precalculus

[viscid thistle]

for my school we used Java programming

Really because my first programming language

Was kotlin

It's a new language

😂

lol

ik terrible choice

barely any YT Tutorials

try java coding its not too bad

yeah I might try to finish out kotlin because i found some playlist which goes with my book i bought

they make you start a baby code and then to complex things

yeah good idea

I should probably start with Java

Ditch this Kotlin nonsense

let me see if I can find my first code

to show you

oh

where you a computer engineer?

or are*

my bad.

biomedical eng

simple code, the engine it called dr java its free

Haha

First output

Was good

xD

yea lol

That's chill dude

I'll show you a more complicated one

: D

sure

when you think it will run and there is a compiler error lol

RIP

its a clock

its looks crappy, I did not set the face colors right, but you can change it to lookbetter

Oh damn

All I could understand were a few snippets

Damn that looks hard

How long did it take to complete this project?

[viscid thistle]

I believe he gave us like a week

so you need some math, there is some equations in there

[thick raptor]

Wait a minimum of 15 minutes before pinging helpers

[viscid thistle]

Oh sorry

[viscid thistle]

No one wants to help me on the step confusion for #17?

[potent pulsar]

@viscid thistle

I can try

[viscid thistle]

I need help finding the angle.

i would move cos(2theta) to the right and then divide both sides by cos(2theta) to get tan(2theta) = -1

For number 2 would it be fine to rearrange the terms before I solve it so that its, x + w - y ?
Instead of x - y + w

[willow bear]

if you're asking whether it is okay to rewrite the expression as $$5 \log(x) + 3 \log(w) - \frac12 \log(y-4),$$ then yes.

@viscid thistle

[obsidian monolith]

Ann:

[steady aspen]

I would greatly appreciate it if someone could do this question for me in steps:
42 is a member of the sequence with the rule tn = 12 + 3n Find which term it is

Sequencing fucks with my head

[willow bear]

you are in essence asked to solve the equation 42 = 12 + 3n

[cyan flame]

We are proving identities

And I feel so damn stupid

Like wtf is this

[willow bear]

"this"?

[cyan flame]

Trig identities

Idk where to start when I look at a problem

[willow bear]

start with the side that looks more complicated, and try to simplify it as best you can

if both sides are equally complicated you can start with either one

[cyan flame]

How would I know when to apply each trig identitie

[willow bear]

experiment

apply what seems applicable. if you find it doesn't work out you can always back up

[cyan flame]

Oh okay. Do you know any good resources to learn the identities? Since that’s prob the part which is hendering me the most.

[willow bear]

hm. i can send you a list i compiled some time ago, though i cannot guarantee that it will be complete

[cyan flame]

That’s fine

I appreciate it

Thanks 🙏

[willow bear]

you'll have to wait until i get home tho

[distant flume]

"A cylindrical tank standing upright (with one circular base on the ground) has radius 20 cm. How fast does the water level in the tank drop when the water is being drained at 25 cm^3/sec?"

I should use related rates here

[limber bone]

V = pi * r^2 * h

[distant flume]

I know that, but not much else

[limber bone]

the water is being drained at 25 cm^3/sec

cm^3/sec is a unit of volume

[distant flume]

Yeah

[limber bone]

how fast does the water level in the tank drop

water level is like height

[willow bear]

$\frac{dV}{dt} = ?$

[limber bone]

nvm let ann help

[willow bear]

shit. bot's still down

okay so you've got two time-dependent quantities here @distant flume

also this should be in #calculus

[distant flume]

Alright

[viscid thistle]

I've got a feeling I screwed up here but can't pinpoint where 🤔

I'm 70% sure logarithms don't work like that

[willow bear]

they do not indeed

log(a) = log(b) - log(c) does not mean a = b - c

you should instead have gone to e^y = x/(1-x)

[viscid thistle]

Well yeah but I just couldn't find a way ti solve that for x

[willow bear]

multiply both sides by 1-x

get e^y - xe^y = x

e^y = xe^y + x = (e^y + 1)x

[viscid thistle]

So x=(e^y)/(e^y+1)?

