help-13

[cedar kiln]

@wide light Has your question been resolved?

[cedar kiln]

@wide light Has your question been resolved?

[cedar kiln]

@wide light Has your question been resolved?

[wide light]

look at this

wth?!?!?!?!?

im soooo confused

[cedar kiln]

@wide light Has your question been resolved?

[wide light]

<@&286206848099549185>

[cedar kiln]

@wide light Has your question been resolved?

[wide light]

ill close it if nobody can help soon

if someone reads this and thinks something like - this is just unclear - do let me know

or maybe this is just a really niche topic

[odd seal]

damn you've been waiting longer than me

[vast isle]

Sorry bro I think that’s the case here

[odd seal]

can you open source it? maybe put it on fdroid https://f-droid.org/

[wide light]

what the hell

@vague moth

i dont think pinging modmail does naything

<@&268886789983436800>

[odd seal]

oh yeah and I bet the moderators would be furious

[rustic coyote]

banned

[wide light]

i think im gonna give up on my question

[odd seal]

noo always believe in yourself

[wide light]

ive even tried asking my teacher but she ghosted me for weeks lol

i saw her at schoool today, i went for a diff exam

she said oh i saw ur email in my junk folder i will reply today

today ended 10 minutes ago

exam is on wednesday (aka tomorrow)

screwed

[odd seal]

what math are you taking?

[versed bone]

@wide light you good?

somehow this channel is still open

[wide light]

I never solved it

But it's closing time :

.close

[cedar kiln]

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[crimson sedge]

what exactly do you need help with?

It appears that this algorithm is traversing an array of numbers and assigning the product of a_{i} and a{j} to a max() function/method

which returns the maximum value between m and a_{i}*a_{j}

alright, so, for the outer loop's iteration, i starts from one and ends at n

so that's always n iterations

but j starts from i+1

so, at value i = 1 j will be 2

or rather, will start at 2

so, it skips the number one

but it stops at n

you with me so far?

i = [1, 2, 3, 4, ..., n]
j = [-, 2, 3, 4, ..., n]

so, i iterates n times

but j iterates n-1 times

j iterates n-1 times on EACH iteration of i

meaning: for every iteration of i, j iterates n-1 times

and there are n iterations of i

so the TOTAL number of iterations is: $$iterations_{i} \cross iterations_{j}$$

[wraith dagger]

Emran & Knuckles

[crimson sedge]

which is why it's n(n-1)

does that make sense, @plain bridge ?

read my replies and tell me if you have any questions

sorry for the ping

goes into?

[cedar kiln]

@plain bridge Has your question been resolved?

[crimson sedge]

yeah?

that's not the way to think about it

the amount of iterations the second loop will do is always going to be $$iterations_{i} - 1$$

[wraith dagger]

Emran & Knuckles

[crimson sedge]

in reality the n from O(n) and the n from i := 1 to n aren't dependent on each other

it just so happens that the amount of iterations i will make (the n from O(n)) is equal to n ( the n from the loop range)

[crimson sedge]

refer to this

i = [1, 2, 3, 4, ..., n]
j = [-, 2, 3, 4, ..., n]

j will always start at i + 1

so when i = 1, j = 2

and on the second iteration, when i = 2, j = 3

suppose n = 5

the iterations would be as follows:
when i = 1 : j = 2, 3, 4, 5 (because it stops at n = 5)
when i = 2 : j = 3, 4, 5
when i = 3 : j = 4, 5
when i = 4 : j = 5
when i = 5 : j = 6 (do nothing because we ordered it to stop at 5)

for each iteration of i the number of iterations of j is iterations of i - 1

so the number of multiplications that take place (a_{i} * a_{j}) is n(n-1)

[cedar kiln]

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[crimson sedge]

and the number of comparisons that take place (max(a_{i} * a_{j}, m)) is also n(n-1)

[cedar kiln]

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[tawdry mist]

i'm fairly new to calc

[sage forge]

You can integrate the function sqrt(r^2-x^2) and do trig-sub

[tawdry mist]

sorry, i am completely new, i don't get what you just said

[sage forge]

You know the equality of a circle? x^2 + y^2 = r^2 ?

[tawdry mist]

never heard of it sorry

[bitter reef]

perhaps x^2 + y^2 = 1

unit circle

[tawdry mist]

hm a unit circle?

is y^2 the curvature

[bitter reef]

the thing you use in trig

[tawdry mist]

oh

okkk i see

[sage forge]

Then solve for y

[tawdry mist]

once you have it, it's y^2 = r^2 - x^2 right

[sage forge]

Yes

[tawdry mist]

and replace y with our r in here?

[sage forge]

r is the radius. You should treat it like a constant

[tawdry mist]

i see

once we solve for y, what do we do with it?

[sage forge]

Integrate it with respect to x

[tawdry mist]

i seeee

ok i think, i get it now

thanks!

[cedar kiln]

@tawdry mist Has your question been resolved?

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[crimson sedge]

hi

[crimson sedge]

did i solve this right?

<@&286206848099549185>

[sage forge]

#❓how-to-get-help

[crimson sedge]

?

[rotund vine]

,rotate

[rotund vine]

I think just find the dot product of the two forces

[crimson sedge]

i tried to find it

is it right?

ig this is physics question?

[rotund vine]

looks good to me for the first step

I mean like mechanics

cause some countries put mechanics in maths

[bitter reef]

$\text{Work} = Fx$ right?

