help-28

[fierce roost]

thanks chat sorry

wait! i have an unrelated question

how is this incorrect

ive done it several times and gotten the same answer each time

[shy stump]

with no negative exponents

look at the question

[fierce roost]

OH

[slate violet]

yeah......

[shy stump]

first thing that popped out to me 😭

[fierce roost]

idk guys i MIGHT be bad at paying attention to little details...

[shy stump]

lowk same bro

[fierce roost]

genuinely 😭 up until recently 90% of the points i get off on tests are just like forgetting dumb stuff

thank you for your help kind sirs and madams and whatnot!

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[feral quest]

i have my sat in 6 days and im trying to find a desmos explanation on how to do these questions, any help would be appreciated

[full forum]

@feral quest Has your question been resolved?

[stone matrix]

@feral quest both equations are redundant so we can deal in any one. The question statement asks us to define the line.

What have you tried so far?

[feral quest]

ive tried considering r as x and isolating

youd get r=5/7 - 6r/7

but clearly thats not the way to go

[stone matrix]

consider the first coordinate to be $x$ in each of the points.

This gives you an expression of $x$ in terms of $r$, we good?

[feral quest]

yea

[stone matrix]

now you have to make the second point, which is $y$ in terms of $x$

[glossy valve]

Adhi

[stone matrix]

do you see why and how?

[feral quest]

oh

yea

so like a system

[stone matrix]

yes. If these equations were not redundant, you would use desmos to ||find the common point and consolidate it in terms of the given options||, but since they are, you have to solve the algebra

[feral quest]

cheers man

appreciate the help

.close

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[neon basin]

!status

[full forum]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[signal solar]

But honestly 1

There's a formula for inverse functions and derivatives it's 1/f'(f-1(a))

But I'm not sure how to apply it here

[neon basin]

Can you show your work

[signal solar]

I don't rly have any work to show

More or less stuck

[neon basin]

Yeah

[signal solar]

I know that I have to solve for f-1'(2) and that 1/f'(f-1(2)

[neon basin]

$(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$

[glossy valve]

USS-Enterprise

[neon basin]

Yes, exactly

First find f(x) = 2

Because we have f^(-1)(2)

[signal solar]

There's a 2 that conveniently cancels and just leaves the polynomial expression

But idk how to solve it afterward

[neon basin]

$-x^3 + 4x^2 - 7x = 0$

[glossy valve]

USS-Enterprise

[neon basin]

Right?

[signal solar]

Yep

[neon basin]

Well first we can multiply both sides by (-1)

$x^3 - 4x^2 + 7x = 0$

[glossy valve]

USS-Enterprise

[neon basin]

Then notice x is in all 3 of the terms

What can we do

[signal solar]

I guess you can factor and solve for 0s

x=0 and x=4/7

[neon basin]

We can factor yes

[neon basin]

Half correct

When we factor x out, we get

$x \cdot (x^2 - 4x + 7) = 0$

[glossy valve]

USS-Enterprise

[neon basin]

Therefore $x = 0 \lor x^2 - 4x + 7 = 0$

[glossy valve]

USS-Enterprise

[neon basin]

Right

[signal solar]

Oh right yeah I forgot about the x^2

Yes

[neon basin]

Yeah

So our first solution is $x_1 = 0$

[glossy valve]

USS-Enterprise

[neon basin]

What about the quadratic

[signal solar]

I don't think you can factor it without manipulating it

[neon basin]

Have you learned the quadratic formula

[signal solar]

Oh

Yeah

It's a bit tedious is there anything easier

[neon basin]

Well we don't have to do the entire thing

But the discriminant tells us everything

[signal solar]

Ok

[neon basin]

You should know if the discriminant > 0, we have 2 real solutions. If it's = 0, we have 1 double real solution. if it's < 0 there are 2 complex solutions

So what's the discriminant for this quadratic

[signal solar]

Ok

[neon basin]

huh?

[signal solar]

Is the b squared minus 4ac the discirminant

[neon basin]

Yes

$\Delta = b^2 - 4ac$

[glossy valve]

USS-Enterprise

[signal solar]

Ok

The discirminant is negative

[neon basin]

Exactly

So this means no real solutions right

[signal solar]

Yep

If there were real solutions does that mean I'd have to solve for them?

[neon basin]

Yes.

Anyway

So we have concluded

The only solution to $f(x) = 2$ is $x = 0$

[glossy valve]

USS-Enterprise

[neon basin]

next, we need to compute f'(x)

$f(x) = -x^3 + 4x^2 - 7x + 2$

[glossy valve]

USS-Enterprise

[signal solar]

-3x^2+8x-7

[neon basin]

$f'(x) = -3x^2 + 8x - 7$

[glossy valve]

USS-Enterprise

[neon basin]

Yes

And then we just need f'(0)

[signal solar]

-7

[neon basin]

Now we can apply the formula

$(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$

[glossy valve]

USS-Enterprise

[neon basin]

$(f^{-1})'(2) = \frac{1}{f'(f^{-1}(2))}$

[glossy valve]

USS-Enterprise

[signal solar]

-1/7

[neon basin]

Correct!

[signal solar]

Ok that's good it works

[neon basin]

Yep

[signal solar]

So in summary set f(x) = a then manipulate into something solvable like solving for zeroes then find the derivative of f(x) and use the formula based on the result of the inverse value we calculate

[neon basin]

Yes

[signal solar]

ok so what if we make it a little harder

[signal solar]

like adding a trig function

[signal solar]

like if i add tan(pix) to this expression how would solving this change

or if i had e^x or some other things like hyperbolic idk other types of terms

[neon basin]

The formula doesn't change

Just how you compute f^(-1)(a), f'

[full forum]

@signal solar Has your question been resolved?