I feel stupid now

Looks so simple and I've been trying to solve it on my own for way too long ;-;

I'd multiply both sides but didn't figure out how to go from there

Thank you so much

[calm tangle]

so what exactly is learnt in precalc?

I'm in algebra 2 and I feel like I already know all of the topics listed in the channel description

[viscid thistle]

Take a precalc test and find out @calm tangle

[calm tangle]

Yeah, seemed to be what I have been learning in algebra 2

[willow bear]

@calm tangle read the channel desc for what this server typically considers precalc content

[calm tangle]

Maybe my school curriculum is different, but yeah I did and thats to be what I am learning in algbera 2

[cyan flame]

So with half angles

I’m not understanding why

When you say for instance sin x = 12/13

In quad 1

How using the half angle formula would give us that angle

I understand settting up the triangle and such

[willow bear]

are you asked to find f(x/2), where f is one of sin, cos or tan?

@cyan flame

[lost talon]

Can someone please help me with this?

[swift glacier]

which one?

it's also a bit blurry

[lost talon]

I don't really understand any of them, but if I could have the first one or two explained, I think I would understand and be able to complete the rest alone.

Usually I can just look at the answer key and figure it out, but there isn't a key for this sheet yet

[potent pulsar]

@lost talon

The 3rd one would be 3tan2theta

Thats the only one i know cuz of what u showed lol

[viscid thistle]

does anyone know what to do with 8b

[willow bear]

your photo is cut off, so it's impossible to tell which one is 8b as opposed to 7b or 9b

also, don't ping everyone.

it's disabled, but it's still rude.

@viscid thistle

[viscid thistle]

the one below @willow bear

sorry about that

lol

[willow bear]

is it the one that begins "Let f(x) = ..."?

[viscid thistle]

yeah

[willow bear]

ok

what have you done so far?

[viscid thistle]

i Have completed the square of 8a

but don;t know where to start for b

[willow bear]

no, what have you done for 8b?

ok so

It is given that the graph of y = f(x) cuts the x-axis at P and Q.

[viscid thistle]

yes

so there are two roots

[willow bear]

yes

so why don't you find them?

[viscid thistle]

yeah but how?

like there's a variable p

[willow bear]

so what?

you'll get an answer in terms of p

[viscid thistle]

ok let me try

right?

[willow bear]

yes, but this can be greatly simplified

to $-p \pm \sqrt{p^2 + 2p + 3}$

[obsidian monolith]

Ann:

[viscid thistle]

oh

yeah

[willow bear]

so then

you've found those two roots

what's the distance between them?

[viscid thistle]

2?

idk

[willow bear]

how did you arrive at that?

[viscid thistle]

the discriminant is inside is 2

as stated in 8a

rihgt?

[willow bear]

no, the discriminant of p^2 + 2p + 3 is not 2. it's -8. but you don't care about that.

[viscid thistle]

oh I mean the minimum value of the whole thing is 2

[willow bear]

yes, but i'm not asking you that

[viscid thistle]

so the minimum value is P+P=1+1=2

[willow bear]

what is the distance between $-p - \sqrt{p^2 + 2p + 3}$ and $-p + \sqrt{p^2 + 2p + 3}$?

[obsidian monolith]

Ann:

[viscid thistle]

P+P

[willow bear]

2p?

how did you arrive at that?

[viscid thistle]

??

[willow bear]

i asked you a question

[viscid thistle]

I add the two roots

[willow bear]

you what?

[viscid thistle]

add the two thingy

[willow bear]

do you know how to find the distance between two points on a number line?

[viscid thistle]

yes

[willow bear]

then tell me

[viscid thistle]

subtract them

[willow bear]

so why don't you do just that?

[viscid thistle]

yeah and then?

[willow bear]

well what do you get?

[viscid thistle]

i get - 2p and the thing

lol

[willow bear]

what?

[viscid thistle]

help me

[willow bear]

$(-p + \sqrt{p^2 + 2p + 3}) - (-p - \sqrt{p^2 + 2p + 3}) = ~ ?$

[obsidian monolith]

Ann:

[viscid thistle]

yeah

this thing

lmao

[willow bear]

how are you getting 2p?

show your work.

