help-26

[stark storm]

PHOTOMATY

[chilly snow]

I USED LIKE

[stark storm]

PHOTOMATH

[chilly snow]

2 AI APPS

AND IT DIDNT

GIVE ME

[stark storm]

Hell nawl

[chilly snow]

ANYTHING

[stark storm]

Nah they lying

Hold on

Stay with me now...

[chilly snow]

😭

its okay thank you anyways 🙏

this math website is so crap

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[ocean terrace]

how did they get the order of the commutator subgroup is 1 or 47?

[wary tulip]

lagrange’s theorem. 47 is prime

[ocean terrace]

i got that H1 is a normal subgroup of G but im not sure what we can say from there?

[wary tulip]

if G’ is a subgroup of H_1, then |G’| divides |H_1|

[ocean terrace]

how do we know G' is a subgroup of H1? I only got that G' is a subgroup of G

and H1 is a normal subgroup of G

[wary tulip]

oh, so you’re wondering about the part right before the “which tells us that’

also H is a typo there right and it should be H_1

[ocean terrace]

oh i did not even read that but yes. i do not understand that part yes

[ocean terrace]

yes. i do not see why G' would be in H1 though

[wary tulip]

:hmmge:

[ocean terrace]

:hmmge:

omg twinsies

fellow nitroless user

[wary tulip]

loll

i am not sure. i am bad at algebra

[ocean terrace]

aw :( okie

algebra is pain

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[ocean terrace]

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[flat quiver]

I can’t figure out this problem at all. Read the textbook and was still confused. It says “the sum of two consecutive even integers is 118. A. Define a variable for the smaller integer.”

[craggy pond]

Alright, well consecutive even integers would be say, 4 and 6, or 18 and 20, or 1000000002 and 1000000004.

They're basically even integers that are only 2 numbers apart from each other

meaning if the smaller integer is say x,

the bigger integer would be x + 2

[flat quiver]

Ok

Thank you. I think I got it from here

Maybe

[craggy pond]

🫡

try it out and tell me what ya see if ya like

[flat quiver]

Ok

Could I do one more problem?

Also the previous one was x=58

Ima open another

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[sweet shard]

Show the full question

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@strong sable Has your question been resolved?

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[echo relic]

@knotty finch

[echo relic]

<@&286206848099549185>

[surreal spoke]

What does H. C. F mean in this context?

[echo relic]

highest common factork

[surreal spoke]

Ah i see

I would try getting the Prime factors of 45 and excluding those of the ones that compose n

(if n is not Prime)

[echo relic]

n belongs to natural number

[surreal spoke]

So i would get the Prime factors of 45 and exclude those and their multiples from a list ranging from 1-200

Then the remaining numbers in this list would be the Primes between 1-200 (excluding the Primes that are factors of 45)

Sorry i didnt read what Set B includes fully

You dont need more than 1-200 since you are supposed to calculate the sum of elements in the intersection

Know what i mean?

[echo relic]

In the problem, two sets ( A ) and ( B ) are defined as follows:

• Set ( A ): Consists of all natural numbers ( n ) such that the highest common factor (HCF) of ( n ) and 45 is 1, i.e., ( \text{HCF}(n, 45) = 1 ). This means ( n ) is coprime with 45, which implies that ( n ) must not be divisible by any of the prime factors of 45. Since ( 45 = 3^2 \times 5 ), the numbers in ( A ) cannot be divisible by 3 or 5.

• Set ( B ): Contains all even numbers between 2 and 100, i.e., ( B = { 2k : k \in {1, 2, ..., 50} } ). This means ( B = { 2, 4, 6, ..., 100 } ).

[thorny flame]

Mr. Aura Himself

[surreal spoke]

Im not sure on the last Part. It says that k can range from 1-100 so 2k could also be 200

[dire trench]

I got 5264 as answer

[echo relic]

This does not mean that ( 2k ) can go up to 200. It just means that for each ( k ) from 1 to 100, the elements of ( B ) are ( 2 \times k ), so the set ( B ) contains:

[thorny flame]

Mr. Aura Himself

[dire trench]

First find the sum of the elements of set B

Which is 10100

[echo relic]

ok

[dire trench]

Then substract those numbers which are multiples of 3 and 5

There are 33 multiples of 3 and 20 multiples of 5 in the set B

[echo relic]

bro mine comes 5050

[dire trench]

Thats the sum of integers from 1 to 100

Multiply it by 2

[echo relic]

reason

[dire trench]

U will get the right ans

[dire trench]

Set B is multiples of 2

[echo relic]

ok

[dire trench]

First term is 2 and last term is 200 with a common difference of 2

So its an Arithmatic sequence

[echo relic]

lasth term is 100

[echo relic]

This does not mean that ( 2k ) can go up to 200. It just means that for each ( k ) from 1 to 100, the elements of ( B ) are ( 2 \times k ), so the set ( B ) contains:

[dire trench]

No its 200 in Set B

[thorny flame]

Mr. Aura Himself

[surreal spoke]

Because B Ranges up to 200 not 100

[echo relic]

ok

[dire trench]

Can you find the sum of an arithmetic sequence? If yes then apply it on set B. You will get 10100

[echo relic]

n/2(2a=(n-1)d)

[dire trench]

Now If we want to get n such that the H.C.F(n,45) = 1 then we have to find the multiple of 3's and 5's

[echo relic]

ok

[dire trench]

