help-13

[modern cairn]

well basically this is what a log really means

its a way of writing a^x=b

now imagine a is any positive integer

lets say a=2

for any value of x b can never suddenly become negative right?

[hoary stone]

yeah

[modern cairn]

same is true for not just 2 but any positive integer

so now do you understand why a log can never take negative values as input?

[hoary stone]

yes

[modern cairn]

great!

[hoary stone]

now my biggest problem

[modern cairn]

now lets get back to our question at hand

[hoary stone]

is trying to plot the points

[modern cairn]

yeah dw we are getting there

[hoary stone]

mb

[modern cairn]

now for this question

we know that whatevers inside the log should always be greater than zero

we just discussed that

so that means -x-1>0

right?

[hoary stone]

yes

[modern cairn]

now we just look for random values of x for which -x-1>0

can you suggest any random number for which this is true?

[hoary stone]

it can be any number as long as it’s greater than 0 right?

[modern cairn]

yeah thats the condition we got from log

but now i am asking for x

btw -x-1>0 just means x<-1 ,that is, x can be any number that is less than -1

do you understand?:

[hoary stone]

yes

[modern cairn]

great

we can actually clearly see this in our graph

can you see how values of x keep coming closer and closer to -1 but never equal to or greater than -1?

[hoary stone]

yes

[hoary stone]

yea i was thinking that

[modern cairn]

so thats basically our graph

if you didnt have a calculator for this, you can simply chug in random values of x(which are less than -1 of course) and get your answer

and voila!

you can easily get your graph

[hoary stone]

im cooked

@modern cairn it was marked as incorrect

[modern cairn]

what were the options?

[hoary stone]

i just the graphing part

i gotta actually draw the graph

[modern cairn]

the graph we discussed should be 100% correct

[hoary stone]

the app is stupid the drawing on the graph has to be 100% accurate for the question to be right

what is this bruh

[stiff moss]

What is the problem?

[hoary stone]

the drawings

look how

still marked incorrect

[stiff moss]

Were you supposed to trace like those arrows?

(like follow their path?)

[hoary stone]

it’s not even there in the first place

it shows you what it’s supposed to look like after you draw your own line

[stiff moss]

Where is this?

[hoary stone]

delta math

[stiff moss]

Huh.

This just seems like simple machine stupidity.

[hoary stone]

exactly

[stiff moss]

I cant tell what it thinks you are.

I dont think we can do anything.

All that glitters isn't gold....

[hoary stone]

Thank God i just got it

[stiff moss]

Or more like all that gives off UV Light isn't smart.

[hoary stone]

out of pure luck

[stiff moss]

Nice.

Could you go ahead and clear the channel?

with .close

[hoary stone]

gotchu

.close

[cedar kiln]

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[cobalt grotto]

I am confused, it seems to skip a few steps

where did sec(4x) come from, and where did 1/4 come from

This could be written alternatively as

$\sin(u)\cos^{-2}(u)$

[earnest geyser]

First take u=4x

[wraith dagger]

Eyesonjune

[cobalt grotto]

I believe

[earnest geyser]

So dx=1/4 du

[cobalt grotto]

the inner part of the integral at the end of the listed steps

[cobalt grotto]

How do you get this?

Integrate 4?

[earnest geyser]

U=4x

du=4dx

[cobalt grotto]

Are you saying that we need the derivative

[earnest geyser]

dx=1/4du

[earnest geyser]

Well that's how substitution works

[cobalt grotto]

I'm bad at u sub

💀

[earnest geyser]

Since we are converting it to u

We need integral of du

Like

One min

[cobalt grotto]

Is this a chain rule thing

[earnest geyser]

$\int tan(u)sec(u) dx$

[wraith dagger]

Monarch of Eternal Night

[earnest geyser]

We got till this after substituting

But now it's in dx but there is no x inside integral

So dx=1/4du

[cobalt grotto]

ok I get that part

[earnest geyser]

Which we got earlier

[cobalt grotto]

So then you can sub in that for dx

[earnest geyser]

Yes

[cobalt grotto]

Which turns it into an integral wrt u

and also adds a constant

Which you can factor out

[earnest geyser]

Yes

[cobalt grotto]

Ok so we are clearly using the 1/cos and sin/cos identities here to get sin/cos^2

But how do we go from that to sec

[earnest geyser]

There are two methods to do now

Directly

Differentiation of sec x is sec(x)tan(x)

So integration of sec(x)tan(x) is sec x

Do you want other method?

@cobalt grotto

[cobalt grotto]

yes

[earnest geyser]

First convert in terms of sin and cos

$\int \frac{sinu}{\cos^2u}du$

[wraith dagger]

Monarch of Eternal Night

[earnest geyser]

Now take cos u =t

Now -sinudu=dt

So it'll be

[cobalt grotto]

So another layer of substitution

[earnest geyser]

Yes

That's why I prefer direct method

[cobalt grotto]

Will this use u sub

I know I can factor out the 3 to begin with

[earnest geyser]

$-\int \frac{1}{t^2}dt$

[wraith dagger]

Monarch of Eternal Night

[cobalt grotto]

$3\int \sin^2 (r) \cos (r)dr$

[wraith dagger]

Eyesonjune

[earnest geyser]

For this since sin has power

You can take sin=u

Since differentiation of sin is cos

[cobalt grotto]

so it becomes cos * -cos?

Kinda confused

[cobalt grotto]

I assume you mean sin(r) right

[earnest geyser]

No?

