

P(x) can be a whole expression, including quantifiers

,, \neg \exists x \forall y (\neg O(x) \lor E(y)) \
\forall x \neg \forall y (\neg O(x) \lor E(y)) \
\forall x \exists y \neg(\neg O(x) \lor E(y)) \
\forall x \exists y (O(x) \land \neg E(y))

[elfin berry]

Nel

[boreal zenith]

okayy i get it now

thanks for your help i gtg now

.close

[marsh citrus]

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[feral lagoon]

can someone check my workings?

[marsh citrus]

@feral lagoon Has your question been resolved?

[opaque agate]

Oof all except 8(b) are wrong

For 8(a) you applied stars and bars wrongly, second of all 5! Is also unnecessary as we already applied stars and bars

For 8(c) you again applied stars and bars wrongly and 3! Isn't necessary since we already applied stars and bars

Also answer of 8(d) will be $\frac{answer of 8(c)}{answer of 8(a)}$

[elfin berry]

Itsuki

[marsh citrus]

@feral lagoon Has your question been resolved?

[feral lagoon]

Lecturer explicitly said no to this

I think it’s because the probabilities for each pattern are different (are they idk)

[crystal lintel]

actually can’t look rn

[feral lagoon]

I think 5! is needed because the boxes are unique

I can’t write shit rn my iPad decided to update and reboot at this moment

[reef fable]

wait is question 8a should be 11C4 only?

[feral lagoon]

i think no cuz distinct boxes

[reef fable]

but i think there are 7 balls and 4 walls so could be 11 positions

[feral lagoon]

yes and then i mult 5!

[reef fable]

depends on the numbers of ways you rearranged them (4 positions) from that 11 posittions so that it would be 11 Choose 4

[feral lagoon]

yes

and then i mult 5!

[reef fable]

Which are the patterns right, the way you rearranged them so 5! is not compulsory

why would you multiply 5!

[feral lagoon]

because 5 distinct boxes?

or does them being distinct not matter

[reef fable]

no i mean you already put them in different boxes, multiply 5! means you take 5 boxes and swap their positions again

[feral lagoon]

i do intend on counting the swapping

is this not necessary

[reef fable]

and this only correct if the question ask the balls are distinct, however the question said the balls are identical which are the same

[feral lagoon]

distinct balls and distinct boxes?

[reef fable]

7 same balls different 5 boxes would be 11C4 only, and if both balls and boxes are different the way you calculate is different too

Suppose 7 identical balls, read from the start of the question

[feral lagoon]

identical balls distinct boxes yes

oh wait

11212 and 11122 is already “swapping boxes”

[reef fable]

yess

[feral lagoon]

shi alr

then new question

[reef fable]

b wrong too

only 5

[feral lagoon]

fair

[reef fable]

120 is way too big

because 4 boxes empty

mean you put 7 balls in 1 box only, and do the same for other 4 boxes

[feral lagoon]

distinct balls and distinct boxes
i dont think itll be 5! at the end
but it cant be 7! neither cuz swapping 2 balls in the same box does nothing
would i have to solve case by case?

[reef fable]

no no you dont need that 5! nor 7! at the end

[feral lagoon]

distinct balls tho (hypothetical)

[reef fable]

you only need the combination formula (k-1)C(n+k-1)

uhm the question never talk about distinct

[feral lagoon]

yes i made it up for better understanding

id then wanna ask about identical balls and identical boxes too lol

[reef fable]

alright if the balls are all different and boxes different

each balls have 5 ways to choose boxes (since 5 boxes), right?

and since you have 7 balls, this would be 5^7

and b c d wrong too since you assume the balls are different

[feral lagoon]

for d, i think the ^7 is necessary

because only by considering them distinct, could i count them as cases of equal chance

[reef fable]

i think you are obssessed with different balls

[feral lagoon]

like, 5555554 and 4555555 looks the same but it might be 2 throws

[reef fable]

the probability in D is using the ways satisfy the condition divide by the way it could happen

which is using answer in C divide by answer in a

[feral lagoon]

this is lecturer’s reply to someone who asked why d isnt just c/a

[reef fable]

so she/he say its not C/A?

[feral lagoon]

its not

and i think its fair cuz flipping (th or ht) is 1/2

[reef fable]

alright c is 150 if you agree

[plain heath]

The axioms of probability only give you for the ordered outcomes:

$\Omega={HH,HT,TH,TT}$

$\mathbb{P}(E)=\frac{|E|}{|\Omega|}$

[elfin berry]

flynger

[plain heath]

$\mathbb{P}({HT,TH})=\frac{|{HT,TH}|}{|\Omega|}=\frac{2}{4}$

[reef fable]

I think the total probabilities of outcomes must be 5^7 first hand

[elfin berry]

flynger

[reef fable]

and the number of ways to have 2 empty boxes is choosing 3 boxes having balls 5C3, as well as choosing stirling prob

[feral lagoon]

bruh

5c3 ways to choose occupied boxes, and 9c7 ways to stars and bars them into partitions

i dont even know what am i not understanding :sob

other than im often overlooking things

[plain heath]

for c be careful as you dont want any of the three boxes to be empty

So a trick you can do is just put one ball in each and calculate the combinations for the rest

[feral lagoon]

oh right

4 free balls in 3 boxes then

[feral lagoon]

man ill settle at this

.close

[marsh citrus]

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[surreal kindle]

Area in terms of a

[jagged relic]

,rccw

[jagged relic]

What did you try?