As 45= 3^2 × 5

In set B there are 33 multiples of 3 and 20 multiples of 5

So the sum of these multiples will be 3366+2100= 5466

[echo relic]

ok

[dire trench]

However we have overcounted here

We over counted the multiple of 3's while counting the multiples of 5's

[echo relic]

ok

[dire trench]

For example 15 is a multiple of 3 and 5

So what we have to do is to find the multiple of 15

[echo relic]

5*3

ok

[dire trench]

Yes thats why we want the multiples of 15

There are 6 multiples in Set B

Adding them we get 630

Now we will substract the sum of multiples of 3 and 5 from 10100

We get 4634

As we overcounted we will add 630

So answer is 5264

I hope you understood

[echo relic]

ok

i understood

[dire trench]

Noiceee

[echo relic]

ar eu an indian bro

[dire trench]

Nah

Im Bangladeshi

[echo relic]

ok bye

[dire trench]

Bye

[echo relic]

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[hallow cape]

can someone help me with some basic chemistry question, I can't find a chemistry help serverf

[hearty turret]

you can find it in #old-network

[hearty turret]

but basically, there'll be several ways of approaching this

but i think it's easiest to think about it like thus:

if we have 4.55g of X2O5 and 2.55g of it is X, then X makes up ~56% of the mass

so if we let its atomic weight be x

$\frac{2x}{2x+5 \cross 16} = \frac{2.55}{4.55}$

[thorny flame]

LY

[hearty turret]

then just solve for x

hope that helped

[hallow cape]

and that formula, where did you get it from?

5 x 16 is the mm

[hearty turret]

well we know that 56% of the mass of our thing is X

and if we had say V2O5

then like we had 100g of it and you wanted to find out the mass of V in that thing

u could do i.e.

$100 \cross \frac{2\cross 51}{2 \cross 51 + 5 \cross 16}$

[thorny flame]

LY

[hearty turret]

so basically we're just using the normal formula but with x as an unknown

hope that helped

[hallow cape]

oKay so x is the atomic number of the unknown element

right?

[hearty turret]

yh

[hallow cape]

thanjs

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[topaz sinew]

Please don't occupy multiple help channels.

[sweet shard]

!1c

[topaz sinew]

Please stick to your channel.

[sweet shard]

#help-32

[bleak sedge]

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[patent quest]

hey calclus question:

How do I know to apply l'hopital here?

So I know the numerator is like 1/0.0000001 = infinity

Denominator ln(sin(0.0000001)) is ...?

How can I calculate this

[whole geode]

ln(x) approaches -infinity as x approaches 0

[patent quest]

so sin(x) equals x?

[whole geode]

sin(x) approaches 0 as x approaches 0

[patent quest]

thanks

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[leaden jungle]

Guys I need help solving this problem I'm stumped on

[leaden jungle]

A total of $16,000 is invested in two corporate bonds that pay 3.5% and 5% simple interest.

The investor wants an annual interest income of $650 from the investments. What amount should be invested in the 3.5% bond?

To which I got

52,900 = x
-36,900 = y

But that... is not right

I feel so argh 😭 I don't know why I'm getting this wrong, I used to do these questions easily before

[night imp]

If interest is 3.5 percent on x, then amount will be 3 5x/100 not just 3.5x

Same for 5 percent

[leaden jungle]

So I should multiply both parts by 100?

[night imp]

Divide by 100

[leaden jungle]

Oh

[night imp]

5 percent of x = 5x/100 not 5x

[leaden jungle]

Okay lemme try

I didn't get it right

I thought I did

I feel sad

Okay so what I did was

350 = x

300 = y

Oh yeah I should've seen thst I was wrong

But where did I make my mistake?

I got too gung-ho at the end and just entered 350 as my answer 😦

[night imp]

-.035y+0.05y ?

[leaden jungle]

Yeah

[night imp]

You did correctly this time

[leaden jungle]

I did?

But I got the wrong answer 😭

[night imp]

Show me

[leaden jungle]

Question 13

I think

I did the problem right but

[night imp]

10000 is the answer

[leaden jungle]

I didn't put in the

Heah

😞

Aaaahhhh

Thank you

[night imp]

No problem

[leaden jungle]

I need to try and get better at math

[night imp]

You will

With practice

[leaden jungle]

Does simple interest always mean go back two decimal places?

[night imp]

Percentage always means /100

[leaden jungle]

Oh okay

So if I see another question with percentage

I should divide by 100?

[night imp]

Not exactly but if it says x percentage of y

Then xy/100

[leaden jungle]

Okay I'll try to remember that

Thank you you guys are so amazing

[night imp]

[leaden jungle]

I'll close rn

🥲🙏

YOU GUYS ARE AWESOME

. Close

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[sterile cedar]

hi hi. please help. I have my first calc 1 exam tmr and i am lost on quite a few things. professor is kinda chill but has his way of doing things.

[tame needle]

Idk why it turns red for the last part when I put (4n + 6)

[sterile cedar]

@tame needle who r u

i alr occupied this channel.

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[iron grove]

I'm really stuck on this intergration question can somebody help me please ?

[neon iron]

make x+2 = u

and distribute

[iron grove]

That's the part I don't understand because if I do that then I get the integral of 2xu^1/2 with respect to du. How do I integrate 2xu^1/2 ?

[neon iron]

recall x= u - 2 too, no?