[earnest geyser]

Yes

[cobalt grotto]

Lemme try rq

[earnest geyser]

So since sin r = u

[cobalt grotto]

$3 \int u\sin (r) \cos (r)dr$

[wraith dagger]

Eyesonjune

[cobalt grotto]

Where

$u=\sin (r)$

[earnest geyser]

Bruh

[wraith dagger]

Eyesonjune

[earnest geyser]

Why is there sin r

It should be u^2

[cobalt grotto]

oh

True

[earnest geyser]

And gtg, I'll be back in 10 mins

[cobalt grotto]

$3 \int u^2\cos (r)dr$

[earnest geyser]

And cos rdr is du

[wraith dagger]

Eyesonjune

[earnest geyser]

Since

U=sin r

[cobalt grotto]

d/dx sinx = cosx?

$u = sin(r)$

[wraith dagger]

Eyesonjune

[cobalt grotto]

$du = cos(r)dr$

[wraith dagger]

Eyesonjune

[cobalt grotto]

$dr = \frac{\cos (r)}{du}$

[wraith dagger]

Eyesonjune

[cobalt grotto]

I feel like this is wrong

I'm a bit confused by the step you did on the previous problem where you said

$u=4x$

[wraith dagger]

Eyesonjune

[cobalt grotto]

$du=4dx$

[wraith dagger]

Eyesonjune

[cobalt grotto]

How did you get that?

Just taking the derivative of 4x?

Where does the dx come from then?

[earnest geyser]

What the

What is this

[earnest geyser]

This is correct

[earnest geyser]

u=4x

now find du/dx

d(4x)dx= 4

Meaning

du=4dx

[cobalt grotto]

Well like how do I get dr by itself otherwise?

oops

divided wrong

It should be

$dr=\frac{du}{\cos(r)}$

[wraith dagger]

Eyesonjune

[cobalt grotto]

Right

Which is easier to just split into

$\sec(r)du$

[wraith dagger]

Eyesonjune

[earnest geyser]

It's correct but

[cobalt grotto]

So we get

[earnest geyser]

$3 \int u^2\cos (r)dr$

[cobalt grotto]

$3\int u^2\cos(r)\sec(r)du$

[wraith dagger]

Monarch of Eternal Night

Eyesonjune

[earnest geyser]

This is correct too

So you'll get

$\int u^2du$

[wraith dagger]

Monarch of Eternal Night

[cobalt grotto]

Because cos/cos right

So we end up with

[cobalt grotto]

well this

I forgot where to go after tho

just sub back in the u and integrate as normal wrt r?

$\int \sin^2(r)dr$?

[wraith dagger]

Eyesonjune

[earnest geyser]

No no

[earnest geyser]

cos r dr=du

So

Just substitute cos r dr

As du

[cobalt grotto]

So we get

[earnest geyser]

You'll get integral of u^2du

[cobalt grotto]

$3 \int \sin^2(r)\cos(r)dr$

[wraith dagger]

Eyesonjune

[cobalt grotto]

Wouldn't that just put us back where we started

[earnest geyser]

No man

[earnest geyser]

.

[cobalt grotto]

Where are we doing the substitution

[earnest geyser]

.

[earnest geyser]

In here

[cobalt grotto]

without substituting the sin(r)s back in yet?

[earnest geyser]

Leave sin^2 r as u^2

We won't change it back until last step

[cobalt grotto]

$3\int u^2\cos(r)dr$

[wraith dagger]

Eyesonjune

[earnest geyser]

Now since cos r dr =du, substitute inside the integral

Or do as you did earlier

$dr= \frac{du}{cos r}$

[wraith dagger]

Monarch of Eternal Night

[cobalt grotto]

If I sub back in dr I just get the same thing

3intu^2du

[earnest geyser]

That's it

$3\int u^2 du$

[cobalt grotto]

Yeah but we already had that step

[wraith dagger]

Monarch of Eternal Night

[cobalt grotto]

Without substituting du a second time

[earnest geyser]

Don't you know to follow through?

[cobalt grotto]

Probably not?

My question is

Why did we take the integral after substituting dr to get it in terms of du

and then re-substitute du in terms of dr

so that we got the exact same thing

[earnest geyser]

We don't need this

Where did we use this

[cobalt grotto]

We went from 3int(u^2)du to 3int(u^2cos(r)dr) right back to 3int(u^2cos(r)sec(r)du) -> 3int(u^2du)

[earnest geyser]

No?

First step

Is all we need

[cobalt grotto]

So what do we do with this?

[earnest geyser]

Integrate

[cobalt grotto]

Integrate u^2?

Too hard 🤣

1/3u^3

[earnest geyser]

Yes

[cobalt grotto]

making it u^3

sin(r)^3

[earnest geyser]

Yep

That's it

[earnest geyser]

So at first step we can already do it

[earnest geyser]

Integrate I mean

[cobalt grotto]

apparently that is wrong?

[earnest geyser]

Remove bracket

[cobalt grotto]

It automatically adds them

I also tried sin(r)^3

no dice

[earnest geyser]

The ans is sin^3 r

Something similar happened too a while bacl

Someone here got correct ans

[royal loom]

you forgot +C

[cobalt grotto]

these mfs

[cobalt grotto]

They always add the +C at the end

and this time they remove it for whatever fucking reason

example

They will pay for this

Thank you for the help @earnest geyser

.close

[cedar kiln]

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[cedar kiln]

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[mellow reef]

adjacency

[harsh comet]

How solve this question

[mellow reef]

!occupied

[cedar kiln]

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[mellow reef]

two graph with adjacency matricies are different

can they be isomorphic

?

[cedar kiln]

@mellow reef Has your question been resolved?