[marsh citrus]

@surreal kindle Has your question been resolved?

[wintry gale]

i think you scared em off

[opaque agate]

💀🙏

[surreal kindle]

Everything

And I mean everything

And nothing works

[marsh citrus]

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[surreal kindle]

.reopen

[marsh citrus]

✅ Original question: #help-33 message

[wintry gale]

its a tedious one for sure, requires A LOT of pythagoras theorem if im thinking this through correctly

[surreal kindle]

I mean

It might be required idek

Like no matter which way i try i always end up with something u cant get the area of

[sacred idol]

i know this is not very straightforward but you can split the red portion into two triangles by a vertical line and find the height of one of them by constructing something like the diagram given below

[surreal kindle]

How do u know triangles r split by 2h and h if u get me

Like how do u know the height of right is twice height of left

Even with that can u proceed?

[sacred idol]

the line segments with lengths b and 2b are parallel

theyre heights of equilateral triangles of sides a/2 and a respectively

[surreal kindle]

Oh nd its a similar

Yeah okay

I don’t see a way to progress if im honest

[sacred idol]

can you find the height h of the smaller triangle from here

[surreal kindle]

A/12?

[sacred idol]

what about the height of the larger triangle then

[surreal kindle]

Omg

Im acc blind

Bro how tf do u come up with this

[sacred idol]

could say i got lucky

i messed up a few times

[surreal kindle]

Like ive worked on this for 2h no troll doing some comparing areas and stuff

[sacred idol]

trig wouldve made it easier

[surreal kindle]

But nothing worked

I mean how do u see something like that

Problems like these just make me feel restarted

[sacred idol]

honestly its not that big of a deal if you dont see a construction

since you can finish it just as easy with later concepts

[surreal kindle]

Okay

Thank you so much

.close

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[magic plaza]

I need explanation about derivation of sine double-angle formula
Source:https://www.onemathematicalcat.org/Math/Precalculus_obj/doubleAngleFormulas.htm

So, how does sinx+sinx over 1 over cosx equal to 2sinxcosx? Where does 1 from 1/cosx disappear?

[magic plaza]

Does it work like x/x = 1, so 1/cosx = cosx?

[glossy flint]

1/cosx is not at all equal to cosx 🤔

[glossy flint]

!xyp

[marsh citrus]

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.

[magic plaza]

Then I'm really confused

[glossy flint]

Me too honestly

[magic plaza]

So it says that, sin(2x)=opp./hyp. = sin(x) + sin(x) / 1/cos(x) = 2sin(x)cos(x).

I'm totally stuck now if it doesn't work this way

[glossy flint]

Can you send a picture of these things?

[magic plaza]

That's the triangle given for example

My bad, I forgot to send it too

[still temple]

Huh?

[magic plaza]

As it says, AFD triangle is used for the sine double-angle formula

[jagged relic]

,, \frac{\sin(x)+\sin(x)}{\frac{1}{\cos(x)}} = 2\sin(x)\cos(x)

[elfin berry]

Nel

[bright pawn]

?

[still temple]

normal fraction

[magic plaza]

But why 1/cos(x) just disappears?

Oh wait

Okay i understood how it works

.close

[marsh citrus]

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[dire basalt]

Find the number of positive integral solutions to the equation $$x+y+z=2010,$$ with $x \leq y \leq z.$

[elfin berry]

robin.dabanc_

[dire basalt]

My solution:

Let $y = x+1+\delta_1, z = x+1+\delta_2$, where $\delta_i \geq 0$. Now note that the equation reduces to $$3x+t = 2008,$$ where $t = \delta_1+\delta_2 \geq 0$. Now I'm setting up a bijection: for every pair $(x,t)$ that is a solution, it can be uniquely decomposed into $\delta_1+\delta_2 = t$, having $t+1$ solutions each. So our final count reduces to the sum $$\sum_{x=0}^{669} \binom{t+2-1}{2-1} = \sum_{x=0}^{669} (2009 - 3x).$$ This evaluates to $$2009(670) - 3(669)(335) = 673,685.$$ Can someone verify the logic steps behind this solution?

[elfin berry]

robin.dabanc_

[lusty pond]

if x,y,z can be equal, then why the x+1+delta and not just x+delta

[dire basalt]

oh right

wait then

[lusty pond]

do you account for delta_2≥delta_1 aswell?

delta_2 + delta_1 = t having t+1 gives all the possible solutions

[dire basalt]

right...

mb

[echo sinew]

take x, x+a, x+a+b
so 3x + 2a + b = 2010

[lusty pond]

you should perhaps take all the possible solutions, then deal with the ones which have equal x,y,z seperately

[sleek lake]

what then?