[wraith dagger]

Doggo

[crimson sedge]

oh

[rotund vine]

yes but dot product for vectors

[crimson sedge]

well if this is maths then it should be fine

[rotund vine]

well tbh
math -> physics -> chemistry -> biology lmao

[bitter reef]

bi 🤮 logy

[crimson sedge]

maths is in everything but if the question is not a topic in math then #old-network -> appropiate server

[rotund vine]

yeah I mean everything is closely related to each other

[crimson sedge]

physics is almost entirely math tho

[rotund vine]

right @bitter reef did we solve your problem

[bitter reef]

I didn't open the channel

I just wanted to make sure I knew the formula right

[rotund vine]

wait sorry

[rotund vine]

it's correct
but for vectors we have dot product and cross product
we need dot product here

$$ W = \vec{F} \cdot \vec{S}$$

[wraith dagger]

||𝓚𝓪𝓼𝓹𝓮𝓻||

[crimson sedge]

answer is correct?

[bitter reef]

this seems correct

didn't plug it in calculator myself

[cedar kiln]

@crimson sedge Has your question been resolved?

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[tulip storm]

Can someone help me understand the difference between the solutions

[cedar kiln]

@tulip storm Has your question been resolved?

[cedar kiln]

@tulip storm Has your question been resolved?

[warped gull]

Hello, to understand the difference between the 2 solutions, you have to manipulate the first one in order to get an expression alike the second one, then see what happens when Ω gets closer to ω

To do so, the first step would be to use the following identity with Ωt and ωt

[cedar kiln]

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[manic acorn]

what is the answer to no. 4?

[dire geode]

What do you get when you Google "arcsin domain"

[manic acorn]

[-1,1]

would it be subset of the range ?

[cedar kiln]

@manic acorn Has your question been resolved?

[manic acorn]

ok its not subset, i just saw the onto part💀

the domain they were talking abt in the activity for the sin function is [-pi/2,pi/2]

which becomes the range of sin inv

but for any other domain that is also one one and onto like [pi/2, 3pi/2]

[manic acorn]

Okay so what i think is that the output of this domain when input in the sin function is the domain of the sin inverse function, but idk if theyre asking this or not

[iron berry]

the domain of arcsin(x) is the range of sin(x)

if that's what you're asking

[manic acorn]

I can understand that from when i did relations..., But inverse trig chapter hasnt been done in school yet,so im not exactly sure what theyre asking

It was like an activity where they used the domain of the sin function from [-pi/2, pi/2], now theyre asking what would the other domains of sin function, for which the function is bijective, will give

These other domains when input into the sin function, they give the domain of sin inverse

So maybe that is the ans?

[cedar kiln]

@manic acorn Has your question been resolved?

[manic acorn]

ok so

if i take the codomain to be [0,1]

then the domain [0,pi/2] for the sin function

is one one and onto

so then if my prev logic was correct

it wud be the subset of the domain of sin inverse?

<@&286206848099549185>

[cedar kiln]

@manic acorn Has your question been resolved?

[manic acorn]

also for question 2 is it more appropriate to use negative or non positive?

seeing as how they have used non negative for question 1

[crimson sedge]

either one should work but just go with negative

[manic acorn]

and what do u think wud be the answer to the 4th question??

[crimson sedge]

uhhh arc sine is sine inverse
defined [-pi/2, pi/2] as range so range

[manic acorn]

what

[manic acorn]

https://ncert.nic.in/pdf/publication/sciencelaboratorymanuals/classXII/mathematics/lelm501.pdf

this is the link

activity 6 if u want context

[manic acorn]

ill just brief you that in the activity they take the domain [-pi/2, pi/2] for sinx, they want us to take other domains for the sinx function for which it is still bijective

[cedar kiln]

@manic acorn Has your question been resolved?

[cedar kiln]

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[idle loom]

Hi all, I am very well versed in sketching the graphs of functions, but I am wondering if there are any established analytical methods (I.e not substituting random points to see where they lie) for sketching implicit curves.

The only thing I can think of other than finding intersects with the coordinate axes is implicit differentiation to find the stationary points. Are there any other useful methods available?

[cerulean onyx]

the gradient helps if its + or -

[foggy merlin]

studying the variations of the functions by seeing the intervals where the function increases or decreases

the sign of the derivative of a function gives us information about its variations

[cedar kiln]

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[cedar kiln]

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[cedar kiln]

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[arctic bluff]

Wait

[arctic bluff]

If I wanted to find the area of the curve x^3 from x= -(sqrt2) till x=0

Could someone please sketch and shade the area it would be

[crimson sedge]

u really can't

find the area

or wait

u can

u cannot exactly shade it and show it I will approx sqrt(2) as 1.41

[upper abyss]

,w integral of x^3 between -sqrt(2) and 0

[wraith dagger]

Results provided by WolframAlpha

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

Requested by Kaynex#7520

[upper abyss]

x^3 is a pretty easy curve. If you can graph it, you can shade the area, right? I'm stuck on where you might be stuck.

[crimson sedge]

the area would be |-1|=1

[cosmic steppe]

I mean

Depends on the context

Generally it would be -1 but unless you want absolute area, it's 1

[cedar kiln]

@arctic bluff Has your question been resolved?

[crimson sedge]

area is always absolute

integral can be negative

[cedar kiln]

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[harsh shadow]

Don’t know how to do this

[cedar kiln]

@harsh shadow Has your question been resolved?

[muted bear]

This forms a sinusoidal wave
Since the car starts at the bottom of tbe wheel we would use a negative cosine function of the form
-acos(bt)+c

We know the amplitude since it has a radius of 8 so a=8
We know the horizontal shrink because it needs to complete a full rotation in 40 seconds so b=pi/20
And we know the vertical shift because the wheel is 12 off the ground so c=12

Bringing it all together you get
-8cos(pi/40×t)+12

b should be easier when you have the formula

[cedar kiln]

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[bold vine]

<@&268886789983436800>

[odd seal]

Write out your question and steps

#❓how-to-get-help

[cedar kiln]

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[cedar kiln]

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[quartz yarrow]

curious what the right answer for this question would be

[quartz yarrow]

.close

[cedar kiln]

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[crimson sedge]

how did the first function become the second function

[dire geode]

Do you know the quadratic formula?