[signal solar]

ok

[full forum]

@signal solar Has your question been resolved?

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[signal solar]

not sure how to get rid of 12 after plugging in and needing to sovle

[fast peak]

there is an easy value x for which f(x)=12

[sand escarp]

do you know inverse function thoerem

[fast peak]

you arent supposed to analytically solve it

[signal solar]

no

[fast peak]

you can nearly guess the solution

[signal solar]

ah

1?

[fast peak]

yes

[full forum]

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[torn jolt]

hi i need help

[torn jolt]

i need help understanding |F₁| and |F₁|

anyone?????

[onyx glen]

Using the same reasoning from the previous example:

can you show us said previous example too?

[onyx glen]

you've waited all of 1-2 minutes and you already yell "ANYONE??? ANYONE HERE???????"

[torn jolt]

yes

[onyx glen]

bit impatient of you innit

[onyx glen]

ok and can you show how they did it in the previous example ie with just one file

are these questions numbered btw

[torn jolt]

this is their continuation of the previous problem

[torn jolt]

no

[onyx glen]

oh ok so they are just part of an infinite string of consciousness.

[onyx glen]

okay, now tell me this: do you understand what they did in this problem, yes or no.

[torn jolt]

no

[onyx glen]

sure is. it's combinatorics.

[onyx glen]

okay then maybe start from there?

!redir

[full forum]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[onyx glen]

look, do you wanna help or do you just wanna yap about how hard math is

<@&268886789983436800> troll

girl just get out of this channel please

1. there are 30 computers on the network.
2. of these computers, 3 particular ones store copies of a file, and the other 27 do not.
3. we select RANDOMLY a SUBSET of 5 computers to kill.
4. there are C(30,5) ways to select which 5 computers die.
5. the file is wiped out if and only if all 3 of the computers storing it are dead.
6. for the file to be wiped out, the set of dead computers must consist of all 3 computers storing the file and 2 of the 27 computers not storing the file.
7. there are C(27,2) ways to select 2 computers among the ones that don't store the file,
8. and therefore there are C(27,2) outcomes in which the file is wiped out.
9. finally, the probability of the file being wiped out is C(27,2)/C(30,5).

@torn jolt can you read this and report to me the EARLIEST line number which confuses you?

before you start to read, reply "understood, will begin reading now"

[torn jolt]

alright

[torn jolt]

Alright ill read it now

[onyx glen]

be aware that if e.g. you tell me that line 5 is the one that confuses you then i WILL assume lines 1-4 make sense to you.

[torn jolt]

line 7-10 confusing

[onyx glen]

there is no tenth line, what 10th line are you talking about

[torn jolt]

i mean 7-9

💀

[onyx glen]

there are C(27,2) ways to select 2 computers among the ones that don't store the file,
this is line 7.

there are 27 computers which dont store the file

among these we need to pick 2

C(27,2) ways to do so

if you understood line 4 then you should understand line 7 is the same shit but different numbers

[torn jolt]

but why are we even doing this

i understand the math behind it but why

[onyx glen]

check line 6

it answers that exact question of why we are doing it

[torn jolt]

Wait

2 of the 27?

[onyx glen]

yes, 2 computers among the 27 computers that dont store the file

3 machines store the file. 5 machines die.

if file is wiped out, then machines dying = 3 that stored the file + 2 others

[torn jolt]

alright I don’t get it tbh

[onyx glen]

okay

lets imagine this

instead of computers on a network there is a bag of balls

30 balls in total

3 red, 27 white

all of them numbered and so pairwise distinct

[torn jolt]

okah

okay

[onyx glen]

you pull 5 balls from the bag without replacement

in how many ways could you have all 3 red balls in hand

[torn jolt]

well

idk tbh

not sure

i know the ways of getting all balls but not specifically the reds

[onyx glen]

to have all 3 red balls in hand, the 5 balls that you pull have to be those 3 reds and 2 whites

do you understand this, yes or no

[torn jolt]

yes

[onyx glen]

ok

[torn jolt]

indeed

[onyx glen]

there are 27 white balls in the bag

how many ways are there to pick 2 of them

[torn jolt]

C(27,2)

[onyx glen]

exactly

thats exactly the same as the amount of selections of RRRWW when pulling 5 balls from the bag

do you understand this, yes or no

[torn jolt]

no tbh can you explain what you mean exactly

i mean i imagine it as RR

[onyx glen]

it's not RR...

no

we pull 5 balls and we want to end up with ALL THREE REDS

as i have repeated like several times i think

ok look

lets put more words to it

lets say this is a game of chance

we reach into the bag, pull 5 balls without replacement, and if we catch all 3 red balls, then we win

[torn jolt]

so with this I selected {WW} ballss

[onyx glen]

what i am saying is that

each winning selection

is characterized ENTIRELY

by which WHITE balls are in it

[torn jolt]

if you only had 2 white balls

[onyx glen]

"had" where.