[compact zodiac]

isolate x from here

and put that in

[iron grove]

Oh ok I'll give that a try. Thank you both I've never encountered a question like this before where you have to isolate the x so this is new for me

So I've gotten to this point but my signs are the wrong way round and I have no idea how

if u sub in u you get the integral of 2(-u+2)u^1/2

oh wait nevermind it's x=u-2

ooops

I'm having trouble with my surds could somebody help me with 2^5/2

I forgot how you deal with fractional powers

[vernal matrix]

Remember that (if $a$ is positive) that $a^{m/n} = \sqrt[n]{a^m} = (\sqrt[n]{a})^m$

[thorny flame]

@vernal matrix

[iron grove]

Oh so is 2^5/2 = (sqrt(2))^5 ?

I ended up getting the right answer. Thanks everybody for all of your help

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[tall kernel]

not sure how to solve this integral

[fallow heart]

Multiply top and bottom by e^x

[tall kernel]

just e^x?

[fallow heart]

Yes

[tall kernel]

k 1 sec

[compact zodiac]

nah

[tall kernel]

sorry, im still not sure how to solve it after this…

[compact zodiac]

look at the last step

[clear crescent]

$e^{x}e^{-x} \neq e^{-2x}$

[compact zodiac]

after you open the parantheses

[tall kernel]

whaaaaaaaat

[compact zodiac]

the powers ADD

[tall kernel]

oh sorry wait i get it

[clear crescent]

$e^{-x}e^{x} = e^{-x + x} = e^{0} = ; ?$

[compact zodiac]

wait isn't there e^x + e^-x in the original equation?

why did you write -

[tall kernel]

ohhhhhh

[tall kernel]

oh yeah sorry, its plus

[clear crescent]

Try to use a fitting u-sub now.

[tall kernel]

u= e^2x?

[clear crescent]

https://tenor.com/view/no-no-gif-26414895

[tall kernel]

oh damn

u=e^x?

[clear crescent]

It might be more clear if you rewrie e²ˣ = (eˣ)².

[clear crescent]

https://tenor.com/view/cat-nod-catnod-yes-vibe-gif-21529329

[tall kernel]

ok got it👍

[clear crescent]

[tall kernel]

lmao thanks

[clear crescent]

[tall kernel]

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[random notch]

what's the sum of all increasing arithmetic sequences of five positive 2-digit integers using between them, the digits 0-9 once each

[random notch]

ik if a = 9, d=9, a=9, d=18 and a=18, d=18 all are valid

but i'm confused how to proceed

oh and '09' is two digits

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[random notch]

.reopen

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✅

[random notch]

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[tight rivet]

Help

What do I do next

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[tight rivet]

<@&286206848099549185>

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[green pike]

oh wait forgot a part

so I'm struggling with creating a automaton for the language B

since I have to create one that only accept that language and rejects all other but I don't know anything about the members of the language other than that they have the prefix 21, 22 and there are about 200 of them, they also have a length of 9

is there another way to prove that B is a regular language other than creating an automaton?

[topaz sinew]

@green pike Has your question been resolved?

[green pike]

<@&286206848099549185>

[crude anchor]

I mean I don’t know if you’re there yet but if you can formulate a regular expression that describes the languages then they are regular

[green pike]

oh i haven't learned about that yet

[crude anchor]

Kleenes theorem it’s called

I’m sure you’ll get there

Do you have any theorems at your disposal?

Because otherwise I’m pretty sure that a finite automaton that recognizes the language is your best bet to show that the languages are regular

[green pike]

I have the one where if languages A and B are regular then, their union, intersection and concatenation are regular

[green pike]

yeah, but I'm not guaranteed to reject words that aren't in the language

[crude anchor]

Ok well maybe you can prove that the language containing a single student ID is regular first

Then using unions you could build up the language B

Since there are a finite number of students in your class you could just union all the single ID languages to get B

[green pike]

oh that's what I did with A

ahh

so I could say {0} is a regular language, {1} is regular and so on

and then union them?

[crude anchor]

Well I don’t think that’s how I would approach it

[green pike]

o

[crude anchor]

So you have the alphabet 0,1,2,…,9 right?

[green pike]

yes

[crude anchor]

Can you prove that the language defined by your ID alone is regular?

It’s a single word language

[green pike]

yes i can

I created a nfa that accepts only my student id

but my student id only contains 0,1,2,3

[crude anchor]

Ok but now consider an arbitrary classmates ID, you could create an nfa in the same manner for them

There’s shouldn’t be anything inherently different about it right?

Then you could in theory do the same for any classmates ID to prove they are all regular languages

And their union is B

I think I would attempt to formulate the proof along that line of reasoning

[green pike]

oh that makes sense

so no matter what id it is I can always create an nfa for it

how do you even put that formally though

[crude anchor]

I guess I would try abstract what you did to determine the nfa for your specific ID

What steps did you take to construct it?

Try to formulate those steps in such a way that it could be done to any ID you are given

[green pike]

ah

so I just have to define a transition function knowing the ID going in

theory of computation

[crude anchor]

Theory of computation

Jinx

[green pike]

guess i owe you a soda

[crude anchor]

Haha

[green pike]

depends

[crude anchor]

Yeah I think maybe try defining 10 variables in the alphabet for example A,B,C,…

And then you can define an arbitrary ID AS ABCDEFGHIJ

the you can define the state changes in terms of the variables

That way you have abstracted the process to be applied to any given ID

And you can from there argue that the union of all these languages is regular since any language containing one ID is also regular

[green pike]

alright, i'll try it out

[crude anchor]

Good luck

[green pike]

👍

[crude anchor]

Well in this math you don’t really work with numbers you work with “alphabets” and “words”

But not in the sense that you’re used to

And alphabet in this sense is any collection of characters tad a word is a combination of elements in the alphabet

[winter garnet]

anyone good at probability?