[crimson sedge]

ohhh

why use quadratic formula

bc its a polynomial?

[crimson sedge]

because it's quadratic

[dire geode]

https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:quadratic-functions-equations/x2f8bb11595b61c86:quadratic-formula-a1/v/using-the-quadratic-formula

Review that

[crimson sedge]

and quadratic formula

is not like

always needed

sometimes u can factor

but in this case u can't

it rather provides a general formula

ohh ok

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

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[hardy crag]

Hi ok so I know this is supposed to be an easy question and stuff but I'm really dumb. How do I know what base to put my log as when I'm log-ing 2 sides of an equation? I'll be focusing on part (a) here.

[hardy crag]

So I tried with bases x and e but it doesn't do anything

(And even base 10 for that matter)

Thanks for any help

(And for future questions like (b) how will I determine what base when I log both sides, because I don't want to keep asking for help lol)

[cedar kiln]

@hardy crag Has your question been resolved?

[hardy crag]

<@&286206848099549185>

Ty

[cedar kiln]

@hardy crag Has your question been resolved?

[cedar kiln]

@hardy crag Has your question been resolved?

[twilit citrus]

you can try using logarithmic properties

and you can take log base e on both sides

[twilit citrus]

i.e., log ^ (something) = something times log

you can try solving now

idk you got that already or not but this is what you need to do

@hardy crag

[hardy crag]

So log base e

Then idk what to do from here though

The question wants me to arrange it into x = a^2 + b

[twilit citrus]

ln x^4 = ? (use property)

[hardy crag]

So rearrange the ln into the exponent form?

[twilit citrus]

the ques wants you to arrange it as a^x+b not a^2+b

[hardy crag]

Oh yeahh

But still, if I arrange it in exponent form I get the original question

[twilit citrus]

ln x^4 = 4 ln x no?

this is the property i was trying to say that you need to use

similarly for other side

[hardy crag]

So 4ln x = (5x+1)ln e

Ln e is one so I just put 5x +1

[twilit citrus]

correct

[hardy crag]

Ah tyty

.close

[cedar kiln]

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[cinder mason]

hi

[cinder mason]

Find the legs of a right triangle when given the hypotenuse which is 7√2

With the angle of 45°

[elder pewter]

draw it out

[elder pewter]

,rotate

[elder pewter]

label sides hypotenuse, opposite and adjacent

[cinder mason]

Alr wait

,rotate

[elder pewter]

so first find opposite

you know sohcahtoa

?

[cinder mason]

No im sorry

[elder pewter]

search it up

[cinder mason]

Ok ok

This was in my previous lesson

hi

@elder pewter

[elder pewter]

I said search it up

[cinder mason]

i did

.close

[cedar kiln]

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[cedar kiln]

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[crimson sedge]

i need help on this pls

[balmy apex]

• Show your work, and if possible, explain where you are stuck.

[crimson sedge]

for the first one i got <-10,20>

second answer i got 22.36

i have no clue for the thrid one

[iron berry]

well you seem to know how to get the norm of a vector

so what's the issue with calculating it for each vector instead of doing it for the added one

[crimson sedge]

the first one and second is correct? in the last attempt i only got the first answer right

[crimson sedge]

how do i calculate it for each vector

[iron berry]

huh how did you do it for the second answer

the first 2 seem correct

[crimson sedge]

sqrt -10^2 + 20^2

[iron berry]

yes

so why not do the same for <-7,24> and <-3,-4>

and then add those values together

[crimson sedge]

25 + 5?

[iron berry]

seems correct

but my mental maths is terrible so better check it again sqrt((-7)^2+24^2) + sqrt((-3)^2+(-4)^2)

[crimson sedge]

30

[iron berry]

yup

do you get how we got to this result?

[crimson sedge]

yeah we just did what we did in question two, but did it twice plus addition

thank you smm

[iron berry]

yes

you're welcome

[crimson sedge]

can you also help me with this pls

is it <7, -40> for the first question?

[iron berry]

seems correct, you multiply numbers on vectors by just multiplying every entry

and then you have the addition as usual

[crimson sedge]

<-0.2, 5> for the second

[iron berry]

you're really testing my mental maths here

[crimson sedge]

nah it okay

[iron berry]

but seems correct

[crimson sedge]

it correct, i submited

and got it right

[iron berry]

nice

[crimson sedge]

anyways thank youu

[iron berry]

you're welcome

[crimson sedge]

.close

[cedar kiln]

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[charred zodiac]

Is limit of this sequence 1?

[cosmic steppe]

Lemme check

It's 0

If you simplify the general term you get $\frac{1}{x(x-2)!}+\frac{1}{(x+1)(x-1)!}$. Evaluating the limit of that gets you $0+0 = 0$

[wraith dagger]

Umbraleviathan

[cosmic steppe]

But like

That's the limit of the nth term

Not the sum itself

@charred zodiac

The infinite series itself will equal to 1

[charred zodiac]

Thank you!

.close

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[light dock]

I know I already answered this but I don’t actually know how to solve it

I swear these questions were never taught to me

[hot bison]

sorry, since english is not my mother language can i ask you what the question is asking is you need to find AC in term of x?

[crimson sedge]

can you help with mine? I can help @light dock

[light dock]

Yeah I’m also a little confused with the question

[crimson sedge]

@light dock lets say the other one dot is gonna be D

[light dock]

It’s asking me to length of one leg

[crimson sedge]

did you get it?