[torn jolt]

Then yeah

if you picked

only 2 white balls then the rest would be red

oh is that why

[onyx glen]

sure if that makes it make sense to you

because i for some reason am not able to explain it to you in a way that sticks

[torn jolt]

but then you can pick 2 white balls and it can count for if you picked other white balls with it

[torn jolt]

can you go over why we did this

[onyx glen]

no obviously i cant

i cant explain it in a way that would make sense to you

i cannot possibly repeat it seven thousand more times and expect it to magically become more clear on the 7001st try

[torn jolt]

why is it the same though

[tawdry heron]

yooooo calm dowwwwwwwn

https://tenor.com/view/chill-pill-chill-rx-oprah-sp0hc-gif-24004119

[onyx glen]

hey, you try explaining this problem to OP

[knotty grail]

hrmrmm

alright lets say there are thirty computers numbered from 1 to 30

and say computer numbers 1, 2, 3 are having the file

now 5 computers are chosen out of the thirty

total ways of choosing is 30C5

[torn jolt]

yes

i agree

[knotty grail]

we need the number of choices where computers 1, 2, 3 are chosen among the five

to ensure the file is completely removed

so three of our choices are already known

[torn jolt]

indeed

[knotty grail]

we know that we are going to choose computers 1, 2, and 3

[torn jolt]

yes

yep

[knotty grail]

and the other two are the doubtful ones

we could choose those two from the remaining 27 ones

[torn jolt]

why does this work

[knotty grail]

numerator shows all possible computer numbers where 1, 2, 3 are present

[torn jolt]

ohh wait so you have 27 to pick the two remaining ones from

[knotty grail]

yes

[torn jolt]

I seeee

[knotty grail]

1, 2, 3, 4, 5
1, 2, 3, 4, 6
1, 2, 3, 5, 10
1, 2, 3, 7, 9
and so on

as only two of the five which are selected from the remaining 27 vary

its 27C2

[torn jolt]

@knotty grail thank you so much

God bless you

[knotty grail]

welcom

are you done then

for the original problem you can once again assume using numbers

[torn jolt]

yes thanks bro

[torn jolt]

yes its the same logic basically

[knotty grail]

you can close using .close

[torn jolt]

.close

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[stuck fiber]

Apparently this converges to 0 but idk how

[rustic frigate]

Try and bound it

By something that clearly converges to 0

[stuck fiber]

surely the power goes 'faster' than the denomnator, no?

[rustic frigate]

I.e estimate |a_n|

[stuck fiber]

So it should go to infinity

[rustic frigate]

The numerator is oscillating between -1 and 1

[stuck fiber]

Ohhhhh

I didnt notice it was 1 lol

[rustic frigate]

Ah lmao

[stuck fiber]

Ok so the denom goes 'faster' to infinity

Which is why its 0

Makes sense

Thanks!

❤️

.close

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[stuck fiber]

Apparently this converges to 1/2

But im confused

If we just ignore the fraction, 1/infinity tends to 0 and sin(0) = 0, so it should be something multiplied by 0 which is just 0, no?

[lime ether]

no

[slate violet]

you should use $\sin(1/n) \approx 1/n$ instead

[glossy valve]

south

[slate violet]

sin x is around x

[stuck fiber]

So the fraction shouldnt matter?

[slate violet]

when x is small

[lime ether]

its n^2 in the numerator

[stuck fiber]

Ah yeye

[lime ether]

not sure where you are getting 1/inf from

[stuck fiber]

Ohhhhhh

True

[slate violet]

yeah cause you would get infinity * 0

which is an indeterminate form (you can't continue like that)

[stuck fiber]

Ok I see

[lime ether]

n^2/(2n - 1) behaves like n/2

sin(1/n) is like 1/n

so you get 1/2

[stuck fiber]

Alright I get it

Thanks guys!

❤️

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[stuck fiber]

In Exercises 101–108, assume that each sequence converges and find its limit.

I thought it would just be like a_n goes to infinity so the answer would be 6/2 = 3, but apparently the answer is 2

I dont see why

[vast fossil]

How would a_n’s limit be infinity and 3 at the same time anyway

Where did 6/2 even come from

[lime ether]

why are you dividing the constants?

[vast fossil]

It’s better to take n -> infty of both sides in the recursion

[stuck fiber]

Where did 3 come from?

[lime ether]

you wrote it?

[stuck fiber]

Oh wait

I got mixed up

[stuck fiber]

Why do we want to take it from both sides

[vast fossil]

You will get an equation with the limit of a_n as an unknown

[stuck fiber]

We get 2 equations, no?

[lime ether]

do you agree that lim a_n = lim a_{n + 1}

[stuck fiber]

Or will they both be the same?

[stuck fiber]

Idk I dont really understand this topic

[vast fossil]

Taking the limits of both sides of an equation yields just one equation

a_n = b_n => lim a_n = lim b_n

[lime ether]

$\lim_{n \to \infty} a_n = \lim_{n \to \infty} a_{n + 1}$

[glossy valve]

knief

[vast fossil]

I’ll let you take over, knief

[lime ether]

😭

[stuck fiber]

Ok sorry...

You dont have to help me next time

.close

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[lime ether]

💀

[somber pike]

@stuck fiber when the sequence settles down, an+1 and an both approach the same value L

I hope that makes intuitive sense

So just set both equal to L

And solve that equation you get

[stuck fiber]

Ah ok!

I think I get it 😄

Thank you!!

❤️

.close

[somber pike]

you're welcome

[stuck fiber]

oh oops lol

already closed

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[stuck fiber]

In Exercises 101–108, assume that each sequence converges and find its limit.

I almost got the right answer

I had L + L^(1/2) - 12

I turned that into x^2 + x - 12

So that gives x = 3 and x = -4

[somber pike]

Can you just take a picture of your work because that'd be a lot easier

[stuck fiber]

Is sooooo messy 😅

[somber pike]

It's okay

We've all been there

[stuck fiber]

A lot of steps are in my head too

[stuck fiber]

Basically I subbed back in L^(1/2) to get L^(1/2) = 3 and L^(1/2) = -4

So L = 9 and L = 16

Which are both greater than 0

So I didnt know which to pick

In the answer they got the same but they picked 9

I dont understand why?

[queen crater]

sqrt is positive, so 12 - sqrt is < 16

[stuck fiber]

Ahhhh I see

Makes sense

Thank you!