[dire basalt]

youd have to do inclusion exclusion

[lusty pond]

yep

[dire basalt]

since $x = y \neq z$ possible too

[elfin berry]

robin.dabanc_

[dire basalt]

thatd be a bit long

[lusty pond]

more like, take all the solutions,
then you can figure out which solutions have x=y≠z or x=y=z

and the remaining solutions have distinct x,y,z

divide the remaining solutions by 6! to satisfy x<y<z

and then deal with the x=y≠z and x=y=z

[dire basalt]

you dont need to subtract x=y=z

equality is allowed

and so you cant just divide by 6! in that case

[lusty pond]

we're isolating the x,y,z solutions where x,y,z are distinct

so we take out rhe x=y=z solutions for that

and then we divide by 6!

[dire basalt]

ah ok

[lusty pond]

3! *

mh

mb

[dire basalt]

and then add back the equal ones?

[lusty pond]

yep

[dire basalt]

ye 3!

alr ima try that

[lusty pond]

been writing 6! like an idioy

bruh

[dire basalt]

isnt it gonna be a hassle to subtract x=y not z

oh not really nvm

[sleek lake]

with this you get like
3x+2a <= 2010
sum 1004, 1003, 1001, 1000...

[echo sinew]

yeah im trying that but idk how to find this summation

[dire basalt]

okay there are only two cases for two equals and one not equals

$x=z \neq y$ is impossible

[elfin berry]

robin.dabanc_

[dire basalt]

(by squeeze)

[echo sinew]

[] is greatest integer function

or floor

[dire basalt]

alright so ive found 1003 solutions where two things are equal and 1 solution where all are equal

and the total solutions is 2009 choose 2

(2017036-1004)/6+1+1003

is this the count you were refering to?

yo it isnt even an integer wth

[echo sinew]

do you know the answer? im getting 336675

[dire basalt]

idk the answer, im getting 337009.333

[dire basalt]

how did you compute it

[echo sinew]

.

for the summation [3k/2] take cases if k is even and odd and the sum is easy

i wrote =2007 to remove the positive integer restriction, now they can be 0 too

[dire basalt]

wait so you wrote x+(x+a)+(x+a+b) = 2010

[echo sinew]

yes

[dire basalt]

what is c

[echo sinew]

also x as 1+c

[dire basalt]

ah fine

[sleek lake]

it's probably like, 1,4,7,10...1003 + 2,5,8,11,1004 so 335 + 3(334)(335)/2 (for the first term)

336675

[echo sinew]

yes

[sleek lake]

i knew there's a direct method like that, but couldn't remember it

so how does this work

[dire basalt]

YES

this is the standard way

[dire basalt]

think of n balls and r bars

note that any solution can be represented by spacing the bars between the bars

for example oo|oo|o represents 2, 2, 1

[sleek lake]

i know stars and bars, how did they make it work here

[dire basalt]

also that calculation is wrong

2009x1004 is not 2017072

[sleek lake]

i see

ignore the screenshot then

[dire basalt]

also witht he sum im getting 200 thousand smth

wheres my error

[echo sinew]

why till 3*167?

first sum goes till 3*334

[dire basalt]

AH RIGHT

such stupid errors

yeah got it

thanks a lot guys

[lusty pond]

you have to subtract the x=z<y solutions too, theyre in the total solutions computed
(2017036-1003-1003-1003-1)/6 gives 335671

and lets say for the case x=y≠z
2x+z=2010
z is 2010-2x
accouting for x<z, x<2010-2x
or x=67
this gives only 669 valid solutions
doing so for the other case gives us 334 since (1003-669 is 334)
335671+669+334+1 gives 336675

[lusty pond]

how does this account for x<=y<=z

[dire basalt]

it doesnt

very faulty

[sleek lake]

it's just fake

[echo sinew]

how did it get the right answer lmao

just copied from somehwre?

[dire basalt]

okay 1003 in each is due to symmetry?

[lusty pond]

yes

[dire basalt]

in the last adding back thing there are only two cases right

[lusty pond]

x=y<z has 669 valid solutions
x<y=z thus has the remaining 334 valid solutions

[dire basalt]

because of this

[lusty pond]

x=z isnt possible

[dire basalt]

got it

[lusty pond]

yes

[sleek lake]

there must be some cooler way

[lusty pond]

this is fast enough if you know what you're doing

[dire basalt]

both methods were pretty neat

alright ima head to bed if this thread doesnt time out and someone finds a cool method pls send it here thanks in advance 🙏

[echo sinew]

nah a+2b+3c+4d+.. = n type equations have no general way to solve

like if it was 4 nunbers instead of 3 itd be much harder to solve

[echo sinew]

i mean there are some but they're too advanced for me idk

its similar to partition function

[lusty pond]

partition yea

[sleek lake]

you can just do (1004+1)/2*670

you probably noticed

[echo sinew]

how?

[sleek lake]

like it's not different enough from a constant difference, i don't know

if you add it to itself reversed it's just 1005, 1005, 1005, 1005...