[hot bison]

let me check your problem

[crimson sedge]

we're going along with how he named them

and the other one is gonna be called D

[light dock]

I know what you mean

[crimson sedge]

degree BAD is 45 right?

[light dock]

D is gonna be when the altitude touches the hypotenuse

[light dock]

I understand how you got that

[crimson sedge]

yup

[crimson sedge]

so AD is x. since ADB=90 and BAD=45, DBA is gonna be 180-90-45=45

right?

[light dock]

Yes, I see

[crimson sedge]

i believe you know how pifagor thing works

[light dock]

I do know the formula for it yes

[crimson sedge]

so let's do pifagor with 🔺 BAD

actually

let me just write them all in one message

would it be ok for u

[light dock]

Of course

Sorry I was drawing the triangle so I won’t have to scroll up every time

[crimson sedge]

We're going along with this picture, and other one is gonna be called D.
$$\text{Triangle BAD} ∡BDA = 90* ∡DAB = 45* -> ∡DBA = 180-90-45=45*$$
Which means Triangle BAD is equalatiral triangle meaning AD = DB. Since AD = x, DB = x too.
$$\text{Triangle CAD} Everything is gonna be same as the one on top, so CD would be x too.$$
So CB = BD + DB = 2x. It is given that CA = AB
$$\text{Triangle ABC} Let's do pifagor on Triangle ABC. AC^2 + AB^2 = CB^2. AC = AB, CB = 2x -> AC^2 + AC^2 = 4 * x^2 -> 2 * AC^2 = 4 * x^2 -> AC^2 = 2 * x^2 - > AC = sqrt(2) * x$$

[light dock]

Did the bot mess something up?

[crimson sedge]

there

[wraith dagger]

Gari

[cold briar]

it's cut off

the right side

[crimson sedge]

im having trouble here kek

[cold briar]

latex
We're going along with this picture, and other one is gonna be called D.
$$\text{Triangle BAD} ∡BDA = 90* ∡DAB = 45* -> ∡DBA = 180-90-45=45$$
Which means Triangle BAD is equalatiral triangle meaning AD = DB. Since AD = x, DB = x too.
$$\text{Triangle CAD} Everything is gonna be same as the one on top, so CD would be x too.$$
So CB = BD + DB = 2x. It is given that CA = AB
$$\text{Triangle ABC} Let's do pifagor on Triangle ABC. AC^2 + AB^2 = CB^2. AC = AB, CB = 2x -> AC^2 + AC^2 = 4 x^2 -> 2 * AC^2 = 4 * x^2 -> AC^2 = 2 * x^2 - > AC = sqrt(2) * x$$

ok doesnt work

[crimson sedge]

We're going along with this picture, and other one is gonna be called D.

{Triangle BAD} ∡BDA = 90* ∡DAB = 45* -> ∡DBA = 180-90-45=45*
Which means 🔺BAD is equalatiral triangle meaning AD = DB. Since AD = x, DB = x too.

{Triangle CAD} Everything is gonna be same as the one on top, so CD would be x too.
So CB = BD + DB = 2x. It is given that CA = AB

{Triangle ABC} Let's do pifagor on 🔺ABC. AC^2 + AB^2 = CB^2. AC = AB, CB = 2x -> AC^2 + AC^2 = 4 * x^2 -> 2 * AC^2 = 4 * x^2 -> AC^2 = 2 * x^2 - > AC = sqrt(2) * x

https://cdn.discordapp.com/attachments/903430334681608232/989174332381077564/unknown.png

@light dock

[cold briar]

you might want to memorize the 45 45 90 triangle

[crimson sedge]

thats exactly what we're trying to prove here xd

[light dock]

Alright so in a right triangle with the other two sides being 45 degrees, the hypotenuse is x times the square root of 2 every time

[crimson sedge]

yuh

[light dock]

I’m still gonna read your explanation in order to fully understand this masterpiece

[crimson sedge]

tell me the part you dont get if you dont get any of them

[light dock]

Ah sorry I’m still reading it

Since CB = 2x, why did CB turn onto 4 * x^2?

Nvm it all makes sense now

Thank you @crimson sedge you helped me understand this masterpiece

[crimson sedge]

yup no worries

[crimson sedge]

it's the square of it

[light dock]

Yeah I figured that out

I looked at a lot of brain.ly explanations but none of them were as clear as yours

[crimson sedge]

you can always ask for help if it's geometry

especially considering that this level of geometry is easy for people who know goemetry

[light dock]

Of course, have a great day

.close

[cedar kiln]

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[desert lotus]

hi, someone could help me out with this limit?

[desert lotus]

i tried to assume t=x-2

And it should goes like this right?

Then like this I think

but i dont know how to go on

the result must be 1/16

ok i actually solved it lol

.close

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[crimson sedge]

How do i find the volume here?

[cold briar]

ok so

have you tried using the equation given

[crimson sedge]

yes

its more that i dont understand how to do the work?

[cosmic steppe]

What's your r and what's your h

[crimson sedge]

3.14 is my r and h is par^2?

[cosmic steppe]

No

r is radius
h is height

[crimson sedge]

Oh wait my bad

mk r is 18.8 ft and 24 is height?

[cosmic steppe]

Yeah

[crimson sedge]

Now I dont understand how to find the volume with those numbers?

[cosmic steppe]

You just plug it in

Replace r with 18.8 and h with 24

[crimson sedge]

this is a new topic for me im sorry lmao

[cosmic steppe]

$$V = πr^2h$$
$$\=π(18.8)^2(24)$$

[wraith dagger]

Umbraleviathan

[cosmic steppe]

Note how the 18.8 replaced the r, and the 24 replaced the h

[crimson sedge]

yes

[cosmic steppe]

So it's just evaluating that number

[lilac totem]

yeah

[crimson sedge]

So evaluate the number like?