[crude anchor]

This course would be like 2nd or 3rd year cpsc

Not very likely to see it if you’re in math

Unless you go to study theoretical cpsc and computation for fun

Cpsc is pretty tough for a lot of people it’s a shock

It’s not really about coding

If you wanna learn to build software do software engineering

[green pike]

yeah there's a lot more proofs and math than i expected

[crude anchor]

Cpsc major is very close to a math major tbh just highly specialized aspects of math

Well it’s not that impossible tbh you have just taken a peek at a random problem from a course that you’re at least 3 years out from

So don’t let this be the thing that dissuades you from cpsc

But do know that cpsc has lots of math

But it’s also very different math from high school

[green pike]

you might actually start liking math too, cause there are some assignments where proofs have clever tricks you're supposed to figure out and it's pretty rewarding when you're able to do it

[crude anchor]

I switched from cpsc to pure math cause of my discrete math course

I disliked math in HS

Too boring

But in uni it becomes neat it’s like puzzle solving

I disliked it cause it’s too brain dead in HS they don’t teach you what stuff is how it works or what’s going on it’s more like this is a logarithm, this is a polynomial, these are what they do now go regurgitate it on the test

It’s really not a good way to teach math

And so many young folk are being misled about what math truly is cause really and teachers who are barely qualified are teaching them that it’s all formulas to memorize and boring as shit

Math pedagogy in the west is a travesty

[green pike]

yeah i felt the same way about calc cause in uni they teach you the how and why and that's way more interesting to me

[crude anchor]

That’s exactly the problem, but the sad thing is that no one wants to be a high school teacher so it’s just a Human Resources issue

Not just, but a lot of it is just a lack of well educated teachers

Haha best of luck with everything

[green pike]

I think I got it

[crude anchor]

It seems reasonable to me

I think that the first phrase needs a tiny change make it so that the claim is: one can construct an nfa such that it accepts the language defined by any single word from B

[green pike]

makes sense

i should probably also state that i'm using the theorem

[crude anchor]

And define the n_i

Also I thing the state changes from q1 and q2 aren’t right

It should be n_i in there as well

[green pike]

oh, that's because all ids start with either 22 or 21

[crude anchor]

But you should define n_1 as 2 and n_2 as either 1 or 2

That should also make the definition of the state changes succinct

Just use a recursive definition

d(q_i, n_i) = q_i+1

Il pretty sure if you just define the domains of the n_i’s then that would work

[green pike]

like the domain of n3 = {0,1,...,9}

[crude anchor]

I mean this is non deterministic so I guess the 2-1 thing is fine too

[crude anchor]

Yeah

I would do something like suppose w is a word in B, then it takes the form n_1 n_2 … n_9 where n_1 = 2 n_2 is in the set {1,2} and n_i for i>=3 is in the alphabet of B

Something like that I guess would work

[green pike]

yeah makes sense

for the recursive definition though, would you also say for i < 9

[crude anchor]

Yeah yeah definitely

Everything needs to be defined

And I guess I wouldn’t even want to reuse i twice

Just not the best practice

Probably not a big deal if you did

[green pike]

yeah my prof said they don't expect the most formal proofs

but it's probably good to do that especially for later math courses I'll have to take

[crude anchor]

Btw do you need to prove that the union of a finite collection of regular languages is regular

Because I assume you only have a pair for now

So maybe you’d want to prove that extension with recursion

Or your prof did already idk

[green pike]

oh you're right

my prof only taught us the proof for a pair

[crude anchor]

Could just do it as a lemma

Instead of using this specific case

Just price it generally for any finite collection of regular languages

Using the pairs thing

It’s easy enough to induct over the number of languages in the collection

[green pike]

yeah I think I should be able to do it by myself, thanks for the help!

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[karmic fox]

When do I use there formulas for Expected Value and Variance instead of the summations?

[karmic fox]

I tried using both on one problem and got a different answer

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[karmic fox]

.reopen

[topaz sinew]

✅

[karmic fox]

<@&286206848099549185>

[tight pasture]

A

[viscid path]

what?

[karmic fox]

huh

i need help with my question

😭😭

[winged belfry]

Hi @karmic fox it looks like you're referencing different types of random variables in your question

[karmic fox]

those are just examples from different distributions

my question is when do i use the summation vs the formula per distribution

[winged belfry]

Right binomial and bernoulli

[karmic fox]

yes and uniform

[winged belfry]

Hold on... I'm new here so I'll try to format LaTeX correctly

[karmic fox]

okay dont worry :)

i was wondering when i use those vs this. If i wasnt clear enough

[winged belfry]

Right OK. Thanks

The first is the formula for the mean of a discrete rv. The second is the variance of a discrete rv.

These can be used to derive the initial formulae you listed

[karmic fox]

yes, but to derive those formulas isnt it a summation to infinity?

[winged belfry]

No

Ah. That's an important question!