[cedar kiln]

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[stoic crystal]

In rectangle $ABCD$, we have $AB=8$, $BC=9$, $H$ is on $\overline{BC}$ with $BH=6$, $E$ is on $AD$ with $DE=4$, line $EC$ intersects line $AH$ at $G$, and $F$ is on line $AD$ with $\overline{GF}\perp \overline{AF}$. Find the length $\overline{GF}$.

[wraith dagger]

938c2cc0dcc05f2b68c4287040cfcf71

[cedar kiln]

@stoic crystal Has your question been resolved?

[cedar kiln]

@stoic crystal Has your question been resolved?

[stoic crystal]

@gleaming path

[potent verge]

Hello

[stoic crystal]

@storm harness

[stoic crystal]

<@&286206848099549185>

[quartz spear]

Let FD = x
Triangle CDE is similar to GFE
Therefore, 8/4 = GF/x + 4
GF = 2x + 8
Let P be the perpendicular intersection point of C to GF
Triangle GFA is similar to GPH
Therefore, (3 + x)/(2x) = (9 + x)/(2x + 8)
You can then solve for x and then substitute it into 2x + 8 to get GF

[stoic crystal]

?

can u do a drowing?

[stoic crystal]

.close

[cedar kiln]

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[desert atlas]

the answer was 12x18, solution is confusing/unclear

[desert atlas]

especially with this part

who is it 150/x +3 ?

shouldn't it be 150/x + 4?

[granite knoll]

no. 1 inch at the top and 2 inches at the bottom

[desert atlas]

can you provide a diagram?

[granite knoll]

no

[desert atlas]

ok

[cedar kiln]

@desert atlas Has your question been resolved?

[cedar kiln]

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[fresh quarry]

Please help

[fresh quarry]

How is it 1

[granite knoll]

it's a trick you have to notice I think

[fresh quarry]

Notice?
How

[crimson sedge]

@fresh quarry

[fresh quarry]

Yes

[crimson sedge]

You can see a lot of common terms right?

[fresh quarry]

Yeah

[crimson sedge]

Did you simplify it then?

[fresh quarry]

I did
And at last I get

(x^2023 + 2024x + 2025)(sinx+2) = 0

So sinx+2 cannot give a value of x
So it must be the other one

Now how to simplify that big one to get the roots

[crimson sedge]

Damn

[livid hound]

approach with rat root

[crimson sedge]

There is a common term In the denominator and the numerator

[livid hound]

you should be able to identify one by insepection

[crimson sedge]

Rat?

[livid hound]

rational

[crimson sedge]

Oh

[crimson sedge]

Mb

[fresh quarry]

It can be done
But is there any efficient way for this

Because you know it came in a competitive exam this year and like you get only 60 minutes to ace 60 marks from such many questions

[livid hound]

yes

identify that one solution

[crimson sedge]

That is sooo wrong

[livid hound]

i'll reveal the steps that follow

[fresh quarry]

Ikr but yk it's fun

[crimson sedge]

^

[fresh quarry]

Got it x=-1

[livid hound]

and the calculus to determine whether that component is monotonous
if it is, that'll be the only solution

[fresh quarry]

But how can I be sure that there are no other roots

[fresh quarry]

Ah ok yes

.solved

[cedar kiln]

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[livid hound]

if it isn't monotonus, you're in a world of hurt

but questions are usually set up to have an elegant approach

[cedar kiln]

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[dull oxide]

<@&268886789983436800> spam

[solar vector]

Stop spamming your form.

Probably your best bet is to ask @vague moth for permission/where you can post this.

[cedar kiln]

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[halcyon falcon]

how is he getting -2 in the numerator?
(value of i^2 is -1)

[livid hound]

simplify
-6-4(-1)

[halcyon falcon]

ookkk

[halcyon falcon]

thanks a lot, I was stuck on this for the past 5 min

.close

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[halcyon falcon]

.close

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[echo kiln]

https://ss.paartx.xyz/3S8MOAhManb0

[idle rain]

Damn. This question is popping up like 10 times already lol

[echo kiln]

Oh lmao

[idle rain]

so what do you need help with?

[echo kiln]

Youve got some previous answers

[idle rain]

you can use the properties of determinant to solve this

[echo kiln]

Im confused because of the t and x in the equation

Im not really familiar with the properties of the determinant yet

[idle rain]

I bet you have not learned about Linear algebra yet.

[echo kiln]

Correct

[idle rain]

It is hard to solve this without prior knowledge

but t here means transpose

not a variable

[echo kiln]

I'd like to know more about it though, was just confused by the 2 unknown values

Ohhh

[silver forge]

they shud use capital T smh

[wraith dagger]

print("NAME")

[silver forge]

makes sense

[idle rain]

it is not that uncommon to see transpose with lowercase

[wraith dagger]

print("NAME")

print("NAME")

print("NAME")

print("NAME")

[frozen spindle]

btw you can prove this very easily from multiplicity and the fact that det(I) = 1

[echo kiln]

Could you explain some more about transpose? Could also watch some videos rq if thats easier for you

[idle rain]

and the determinant of $2\times 2$ matrix $\begin{bmatrix}a & b\ c & d\end{bmatrix}$ is $ad - bc$

[wraith dagger]

print("NAME")

[idle rain]

transpose in short is just you change row with column. For example $\begin{bmatrix}a & b\ c & d\end{bmatrix}^\top = \begin{bmatrix}a & c\ b & d\end{bmatrix}$

[wraith dagger]

print("NAME")

[echo kiln]

Ahh

[idle rain]

I know this is game question so I want to ease you up a bit but if you want to understand all of these, you probably need to learn Linear Algebra first

[echo kiln]

Ill try some more stuff out then and research some more. Thanks for your time!

[frozen spindle]

game?