[lilac totem]

you could just keep pi like a variable or substitute the value of pi

[crimson sedge]

like sub

ohh

[cosmic steppe]

Well π is a number

$$π ≈ 3.14$$

[wraith dagger]

Umbraleviathan

[cosmic steppe]

We just rounded it

[lilac totem]

but yeah, you treat pi like a variable

[crimson sedge]

so i just add 3.14 to the equation?

[lilac totem]

i don't get you.

[crimson sedge]

I don't get this stuff man

[lilac totem]

you mean substitute $\pi = 3.14$?

[wraith dagger]

ThickduckPlayz

[crimson sedge]

yes?

[lilac totem]

oh yeah you could do that

else, you could just pi as it is to reduce calculations

you had $\pi(18.8)^2(24)$

[wraith dagger]

ThickduckPlayz

[crimson sedge]

ehm

[lilac totem]

nvm

[cosmic steppe]

They want them to use 3.14 in lieu of pi

Which is dumb lol

[lilac totem]

oh

i didnt note that.

substitute $\pi = 3.14$ and calculate

[wraith dagger]

ThickduckPlayz

[crimson sedge]

is that final?

[lilac totem]

yes.

[crimson sedge]

okay

[lilac totem]

in final, you will have a number

if you substitute $\pi = 3.14$, you will get :
\newline $(3.14)(18.8)^2(24)$

[wraith dagger]

ThickduckPlayz

[crimson sedge]

Wait okay so is there an easier way to calculate this?

As in is there an easier way to get the answer?

.close

[cedar kiln]

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[ashen stone]

I'm looking for equation(s) that would return an X,Y,Z point on a Circle in 3D space, given theta, radius, and a rotation of one specific axis (let's call it spin)

I believe the axis I need to spin the circle around is Z, (so that I could spin the circle as a coin would spin on a table, see pic)

So Given:
theta (radians)
spin (radians)
radius
centerpoint can be assumed to be 0,0,0

Return:
X,Y,Z

https://math.stackexchange.com/questions/73237/parametric-equation-of-a-circle-in-3d-space
This top answer here feels really close to what I want. Though I'm confused about how to implement a and b to respect the rotation ("spin") I am looking for (and whether or not that's even feasible)

Thanks

[undone ore]

wtf r u serious

[ashen stone]

?

[undone ore]

pls help me with ass 2

pls

[ashen stone]

Oh. I've asked a question here. Could you please visit one of the other unused help channels and post there? #❓how-to-get-help

[undone ore]

ok

[cedar kiln]

@ashen stone Has your question been resolved?

[vernal shell]

Hello, are you assuming any position for the circle? I mean, we need at least the axis perpendicular to the circle that crosses its center

[ashen stone]

Hmm. I believe I need this assuming that the center point is always 0,0,0. Does that answer?

[vernal shell]

If you want to rotate the circle in different ways, you can start with the regular parametric equation of a circle, and then apply the rotations matrices as you wish

[vernal shell]

I mean what's the "initial" orientation of the circle? You want a circle depending on its spin, but the result of the spin depends of the initial position of this circle

[ashen stone]

Oh I see. I believe the initial rotation can be arbitrary for what I need it for. So parallel along either the X or Y axis is fine with me

[vernal shell]

These are the rotation matrices in case you haven't seen them before

So you can start with

$(x,y,z)=(R\cos\theta, R\sin\theta, 0)$

[wraith dagger]

leonardogtf

[vernal shell]

This is flat on XY plane

Then use the rotation matrices to rotate around the axis you want through matrix-vector multiplication

If you want the circle's orientation to depend on the axis v just like in the math stackexchange answer, you can take a = (1,0,0). Then b = v x a, just make sure v is an unit vector

[ashen stone]

I'm a game developer with unfortunately very little math background. Very new to matrices.

The third image describes the rotation I seek

Question...

Does theta in the image you sent represent the "spin" that I described

While theta in the equation represents where on the circle I'm getting the point from?

[vernal shell]

Yes, it represents the spin you described. If you want it to spin around the z axis, then use the matrix related to this axis which is Pz. Also what you said about the theta in the equation is correct

[vernal shell]

Sorry forget what I said about a=(1,0,0) here, it's wrong

[ashen stone]

Ok nice. So it's just a matter of multiplying that Pz Matrix with the Equation. I will need to go learn Matrix-Vector multiplication then

One more Q, what exactly is that equation you sent?

[vernal shell]

This is a circle flat on XY plane

[ashen stone]

Ohh. Because I feel like Z shouldn't always be 0?

[vernal shell]

If you want it to initially be flat on a different plane like YZ, just change the position of the components:

$(x,y,z)=(0,R\cos\theta, R\sin\theta)$

[wraith dagger]

leonardogtf

[ashen stone]

Riiiight. Okay that clears things up. Great explaining thank you so much.

.close

[cedar kiln]

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[cedar kiln]

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[prisma ether]

Considering the following quadratic function f(x) = -x²+7x-18
For what values of d, the range of the quadratic function g(x) = f(x+d) is [-20, +∞)

[prisma ether]

So I tried replacing x+d in f(x) and I got this

x² + 2xd + 7x + 7d - 18

And I think what I did is useless lol

[cedar adder]

Chuti please show your work as then I can see what went wrong.

[rugged palm]

how did you get this?

[prisma ether]

1 sec

[dawn junco]

think about it, what does transforming f(x) into f(x+d) do to the graph of your function ?

[prisma ether]

uh

it moves d in Y

?

sorry idk

[dawn junco]

nah

[prisma ether]

1 sec

[dawn junco]

there's not a lot of other choices

[prisma ether]

moves in X ?