[viscid thistle]

oh no

I got it wrong

so its 2times that thingamijig

[willow bear]

yes, the distance is $2\sqrt{p^2 + 2p + 3}$.

[obsidian monolith]

Ann:

[viscid thistle]

yas

[willow bear]

so what's the smallest possible value of that?

[viscid thistle]

so the distance is 2 times square root of 2 !!!

rihgt?

RIGHT

[willow bear]

yes

[viscid thistle]

OH

I THINK ALL MY CLASSMATES INCLUDING ME GOT IT WRONG

:))))

::)

[willow bear]

what was y'all's answer

[viscid thistle]

SOME SAID 4 AND SOME SAID 2

LOL

actually I was quite close

I just didn't know what to do after I found the two rrooots using the quadratic formula

like i coludn't see it could be simplified to the form in 8a

and then I got stuck

anyway thx for helping 😃

[fast cape]

Any idea how to simplify (1+i)^2012+(1-i)^2012

[serene heath]

whats (1+i)^2?

[fast cape]

Ah I see. Thanks

[serene heath]

np

[limber bone]

DEJAVU?

WHAAT

[distant flume]

Can I post something that goes into Calculus since it's busy?

[warped anvil]

@distant flume you can try one of the question channels

[distant flume]

Ok thanks

[weak helm]

if you had sqroot(50^2 + (20-x)^2) wouldnt the square roots cancel with the squares and would just be left with 50 + 20 - x?

[patent beacon]

Square roots don't distribute over sums

√[a + b] ≠ √a + √b

[willow bear]

$5 = \sqrt{25} = \sqrt{9+16} \neq \sqrt{9}+\sqrt{16} = 3 + 4 = 7$

[obsidian monolith]

Ann:

[weak helm]

oh thanks

so wait could i square root one though?

and make it like 50 + (20 -x)^2 ?

[patent beacon]

You've got to point out to yourself which rule you're using at any given moment. You are saying:
√[a + b] = √[a] + b

Which isn't true

In general, there's nothing nice you can do with square roots over sums

[cyan flame]

Wait I’m sorry

I’m just seeing your response

[serene heath]

plug in 2 for x solve for f(4)

then plug in 4 and solve for f(16)

[broken minnow]

Plug in 2 where

[serene heath]

for x

[broken minnow]

f(2)^2-f(4) = 7

??

[serene heath]

yea

now do what i said

plug in 2 for x solve for f(4)
then plug in 4 and solve for f(16)

^

[broken minnow]

I got

f(4)=2

k ig ot it

thx

[viscid thistle]

cuz what is cos(pi/2)/1

[short sorrel]

well, it's just cos(pi/2)

do you know what that would correspond to on the unit circle?

[viscid thistle]

90 degs

[short sorrel]

The angle pi/2 corresponds to 90 degrees, yes

But i was asking for its cosine

(as a leading question, mind)

Or, to ask a similar question: what does cosine mean on a unit circle, anyway?

[gilded bolt]

Cosine is the x coordinate of a point on the unit circle iirc

cosine of 90 degrees will return 0

@short sorrel I hope that helps :+1:

(im worried that I mightve answered it too late considering that you asked that question last night <<;;)

[serene heath]

lol

[short sorrel]

I wasn't the one asking the question

Lmao

I was trying to lead someone else to the answer

.>

I appreciate the sentiment, though @gilded bolt

[gilded bolt]

yeye! I need to read more of the chat history before saying something

[viscid thistle]

hahahaha namington

[viscid thistle]

Anyone on right now

And down to help?

Depends on what it is

@viscid thistle just post your question and one of our helpers will be with your shortly
Or in a few weeks
They don't get paid

[viscid thistle]

@viscid thistle

YOU THERE?

You there*

What you need fren

can you go to #questionk

[viscid thistle]

Ew double-posting

[half axle]

whats the correct fill in?

[viscid thistle]

M

log and exponentiation are inverses

[heady mango]

How woupd i solve these two problems

Would i solve

[bright oak]

cos x = x/2 ?