[robust harness]

it is roblox game actually

[frozen spindle]

how about that

[echo kiln]

Yep, just testing my knowledge, never had this though and I'm very interested in learning more

Dont want it easy or something, rather wanna understand it

[idle rain]

well, determinant concept alone would take the entire 3 hours lecture to explain lol

there is a nice intuition for 2x2 or 3x3 matrices but not the generalized version of determinant

like for 2x2, it can be thought as the area of parallelogram. here is the wikipedia image:

and as you might think, if this is a shape for each side of the 3d object (parallelepiped), then you can get the determinant of 3x3

[echo kiln]

Basically a 3d parallelogram?

[idle rain]

like this, again from Wikipedia:

[echo kiln]

Yea right

[idle rain]

and the rest is just use the properties of determinant to solve it. you will have linear equation (I think) to solve for $x$

[wraith dagger]

print("NAME")

[echo kiln]

Ill look some more into the specific properties of the determinant then.

Thanks a lot for your time! Ill see how far I can come with my limited knowledge atm.

.close

[cedar kiln]

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[barren path]

yo

I need help with that question ☝️

Let's say I choose a Ferris wheel for example

I'm not sure how to set up the question

Let's say it takes 15 minutes to make a full circle

It's like 10 meters high

[cedar kiln]

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[lone canopy]

hello

[lone canopy]

how is the integral of 2y : y^2 i thought it was y^2 / 2

[idle rain]

you can try differentiate the $\frac{y^2}{2}$ with respect to $y$ to see that is integral of $y$ not $2y$

[wraith dagger]

print("NAME")

[lone canopy]

oh

thanks

[barren path]

I wasn't done my question 💀

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@lone canopy Has your question been resolved?

[crimson sedge]

Integration of y is y^2/2. If you multiply this result with 2 then the 2s get cancelled and you have y^2

[cedar kiln]

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[tepid sonnet]

hi

[tepid sonnet]

i have a problem

[fervent jackal]

??

[tepid sonnet]

how did they go from the top line

to the bottom line

[fervent jackal]

try factoring out 30

[tepid sonnet]

i did it but i dont understand

how u get 9x-1

[fervent jackal]

$30((3x-1)^4 + 6x(3x-1)^3)$

wait let me try

i dont know yet

[wraith dagger]

N(μ,σ²)

[fervent jackal]

$30((3x-1)^{3}((3x-1) + 6x))$

[wraith dagger]

N(μ,σ²)

[fervent jackal]

yea

then u get 9x -1

any questions @tepid sonnet

[tepid sonnet]

but isnt it 6 times 3x-1

[fervent jackal]

no

u factor out (3x-1)^3

to make it easier

let y = (3x-1)

then u have $30y^{4} + 180xy^{3}$

[wraith dagger]

N(μ,σ²)

[fervent jackal]

first you factor out 30

you get $30(y^{4} + 6xy^{3})$

[tepid sonnet]

yh

[wraith dagger]

N(μ,σ²)

[tepid sonnet]

ok

[fervent jackal]

then you factor out y^3

you get $30(y^{3} (y + 6x))$

[wraith dagger]

N(μ,σ²)

[tepid sonnet]

u mean 3?

not 4

oh nvm

[fervent jackal]

yea my bad

then u put the values back in

you get $30((3x-1)^{3}((3x-1) + 6x))$

[wraith dagger]

N(μ,σ²)

[tepid sonnet]

i get it bro thx alot

[fervent jackal]

which is equal to $30(3x-1)^{3}(9x-1)$

[wraith dagger]

N(μ,σ²)

[fervent jackal]

no problem

can u do .close

??

[tepid sonnet]

.close

[cedar kiln]

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[cedar kiln]

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[rustic storm]

Hi, how would one come to solve such integral? I did find a (imo) complicated way tha uses Grouping to solve it but i was hoping there is a simpler way to calculate it

[soft lark]

ty for making a new one

we just must have been typing at the same time

[rustic storm]

Ye that was an unfortunate timing lmao

[round pagoda]

I think you should first notice, that (x^2-4x+9)'=2x-4, so you can do: 6x-10=1.5(2x-4)-4

so its 1.5*(2x-4)/(x^2-4x+9) (which integral simply to solve) -4/(x^2-4x+9)

[rustic storm]

.rcwu

Without the minus in the beginning. Aka in front of the 3/2.

[round pagoda]

yup

[rustic storm]

Noo idea what's next tho, how do i make the x^2-4x+9 part easier to integrate?

[round pagoda]

x^2-4x+9=(x^2-4x+4)+5=(x-2)^2+5

and put t=x-2

its easier cause there is just a number in the counter

[rustic storm]

Would be it wise to do it for both of equations?

[round pagoda]

no

the first part is just 3/2*ln(x^2-4x+9)

[rustic storm]

What happened to the 2x-4 what was above it?

[round pagoda]

Calculate the derivative of this

and youll see

[rustic storm]

Okay yeah

So that's the right way now?

[round pagoda]

how u get that 1/2x

put t on x-2

[round pagoda]

no no

x^2-4x+9=(x-2)^2+5

and u put t on x-2

so u get dx=dt

and its 1/(t^2+5)dt

and u just use it

[rustic storm]

Oh okay, why isn't my way correct tho? Did i break some math rules?

[round pagoda]

it just doens't help u

u got x and t at the same time

xd

[rustic storm]

Lmao

Okay shi lol

Hehe first time in a while that i didn't forget the + C

[round pagoda]

looks okay I guess

[rustic storm]

Calc gave me back this

[round pagoda]

oh yeah

my bad...