❤️

.close

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[queen crater]

You could also just say 12-sqrt(16) = 8, which is not 16, so it's not a fixed point

12-sqrt(9) = 9, that's a fixed point

[grave elm]

L^(1/2) = -4

L = 16

This isn't even right

no square root is ever negative

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[deep flame]

guys

[deep flame]

i need help

!

[somber pike]

With

[deep flame]

i asked chat gpt and gauth and they were both wrong

and the video doesnt help

and its due tmr

😔

[edgy glade]

You know what the area under a velocity time graph represents?

[deep flame]

is it ok if you just tell me the answer

:v

[edgy glade]

I think the spirit of help channels is to get you to understand it

Not spoonfeed the answers

!noans

[full forum]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[deep flame]

hmm

alr

[deep flame]

nah

[foggy vapor]

knowing this is admittedly paramount to solving the question, and telling them doesn't actually answer the question either, so I believe it's fine

[deep flame]

💀

[edgy glade]

I mean I was gonna tell them what the area under the graph means

But not hand the numerical answer

[foggy vapor]

oh ok

carry on

[deep flame]

can i just get my help

🥺

what do i do

[edgy glade]

So think if the velocity time curve was a straight line

Parallel ot x axis

To*

What would area under it be calculated using

What would be the formula

[deep flame]

😭

im 14 years old

i cant even comprehend what u just said

[edgy glade]

What's the formula for displacement

[deep flame]

isnt that physics

[edgy glade]

Yeah.

[deep flame]

substitute?

🤔

genuinly dont know

this is like college stuff

[silent dove]

Remember velocity is meters per second. So in the. First 250 seconds, it went from 0-14 meters per second at a steady pace. Have you learned how to calculate that?

[deep flame]

yeye

isnt it 14 divided by the number of seconds

https://tenor.com/view/jgmm-monkey-think-monkey-meme-ponder-monkey-idea-gif-6401133862108294696

[silent dove]

That doesn't sound right. But if you know how to figure out how many meters it went during the first 250 seconds, the rest should be easy enough.

[deep flame]

thats so smart

[silent dove]

Id guess it'd be travelling at an average of 7

Because that's half of 14 and it's constant and symmetry stuff

But idk. So for the first 250 seconds, if it's travelling at an average of 7 m/s, how many meters does it go?

[deep flame]

uhh

7 x 250

1750!

waitr

different question now

[silent dove]

?

[silent dove]

Did you figure this one out?

[deep flame]

it told me to seek help

[deep flame]

i thought i was onto smt

but i was wrong

[silent dove]

Ok. This has the same concept

First 250 seconds it seemed to travel at an average of 14 m/s

[deep flame]

ur so smart

🥹

[silent dove]

No it's 7

[deep flame]

💔

[silent dove]

I forgot to divide by 2

You need to understand what I'm saying and correct me lol

[deep flame]

why do u need to divide by 2

[silent dove]

I'm just saying that because it starts at 0 and goes to 14, so it seems average in the middle

Then, for the rest of it, it's at 14

So the first 250 seconds go 1750, but the rest go 3500 per 250 seconds

So 9800-1750=...?

[deep flame]

uhh

8050

[silent dove]

Correct

[deep flame]

😎

[silent dove]

Now you need to divide that by 14 I think to find how many seconds more

Because it's 14 m/s for the rest

[deep flame]

575

[silent dove]

So the answer should be 250+575? I think?

I'm not completely sure. Does your homework program tell you if that's correct or not?

[deep flame]

ur right

🥹

tysm

i wont get detention

[silent dove]

You're welcome!

You understand how to do the problem on your own now?

[rigid whale]

hi

can i get some help

[gritty rose]

#❓how-to-get-help

[silent dove]

Oh glad should probably close this channel

.close

😢

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[deep flame]

idk how

mb

[deep flame]

ye

[silent dove]

It's closed now. You just type ".close" when your question is resolved

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[gilded bluff]

hai, I've been trying to self-study linear algebra for a few months now. I did the same thing with calculus, but at some point, things came "full circle", it kinda feels like im just learning and proving random facts. How do i actually learn this? (sry if thats a bad question)

[foggy vapor]

You probably need a structured curriculum to follow

Either pick out a good book or find a good series online to follow with problem sets

OCW has some nice lin alg lectures I followed. Would recommend!

Also, this is more personal advice, but I'd say try to anchor everything you learn around transformations

Matrices are just the representation of these transformations under a chosen basis, for example

[fickle adder]

for actual personal interest and genuine desire to grasp, 3blue1brown has a famously helpful visualization series of linear algebra topics that youd find covered in your average one-semester course. very valuable for lin alg in particular IMO because unlike calculus or other math, i feel like without visual aid lin alg is very hard to understand what all the different facets physically represent

it probably wont make you super great at solving problems but if you watch it in junction with something else + problem sets i think its a very valuable supplement

for me at least, determinants were a purely symbolic construct before watching that series, and even though i thought i had a conceptual understanding of how bases work, i realized my understanding was very primitive and watching it helped me get a deeper grasp of both

https://www.youtube.com/watch?v=fNk_zzaMoSs&list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab

[gilded bluff]

thank you! i think ill just look at a lot of resources and hope to find understanding there. how long does it take to get a solid grasp of the subject?

[full forum]

@gilded bluff Has your question been resolved?

[prime pier]

once I really, properly, understood the idea of span and linear (in)dependence, i felt like i could start to put pieces together. a good metric for this is being able to do proofs relating to basis vectors, dimension, span, etc. very easily, or at least being able to tell if theorems are true or false

after that, getting to eigenvalues and orthogonalization was very helpful, because at that point you start to treat matrices as what they are: transformations of n-dimensional space. once you've grasped that, you're pretty much set until abstract algebra

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[gilded bluff]

thank you

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[upbeat skiff]

I'm really struggling with this:

[upbeat skiff]

I am up to the part where I got to L = sqrt(2) int (from 0 to 1) of 1/(sqrt(2-x^2))

[leaden ermine]

how did u get there

[upbeat skiff]

Let me see if I can just upload my tablet writing standby

That is the nice thing about ReMarkable.