[willow bear]

[viscid thistle]

No complex exact analytic answer

[hollow estuary]

I can’t figure out how to prove that -2 to > 0 doesn’t work without using testing point.

[patent beacon]

@hollow estuary
What? Do you mean you're trying to prove that f(-2) > 0?

[hollow estuary]

(-5,-3] U [0,1) is the part I want to prove

[serene heath]

what exactly are u trying to prove

[hollow estuary]

Why -2 to 0 doesn’t work

For the domain

[patent beacon]

I'm guessing that you're trying to say "I want to prove that (-2, 0) isn't in the domain of f(x)"

Well, let's look at f(x).
There's a square root, so anything under it must be greater than or equal to zero:
x² + 3x ≥ 0

x² + 3x is a parabola that has roots at x = 0, x = -3. It opens upwards, which means that everything between those roots is negative. (-3, 0) is not in the domain

@hollow estuary

[hollow estuary]

Thank you, sorry for being so unclear.

[quasi dirge]

Can anyone explain how I would go about finding the right multiples of 2 pi for coterminals? I'm not sure I understand where they come from;;

[solemn ore]

4pi/3 is answer

just take it mod 2pi

[quasi dirge]

What do you mean?

[solemn ore]

You Know Like How 5pi is coterminal with 1pi

[quasi dirge]

Yeah

[solemn ore]

Just subtract 2pi until you get a value less than 2pi

same for negative

add 2pi until you get an absolute value less than 2pi

theres your Answer.

-2pi/3

[quasi dirge]

Oooh

[solemn ore]

add 2pi and u get 4pi/3

[quasi dirge]

I see...thank you very much for your help 😄

[solemn ore]

np men

[stable kernel]

hey can anyone help me out ?

confused about vectors

sorry if this isnt very visible but confused about TU + UP + PS

[pliant niche]

d)

The Red ones cancel out.

[stable kernel]

Ah Thank YOU!

[pliant niche]

Think of it like driving

You drive to school, take a left and drive a block, and then another left and drive back toward home

Would be the same as just taking a left at home

Or going the distance from UP

[stable kernel]

I see, thanks for the analogy!

[pliant niche]

Also Vectors don't really have a position(i think?) but it's good enough. ¯_(ツ)_/¯

🤔 wait, they don't have positions

Wouldn't the one above just be twice UT and then just UP?

Wait

Gosh darn it I'm confusing myself.

How are you calculating the Magnitudes?

Because PS might = TU

[stable kernel]

What do you mean by that, like the lengths?

[pliant niche]

And not PS = -TU

Yeah

ah, I should go to bed and stop making this worse

[stable kernel]

Oh for this section we are just completing the statements so finding the vector that completed it

Np now I'm confused on SR+ TU lol

Because those vectors are really far apart

[pliant niche]

oh so it wouldn't be anything as fancy as the second thing I said. Neat

[pliant niche]

Maybe you can use the idea that some other lengths are known to be the same as TU

[stable kernel]

Yeah these are the answer ive got so far

Yeah that's what I was thinking so TU is SP

[pliant niche]

yep

[stable kernel]

So would that be like RP??

[pliant niche]

Yeah

[analog tinsel]

Write an equation of the plane that contains the points P(2,-3,6) and Q(4,1,-2) and is parallel to the line [x,y,z] = [3,3,-2]+t[1,2,-3]
https://cdn.discordapp.com/attachments/238956921581862913/562463456670777344/477e7edb6777b7254e707281399edb9e.png

Thats the answer but I dont get it

[pliant niche]

So first of all you get to ignore the position of the line you’re trying to make it parallel to. You only need the information about the 3D slope. (It helps to simplify the information) because of this, you might be able to picture that since the plane will be parallel to the line, one of the direction vectors will BE the line. (See the answers for that)

[analog tinsel]

Oh

so cant i

just find a vector given the two points

then take the cross product between two vectors

@pliant niche ?

[pliant niche]

Yeah that too. I was having a hard remembering if it was the dot or cross product

Lol

[analog tinsel]

yeah lol

[frosty heart]

why is 1-sinA-cos2A/cosA+sin2A equal to tanA ??