That's rarely the case. We sum over the values for which $P(x) \neq 0$

[thorny flame]

Stefano

[winged belfry]

Oh snap! LaTeX just works!

More precisely, the support of our random variable.

Does that make sense or might an example help?

[karmic fox]

it makes sense. I did a problem where I used the provided formulas and then did it with the summation but got different results. Did I make an error or should that be the case?

[winged belfry]

You should not get different results.

May I see the formula you used and the summation you used?

[karmic fox]

yes one second

[winged belfry]

(In particular, the summation with the bounds)

[karmic fox]

this was for a binomial distribution

4b is with the summations and 4c is with the formula

[winged belfry]

Ok. Thanks. Lemme take a look

[karmic fox]

i did a lot in my calculator so the math might be missing

[winged belfry]

It's all right, just gimme a few minutes to digest this

[vernal matrix]

Start summation at 0

(actually that shouldn't affect it, nvm-)

[winged belfry]

It looks like they are starting at 0. $\frac{2}{3}^0$

[thorny flame]

Stefano

[karmic fox]

it's not. It's a binomial so that is (2/3)^(n-x) :(

wups sorry replied to wrong person

[vernal matrix]

Actually wait, how are you doing these

Remember for the binomial (with $n$ trials, "success" probabiliy $p$) you have each $P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}$

[thorny flame]

@vernal matrix

[winged belfry]

So $p = \frac{1}{3}$

[thorny flame]

Stefano

[karmic fox]

okay ngl, I accidentally wrote the start of binomial then did geometric

[winged belfry]

Yeah

[karmic fox]

i wrote the formula wrong i think

😭😭😭

[winged belfry]

$q^{(n-k)}$

[karmic fox]

yeah 😞😞 okay thank you

[thorny flame]

Stefano

[karmic fox]

yeah i got it now thank you 😭😭 my mistake

[winged belfry]

It happens to all of us!

[karmic fox]

Ill retry and hopefully it goes better, thank you!

.close

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[winged belfry]

And it looks like you used that mean in the variance formula. Hopefully, your correction will smooth everything out

[topaz sinew]

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[tacit turret]

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[tacit turret]

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[tacit turret]

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[tacit turret]

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[agile harness]

my fault

so what condition must be met for a vertical asymptote to occur

[tacit turret]

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[agile harness]

two things really

[tacit turret]

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[agile harness]

the denominator must be zero

[tacit turret]

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[agile harness]

with a nonzero numerator

[tacit turret]

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[agile harness]

so let x=1

and set the denominator equal to zero

then set x=-2

for removable discontinuity

what must be satisfied to have a removable discontinuity

[tacit turret]

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[agile harness]

nope

0/0

[tacit turret]

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[agile harness]

yes

that is a removable hole discontinuity

[tacit turret]

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[agile harness]

set both the numerator and denominator equal to zero

with x=-2

[agile harness]

there is a removable hole when it’s 0/0

[tacit turret]

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[agile harness]

but x=-2

[tacit turret]

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[agile harness]

-2

[tacit turret]

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[agile harness]

yea

but also the denominator

=0

[tacit turret]

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[agile harness]

it’s +ax

[tacit turret]

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[agile harness]

and +b

[tacit turret]

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[agile harness]

yup

[tacit turret]

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[agile harness]

mhm

[tacit turret]

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[agile harness]

^

you need the other equation

[tacit turret]

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[agile harness]

you have two unknowns

[agile harness]

yea

and now you know c

[tacit turret]

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[waxen timber]

i have to claim if this is a vector space or not. i am not sure where to start

[vernal matrix]

,rccw

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[topaz sinew]

Please don't occupy multiple help channels.

[rigid prawn]

Oops

Can someone please explain to me why this log function curves like that

Like downward towards the asymptote

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[shy lance]

is i correct?

how to start ii?

[vernal matrix]

From your first part, you know that $f'(x) = 3 - \frac{14}{x^3}$, what happens when $x$ gets large in size?

[thorny flame]

@vernal matrix

[shy lance]

ok will try

[shy lance]

ahhh 14/x^3 gets smaller so like 3 - 0 (approx) when x gets larger

how do we show that

like do we just sub

@vernal matrix

[vernal matrix]

Yep, basically that

[shy lance]

wait sub what

like explain only?

[vernal matrix]

Well, I'd just argue that 14/x^3 gets smaller as x gets larger in size ("bigger denominator, larger fraction"), you can't really "sub infinity" into the function

[shy lance]

or need solns

[shy lance]

okay i see

[vernal matrix]

I would suspect you just need to explain it really,

[shy lance]

erm what abt like getting increasinly small

[vernal matrix]

Well, by "increasingly small", they probs mean "large negative", so like as you go to -inf

[shy lance]

but the statement says approaching 3 for both right?

or is it asking to check one of those only

[vernal matrix]

Well, yep, approaches 3 for both, but, like, as x goes to either +inf or -inf, you get f'(x) going to 3

"x goes to +inf" is the "x becomes increasingly large", and "x goes to -inf" is the "x becomes increasingly small"

[shy lance]

idont get the neg inf part

[shy lance]

is there a way i can confirm when the x goes -inf

coz when i tried computing only the +inf approach 3 for f'(x)

(direct sub, random large num)

[vernal matrix]

Well, you could either e.g. do it graphically, or choose a really large negative number (I really dislike their choice to say "increasingly small", it's an ambiguous term)

So e.g. try x = -1000000, x = -1000000000 etc whatever

Or, even-

,w lim 3 - 14/x^3, x to -inf

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[vernal matrix]

Just as-

,w lim 3 - 14/x^3, x to inf

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[shy lance]

should i just say that when x becomes inc small/big, 14/x^3 approaches zero
thus the func approaches 3?