Pls forgive how jumbled it is

[leaden ermine]

oh i see now

welp, a trig sub might fit here

[upbeat skiff]

Is that the trig sub?

[leaden ermine]

yes

[upbeat skiff]

Ok. Standby

How did you get that?

[leaden ermine]

well i am abit familiar with trig subs but it comes from the identity sin²(x)+cos²(x)=1

[upbeat skiff]

(Can you tell Cal2 has been killing me?)

Ohhhhh

[leaden ermine]

so we can rearrange that as cos²(x)=1-sin²(x)

alternatively, there is a method, where you draw a right triangle and derive its lengths

[upbeat skiff]

I wound up trying to keep going until I got to a 1/x situation but ln(0) is undefined and that's where I kept getting stuck. Let me try to run through it with the trig sub

I truly have no idea how I'm supposed to memorize every single one of these trig identities

Ok so I'm almost there.

Just gotta convert the bounds

[upbeat skiff]

I'm at sqrt(2) * [arcsin(1/sqrt(2)) - arcsin(0/sqrt(2))]

so it should be sqrt(2) * arcsin(1/sqrt(2))

[leaden ermine]

so you can find the value for alpha maybe easier

[upbeat skiff]

so the answer is (pi*sqrt(2))/4

[leaden ermine]

yes

[upbeat skiff]

TYSM! I'm all good.

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[potent loom]

helo how does this looks like

[potent loom]

Help

[foggy vapor]

What do you need help with exactly?

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[leaden ermine]

What have you tried (also ty for not using help 47)

[torpid perch]

(x,y,z) = (cos(θ), sin(θ), z)

trying to use cylindrical coorsinates

[leaden ermine]

okok

[torpid perch]

0 <= z <= r

cos(θ) >= |sin(θ)|

r^2 + z^2 <= 4

z ∈ [0, r]

r^2 + r^2 = 4
2r^2 = 4
r^2 = 2
r = sqrt(2)

z ∈ [0, sqrt(2)]

so I found the bounds for the integral wrt z

but I am having trouble finding bounds for the integral wrt theta

this shit is too trig pilled: cos(θ) >= |sin(θ)|

also, we can find the bounds for r I think

x^2 + y^2 + z^2 <= 4

r^2 + z^2 <= 4

[leaden ermine]

geometric intution is a cheat code with these

[torpid perch]

r^2 < = 4 - z^2
z ∈ [0, sqrt(2)]

[torpid perch]

is not easy to visualize

[leaden ermine]

whats the first shape

[torpid perch]

a sphere

[leaden ermine]

second

[torpid perch]

like a ball of radius 4

[leaden ermine]

ya

[torpid perch]

another ball?

[leaden ermine]

no

[torpid perch]

help

[leaden ermine]

it's a cone

[torpid perch]

yeah I guess, yeah

only the positive part of it

not a double cone or anything

[leaden ermine]

you know z^2=x^2+y^2 is pythagoras

[leaden ermine]

so theoretically you have the distance from a point to the origin, and visually a triangle

[torpid perch]

the z is squared

[leaden ermine]

ya

[torpid perch]

@leaden ermine

[leaden ermine]

@torpid perch

[torpid perch]

your point is

r^2 < = 4 - z^2
z ∈ [0, sqrt(2)]

r = 2

r ∈ [sqrt(2), 2]

with the sphere r has interval sqrt(2) <= r <= 2

with the cone... however

0 <= z <= r

oh but we already found z ∈ [0, sqrt2]

this shit is too hard

let me start from scratch

and let me draw a visualization

[leaden ermine]

now we are talking

this is also how inequalities will mean to you something

instead of just being inequalities

[torpid perch]

I gueeeess

though I suck at drowings

[leaden ermine]

u can improve

[torpid perch]

life sucks

[torpid perch]

https://www.geogebra.org/3d/xcvgtrhm

you here @leaden ermine ?

.close

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[leaden ermine]

.i am about to go off anyway, but noticed, you plotted z>=|x| instead x>=|y|

[jagged relic]

can anyone help me with the question i cant solve it

[placid oak]

start with simplifying the diagram by drawing some triangles

[jagged relic]

ai

[jagged relic]

now what

i got one that looks like this

[placid oak]

that doesn't look like a triangle to me

[jagged relic]

and one like this

[jagged relic]

i drew it on paper

[placid oak]

ok cool

[jagged relic]

iont use my computer to draw

[placid oak]

so, you've kept the top angles?

[jagged relic]

yeah

the 45 and the thirty five

[placid oak]

Ok, you have these two?

[jagged relic]

yep

[placid oak]

nice, so fill out the lengths and angles

[jagged relic]

ok

[placid oak]

Hint: If you let the plane be O, then you can work out OA from the first triangle and OB from the second.

[jagged relic]

now what

[placid oak]

What do you have for OA?

[jagged relic]

six eight nine four

[placid oak]

that doesn't look right. how did you calculate that?

[jagged relic]

sin 45 times nine seven fifty

[placid oak]

why sin?

[jagged relic]

opposite over hypotonuse

[placid oak]

which is the hypotenuse?