[eternal lotus]

$\frac{1-\sin(A)-\cos(2A)}{\cos(A)+\sin(2A)}$?

[obsidian monolith]

stephen:

[willow bear]

[echo plaza]

double angle

[valid vector]

Can someone explain to me why it's the second graph?

When I did the bx-c thing I got π/3 (0) and 5π/3 (2π)

[frozen needle]

Well it's the only one that's got the correct amplitude

[valid vector]

But what if I were to graph it myself, I would have gotten it wrong

Cause when I equaled it to 0 and to 2π I got π/3 and 5π/3

That's not what any of them say

[mellow iris]

cos(x + pi/3) is different from cos(x) + pi/3

[valid vector]

?

It's all inside the function

This is what I did

[mellow iris]

the one you did on the left should be
bx - c = 0
x + pi/3 = 0 (instead of x - pi/3 = 0)
x = -pi/3

plugging -pi/3 and 5pi/3 into y=4cos(x + pi/3) gives 4 for both, the max on the second graph lines up with that

[valid vector]

I know, but where do the -2π/3 to 2π/3 in the second graph come from?

[mellow iris]

I guess that's just to give you a reference to figure out for what prominent pts lines up with what x values?

though I guess you can plug in 2pi/3 into the given expression and find you'll get -4, which lines up on the graph

[fiery spruce]

How would you get the horizontal asymtote from the equation?

[frozen needle]

you could calculate the limit at ±infinity

[short sorrel]

If you haven't learned limits get, the "trick" is to divide the leading coefficients of each polynomial, assuming they are of the same degree

In this case, 3/1 = 3, so the h. asymptote is as y=3

Explaining why this "trick" works is best done with an understanding of limits, I'm afraid

[serene heath]

[short sorrel]

I say this because some curricula introduce this concept without formally introducing limits

Which is dumb imo but whatever

[analog tinsel]

lets say line = [x,y] = [3,2] + t[4,2]

find the scalar equation of a plane that intersects that line

[frigid vale]

guys i managed to make desmos depressed

[grave furnace]

Idk if this is precalc but I need help on #9

[grave furnace]

<@&286206848099549185>

[fading token]

,rotate

[fading token]

I suppose this fits in precalc

[willow bear]

@grave furnace what's holding you up?

[valid vector]

What's the period of this function?

@willow bear Can you help me? It's urgent, sorry for the tag.

[willow bear]

okay well

you know the period of tan is π, right

@valid vector

[ruby otter]

We can multipy both sides or bring smtng from one side to another if the value/entity is positive right??

[serene heath]

what

[ruby otter]

wait im not good at latex

can u tex it??

[serene heath]

tex what?

[ruby otter]

$ |\frac{2x}{x-2}| + |x| = \frac{x^2}{|x-2|}$

[obsidian monolith]

Radical Ninja:

Radical Ninja:

[ruby otter]

@serene heath u der??

[ruby otter]

Fine at least some one else help

<@&286206848099549185>

[hasty vale]

no: |x||x-2| = |x(x-2)|

[keen heath]

I think I should go here

Hello

[deep glacier]

does anyone know how to find the interval such that the polar graph is traversed once?

[viscid thistle]

Yes

[deep glacier]

is there a way to do with algebraically

without a grph

[viscid thistle]

Yes

[prisma prairie]

what function is eqivalent ot log(logx)?

if there is any

like how can i write it differently

[short sorrel]

there isnt really a way to write it in simpler form

is this for classwork @prisma prairie? if so, mind giving the full question?

[patent beacon]

@vestal plaza
Still want it?

[vestal plaza]

@patent beacon yes

how do I do this

I know we have -1, and 3

so should be something like

(x+1)(x-3)

[ruby otter]

ya

but remember at x=3,-1 it is not defined so
x cant be = 3 , -1

change accordingly !

@vestal plaza you got it now??