[echo sinew]

i mesn this is sum of AP with first term 1 last term 1004 and total terms 670 right? how did you get that

[marsh citrus]

@dire basalt Has your question been resolved?

[sleek lake]

i just realzied it's partitions lol

[marsh citrus]

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[tame onyx]

Given a natural even number n, prove that $n^{3}+20n$ is divisible by 48.

[tame onyx]

Having reached that moment, can I stack induction proof and prove that for k=1 the third factor is divisble by 6 and for every k+1?

[still light]

,rotate ccw

[still light]

did you mean to say divisible by 48

[tame onyx]

I already have an 8 at the beginning, so I only need to multiply it by 6

[still light]

right

[tame onyx]

its $8(k+1)(k^{2}+2k+6)$

[elfin berry]

EntuzjastaPKP

[still light]

here's an idea $k^2 + 2k + 6 \equiv k^2 + 2k \pmod{6}$

[elfin berry]

Ari

[still light]

and $k^2 + 2k = k(k+2)$

[elfin berry]

Ari

[tame onyx]

and you get k(k+1)(k+2)+6(k+1), first element has numbers divisible by 2 and 3 therefore is divisble by 6, second is obv divisible by 6 meaning the whole is divisible by 6

but I also kind of wanted to spam induction for fun

but ig that question would better be suited for #math-discussion

[tough raft]

Are you sure its 40

?

[tame onyx]

where

[tough raft]

Here

[tame onyx]

yeah Im sorry its late

[tough raft]

Ig it should be 48

[tame onyx]

.close

[marsh citrus]

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[boreal zenith]

i dont know how to start

[quiet anvil]

So do you understand the idea of proof by contrapositive?

[boreal zenith]

I have to show notQ then NotP

thats all i know about it

[quiet anvil]

alright, so not Q => not P is what statement in this case?

[boreal zenith]

either a or b is even is notq

[quiet anvil]

not quite!

what you have is not (a or b is odd)

which is not the same as (a or b is even)

instead it's (a and b are even)

do you see why?

[boreal zenith]

i think

cause you distribute the not?

[quiet anvil]

~(r or s) = ~r and ~s, de morgan's law.

[boreal zenith]

oh okay

[quiet anvil]

https://en.wikipedia.org/wiki/De_Morgan's_laws

[boreal zenith]

so a and b are even

[quiet anvil]

yes, because if either one of them were odd, then a or b would be true so not(a or b) would be false.

so the only way not(a or b) is true is if they're both even.

[boreal zenith]

hmm okay

[quiet anvil]

not(1 is odd or 2 is odd) = not(true or false) = not(true) = false

for a specific example.

[boreal zenith]

okay

[quiet anvil]

so can you prove the contrapositive now?

[boreal zenith]

let a and b be integers, and assume both are even

i dont know thats the only one I know

A +B is even?

so there would only be 2 blocks?

[quiet anvil]

o7 that sounds correct to me.

[boreal zenith]

it says its too short..

[quiet anvil]

really?

[boreal zenith]

yeah

[quiet anvil]

oh 3 blocks

sum of two even integers is also even

[boreal zenith]

yep thats right

im gonna look at the next one and try that

nvm

.close

[marsh citrus]

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[untold spruce]

Is Axiom of Choice being used here to choose these vectors?

[untold spruce]

I don't think so, since this is a finite number of choices

and not infinitely many choices

so it seems to be allowed?

[devout mauve]

well you need some form of choice but not the full version, yes

cause its just a finite choice

[marsh citrus]

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[marsh citrus]

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[blazing torrent]

can someone solve this for me

[blazing torrent]

@topaz escarp need u boss

i m going crazy

[quiet anvil]

!noans

[marsh citrus]

The purpose of this server is to help you learn; please don't ask for direct answers. Ask for guidance, explanations, or feedback instead.

[quiet anvil]

!noping

[marsh citrus]

Please do not ping individual helpers unprompted.

[quiet anvil]

@blazing torrent ^

What have you tried so far?

[marsh citrus]

@blazing torrent Has your question been resolved?

[marsh citrus]

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[sacred totem]

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

\begin{Theorem}
Assume $r : (-\infty, 0) \to (-\infty, 0)$ where $r(x) = -|x + 4|$. Then, $r$ is a surjection.
\end{Theorem}

\begin{proof}
Let $x = b - 4$ where $b \in (-\infty, 0)$.
Since $b - 4 \in (-\infty, 0)$, $x$ is also in $(-\infty, 0)$.
Observe that
\begin{align*}
     -|x + 4| &= -|b - 4 + 4| \\
              &= -|b| \\
              &= -(-b) \\
              &= b 
\end{align*}
Since $b$ was chosen arbitrarily, we have shown that for every $b \in (-\infty, 0)$, 
there exists an $x \in (-\infty, 0)$ such that
$r(x) = b$.
Therefore, by the definition of surjection, $r$ is a surjection.
\end{proof}

[elfin berry]

Mor Bras

[cunning fiber]

um

,w -|x+4|, x=-4

[elfin berry]

Results provided by WolframAlpha

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

Requested by civil_service_pigeon

[cunning fiber]

is the question written correctly

[crystal lintel]

also as an aside... labeling this a "theorem" irks me lol

not saying it's against the law, just that it irks me

[delicate moat]

Observation 1

[static quarry]

the famous Theorem 1, i cite it often

[marsh citrus]

@sacred totem Has your question been resolved?