[jagged relic]

yo iont mean to be rude but can you just show me how to solve it i have like 6 more questions like this and i dont have time like that

i just want to know how to solve these

[placid oak]

it's not sin

have another look at which sides you have

[jagged relic]

i did tan 45 and its just the same answer

[jagged relic]

and this question is supposed to use the sin law

[glossy valve]

riemann

[placid oak]

oooh nice

[jagged relic]

ok

i used tan and its the same as the height

[placid oak]

yep

now we do the same for the second triangle

[jagged relic]

ok

six sero nine two

[placid oak]

nope. which sides do you have?

[jagged relic]

o and h

[placid oak]

That's the problem. You should have o and a

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[rapid olive]

curious

[rapid olive]

for this SVD, is the professor wrong

oh wait im tripping

.close

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[astral brook]

Might be a stupid question but what does this evaluate to? $\int \sec(x) \left|\sec(x)\right| , \mathrm{d}x$

[glossy valve]

impract1cal

[astral brook]

i mean i guess i can evaluate it case by case? My instinct is that you can just turn this into the integral of sgn(x) sec(x)^2 and i think u can pull sgn(x) out of the integral because its a locally constant function

which then evaluates to sgn(x) tan(x) i guess??

idfk

[narrow ermine]

You can do something like to make it work $\int \sec(x) |\sec(x)| \dd{x} = \frac{\sec(x)}{|\sec(x)|} \int \sec^2(x) \dd{x}$

[glossy valve]

Azyrashacorki

[narrow ermine]

Then you get a "nicer" expression in the end

[astral brook]

how did that turn into that

hm and can i turn sec(x)/abs(sec(x)) into sgn(sec(x))

[narrow ermine]

Well if anything then you get the antiderivative given by $$\frac{\sec(x)}{|\sec(x)|} \tan(x) + C = \frac{\sin(x) |\cos(x)|}{\cos^2(x)} + C = \frac{\sin(x)}{|\cos(x)|} + C$$

[glossy valve]

Azyrashacorki

[narrow ermine]

And it turns into that because sec(x)/|sec(x)| is sgn(sec(x))

So you can pull it out

[astral brook]

now hold on i stilld ont get how u got this

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[hoary current]

I need help with logarithms

[hoary current]

For the first 2 images im wondering why we solve for y while the other we dont

And the 3rd image i am confused on what the regular number is to the logarithm. What i mean is when its log 180 or whatever number for that matter what actually is it? Is it the parenthesis so like log(180)? Or what!?

[onyx glen]

aight hold on

for the first image, is the left hand side meant to be (52/10)^x, or is it meant to be 5*(2^x)?

[onyx glen]

i mean, for starters there are two different goals here:

in 1 and 2 you want to verify an identity (ie your answer is "yes this is true for all x" or "no this isn't true for all x")
in 3 you are instead solving an equation (ie your answer is a number that, if plugged into x, makes the original equation true)

i am confused on what the regular number is to the logarithm. What i mean is when its log 180 or whatever number for that matter what actually is it? Is it the parenthesis so like log(180)? Or what!?

even after you tried to explain what you mean by "what the regular number is to the logarithm", this is still confusing.

[full forum]

@hoary current Has your question been resolved?

[hoary current]

Sorry for the late response

[hoary current]

Its the second one

[hoary current]

For this i mean when we do the power property its x times log with a number after. However when were writing it the space in between the log and number, what is it? Is it multiplying? Subtracting? Adding? Is there parenthesis?

[onyx glen]

if it is the second one then the equation is solved incorrectly.

the correct solution would be to first divide both sides by 5 to get 2^x = 36 and then take the binary log on both sides to get x = log_2(36).

[onyx glen]

log is a function. log(180) is "the logarithm of 180".

people sometimes do not write the parentheses but imo that convention can be & often is misleading.

[hoary current]

Wait are you talking about the one thats the same or the one thats not

[onyx glen]

... i have no idea how to answer that

i guess i am "talking about the one that ISN'T the same"

the 3rd image from your original post

5 * 2^x = 180
this one. i am talking about this equation.

oh, i said "first" and not "third".

rather a big blunder on my part, communication-wise.

my bad.

[hoary current]

Ok cus what i thought it was is parenthesis so what you mean is in the formula for logarithms that number is the equals to in exponential form?

[onyx glen]

christ. that sentence is so damn hard to parse.

[hoary current]

Lemme make it easier

When we have log 180

Its log(180)

So

Lets say we were to inverse it so its an exponent

Then that 180 would be equals to the whole other side?

[onyx glen]

log(180) is the number that answers the question of "10^what = 180"

[hoary current]

Yes

Ok

[onyx glen]

so i guess i have to answer yes to you, but i sure don't like the way you worded it

[hoary current]

Its fine its just i understand it in an exponent way but how would the 3rd image be wrong?

My teacher told us to like put log on both sides of the equation first

Then use the power property then divide by log whatever to get x

[onyx glen]

$\log(5 \cdot 2^x) \neq x \log(5 \cdot 2)$

[glossy valve]

Ann

[onyx glen]

\textbf{if} you were to log both sides of the equation immediately (presumably because the teacher can and will give you zero points as punishment for direct-instruction disobedience), then the correct simplification from $$\log(5 \cdot 2^x) = \log(180)$$ would be $$\log(5) + x \log(2) = \log(180)$$

[glossy valve]

Ann

[hoary current]

Its 5.2

Sorry if my writing is terrible

[onyx glen]

mate

i literally asked you

is the left hand side (52/10)^x or is it 5 * (2^x)

you said, and i quote, "the second one"

and now you are revealing to me that you lied to me earlier?

[hoary current]

Im sorry i didnt relise you put 5.2 as a fraction😭

[onyx glen]

i specifically put it as a fraction to make it easier to tell apart from 2^x multiplied by 5 !!!

but this still somehow resulted in a failure of communication.

i ask again for the SECOND time

is it

(5.2)^x

or is it

5(2^x)

tell me once and for all now, WHAT WAS MEANT?