[vestal plaza]

no

[ruby otter]

fine atleast say do you know wavy curve method

[vestal plaza]

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

idk what number goes in the front

or how to figure that out

[ruby otter]

its wrong

[vestal plaza]

I know its wrong

idk what number goes in the front

[ruby otter]

just say i will say how to figure out this

do u know wavy curve??

[vestal plaza]

no

[ruby otter]

fine no probs

you know 1/0 = infinity right?? (or) 1/0 -> infinity

you can see at points 3 , and -1 the graph is not defined or in other words it tends to infinity

thus you can say that at that point/s the function isn't defined

so it must be
$ \frac{1}{(x-3)^n(x+1)^m} $

[obsidian monolith]

Radical Ninja:

[vestal plaza]

I have no clue what you are saying :x

[ruby otter]

fine remember this

anything by zero is not defined

[vestal plaza]

yes

[ruby otter]

in the graph you can it goes straight up which means it tends to infinity or its not defined

and at two points its not defined

they are 3 , -1

[vestal plaza]

yea

[ruby otter]

so if you put
$ \frac{1}{(x-3)^n(x+1)^m} $
as a trial and error

[vestal plaza]

so the bottom part is (x-3)(x+1)^2?

[obsidian monolith]

Radical Ninja:

[ruby otter]

try to put x = 3 or -1

it will not be defined it will give 1/0

now that you have found the function what is remaining m,n the powers of each term

for that remember

got it??

[vestal plaza]

no

lol

[ruby otter]

-_-

which part

[vestal plaza]

english is my second language

unless you visually show me

I wont get it

[ruby otter]

fine say this

can u spend some time to learn a concept

its very easy

[vestal plaza]

I dont got time I gotta turn this in

in like 2 hours

its my last exercise

[ruby otter]

you have 2 hours -_-

it will take 10-15min

[vestal plaza]

and it doesnt tell me the answer

I've been doing this for the past 6 hours

[ruby otter]

no problem the problem you guys are taught some shit in a hard way

[vestal plaza]

doing math aint everything in my life and what is annoying that it wont even tell me the answer if I get it wrong, cause garbage program

[ruby otter]

there is a easy way out there

just study this concept
Wavy curve

after that just by seeing that you can answer

[vestal plaza]

I never heard or the teacher ever said wavy curve

[ruby otter]

you dont even have to use pen/pencil

so dont say it to the teacher that you used it

[vestal plaza]

I just wanna know wtf the answer is

[ruby otter]

$\frac{1}{\left(x-3\right)^2\left(x+1\right)^1}$

[vestal plaza]

at this point, or I am just gonna leave it blank and get a 93

[obsidian monolith]

Radical Ninja:

[ruby otter]

try this

wait

[vestal plaza]

that isnt

correct

the graph is different

[ruby otter]

i said wait

[vestal plaza]

yea I put it in desmo

[ruby otter]

i forgot to do one thing

[vestal plaza]

should be something positive multiplied by the polynomial fraction

[ruby otter]

0.36

try now

$\frac{0.36}{\left(x-3\right)^2\left(x+1\right)^1}$

[obsidian monolith]

Radical Ninja:

[vestal plaza]

sec

nop

its not it

[ruby otter]

wut???

[vestal plaza]

put it in desmos

and compare the two graphs

[ruby otter]

sorry its not 0.36 its 3.6...... i made mistake in decimals

lol i know desmos

[vestal plaza]

its not it

it touches the x axis at 1

[ruby otter]

not this time its definitely correct

[vestal plaza]

I am sure its not

if you put it in desmos

it doesnt touch the x-axis

top should be (x-1)^2

[ruby otter]

wait does it touch the x-axis in the question??