[rustic storm]

Yeah okay its correct, i just had to get the whole equation time two and then get the roots from the bottom of a fraction

[round pagoda]

at the beggining

[rustic storm]

No wait

[round pagoda]

no no, I've made a mistake at the beggining

[rustic storm]

4 times two aint 2

[round pagoda]

6x-10=3(2x-4)+2

[rustic storm]

Ye mb, go on teacj me sire

[round pagoda]

at the beggining it should look like this

so the whole idea is okay, u just need to correct the coefficients

[rustic storm]

Oooo yea mb i should have fact checked too and not mindlessly write it down

Thanks tho homie, your a big one 🙏

[round pagoda]

good luck with next integrals!

[rustic storm]

Thanks handsome!!!

.close

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[lone burrow]

I sort of moved on from this question a while ago but I read something that peaked my interest. So a definite integral gives you the area between two bounds a and b. We can evaluate the area to be F(b)-F(a) using the fundamental theorem of calculus. But, looking at an indefinite integral, it will return a function F(x)+C. If I assign x a value of b, does it give me all conceivable area up to the point b, + a constant C? Or is it truly meaningless on its own if not being subtracted by F(a)

[dull oxide]

If x=b, then area would be solved with C=-f(a)

[lone burrow]

So its just meaningless

[dull oxide]

basically

[lone burrow]

yeah alright

.close

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[cobalt grotto]

Trying to substitute but Idk where I am going wrong

$\int\frac{-23}{x^2\sqrt{x^2+25}}dx$

[wraith dagger]

Eyesonjune

[cobalt grotto]

$x=5\tan\theta$

[wraith dagger]

Eyesonjune

[cobalt grotto]

$dx=5\sec^2\theta d\theta$

[wraith dagger]

Eyesonjune

[cobalt grotto]

$\int\frac{-23(5sec^2\theta)}{25\tan^2\theta\sqrt{25\tan^2\theta+25}}d\theta$

[wraith dagger]

Eyesonjune

[cobalt grotto]

$\int\frac{-23\sec^2\theta}{5\tan^2\theta\sqrt{25(\tan^2\theta+1)}}d\theta$

[wraith dagger]

Eyesonjune

[cobalt grotto]

$-\frac{23}{5}\int\frac{\sec^2\theta}{\tan^2\theta\sqrt{25(\tan^2\theta+1)}}d\theta$

[wraith dagger]

Eyesonjune

[cobalt grotto]

$-\frac{23}{25}\int\frac{\sec^2\theta}{\tan^2\theta\sqrt{\tan^2\theta+1}}d\theta$

[wraith dagger]

Eyesonjune

[cobalt grotto]

$-\frac{23}{25}\int\frac{\sec^2\theta}{\tan^2\theta\sec\theta}d\theta$

[wraith dagger]

Eyesonjune

[cobalt grotto]

$-\frac{23}{25}\int\frac{\sec\theta}{\tan^2\theta}d\theta$

[wraith dagger]

Eyesonjune

[cobalt grotto]

$-\frac{23}{25}\int\frac{\cos\theta}{\sin^2\theta}d\theta$

[wraith dagger]

Eyesonjune

[cobalt grotto]

This would be the "fully simplified" integral right?

[grim rain]

Yes

[grim rain]

But here u wrote 23/5 instead of 23/25

[cobalt grotto]

oh

thx

.close

[cedar kiln]

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[tribal sonnet]

how to decide the start state of the converted DFA if the corresponding NFA has multiple start states?

[humble karma]

If the NFA has multiple starting states, you can create an equivalent NFA with a single start state from which you have epsilon transitions to every start state in the original NFA.

Then you can proceed as usual.

[cedar kiln]

@tribal sonnet Has your question been resolved?

[tribal sonnet]

but i converted the epsilon NFA to NFA 🥲

and it ended up with multiple start states

[humble karma]

I don't think this is supposed to happen

Having multiple start states is implicitly hiding multiple epsilons, so something went wrong

[tribal sonnet]

Step-3: See that if the vertex v1 is a start state or not. If vertex v1 is a start state, then we will also make vertex v2 as a start state. If vertex v1 is not a start state, then there will not be any change.

so is this wrong?

[humble karma]

It could be right, depending on how they explain how to do the NFA to DFA construction afterwards, but it is certainly not conventional.

[cedar kiln]

Channel closed

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[tribal sonnet]

.reopen

[cedar kiln]

✅

[tribal sonnet]

i have one more doubt

[humble karma]

Usually you check the eps-closures of every node and then try to "fix" the epsilons away. I can't see a reason to introduce more start states, but it could work with the method you're learning

[tribal sonnet]

i'm trying to convert this to DFA using subset construction

{q0,q2} ---> a ----> {q0,q2} but not q0

[tribal sonnet]

https://www.geeksforgeeks.org/conversion-of-epsilon-nfa-to-nfa/

it was just for epsilon NFA to NFA

[humble karma]

Yeah maybe they mean accepting states?

Or not

[tribal sonnet]

shouldnt it be just q0
q0 on a --> no transition
q2 on a --> q0
hence {q0,q2} on a ----> q0

[humble karma]

Anyways it's explained semi-clearly, and it looks like their NFA -> DFA doesn't show what to do with multiple initial states.

[tribal sonnet]

why is it {q0,q2}

[humble karma]

You have to do like

closure(transition(closure))

In the sense that

You start from q0, then see if you can go anywhere for free. Here you can go to q2 for free, so we consider {q0, q2}. Then we apply the "a" transition to that set, which yields {q0}, but then you have to check where you can go from q0 from there for free. This gives back {q0, q2}

[tribal sonnet]

q0 on a --> no transition
q2 on a --> q0 and also q2 following epsilon route from q0 hence {q0,q2}

like this? or this is wrong

[humble karma]

Is that meant to be the transition function after the conversion to NFA or just your train of thought?