[vernal matrix]

Yep, that'll do when x gets large positive/large negative, the fraction gets smaller and smaller

[shy lance]

thanks so much!

[vernal matrix]

Awwww

[shy lance]

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[hybrid bear]

i used 5, -4 and -5 -6 btw

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[hybrid bear]

UFN

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[tight shale]

hey just real quick check, i im taking ap calc right now but i didnt take pre calc, and i think im starting to grasp piecewise functions a bit but wanted to make sure, does this look right?

[rocky salmon]

Yup

[tight shale]

oh great :D

thanks for the clarification

[agile harness]

you’re welcome

[tight shale]

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[strange mortar]

need help with phrasing this

[strange mortar]

i got this into the form of a homogeneous system but idk if it helps

[odd pagoda]

how did you get it into that form

[strange mortar]

just substituting

cuz T(x) = 0 and T(x) = ax

so ax=0

the idea is just that a is a zero vector or smth ig

[odd pagoda]

from where do you have T(x)=ax

[strange mortar]

i mean T(vi)=0 then

[odd pagoda]

aka something very different

[strange mortar]

hm

so ig all i can say is that Avi = 0

[odd pagoda]

what is A

[strange mortar]

doesnt say a lot

[odd pagoda]

how can you relate the v back to the vi

[strange mortar]

its saying that all values of v when plugged into T produce 0

[odd pagoda]

well thats what you want to show

you dont know that yet

what do you know

[strange mortar]

v1....vp span R^n

domain and codomain are R^n

n = m so A is a square

[odd pagoda]

so, we know that T(vi)=0 for all i. and we know that the vi span R^n

we want to show T(x)=0

so roughly speaking, we want to show something for x and we know something about the vi

how can we relate the x and the vi

[strange mortar]

im unsure

im guessing the way to do so is to say that vi cant always produce the zero vector in T unless A is the zero vector

[odd pagoda]

we know two things about the vi. one of those things allows us to relate the x to the vi

which one

[strange mortar]

ig vi spanning R^n

cuz T(vi)=0 for all i only really says anything ab vi

[odd pagoda]

yes

good

so what does vi spanning R^n mean and how exactly does it relate the x to the vi

[strange mortar]

all possible linear combinations of v1....vp are equal to 0

[odd pagoda]

not at all

thats like basically the opposite of spanning

spanning means that for every x in R^n you find some linear combination of the vi which equals that x

for example x=a1v1+...+apvp

[strange mortar]

oops

[odd pagoda]

so then T(x)=T(a1v1+...+apvp)

what now

[strange mortar]

factor the constants

so its like T(x)=a1T(v1)+...apT(vp) ig?

[odd pagoda]

why can you do that?

[strange mortar]

this thing

[odd pagoda]

yes

[strange mortar]

oh

so we can substitute T(v1) to T(vp) with the zero vector

[odd pagoda]

yes

[strange mortar]

then that would mean T(x) = the zero vector

[odd pagoda]

yes

[strange mortar]

wow that was really helpful. thanks for ur patience dude

ill write smth formal rq

ok i wrote smth formal but i dont feel like taking an actual picture. if ur still here, i would like to know a little bit more of the formal reasoning for the span linear combination part, which i doofed up on.

so im guessing since we knew x was in the span of R^n we were able to write that part out

ok nvm i think i got it. will close this now. again thanks for all your patience.

def gonna study all of the conceptual stuff really hard 🫡

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[strong rain]

should i substitute lnx with u?

[strong rain]

I am not getting it.

[golden blade]

what exactly

[strong rain]

I cant solve the problem with substituting lnx with u!!!

[golden blade]

have you tried finding f'(x)

[strong rain]

yes!!. but for that i have to proceed with some method of differentiation!!

[golden blade]

yes

[strong rain]

but i am not getting how should I do it

[golden blade]

product rule?

[strong rain]

ooo. ok wait lemme me try it again

[golden blade]

you are having x and ln(x)²

[strong rain]

yes

[golden blade]

recall how was the product rule defined?

[strong rain]

uv'+u'v

[golden blade]

apply it

u = x
v = ln²(x)

[strong rain]

ok

@golden blade how can I find x from here?

[golden blade]

if you look closely, you notice it's a quadratic equation in disguise

[strong rain]

Oh right right

[golden blade]

so you might wanna do a mini substitution like z = ln(x)

[strong rain]

Right. Doing it

?

[strong rain]

Is this correct @golden blade

[golden blade]

yes

[strong rain]

thanks for helping😊

my ans is not matching. i cant find f'(t)

[golden blade]

why did you do?

[strong rain]

I did this only

[golden blade]

and what is the answer?

[golden blade]

this doesnt make sense

[strong rain]

am i wrong or the ans is wrong?

which one?