[sacred totem]

Surely this is an error on the part of the author since $r$ is a function. Changing $r$ to be $r : (-\infty, 0) \to (-\infty, 0]$, then the proof would be OK?

[elfin berry]

Mor Bras

[crystal lintel]

sure

[sacred totem]

.close

[marsh citrus]

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[devout sorrel]

help

[devout sorrel]

idk how to factor this

like im supposed to do (5x____)(x_)

right?

but then what nex

[exotic rampart]

you can use the quadratic formula

but theres an easier way to do it

[devout sorrel]

i know but im trying to use that method

cause thats the method we are supposed to use

[exotic rampart]

okay so like

multiply 5*8 and find which two numbers multiply to 40 and add to 14

which are.....

[devout sorrel]

10 4

[exotic rampart]

right

so we can rewrite this as

5x^2+10x+4x+8

agree?

[devout sorrel]

yeah

[clear latch]

Or you can also use a simple cross.

[exotic rampart]

okay so

look at the first two terms and the second two

find the common factors

for the first is

5x

so you can do

and for the second is 4

so

5x(x+2)+4(x+2)

right?

[devout sorrel]

yeah

so 5x+4

and x+2

so zeros are -2, -4/5?

[exotic rampart]

yep

[devout sorrel]

ok ty

[exotic rampart]

np

[devout sorrel]

.close

[marsh citrus]

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[true creek]

integral of x^3/3^x

[pearl nacelle]

need help to proceed man

[wintry gale]

!occupied

[marsh citrus]

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).

[trail sleet]

Wow.

[true creek]

hello again

[trail sleet]

Integration by parts.

[true creek]

whats that

[true creek]

what the hell

[trail sleet]

Here's the formula.

[true creek]

okay

how do i know when this formula is supposed to be used

[trail sleet]

What?

Don't be scared.

There's even crazier one in the future.

[trail sleet]

When you see two functions.

[trail sleet]

Basically, you can split this into x³ and 3^(-x).

[true creek]

okay

wait so when is usub used

like over this

[trail sleet]

U sub failed here.

You have nothing to be substituted.

[kind mulch]

factor out x no?

then its just x^2/3

[plain marsh]

in the denominator its 3^x not 3x

[kind mulch]

oh mb

[wintry gale]

It's alright

[trail sleet]

Hehehe.

[wintry gale]

Weird giggle

[marsh citrus]

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[tardy oracle]

Let AB |||| GH|| EF || CD

[obtuse umbra]

Why spoilers on GH?

[tardy oracle]

Let the trapeziums ABHG, GHFE and EFCD have the same height. If the ratio of AB and CD is R, find area(GHFE):area(EFCD).

[tardy oracle]

suspense

[obtuse umbra]

So probably it's best to draw the heights

[tardy oracle]

The answer needs to be expressed in terms of R = AB/CD

[obtuse umbra]

Try drawing the heights, like without it you probably can't do anything

[tardy oracle]

AJ = JK = KI

[obtuse umbra]

Idk if thales could help here

[obtuse umbra]

I don't think JK and KI are the heights here

[tardy oracle]

I constructed AI perpendicular to CD.

By the corresponding angle axiom, we get angle JKF = 90 = angle AJH

= angle JAB

[obtuse umbra]

@tardy oracle

Should be GK and EI where GK and EI is perpendicular to the bases

[tardy oracle]

wth

[obtuse umbra]

Ah i see your point here

Mb

[tardy oracle]

yes

so AJ, JK and KI are the heights

[obtuse umbra]

Did you learn Thales by any chance?

[tardy oracle]

yes

[obtuse umbra]

So by Thales, EK = 2GJ and DI = 3GJ

Can you see it?

[tardy oracle]

yes

[obtuse umbra]

So if you draw the heights again, this time at the right side

You should see 3 rectangles forming

[obtuse umbra]

You can notice that ABMJ, JMNK, and KNLI are both rectangles

[tardy oracle]

yea

[obtuse umbra]

And also that NF = 2MH and LC = 3MH

So you can find the area of the 2 trapeziums that you mentioned above using only AB, AJ, GJ and MH

[tardy oracle]

i don't thik that's true

[obtuse umbra]

Why?

[tardy oracle]

it would imply thatABCD is isosceles

which it isn't; it's just a regular trapezium inside which are drawn two line segments

[obtuse umbra]

AGJ = BHM
AJG = BMH

So these 2 triangles are similar
And AJ = BM so they are congruent

[tardy oracle]

how AGJ = BHM

[obtuse umbra]

ABHG is a trapezium

Oh wait hold up

[tardy oracle]

yes, but does that mean it has to be isosceles?

[obtuse umbra]

My bad there

[tardy oracle]

let IL = x and GJ = y and MH = z
the ratio of the areas is equal to (GH+EF)/(EF+DC) = (x + y + z + 2x + y + 2z)/(2x + y + 2z + 3x + y + 3z)

[obtuse umbra]

.