The final NFA will have an "a" transition. To itself and to q2

[tribal sonnet]

conversion to DFA from epsilon NFA

[tribal sonnet]

what if instead of {q0} we had a combination of states

, which yields {q0,q4},
something like this

like {q0, q4}
just for example

what do we do then

consider the epsilon closure of both q0 and q4

and then add them in {q0,q4}

that will be the final ansswer?

[humble karma]

Yeah ok you're trying to go straight to DFA

You can shortcut your way through it like that yes. in the DFA, the starting state is {q0, q2}, but that doesn't mean it has multiple starting states. You just follow through the diagram from there

Like : starting, we can be in {q0,q2}. Then if we pick "a", we stay in {q0,q2}, if we pick "b", we go to {q1}.
Then you go from {q1} : picking "a", we get {q1,q2}. picking "b", we get {q2}.

And so on. Eventually you'll fill all the transitions.

Doing the eNFA -> NFA transition just makes this more verbose because you already have all the transitions to the new DFA "combined" states.

[cedar kiln]

@tribal sonnet Has your question been resolved?

Channel closed

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[tribal sonnet]

.reopen

[cedar kiln]

✅

[tribal sonnet]

ok thnx

[humble karma]

Oh and also now that I think of it, the multiple initial states just get bundled together if you have them, so you would just have them as one state in the DFA

So the link you sent was ok

[tribal sonnet]

bundled together?

[humble karma]

The final DFA's states are like bundles of the states of the NFA.

{q0,q2} becomes a state in the DFA

a single state

[tribal sonnet]

what if there's a state in DFA in which only one start state of NFA is present

[humble karma]

{q2} is still a state for instance

[tribal sonnet]

{q0q1q2}

and NFA has q2 q3 has start states

for example

so do we consider {q0q1q2} as start or not

in DFA

[humble karma]

For any state in the DFA, like {q0, q1, q2}, what we're encoding in this is that if we're in this state, the NFA could've found a way to be in q0, q1 or q2. But the NFA always makes the right guess, so if one of those is a final state, we consider the "composed" state to be a final state also.

There will only be one starting state, which will be the epsilon closure of the initial state in the original NFA>

[tribal sonnet]

so it doesnt matter to the DFA if the NFA has multiple start states

[tribal sonnet]

epsilon closure of the initial state in the original NFA>
initial state or initial states?

[humble karma]

Most textbooks will probably explain eNFA -> NFA without introducing multiple start states, but you can see it as taking the epsilon closure of all the initial states yes.

And especially if you're going to go straight from eNFA to DFA, then you probably won't need these starting states to begin with

[tribal sonnet]

yes okay

thnx

[humble karma]

No worries!

Sorry it took so long I could've explained better but it's been a year since I've done automata

[tribal sonnet]

man this course sucks i hate it lol

so tough

[humble karma]

Yeah it's very different compared to other comp classes

I major in maths too though so I guess I liked the proofy comp courses

[cedar kiln]

@tribal sonnet Has your question been resolved?

[cedar kiln]

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[frank kestrel]

[ \int 12x \cos(x^2) , dx ]

[wraith dagger]

qwerty

[naive tree]

take x^2 = t

[frank kestrel]

how do i determine the indefinite integral using analytical methods

[slate lintel]

u-substitution

[austere ember]

^ notice that the x term multiplied in front of the cos(x²) term looks a lot like the chain rule done for the inner function x²

[frank kestrel]

thanks i got it

🙏

for this i would just change C right

yea i think it is

[cedar kiln]

@frank kestrel Has your question been resolved?

[frank kestrel]

.close

[cedar kiln]

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[tepid sonnet]

hi

[tepid sonnet]

for ln(3x+1)

do you use chain rule to differentiate?

[upper ruin]

Yes

[tepid sonnet]

but wb for ln(3x)

[upper ruin]

This one can be done with the chain rule as well

Or if you want you can use log rules

[tepid sonnet]

can you show me please?

[upper ruin]

ln(3x) = ln(3) + ln(x)
=> D(ln(3x)) = D(ln(3)) + D(ln(x))
=> D(ln(3x)) = 1/x

[tepid sonnet]

what would you get for ln(3x+1)

[upper ruin]

This one can't be splitted

Because the inside of the log is not a product

Therefore, you are forced to use the chain rule

[tepid sonnet]

so if you use the chain rule what do you get?

[upper ruin]

( ln(3x+1) )' = 1/(3x + 1) • (3x+1)'

Hence you get 3/(3x+1)

[tepid sonnet]

how do you get 1/3x+1?

i get the other bit

oh wait

u would equal 3x+1

so when you do dy over du you get lnu

which would be 1/3x+1

.du/dx which would be 3

[upper ruin]

Yep

[tepid sonnet]

so u would get 3/3x+1

thanks alot bro

[upper ruin]

You're welcome 🤗

[tepid sonnet]

.close

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[lilac rune]

i need the formula for this

[lilac rune]

<@&286206848099549185>

[pseudo idol]

hi what's Sice and Sw called in english?

[pseudo idol]

.

[lilac rune]

ye?

[lilac rune]

its ice

and w is water

[pseudo idol]

no what S is reefering to, what physical magnitude

[pseudo idol]

.

[lilac rune]

oh its specific heat

[lilac rune]

.

[pseudo idol]

c is the same as S

you're gonna do some calculus to get teta f wich is the dinal temperture of the system

[lilac rune]

nah there is no need of calculus in this particular question

[lilac rune]

.

or ok thanks just needed formula

.close

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[fathom zealot]

<@&286206848099549185> may I get some help in this question?