[golden blade]

yours is right, so much i can tell

[strong rain]

ohh. ok then. answers might be wrong sometimes dute to typing errors

[golden blade]

,w (1-ln(t))/t^2 = 1/t^2 + 2*ln(t)/t

[thorny flame]

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[golden blade]

the expressions are not identical

so their solution is wrong

but i wonder, it asks you to find f' and f''

yet there is just some random expression

[strong rain]

This is what I am getting while finding f"(t)

[golden blade]

looks also right

except

-2tln(t) should be +2tln(t)

[strong rain]

O thanks

I simplified this

[golden blade]

nice

[strong rain]

this question. should I use both product and quotient rule?

[golden blade]

yes

[strong rain]

i am out of mind. i am not understanding that how can i even use product and quotient rule both at the same time

[golden blade]

in the numerator you have a product

so when you use quotient rule, then at some point you will have to diff. xln(x)

for which you then need product rule

[thorny flame]

bacc

[golden blade]

This is how you can look at it

You apply as usually quotient rule

and when you do u'(x)

that's a'b + ab'

[strong rain]

oh u made it easier to think off. thaks

How can I find x now?

@golden blade

[golden blade]

I think you did a mistake

(1+ln(x))(1+x)

[inner crow]

Yeah that's what they show

They did write it in an incredibly strange way, but x * 1/x = 1, and lnx * 1 = lnx

[golden blade]

You are basically missing a (1+x)

in your final solution

[inner crow]

They have it to the right of the bracket that you put in

OHH you mean that they didn't actually put the brackets, you drew that yourself

Then yeah you're right, mb

[strong rain]

@golden blade is it okay now?

[golden blade]

no

[strong rain]

Now how can o find x

[golden blade]

(1+x)^2

not 1+x^2

[strong rain]

Sorry

Sorry

I am so dumb man

Now how to solve x

[golden blade]

weird task

You are basically having ln(x) = polynomial

btw

your first derivative has a maximum at y = 1/2

so there is not even a solution

[topaz sinew]

@strong rain Has your question been resolved?

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[neon iron]

g

[neon iron]

Vikas has Rs.
(
x
3
+
2
a
x
+
b
)
, with this money he can buy exactly (x-1) jeans or (x+1) shirts with no money left. How much money Vikas has if x=4?

[vestal totem]

Well that's not readable.

[vestal totem]

That's better.

you're supposed to have costs?

[neon iron]

total money

i have to find the total money

he has

@vestal totem

[vestal totem]

that's all the data?

[neon iron]

yea

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[keen solar]

why do all the signs switch

[vestal totem]

distributive property

[keen solar]

O

thanks

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[scenic coral]

You could also explain this because we are subtracting the whole generally the whole is added then subtracted since we have variables we can’t add them but turn them negative since only the sign changes not the magnitude

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[mortal heart]

Guys, for this question, am i allowed to cross the cos x like that?

[golden blade]

yes

[timid flame]

help me check if this is correct

[shut garden]

What is cos^2(x) in terms of sin^2(x)

[timid flame]

bruh sorry my internet so slow

[mortal heart]

1-sin^2x

[mortal heart]

Okay thanksss

[mortal heart]

Thank youu

[mortal steeple]

i would recommend remembering that you cancelled it

or write something like cosx != 0
so x!= 90, 270, ....

[shut garden]

$\cos{x} \neq 0$

[thorny flame]

Roman_Garland

[mortal steeple]

thnx

[mortal heart]

What does the ! mean?

[mortal steeple]

!= is not equal to

or else you would be cancelling 0 by 0

[mortal heart]

Ohh

Btw guys

How to solve 4tanx=3cosx?

[mortal heart]

Thank youuu

[mortal steeple]

4sinx / cosx = 3cosx
take cosx to the other side (cosx != 0)
4sinx = 3cos^2 x

[mortal heart]

Ohhhhh

[mortal heart]

Okayyy thank you so much

I can’t do inverse sin

My calculator says error

Btw i think it’s in radian bcs the domain is 0-2π

[untold shoal]

Just tell me range of sinx

[mortal heart]

-1 to 1

[mortal steeple]

Bro that's not how you solve a quadratic

@mortal heart

Let t = sinx

Solve for t

Then solve for x

[mortal heart]

Ohhhhhhh that is a quadratic? 😭

[mortal heart]

Ohh

I thought i can just factorise

Ive been doing it wrong

[mortal heart]

Btw thankyouuu smm

I’ll try it

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[cinder oxide]

With association law can we remove brackets entirely?

It doesn’t seem to make any difference with or without brackets, if all operators are the same

Even for the original expression using “if then” could we move the brackets or remove them?

p -> (q -> r)
(p -> q) -> r
p -> q -> r

In other words, all 3 of these are equivalent?

(Note: I am not using commutative law with “if then”, only associative)

[ruby tree]

I don't think implication is associative

[ruby tree]

Yes but not really

[outer salmon]

With p: F and r: F you have different truth values for the two

[glass maple]

u can draw a truth table do check as well

(if u want to suffer)

[ruby tree]

You can think of it like addition or multiplication, (a+b)+c = a+b+c = a+(b+c) because addition is associative

[cinder oxide]

So we are never allowed to remove brackets entirely?

Let’s say all operations are AND or V

[ruby tree]

But formally, the given rules don't work when you remove parentheses

Or rather, removing parentheses is simply disallowed

[outer salmon]

This does depend though, some do allow removing brackets

[cinder oxide]

If it’s entirely associative with same operation I don’t really see the harm in removing brackets

[ruby tree]

There is none, like I said it's similar to addition or multiplication

[cinder oxide]

It sounds like “if then” is the exception to this tho

[ruby tree]

AND and OR are binary operators, they take two inputs and produce one output, so formally you need to know exactly what the inputs are to every operator

[cinder oxide]

You mean if every operation is AND or OR?