😄

[dawn junco]

yeah

if you shift your function horizontally, the range doesn't change

[prisma ether]

ooooh

loool

haha

[dawn junco]

so you just have to look at f(x)

[prisma ether]

how are you guys so smart 😄

so basically

for all values of D

[dawn junco]

well depends on what happens with f(x)

if the range of f(x) is [-20, +inf) then any value of d will do

[prisma ether]

right so I just need to get that for f(x)

no need of adding x+d

[dawn junco]

if it's not, then no d will work

[prisma ether]

thats what you mean right?

[dawn junco]

yeah

[prisma ether]

amazing!

[dawn junco]

don't forget to close

[prisma ether]

checking something b4 closing 😛

[dawn junco]

okok

i mean also your f(x) faces downwards

[prisma ether]

so its a decreasing function

plus

[dawn junco]

the coefficient of x² is negative

[prisma ether]

yyyup

Xv 3,5
Yv 3

So its not possible to go above 3 in the range

[dawn junco]

yeah

image is ok also lol

[prisma ether]

haha okok

so the range is (-∞, 3]

[dawn junco]

yah

(very weird question but whatever)

[prisma ether]

meaning its impossible to have that range [−20, +∞)

so no value of D

[dawn junco]

yup

[prisma ether]

😮

you are very smart

tysm

💙

.close

[cedar kiln]

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[cedar adder]

last one i swear @crimson sedge

lmao im sorry

anyone can reply

[crimson sedge]

notice IC and HD are parallel

[cedar adder]

yeah?

[crimson sedge]

and JF intersects them

[cedar adder]

yes.

[crimson sedge]

wait what's the problem asking

[cedar adder]

the intersection of eg and ij is

......

[crimson sedge]

oh uhhhh look at the line that EG is on

and look at the line IJ is on

[cedar adder]

yeah?

[crimson sedge]

when does the two lines intersect

[cedar adder]

on line f?

[crimson sedge]

nope

[cedar adder]

then where

[crimson sedge]

line IJ is part of B and C right

so IJBC is all on one line

[cedar adder]

yes

i see that

[crimson sedge]

EG is part of A and B so EGAB are all on the same line

[cedar adder]

yeah

[crimson sedge]

however there is only one point that IJBC and EGAB intersect at

[cedar adder]

OH

point b

[crimson sedge]

yup

[cedar adder]

alright thanks man

ur the best

[cedar kiln]

@cedar adder Has your question been resolved?

[cedar kiln]

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[coral garden]

what is the variance of T?

given that n = 6

i tryied e[T^2] - e[T]^2, but got stuck calculating E[T^2]

[cedar kiln]

@coral garden Has your question been resolved?

[cedar kiln]

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[amber plover]

Hey can anyone solve this?
Determine the plane in which the lines are contained:
l:{ x=5+3t, y=-1-2t, z=3-5t and k: { x=2-4t, y=-1-t, z=3+2t

Determine the equation of the plane defined by points A, B, C, and verify that point D is contained in the plane. A(0, -2, 3), B(2, -7, 3), C(-1, 4, 4) and D(-1, 2, 9)

[long arrow]

ye

[amber plover]

can't solve the first one at all, in the second task I determined the equation (don't know if right) but I don't know to to check if D is contained in the plane

[long arrow]

for the 1st one firstly find directional vectors of the lines

then take their cross product to get normal vector of the plane

then use formula (choose some point on some line)

[amber plover]

so for l: (5,-1,3) and for k (2, -1,3)?

[long arrow]

coords stand next to the parameter t, u know?

[amber plover]

oh sorry my bad, thought it was the first one (im really bad)

[long arrow]

np.

[amber plover]

so it's gonna be (3, -2, -5) and (-4, -1, 2)

[long arrow]

yep

do cross product

[amber plover]

ok sec

(-1, 26, -11) if i did it right 🤔

nvm did it wrong

[long arrow]

do u know the method?

[amber plover]

yea

[long arrow]

I guess u use matrix

[amber plover]

im using this one

but i wrote it wrong in my notebook

[long arrow]

aa

ok

[amber plover]

(-9, 14, -11)

this one should be right

if i can count

yep it's right

[amber plover]

and i don't get this

like im supposed to choose a random point?

[long arrow]

Now I look at these lines and they don't form the plane cuz they don't intersect, can u check if you rewrite their equations correctly?

[amber plover]

I think I wrote it good, I had some trouble with reading the last z but i think it's written good

(and it's in Polish so sorry for that)

[long arrow]

it seems that plane you're looking for doesn't exist, if you have answer u can check if it's possible lmao

[amber plover]

unfortunately I dont have it, but now at least I know how to solve a task like this

but I still don't get how im supposed to use a random point

[long arrow]

if plane contains the line it means it contains every point which lies on that line also

Equation of the plane which is perpendicular to the vector $\vec{u}=[A,B,C]$ and passes through the point $P=(x_{0},y_{0},z_{0})$ is:
$$\pi : A(x-x_{0})+B(y-y_{0})+C(z-z_{0})=0$$

The easiest way it so plug t = 0 and get some point

from l or k

it doesn't matter

[amber plover]

so I could choose like a=(0, 2, 3)

and than just use those coordinates for xo, yo, zo?

[wraith dagger]

Modus

[amber plover]

and for A, B, C i have to use the normal vector plane

[long arrow]

yes, if a belongs to l or k

(in this case)

[amber plover]

and how im supposed to know if it belongs?

it has to equal 0?

0=0?

[long arrow]

you have the equation, plug t = 0 or any t, then take x,y,z

l:{ x=5+3t, y=-1-2t, z=3-5t
t = 0:
P = (5, -1, 3)

[amber plover]

oh yeah right...

forgot about that and was thinking from where im supposed to take the t lol

[long arrow]

you can solve 2nd?