[brittle wind]

which one

[fathom zealot]

@brittle wind

[brittle wind]

i can'tdo that :cri:

sry

also wait 15 minutes before pinginghelpers

[fathom zealot]

oh its alright 🙂

[fathom zealot]

👍

[earnest geyser]

Amplitude 4 means it changes 4 units from mean position

[short stratus]

so a+b = 4

[earnest geyser]

bsin(cx) maximum and minimum value are b and -b

[earnest geyser]

No

b=4

[short stratus]

oh yeah mb

[earnest geyser]

Can u understand?

How this comes?

[short stratus]

its not my qustion

[earnest geyser]

a is mean position

[short stratus]

i was trying to help

[fathom zealot]

I see

[earnest geyser]

Ohh

[fathom zealot]

umm I dont understand this?

[earnest geyser]

Do you know sin x maximum and minimum value?

[fathom zealot]

oh, may I know why is a +b is 4

[fathom zealot]

yes

[short stratus]

im wrong, ignore this

[fathom zealot]

oh alright

[earnest geyser]

Yeah so b(sin(cx)) maximum and minimum are respectively b(1) and b(-1)

[fathom zealot]

oh I see

[earnest geyser]

Meaning the graph moves from mean position (a) to a maximum of b on both sides

So amplitude is b

[fathom zealot]

I see

[earnest geyser]

Maximum units it moves is amplitude

Which here is b

We ignore a since it's just a position

[fathom zealot]

i see

[earnest geyser]

And then time period is pi/3

[fathom zealot]

I see

[earnest geyser]

2pi/c=pi/3

[fathom zealot]

Does that mean pi/3 times 2pi?

[earnest geyser]

$\frac{2\times\pi}c=\frac{\pi}3$

[wraith dagger]

Monarch of Eternal Night

[fathom zealot]

oh I see

[earnest geyser]

So now what is c?

[fathom zealot]

is it 6?

[earnest geyser]

Now we know both c and b

How to find a?

[fathom zealot]

Do i need to the substitude the (pi/2,2) in the equation?

[earnest geyser]

Yes

Then that's all

[fathom zealot]

Oh I see

Thank you

.close

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[fleet remnant]

can someone explain me what happened here?

[earnest geyser]

They are doing (a+b)(a-b)

To make denominator real number

[fleet remnant]

no not the red part

how they got this

[earnest geyser]

(1+i)^2

Equals 1-1+2i

So the inside part of bracket becomes i

2i/2

i^2015 is -i

Can u follow?

[fleet remnant]

yeah i get that part

i dont see what he did to get 5/3 k + (1/3 - 1) i

[earnest geyser]

That's it then?

Then (...)^2015 is -i

So -i+i/3 +5k/3-1

They just wrote it in different order

@fleet remnant

[fleet remnant]

okay okay i get that but how does he group it to ( 1/3 - 1 ) i

[earnest geyser]

It should be -2/3i

Idk why they didn't subtract

-i+i/3

Same as i(1/3-1)

[fleet remnant]

ohh now i see it it

should it be

-1 + 5/3 k - 2/3 i

[opaque root]

It is the same

[fleet remnant]

yeah i see it now i didnt get what he did with i for some reason now i feel dumb

thanks

.close

[cedar kiln]

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[crimson sedge]

Doing summer school right now and i can mostly do it but im stuck on this one right now

[digital cliff]

what have you got so far, any ideas?

[crimson sedge]

Fuckinnn nope

[digital cliff]

do you understand the function notation?

[crimson sedge]

Its algebra 2 with stat from when i was in 10th grade and im going into the 12th i dont remember any of this

[digital cliff]

is 300<=200 or >200

[crimson sedge]

Latter

Do i just plug in 300 on the x in the bottom equation

[digital cliff]

you do, yes

[crimson sedge]

Ohhh shit thanks a lot

i get it now

[crimson sedge]

So is the correct answer here C

Ok got it

Im thinking this is B here am i correct

Also dont think ive forgotten to close this if i havent typed in a bit i just got a lot of this to go thru and ill be in and out all day most likely

Ok it was indeed B

Ok ill close this for now im thinking im ok for now

Ige moved on to factors ans those are piss easy

.close

[cedar kiln]

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[cedar kiln]

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[sharp pike]

Would the interior of the 3rd set be the empty set?

[elfin hemlock]

Assuming standard (euclidean) topology, yes

[sharp pike]

Ok, so would the boundary then be the set itself and the exterior be the complement

[elfin hemlock]

Si

[sharp pike]

thanks ❤️

.close

[cedar kiln]

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[cedar kiln]

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[halcyon jackal]

There's a question that goes ike: What is s19 in the series: 8-16 + 32-64 +...

I got 4096 for that but apparently that's not one of the choices. What's the answer here?

[mighty shuttle]

is this an infinite series?

why would it even converge ?

[solid juniper]

i don't think it matters

[halcyon jackal]

why is it infinite?

[solid juniper]

s19 is prolly supposed to be the 19th term

or something

[mighty shuttle]

ah

[halcyon jackal]

yeah

[mighty shuttle]

makes sense

[solid juniper]

or like

19th partial sum i should say

[mighty shuttle]

the sum to the 19th term

right

ok break it into two sums

[halcyon jackal]

so -8 + -32...

[elfin hemlock]

So for a start how did you get a positive number as answer when all the terms are negative?

[halcyon jackal]

good question.

I think I used r=4, t1=-8, and n=19 and plugged that in

so like s19=-8(1-4^19) / 1-4

but that gives me -7.33 x 10^11

[elfin hemlock]

This is likely 4 terms, not 2

So when you do -8 + -32 + -128 + ...