[obtuse umbra]

Yep i think so

And AB/CD is just x/(x + 3y + 3z)

And CD is just x + 3y + 3z

So probably try to do some algebraic manipulations to get the ratios

[tardy oracle]

= (3x + 2y + 3z)/(5x + 2y + 5z) = (3x + 3z + AB + AB)/(3x + 3z + AB + 2x + 2z + AB) = (AB+CD)/(CD+AB+2/3(3x+3z))) = (AB+CD)/(AB + CD + 2/3(CD-AB)) = (3AB+3CD)/(3AB + 3CD + 2CD - 2AB) = (3AB + 3CD)/(AB + 5CD) = (3R + 3)/(R + 5) = 3(R+1)/(R+5)

is this correct

[obtuse umbra]

GJ is y tho

Not AB

[tardy oracle]

probably, because putting R = 1 (which means AB = CD) gives the ratio as 1, which is consistent with the area ratio

huh

[obtuse umbra]

What if R is smth different

Like in your diagram

[tardy oracle]

wdym

R = AB/CD

[obtuse umbra]

R might not be 1 tho

[obtuse umbra]

Like your diagram says

[tardy oracle]

i know, that was just a verification for R = 1, and since it worked, the final expression is likely to be correct

so i might not have made a mistake

[obtuse umbra]

The general expression might have something different

[tardy oracle]

i k r

we just need to find the error if there is one

And if we must find area(ABHG):area(GHFE), we may just replace R by 1/R in the final expression

[marsh citrus]

@tardy oracle Has your question been resolved?

[marsh citrus]

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[echo scaffold]

Could anybody please check these for me I don’t have an answer key and I am not confident at all

[fickle vessel]

which question?

[echo scaffold]

All of them… a lot of them are true or false tho

Yea I’m trying to redo it all because it’s from before break b but I forget everything so now I’m just stuck in a

[fickle vessel]

where you stuck at a?

[sharp viper]

yes, you're right

[echo scaffold]

Yea but now I don’t know how to do it anymore

[sharp viper]

your answer here (approx .11.817) is correct

[sharp viper]

so, you can find out what sin 80° is from a calculator

YH = 12sin 80°

$\approx 11.817$

[elfin berry]

Tillman

[echo scaffold]

Do I multiply the answer by 12

The .984

[sharp viper]

where did .984 come from?

oh hang on

that's sin 80°, alright

yes, multiply that by 12 to get the length of YH

[sharp viper]

you may want to rephrase your answer here for a
how you've written, it looks like sin 80° = 11.817

[echo scaffold]

Oh yea

How should I write it

[sharp viper]

simple, $\sin \ang{80} = \frac{YH}{12}$, so $YH = 12\sin \ang{80}$.

[elfin berry]

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

[sharp viper]

ugh how do i write degrees in latex

[sharp viper]

everywhere there's 80 here, assume that means 80°

[sharp viper]

for b, your idea is fine
note you need 1/2 * (two adjacent sides) * sin (angle between them)
here, you have YZ, ZX and the angle between them

make sure to substitute them right

[echo scaffold]

Wait I’m so confused

[sharp viper]

okay, what have you learnt?

[echo scaffold]

nothing really this is preview homework

[sharp viper]

what did you do to find the area of XYZ?

[echo scaffold]

1/2 • base • height

[sharp viper]

ah okay

well, then it's fine your ans is right, sorry

[echo scaffold]

What’s ans? Area?

[sharp viper]

your answer for the area is right

sorry for the confusion

[marsh citrus]

@echo scaffold Has your question been resolved?

[marsh citrus]

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[marsh citrus]

Please don't occupy multiple help channels.

[main idol]

.close

[marsh citrus]

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[frigid dust]

b

[main idol]

do you know parallelepiped formula

[frigid dust]

yes

i jst dont know

how the C works

like u dont know cords of C

ok

CD is 5

everything else

like CA and CB which are needed for formula

we dont know

[marsh citrus]

@frigid dust Has your question been resolved?

[main idol]

try picking arbitrary point for D, say lambda = 0. then solve for lambda using CD = 5 and that'll get you C

or if you want a geometric approach: https://interactivetextbooks.tudelft.nl/linear-algebra/Chapter5/DeterminantsGeometric.html has some nice formulas

[marsh citrus]

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[ashen lantern]

Hello, I need some help with my statistics assignment. The question was to find how many independent events I can generate if

[ashen lantern]

$\Gamma = {a,p,p,l,e},\ \Psi={b,a,n,a,n,a}$

[elfin berry]

letmesleep

[ashen lantern]

And each events cannot be empty set or the whole sample space

[signal stream]

Are the different letters outcomes? And is each p different or consider the same

[ashen lantern]

Example of outcome is (a,b) and I'm not sure if I understood correctly, but each p in \Gamma is considered different outcome

[indigo nest]

Seems like it'd be helpful to show us the original question since you're not sure

!original please

[marsh citrus]

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.