[amber plover]

I got the equation, but don't know if i did it right, and I dont know how to check if the D is contained in the plane

[long arrow]

what's your equation?

[amber plover]

first of all I did AB, BC than I got the normal vector from AB x BC and then i used it in the equation while using xo, yo, zo from the A point and i got something like this:
67(x-0)+37(y+2)-43(z-3)=0

[long arrow]

strategy is good

you can check calculations

[amber plover]

so i calculated it wrong? fck

bruh im dumb lol

[long arrow]

I've got

-5x - 2y + 7z - 25 = 0

[amber plover]

yeah i already see what i did wrong

but i dont know how i did that lol

like what was in my mind

[long arrow]

I generally would recommend you use another method of finding cross product

it's hard to memorize that formula

[amber plover]

this?

[long arrow]

yes

[amber plover]

the good part is that i dont have to memorize it

i can have it on paper

[long arrow]

aa

then it's fine np

[amber plover]

but the bad part is my brain lol

give me a sec im calculating once again

[long arrow]

AB = [2, -5, 0]

[amber plover]

yep

[long arrow]

BC = [-3, 11, 1]

[amber plover]

yep

[long arrow]

and AB x BC should equal [-5, -2, 7]

[amber plover]

yep it equals

i dont know how i did that but i got AB=[7,-15,-2] and BC=[-8,11,-3] on the first try 😭

[amber plover]

from where do we take the D?

D=25 i mean

from the equation

[long arrow]

if point lies on the plane = it means its coords satisfies plane's equation

so you can plug D = (x,y,z) = (-1, 2, 9) into plane's equation and check

[amber plover]

but from where did u get 25?

[long arrow]

after simplification

[amber plover]

which point did u use for xo, yo, zo?

cause I don't fully understand the simplification

[long arrow]

doesn't matter, let's say we take A as you did

[amber plover]

like we got -5, -2, 7 as the normal vector and we have to use it in the equation so it will be -5(x-x0)-2(y-y0)+7(z-z0)

[long arrow]

u = [-5, -2, 7]
A = (0, -2, 3)

-5(x - 0) - 2(y + 2) + 7(z - 3) = 0

-5x - 2y - 4 + 7z - 21 = 0

-5x - 2y + 7z - 25 = 0

[amber plover]

yep

[long arrow]

ye

[amber plover]

got it

[long arrow]

so now

check D

[amber plover]

but I have to use D coordinates for x, y,z?

[long arrow]

sure

it's like checking if point lies on the line in R2 (Cartesian plane)

you can imagine it to better understanding

[amber plover]

i dont think 39=0

so D isnt in the plane

[long arrow]

exactly

gj

[amber plover]

ok now i get it

but i still have one question to the first task

[long arrow]

ye?

[amber plover]

let me think about it for a sec

how to form it

we do the normal vector, write it in the equation for the A, B, C and than we have to use something for the x0, y0, z0 and you wrote that we can use it from one of the lines

[long arrow]

yes

[amber plover]

l:{ x=5+3t, y=-1-2t, z=3-5t
t = 0:
P = (5, -1, 3)

like this

and than we just use the (5, -1, 3) for x0, y0, z0?

[long arrow]

yee

maybe I've messed something up, but you know - if the lines don't intersect and aren't parallel (skew lines) then they don't form a plane

[amber plover]

and that's literally all?

[long arrow]

it's easier than you think it is, the fundamental thing is to associate the line with the directional vector and the plane with the normal

then it's just illustrating it in the space

[amber plover]

yea it seems pretty easy now after you told me how to do it

i've got one more task, but I dont know if you have the time

[long arrow]

np., if I can help I'll

[amber plover]

Find the equation of the line which is the common part of the two planes of pi1: -x-2y+5z-3=0 and pi2: x+3y-z3=0

and I did it

i can take a picture

[long arrow]

not needed, strategy can be:
find directional vectors
take cross product
find some point
use equation of the line

[amber plover]

looks like this

[long arrow]

I'll check

[amber plover]

i was trying to do it with normal vector, got almost the same stuff but it was -13t, 4t, -1t

[long arrow]

wait

you've assumed z = t?

or why is it t

[amber plover]

yea

they teached us like this

[long arrow]

😮

ah

it's z + 3

wait, pi1: -x-2y+5z-3=0 and pi2: x+3y-z + 3=0 *

[amber plover]

yea

[long arrow]

then it's ok

[amber plover]

great

i've been siting on this for the past 6 hours

[long arrow]

sorry, I thought it was 3z

😄

[amber plover]

haven't spent so much time for math or any thing else like for the past 7 months 😭

[amber plover]

thank you for the help, I'll let you know how my exam went tomorrow

[long arrow]

np, all the best tommorow

[amber plover]

i really appreciate your help

thanks once again

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[cedar kiln]

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[muted bear]

This is the classic problem

Lemme explain rq

[gray blade]

alr yeah I'm just confused

[muted bear]

The intersection of two sets (M for mustache, B for beard) can be represent as the sum of the cardinality of the two sets minus the union

Do you want that explained more simply?

In other words
|M|+|B|-|M U B|= |M intersect B|

[gray blade]

uhh yeah those are some confusing words..