[hoary current]

5.2

[onyx glen]

ok in that case the solution you wrote is correct.

[hoary current]

Ok and i have 1 final question

Why is it that we apply log to both sides, and is it multiply by log or put it at the exponent

[onyx glen]

we do not "multiply by log" nor do we "put it at the exponent".

[hoary current]

Are we simply rewriting it?

[onyx glen]

when we write $\log 180$ or other such things, it \textbf{does not, and never has, and never will} mean the multiplication of 180 by some magical object called ``log''.

as i said before, and as bears repeating, log is a \textbf{function}. we are \textbf{applying} the logarithm function to both sides.

[glossy valve]

Ann

[onyx glen]

Why is it that we apply log to both sides
the whole thing of logarithms is that they are the inverse function to exponentials

that's how we get from e.g. 2^x back to x. the function that does this is called the (in this case) binary logarithm, a.k.a. log_2

[hoary current]

Ok so your prob gonna get mad at me but when your doing this log thing why arnt we doing the inverse thing where we change the exponent into a log but instead this

[dry arch]

a function takes an input and gives an output
if the inputs are the same, then the outputs must be the same
in other words, if a = b then f(a) = f(b)
that goes for any function, including log
if a = b then log(a) = log(b)

[onyx glen]

what "inverse thing"...

[hoary current]

The inverse of an exponent is a log and vyse versa

[onyx glen]

but you're speaking specifically about some sort of procedure? or what.

[hoary current]

Like why are we applying log to both sides? Why arnt we changing the exponent into a log? Is it because when doing inverse its not finding x?

So what your saying is we have an exponent which is x, and we cant find x when its an exponent. So we must change it to a log so x isnt an expoenent. However to do this we must do it to both sides?

[onyx glen]

where did you get this idea from that you somehow have to "change" something into something before doing the algebraic move appropriate for the problem

this is like if you were saying "When we solve the equation 6x = 24, why aren't we changing the multiplication into a division?"

[hoary current]

Because x is an exponent

If x wasnt an exponent we wouldnt need to do this log stuff

So we are using this log stuff to change that x to be do able

So we use that log with the power property to make x not an exponent to solve as normal

[onyx glen]

not really, i think that's kind of a crooked way to think about things

the relationship between logarithmic and exponential functions
is the same as
between addition and subtraction,
or between multiplication and division,
or (with some wrinkles) between squaring and square-roots.

[hoary current]

Ok i understand it for what i need to for now sorry for the trouble but tysm @onyx glen

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[sullen vessel]

hi guys quick question, how to find normal vector given an equation

[sullen vessel]

not ur usual linear equation. but like circle for example

[onyx glen]

equation of what? a curve in the plane?

and you want the normal to it at a specific point?

[sullen vessel]

that n

I totally get n would be <0,0,1> or <0,0,-1> if it is lying on a the axis plane, but im afraid my prof would be devious enough to ask us this kind of thing

[onyx glen]

normal to a surface then

[sullen vessel]

yes vyes

[onyx glen]

is your surface given implicitly or parametrically

[sullen vessel]

parametrically

[onyx glen]

right ok

then you take the partial derivatives of (x,y,z) wrt u and wrt v, and cross product them

[sullen vessel]

OHH like gradient vector?

[onyx glen]

gradient would be more visible if you were given an implicit equation ie F(x,y,z)=0

then your normal really would be just grad F

[sullen vessel]

dang ok, so wym by wrt u and v? its 3d

[onyx glen]

you have a surface defined via 2 parameters right

maybe show the question?

[sullen vessel]

um its not quite a question but ill show u hehe

alr so originally, the circle equation is only x^2 v+ y^2 <= 2

but im just considering other cases where its normal to a surface

[onyx glen]

ok so your parameters are called r and theta by the looks of it

[sullen vessel]

and how would I find it

yes r, theta, and z

the question just put the circle when z = 0

that why it zero

[trim yew]

uhh guys

can someone teach me quadratic equations class 10?

[hot kraken]

Go to a free help channel

[trim yew]

okay

[sullen vessel]

sooo any idea?

[onyx glen]

this is a little too disorienting for me to confidently answer with anything

[sullen vessel]

ah ok ok no worries

[full forum]

@sullen vessel Has your question been resolved?

[sullen vessel]

help help <@&286206848099549185>

[devout inlet]

I'm not sure what you're trying to do really. The image above seems to have "3)" which makes me think you've left some details out

[sullen vessel]

thats actually not the point

[sullen vessel]

im just asking how to find this n

[devout inlet]

Ok. I saw it discussed about it being paramaterised in 2 variables and using the cross product and you understood that.
Is the question about which of the two normals to pick?

[sullen vessel]

nope, im asking given a parametrization of the circle from the pic I drew, how can I find the n

[sullen vessel]

u see, the ones here is just a circle lying on xy-plane

so n is just<0,0,-1>

but im afraid of cases like the picture I dew

drew

[devout inlet]

Ok I see

Do you know the equation of the circle you drew?

Like it looks like it lies in a plane, maybe z = y?

[sullen vessel]

probably...? hehe not sure its just a thought from my head

[devout inlet]

If it's z = y you could use a similar trick and say the position vector for all vectors of that plane is r(x, y) = <x, y, y> and use the cross product r_x croxx r_y

[sullen vessel]

so are u saying find two vectors lying on the cirlce with zero dot product and find the cross product of those two?

[devout inlet]

If the surface lies in a plane you can use this trick is what I'm saying but it always depends on the surface and how it's represented when given to you (it could be given as F(x, y, z) = 0 or x = f(u, v), y = g(u, v), h (u, v) or as z = f(x, y), etc.)