[ashen lantern]

The original problem is on Norwegian, but I'll try to translate it to English

The problem is question e, "How many independent events is it possible to construct on this probability space?"

[deep tide]

oh you translated gamma and psi

this would change the solution since {a, p, p, l, e} = {a, p, l, e} and {b, a, n, a, n, a} = {b, a, n}

[ashen lantern]

Yep, I tried to change gamma and psi since it was Norwegian word, but now I see it wasn't a good choice since there where many duplicate words

[deep tide]

in retrospect you could've just done it as {a, b, c, d, e, f} and {a, b, c, d, e, f, g} if you wanted to, since the elements themselves don't really matter as (ω₁, ω₂) is an ordered set

[ashen lantern]

.close

[marsh citrus]

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[exotic rampart]

sorry guys i need help with another problem 😭 🙏

im not sure why my answer is wron

wrong

[lavish pelican]

Like the other one u have to split it again into 2 summations

[exotic rampart]

right

[lavish pelican]

Is that what u did

[exotic rampart]

so 3^n/11^n

yes

[lavish pelican]

Could u send ur working then

Or type it out

[exotic rampart]

yeah gimme a sec

wait

nvm it worked 😭 🙏

im a little slow my bad guys

sorry for wasting ur time

[lavish pelican]

All good

[exotic rampart]

ty for ur time bro

[lavish pelican]

Not wasting time btw ur learning

[exotic rampart]

broo ty bro

.close

[marsh citrus]

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[exotic rampart]

i hate to keep disturbing ppl but

im lowk lost

i rewrote this as

((x-9)/7)^n

[wintry gale]

Uh huh

[exotic rampart]

i think it converges from

[wintry gale]

You're just trying to find the radius of convergence

[exotic rampart]

(0,infinity)

[wintry gale]

Do you know how to?

[exotic rampart]

I do not 😭

[wintry gale]

My bad interval of convergence

[wintry gale]

Try using the ratio test

[exotic rampart]

right

okay

so the ratio isss

lemme find it

[wintry gale]

u_n+1/u_n

[wintry gale]

Go for it

[exotic rampart]

okay i got

-9/8

for the ratio

[wintry gale]

That was fast lol

[exotic rampart]

wait isnt the ratio

a2/a1

a2 we plug 2 for n

[wintry gale]

Yeah as lim n tends to infinity

[wintry gale]

Nooo

[exotic rampart]

oh

😭

[wintry gale]

just plug n+1 for numerator and n for denominator

[exotic rampart]

ohhh

wait what

[wintry gale]

Google the ratio test

[exotic rampart]

im super confused 😭

ohhh

yeah thats what i did

i used a2 and a1

[wintry gale]

Can you show your working?

[exotic rampart]

yes

[exotic rampart]

i got a diff asnwer this time im dead

so the ratio is x-9/7

[wintry gale]

Eh

[exotic rampart]

i think

is that wrong 😭

[wintry gale]

You're plugged integral values

[exotic rampart]

right

oh wait do you not want me to plug in the values for n

[wintry gale]

You have plug n+1 and n

[exotic rampart]

you want me to keep it n and n+1?

ohhhhhhh

[wintry gale]

Yes

And lim n tends to inf

[exotic rampart]

okay so i got this then

i have to find the limit of that?

[wintry gale]

Yes after simplifying

[exotic rampart]

okay

uhhhh

how do i simplify that

[wintry gale]

See how you write a^n+1 as a^n times a^1

Btw i have to go too uni now for class, someone else will take over

Probably

[exotic rampart]

bro i got

like

no problem

i got

x-9/7

as the ratio

😭

[wintry gale]

Good

Now you got | x-9/7 | < 1

For it to converge

Adios lmfao

[exotic rampart]

byee

so is the interval

(-infinity, 7/9)

if someone comes pls tag me

[slim sparrow]

@exotic rampart

[exotic rampart]

Hi

[slim sparrow]

hey

[exotic rampart]

Hii

Can you help me with this problem kind sir 😭 🙏

Uhhh

[wintry gale]

How did you find this interval?

[exotic rampart]

I solved the interval by isolating x

[wintry gale]

Okay so you had this

[exotic rampart]

Yes

And I solved for x

[wintry gale]

So do you know how find intervals from absolute values?

[exotic rampart]

Also if you’re in class pls don’t worry abt helping me

[wintry gale]

Prof hasn't arrived yet.

[exotic rampart]

Isn’t it like find out where it’s negative and solve that differently

Negative of the equation

[wintry gale]

If you had | x | < 2

Then it will be -2 < x < 2 right?

[exotic rampart]

Yes

[wintry gale]

Do the same thing here

[exotic rampart]

Ohhh so would it be

(-7/9,7/9)

[dire basalt]

A good thing to do is plugin some value from your interval to check if it satisfies

[wintry gale]

Yes check inclusivity

You wanna take over robin?

[dire basalt]

Your initial interval with -infty doesn't

[dire basalt]

I gotchu

[exotic rampart]

Ohh right

That sounds like a good plan

[wintry gale]

Thanks!