[gray blade]

I don't really get this either

wdym intersect

[muted bear]

Oh ok

If you dont mind me asking are you above or below college level

[gray blade]

below, I've finished precalculus

this is an sat practice problem btw

[muted bear]

Ah ok

Then im using crazy af words

Lemme make a lil diagram rq

[gray blade]

Lol

alright

[muted bear]

I have something like this

Does that make sense

[gray blade]

ohh yes

wait so that means you can take 5 away from both?

maybe

[muted bear]

Youre close

So youre trying to find the overlap

[gray blade]

ohh okay so I think 5 beard men are actually in the "both" section which leaves 10 with only beards?

or no.. 10 are in the both section

hm

[muted bear]

Yes 10 are in the both section

[gray blade]

ohh okay

so 5 have only beards

[muted bear]

Yes

[gray blade]

thanks

.close

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[crude scroll]

Is this correct

[dawn junco]

little typo at the end

otherwise it's alright

you didn't necessarily have to introduce y for your last point (stability by inverses)

[crude scroll]

make sense..

thanks

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[sharp vine]

Hello, I'm having trouble with this expression; the answer that I got was 8x^7/4y^6, and the answer was 32x^7/y^6

The topic that I'm dealing with has something to do with properties of exponents; which I'm sort of familiar with.

[sharp vine]

^^^ the expression I'm talking about is the bottom one.

[rugged palm]

can you show your work?

[sharp vine]

Could I detail it in steps, instead

[rugged palm]

of course

[sharp vine]

What I did was I flipped the numerator and denominator,

thus, removing the negative signs ( i think)

then, I got 2^2 (4), y^6/ 8, x^7

Sorry, I'm not too familiar with properties of exponents lol

[rugged palm]

if you can, use parenthesis to to type fractions, by "," do you mean multiplication?

[sharp vine]

could you go into detail

im confused, sorry

I switched the positions of the numerator and denominator, and from there, I removed the negative signs, and simplified the integers

Is there anything extra that I should of done?

[rugged palm]

wait, let's do it step by step

[sharp vine]

ok sure

[rugged palm]

$\frac{4y^{-6}}{8^{-1}z^0x^{-7}}$

[wraith dagger]

Nonna

[sharp vine]

ok so you simplified 2^2,
is that all that changed?

[rugged palm]

yes

z^0 = 1

[sharp vine]

ok 👍

[rugged palm]

$\frac{4y^{-6}}{8^{-1}x^{-7}}$

[wraith dagger]

Nonna

[sharp vine]

wait what happened to the 1?

whyd it disappear

[rugged palm]

1 * n = n

we are multiplicating

[sharp vine]

what is n?

[rugged palm]

whatever number you want (not 0)

if you elevate something by 0 you always get one (if the base is not 0)

[sharp vine]

ok yeah z^0 = 1 (assuming z doesnt equal 0)

[rugged palm]

yes

[sharp vine]

i still dont get how 1 disappeared

[rugged palm]

Oh I'm sorry

I misunderstood what you meant

You can multiply what you want by 1, you will get what you started with

[sharp vine]

ohhh so we're multiplying 1 to the denominator?

[rugged palm]

Yes

[sharp vine]

are there invisible multiplication signs?

[rugged palm]

there's no reason to leave it

[sharp vine]

oh ok

[rugged palm]

[sharp vine]

i think i get it lol

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[cedar kiln]

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[rugged palm]

$\frac{4y^{-6}}{8^{-1}\cdot z^0 \cdot x^{-7}}$

[wraith dagger]

Nonna

[rugged palm]

oh ok

[cedar kiln]

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[proper plaza]

Where did I go wrong? I think it was in my distribution of the exponent 2.

[proper plaza]

Ignore 16+6i, I mistakenly put that in there

[livid hound]

how are you getting from
(1+3i)^2 to (1^2 +4i^2)

[proper plaza]

@livid hound it is 9i^2

Distribution of the 2 exponent

[livid hound]

$\fdream$

[wraith dagger]

ℝamonov

[proper plaza]

How does it work?

[livid hound]

(1+3i)^2 = (1+3i)(1+3i)

[proper plaza]

Oh shit lol

I remember

Sorry

Thank you very much for your help

[cedar kiln]

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[distant delta]

Help

[distant delta]

why isnt -7 =x?

[ocean badger]

Try to cross multiply in the beginning, then distribute, then get all the terms to one side so that you end up with a quadratic equation.

Also if x = -7, in your original question you would get 3/0 on the right side which is undefined and not equal to 7/-8

[dusty hazel]

What you forgot is all of this is set equal to zero.

And (x-1)(x+7) is the denominator.

[ocean badger]

If x = -7 then (x-1)(x+7) = (-8)(0)

Which is 0 in the denominator still

What you did was somewhat correct except you only look at the numerator for zeroes, not the denominatoror.

So leave out -7 and you’re fine.

[distant delta]

oh

you take the denominator for restrictions

i mixed it up

[ocean badger]

Yes.

[distant delta]

ok thanks

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[sand shore]

Hello, I've missed out in the discussion and idk how to solve the whole thing.

[cedar kiln]

@sand shore Has your question been resolved?

[sand shore]

<@&286206848099549185>

[cosmic steppe]

The formatting of your homework is dogshit

What am I supposed to do?

The only thing you can do is find the other angle

[glossy canopy]

The other angle is 58, by the way.

[glossy canopy]

'Tis impossible!

[cedar kiln]

@sand shore Has your question been resolved?

[cedar kiln]

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[cedar kiln]

@crimson sedge Has your question been resolved?

[hexed ocean]

Do you know what it means for a point to be the local min of a function?

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

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[craggy anchor]

how do i prove similarity or am i assuming

[leaden snow]

@craggy anchor There's a common angle. Now you just have to show the side ratios match. But I guess you could also assume it in such a case

[craggy anchor]

(1) Angle CAB is common (2) AE/AC = 6/11, AD/AB = 6/11 (3) Therefore Triangle AED is congruent to Triangle ABC (SAS)?

is that the proof?

[leaden snow]

Yes

[craggy anchor]

tysm for your help

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[tall dome]

would #1 be angle Y? pretty sure im right but am debating a friend abt it