[ruby tree]

If you have a AND b AND c, you don't know whether the first AND receives the inputs a and b, or the inputs a and (b AND c)

[outer salmon]

This is because P and Q or S

Is ambiguous (there’s two different interpretations based on how to set the parentheses)

[cinder oxide]

Everything would need to be true regardless. If AND is connecting all brackets and AND is the only operation. I don’t see how brackets would matter

[outer salmon]

Wouldn’t + be the same though

[ruby tree]

If you're making a circuit with logic gates, you have two ways to make this one

Either a AND b for the first gate, and then the result of that AND c

or the other order

It may physically matter

For the end value, it doesn't

[cinder oxide]

Doesn’t it just go left to right? Only 2 compared at any given time?

[ruby tree]

That would be a convention

[cinder oxide]

With no brackets it’s kinda implied.. first two have brackets

[ruby tree]

You could also just have an AND gate with three inputs

[cinder oxide]

That would be like first 2 with brackets right?

[ruby tree]

No, it would be (a AND b AND c)

[cinder oxide]

I haven’t done logic gates yet. Maybe this is where brackets are still important?

[ruby tree]

Or maybe written differently like AND(a,b,c)

[cinder oxide]

No lol I haven’t gotten there yet

[ruby tree]

I'm just inventing notation btw, just trying to get an idea of why it could matter

[outer salmon]

You can think of it like arithmetic. 1 * 2 + 3 has a meaning, but only because we defined that * comes before +. This means you can’t just omit the brackets for 2 * (3 + 4). It’s similar for logic, some prioritize AND over OR for example. But if all operators are only + or only * you can omit the brackets, and some do allow you to omit the brackets if all operators are OR or all operators are AND

[cinder oxide]

Yeah that’s how I would assume it would be interpreted without brackets

[ruby tree]

In practice I doubt the order of two AND gates would ever matter

[cinder oxide]

So brackets don’t matter, if all operations are exactly the same, is what you are saying?

[ruby tree]

If all you have in an expression is one associative operator repeated a number of times, brackets don't affect the end result

[outer salmon]

No, not at all. For implication it does matter

[cinder oxide]

OK TY

[outer salmon]

And depending on who you’re doing the task for, they may always matter

Just do as your prof tells you, or whoever you are writing logical expressions for

[cinder oxide]

Lol full circle back to beginning

[outer salmon]

If you’re uncertain if you’re allowed to omit from the instructor, just use brackets. That’s never actually wrong

[ruby tree]

It's nice to know about it because it can be a shortcut in your mind, but when doing an exercise you should conform to whatever the course uses

[cinder oxide]

But if it’s equivalent I don’t see the harm in doing so or how it could be marked as “wrong”

[outer salmon]

Because if it’s not defined then it’s undefined

[cinder oxide]

As long as your not removing brackets for -> if then statements

[ruby tree]

It could be marked as wrong because the material says so, that's all

[outer salmon]

And if what you’re writing is undefined you’re getting it marked wrong

[cinder oxide]

Haha that’s kinda absurd 🤣 maybe the textbook is wrong for saying so

[ruby tree]

Doesn't matter for whoever is grading you

[cinder oxide]

Wow so it’s just like “follow instructions and don’t ask questions”

Lol this is how a big power came to be in WW2

[ruby tree]

You should usually think of (a and (b and c)) as (a and b and c), but you should also remember that there is a formalism behind where parentheses are necessary

[cinder oxide]

OK

[ruby tree]

That's not really how it works... follow instructions, yes, to have the best chances at a higher grade, but you should ask questions

Ask you teacher if you'll get points off for omitting unnecessary parentheses

[ruby tree]

The thing is, if you can omit parentheses, then the "association" rule becomes useless in this kind of exercise

[cinder oxide]

Rather than “don’t ask questions” maybe “don’t object” would have been a better way to phrase it.

[outer salmon]

There’s nothing to object though

There’s nothing wrong with demanding brackets

[cinder oxide]

True but if your way is also correct that’s where you could object

Maybe it’s just a preference thing

[outer salmon]

But it’s not also correct if you haven’t defined it

[cinder oxide]

Countries teach maths slightly differently

[ruby tree]

I feel like you didn't quite understand that it depends on context

[outer salmon]

In maths you have to define EVERYTHING. If you write something you didn’t previously define, it’s wrong. If you haven’t defined how A or B or C is to be interpreted, it’s wrong. It’s quite simple in that way

[ruby tree]

In a very formal context, like in the picture you posted, you need parentheses, otherwise the solution would be different (at least one step would be removed, and the associated rule made useless)

In a more "informal" context, like when you just want to simplify a logic expression and the only thing that matters is that you get to the correct result, of course you can omit parentheses

[cinder oxide]

It’s funny where math seems to get “lazy” in some areas with adding parenthesis and other areas it can be super rigid (for no added benefit)

Look at tanx for example it’s almost never written as tan(x) and imo it would be helpful to always see that, x is the argument of the tan function. We don’t write fx we always write f(x)

With discrete math I guess we tend to write brackets even if not necessary

[ruby tree]

I always write tan(x)

[outer salmon]

If you write tanx and you haven’t defined it it’s wrong