[exotic rampart]

Ty for ur help mistav

[dire basalt]

So what would be your new interval?

[exotic rampart]

Is it not this

[dire basalt]

Do this

[exotic rampart]

Oh

[dire basalt]

-7/9 doesn't work

[dire basalt]

It's just like this with $t=x-\dfrac 7 9$

[exotic rampart]

Ohh wait

It’s gonna be

[elfin berry]

robin.dabanc_

[exotic rampart]

(2,7/9)

I lied again

It’s (2,16)

[dire basalt]

You're solving |t|<1.

[exotic rampart]

Right

-1<t<1

[dire basalt]

Now plug t back

[exotic rampart]

Okay so

-1<x-9/7<1

[dire basalt]

Yes

[exotic rampart]

-7<x-9<7

[dire basalt]

You mean 7x-9

A simple way to do it is adding 7/9 everywhere

[dire basalt]

From here

That immediately isolates x

[dire basalt]

16-7/9 is not less than 1

[exotic rampart]

Wait you lost me

[exotic rampart]

How’d you get this

[dire basalt]

You multiplied 7 throughout right?

[exotic rampart]

Yes

[dire basalt]

Yea so x becomes 7x

[exotic rampart]

Oh it’s mb I’m writing it wrong

It’s (x-9)/7

[dire basalt]

OHHHHH

I'm so sorry then -2,16 is right

[exotic rampart]

No it’s my fault 😭 🙏

Right okay

So the interval is -2,16

[dire basalt]

Quick check: end points give 1

That's how you know you're right

[exotic rampart]

Ohhh

Alright

So for the sum…

[dire basalt]

It's an infinite GP

For a given x within the radius of convergence

Can you recall the formula for the sum of an infinite GP?

[exotic rampart]

Uhh

Isn’t it

Wait lemme see

A/(1-r)

[dire basalt]

Yup

[exotic rampart]

Alr

Okay lemme try that

But wait

What abt the x

Would the sum answer have the x in it

uhhh

im ngl i kinda needa to finish this

if someone comes plez ping

and also my hw said the interval was wrong

brooo

if someone comes kindly ping me

@tight lance

sorry for disturbing but uh

ive been stuck on this problem for a while

<@&286206848099549185>

wrong one whoops

[lofty gyro]

ah, hi um... i know the post is pinned, but do you mind reposting here for easier reading?

[exotic rampart]

sure

[lofty gyro]

okay, so

its ((x-9)/7)^n

[exotic rampart]

yes

[lofty gyro]

can you tell what is the common ratio? r=?

[exotic rampart]

(x-9)/7

[lofty gyro]

and we need this to be <1 right?

[exotic rampart]

right

so it would be

(x-9)/7 < 1

[lofty gyro]

yea

and finally

A/(1-r)

[exotic rampart]

wait but

for the interval

[lofty gyro]

yes?

[exotic rampart]

wouldnt it be th absolute value or

[lofty gyro]

nah, ratio itself can be negative too, so you dont need to add absolute

but just this question is +

[exotic rampart]

ah okay

so we solve for x right?

it would be

0<(x-9)/7<1

well maybe not the 0

[lofty gyro]

oh dear, my bad

you are right

abs it is lol

so x should be within 2 to 16 right?

[exotic rampart]

oh lol

right

[lofty gyro]

next

[exotic rampart]

okay

[lofty gyro]

with the above range

we dont have to worry for the next question to be out of range

[exotic rampart]

right

[lofty gyro]

just plug in 1/(1-((x-9)/7))

and you'll get the final answer

[wintry gale]

Oh you still cooking? Nice

[exotic rampart]

but what would i plug in for x

[lofty gyro]

just x

[exotic rampart]

oh

[lofty gyro]

need to keep x

[exotic rampart]

oh shooot ur right

broo

[lofty gyro]

well done

[exotic rampart]

tysm bro

i appreciate you

a lot

[lofty gyro]

np

[exotic rampart]

wait if ur not too busy

can i get ur help with another problem..

[lofty gyro]

do post

[lofty gyro]

ah

similar similar

[exotic rampart]

kind of

but the n is negative

[lofty gyro]

1. whats the common ratio

oh

[exotic rampart]

uhhhhh

not entirely sure

[lofty gyro]

it's asking lol

sorry

[exotic rampart]

youre good

[lofty gyro]

i thought it need an expression

[exotic rampart]

it might ask that next but

[lofty gyro]

well, recall:
common ratio r = t_(n+1) ÷ t_n

[exotic rampart]

for now just the ratio

i can try to do the rest myself

[exotic rampart]

right

[lofty gyro]

so, just find any n that's in the sum

[exotic rampart]

right

[lofty gyro]

try to work out t_(n+1) and t_(n)

[lofty gyro]

and that

[exotic rampart]

what abt like

(1+c)^-3/(1+c)-2

[lofty gyro]

yea

so it's just 1/(1+c)

[exotic rampart]

right

[lofty gyro]

that'll be the common ratio

[exotic rampart]

ohhhh

im slow bro

im crine

[lofty gyro]

no worries

it's a good start