It isn't the 19'th term of the partial sum you need anymore

[halcyon jackal]

so what would I do instead?

[elfin hemlock]

You can just define r correctly for the initial series you had and apply formula

[halcyon jackal]

So r=2?

[elfin hemlock]

Write out what the 4 first terms will be if r=2

to see if it matches

[halcyon jackal]

-8, -16, -32, -64?

[elfin hemlock]

So not that

[halcyon jackal]

why?

Idk what r would be other then 4 or 2?

[elfin hemlock]

what is (-1)^1 and (-1)^2?

[elfin hemlock]

Because the question had -8+16 and your now give -8-16?

[halcyon jackal]

I thought it was subtraction

Your saying it's 4 terms?

[elfin hemlock]

What?

You want to find the sum of the first 19 terms of 8-16+32-64+...

This can be written like a geometric series and then applying formula

But you need to define it correctly or you aren't finding the sum of the correct thing obviously

[cedar kiln]

@halcyon jackal Has your question been resolved?

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[calm bobcat]

for the function h=60sin(2πt/3), t is bigger or equal to 0

[calm bobcat]

assuming t is the period would be multiply that by 2pi on 3 or inverse 2pi on 3

[digital cliff]

the period would be 2pi * 3/2pi

t isnt the period though

[calm bobcat]

oh yeah mb

i thought for horizontal dilation you put one over the sf

[digital cliff]

thats what i did

1/(2pi/3)=3/2pi

[round dawn]

Let $ABC$ be an equilateral triangle whose side has length 3. There is a square inscribed in the triangle and one of its sides is on the side $AB$. What is the area of the square?

5.196
9sqrt(3)
3.897
3.333
3.2
3.1
63-36sqrt(3)
2.411
2.25
2.163
2
1.939
1.333
1.299
1
0.646
0.341
0
-1
2.598

is someone know how to solve this?

[wraith dagger]

vrare

[digital cliff]

!occupied

[cedar kiln]

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).

@calm bobcat Has your question been resolved?

[cedar kiln]

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[kindred hornet]

how do i do this?

[kindred hornet]

Anyone?

[fair geyser]

i don't know

you can do this

[limber marsh]

how many 2x2 little squares are in the big square?

we’re gonna use the pigeonhole principle

[fair geyser]

count how many sqaures you can cut out in this pattern

supposedly it would be more than 99

which is convincing enough

[limber marsh]

it’s not convincing at all

[fair geyser]

it is not not convincing at all

[raven shard]

amoung us

[kindred hornet]

I don’t understand

[kindred hornet]

14^2?

[limber marsh]

even more than that, think of the fact that a 2x2 square is determined by its top left corner

[abstract tinsel]

Hello

May i ask a question

[limber marsh]

!help

[cedar kiln]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[limber marsh]

ask in an open help channel!

it’ll be better

[kindred hornet]

im not sure how there could be more

[limber marsh]

they can overlap with other squares (as long as they’re not the EXACT same square we’re good)

[kindred hornet]

then its 14^2+28+27

[limber marsh]

try something like 28^2 (this might be a bit shocking at first, but for every tile on the 29x29 grid that’s not on the edge you get a unique square with that tile as the top left corner)

[kindred hornet]

oh right

[limber marsh]

okay great, so now, how many tiles are in 99 2x2 squares?

[kindred hornet]

4*99

[limber marsh]

yeah, okay so we have 396 tiles. we can suppose that when we put a tile down it will go in 4 squares (because of the overlap)

like forget about the 2x2 square pattern, when we put one (single) tile down this will happen

and the goal (where we’re going with this) is to suppose that we do the meanest possible arrangement of 396 tiles, getting 4 squares per tile, how many squares are used up now?

[kindred hornet]

1584?

[limber marsh]

..oh uh, yeah. i guess we should instead think about how many 2x2 squares are used up by placing a 2x2 tile

because the meanest possible arrangement of single tiles doesn’t let us say there’s a 2x2 tile left

but that’s no big deal.

[kindred hornet]

wwdym?

You there?

[solid juniper]

frownyfrog's pic is actually pretty convincing to me but probably annoying to prove that it's a "worst way to cut 2x2 squares out"

from the point of view of trying to construct 99 little squares to cut out to make no 2x2 square remain: some 2x2 cut out needs to use at least one of these squares and there are only 4 options for it. the one that occupies all 4 of them at the same time is clearly worse than the the one in the pic. the other three force "best" ones to cut out (best in the sense of removing eligible 2x2 squares)

but fwiw i think this is a bad way to go about the problem

[cedar kiln]

@kindred hornet Has your question been resolved?

[cedar kiln]

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[solid juniper]

is this problem even right

[cedar kiln]

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[peak salmon]

find a and b (both real numbers)

i cant think of any way to this other than to use the binomial theorem and rearange the terms

[ebon juniper]

do you know about DeMoivre?

[peak salmon]

but there has to be a better method

[ebon juniper]

there is

[peak salmon]

ive heard of it

could you remind me what it was again tho?

[ebon juniper]

u can rewrite the interior of the ^10 as a ei3pi/10

and then use exponents properties

[peak salmon]

alright

thanks

.close

[cedar kiln]

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[cedar kiln]

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[snow hare]

I solved it and got 4 but it says it is wrong

[snow hare]

Domain is x - 4 /geq 0 which gives x /geq 4

[crimson sedge]

you're looking for the domain of g(h(x)) not g(x)

[snow hare]

So how would that look like?

[crimson sedge]

whats the formula for g(h(x))?

find that, simplify it, and determine the domain