[sullen vessel]

aw dang it, wait lemme try giving u a concrete circle equation that looks like that

[devout inlet]

Ok 🙂

[sullen vessel]

pls wait, might take a while hehe

[devout inlet]

No problem 🙂

[sullen vessel]

me experimenting

fahhh i dont think I can find

wait a moment, ill try my last attempt

u know what, z = x^2 + y^2

[devout inlet]

For a function represented as an implicit function like that which you have no idea what it would look like visually you'd probably use the gradient

[sullen vessel]

gradient vector?

[devout inlet]

This could be treated as r(x, y) = <x, y, x^2 + y^2> and you can use the cross product trick

[devout inlet]

I'll try and find a source as you might have come across it before

[sullen vessel]

shouldn't we change x = rcostheta

and the others

[devout inlet]

You can use any paramaterisation you wish

[sullen vessel]

ahh i see ok ok, and can u tell me what is this cross product trick

i think i forgot

[devout inlet]

You are essentially finding two vectors on the tangent plane at a point and using the cross product to find a vector perpendicular to them both and therefore obtain a vector parallel to the normal at that point

[devout inlet]

https://tutorial.math.lamar.edu/Classes/CalcIII/GradientVectorTangentPlane.aspx

[sullen vessel]

ohh I see I see

[sullen vessel]

alr tysm

i think i can close this now

cya

.solved

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[tepid needle]

How can i do this?

[brittle steeple]

surely there's not enough information there

[whole lily]

i'm pretty sure there is

[knotty grail]

guessing this is linear equation in one variable?

you could take speed of river to be x

so when the motor boat moves upstream

it moves against the river and has velocity 20 - x

[tepid needle]

Yep

Alr

Okey

Down stream means 20 plus x

[knotty grail]

yes

like a slide

[tepid needle]

Yeah

[knotty grail]

and time = distance/speed

[tepid needle]

Okey

They didn't give distance bro how is this possible?

[knotty grail]

hmyeah

[knotty grail]

the question when i search it up has an extra "48 km" phrase

[tepid needle]

Oh

[knotty grail]

"takes 1 hour more to go 48 km upstream"

[tepid needle]

Yeahh it had the same for me

Is there some kind of other method exclusively to do this or smth?

[knotty grail]

hmm i dont think so

[tepid needle]

O alr then

[knotty grail]

you'll get a quadratic in terms of x

and d is necessary to get all constants

[tepid needle]

Yep

We need it for lcm

Then alr

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[silver dagger]

How can I calculate the area of this shape?

[keen vector]

square minus quarter-circle

[graceful nova]

I will add by saying that no assumptions are needed, in case you're wondering.

[silver dagger]

O like the black spot

No*

[jaunty narwhal]

yes, thats what hes talking about

[silver dagger]

I translated it

He said the area of the square - 1/4 of the circle

[graceful nova]

you want the area of the shaded region, correct?

[silver dagger]

Yeah

[graceful nova]

that's exactly what you should do.

[silver dagger]

Yes I tried that and it’s incorrect

[graceful nova]

please show your working.

[silver dagger]

So that’s why I need to do correction

Alr

16x16divided by 4

[graceful nova]

divided by 4?

that's not how you factor in the quadrant.

16^2 is indeed the correct area of the square, however. you're just missing a subtraction by the area of that quadrant.

[silver dagger]

Ik now

I gotta learn basic math again ig 🥀

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[sturdy valve]

i got N belongs to {2,3,...,12}

and favourable cases are {4,5,...12}

so shouldnt it be 9/11

[delicate torrent]

4.9-3.11=3

Where . Is the mult sign

[sturdy valve]

yes

[delicate torrent]

Idk HOW it's not in the answer choice

[sturdy valve]

even idk thats why im asking

[full birch]

the question was 4m - 3n

[sturdy valve]

yes

[full birch]

and m and n here, are not exactly the n that you computed. it's kinda misleading but the probability was to be represented in m/n (Or p/q) form, and then you calculate it based on that

[delicate torrent]

m, n is coprime

[full birch]

yes

[delicate torrent]

So gcd(m, n) = 1

So isn't m = 9 and n = 11?

[sturdy valve]

yes..we represented it in m/n form..

9/11

[full birch]

have you checked what the "probability" of 2 to the power N being smaller than N factorial would be?

[delicate torrent]

Yep

[delicate torrent]

9/11

No way the twin towers is here 💀

[full birch]

alright, then 4m - 3n would be 4 times 9 - 3 times 11

36 - 33

[sturdy valve]

yes 😭 we started with that

[delicate torrent]

You see my point?

[delicate torrent]

Lmao

[full birch]

mmmmmm, you might have made a mistake in calculating the 9/11 part

[sturdy valve]

bro laughed at his own joke

[delicate torrent]

Calculate it yourself

[sturdy valve]

what is the mistake..that is what im here to find out

[somber pike]

Each sum N doesn't happen equally often. You need to list all ways two dice can add up to a given number. sum = 4 --> 3 ways

Do that for all N and add up

[sturdy valve]

ohhh

[somber pike]

Then divide by amount of favorable outcomes

[sturdy valve]

ok let me try

[delicate torrent]

Didn't describe it correctly

[sturdy valve]

i did 1-(P(X=2) + P(X=3) = 1-(1/36+2/36) = 11/12 is this correct?

[somber pike]

Yeah

[sturdy valve]

okay thanks a lot

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[spice grail]

How do we know that the rocket C is downwards

[spice grail]

using the right hand rule?

[burnt grotto]

vector format the boats

Take east as i and north as k

[spice grail]

and j?

[burnt grotto]

upwards/downwards

now what are the vectors of v1 and v2?

[spice grail]

oh okay i think i got it

.close