[vestal plaza]

yes

lol

[ruby otter]

fine my brain isn't functioning

just add what you said

i haven't noticed 0.4 and this

thats why i made a mess

[vestal plaza]

3.6*((x-1)^2)/((x-3)^2(x+1))

something like that

[ruby otter]

exactly that

[vestal plaza]

yup its right

[ruby otter]

[vestal plaza]

only part I couldnt figure out is the 3.6

I can figure out the rest

[ruby otter]

put x=0

[vestal plaza]

thanks for the help

[ruby otter]

put y=0.4

(in LHS put x=0 ) = (RHS is 0.4)

just put some unknown like K ,N

$\frac{n(x-1)^2}{\left(x-3\right)^2\left(x+1\right)^1} = 0.4$

after x= 0 you get
$\frac{n (0-1)^2}{\left(0-3\right)^2\left(0+1\right)^1} = 0.4$

[obsidian monolith]

Radical Ninja:

Radical Ninja:

[ruby otter]

find n

@vestal plaza

[vestal plaza]

oh thats all I had to do

wtf

[ruby otter]

ya

thats why i said you to learn that concept

[vestal plaza]

complicated words = no comprende

[ruby otter]

me??

comprende??

[viscid thistle]

yo tango grande pito

[slender river]

t a n g o

lol

[stable kernel]

How to find speed?

[stable kernel]

please someone

[keen bloom]

@stable kernel the magnitude of the velocity vector gives you speed

[stable kernel]

oh sorry forgot to update I understand it now

Thanks though

[keen bloom]

o7

[half axle]

Anyone have any idea? Not sure how to solve this one. Thank you ^.^

[eternal lotus]

tan(750)=tan(2(360)+30)

[woeful wedge]

How do I do this?

[hexed ermine]

a(cis(theta))

10cos(210)+10sin(210)i

[woeful wedge]

oh really? I was about to do that but doesn't that give me x and y as points not as an equation?

also it assumes I can do this without a calculator oddly enough, is that a reasonable assumption or should I pull one out

it says "express in exact terms" does that mean -8.8387blah blah blah or 10cos(210)+10sin(210)i

[hexed ermine]

No you should get it into radicals

Itll involve 0.5 and sqrt(3)/2

[woeful wedge]

Oh alright, thanks

How do I get it into radicals xD

[hexed ermine]

Do you know cos(210)

And sin(210)

[woeful wedge]

generally, this one I have here is 281

[hexed ermine]

Wut

[woeful wedge]

which confuses me

the angle is 281

[hexed ermine]

What angle

[woeful wedge]

oh

also im stupid

this one asks for nearest thousandth

so no rads

🤦

[swift glacier]

Can a polynomial in intercept forms have imarginary roots?

[rocky bison]

ofc

,$ y=\left(x+i\right)\left(x-i\right)

[obsidian monolith]

Pseudo:

[rocky bison]

@swift glacier

[valid vector]

Can someone tell me how do I find the bottom part of the chart?

The y part of the chart

Tag me pls

[grand snow]

@valid vector y=2csc(x+π/4). csc(x) is the reciprocal of sin(x), so I would rewrite it as y=2/sin(x+π/4).

From there, you can plug in the θ values and figure out the the sine.

[valid vector]

And then for the exact points of the csc graph?

[grand snow]

Sorry. Can you rephrase that?

[valid vector]

Well if I do what you say I'd end up finding the Sine function graph, right? But I need the exact coordinates for the csc graph, so how would I find those points if I turn it into a Sine function?

[grand snow]

Oh. Let me see...

y=2/sin(x+π/4)

For example, if you plug in π/4, you get sin(π/2), which is 1.

2/1 is 2

So, for you original equation, y=2csc(x+π/4), plugging in π/4 gives a y value of 2.

Sorry. Made a dumb mistake there.😫

Does that help?

[valid vector]

Oh yeah, it gives me an idea at least. Thank you so much!

[grand snow]

I want to say one more thing. For θ=3π/4, you would get csc(π), which is 1/sin(π), which gives you an undefined value. Just a heads up.

csc(x) has vertical asymptotes at x=kπ, where k is an integer.

[valid vector]

Ah thanks! That had me confused!

[grand snow]

Does that clear up anything? I sometimes get confused when evaluating reciprocal trigs.

[valid vector]

It's confusing yeah, but it helped clear some doubts! Thanks again!

[woeful wedge]

@fringe stream What are the rules about pinging people for help?

[ocean plover]

Hello there. My first post here, so please bare with me. I would like if someone could check my proof and see if it is acceptable for Q92. I'd appreciate it very much. Thanks!

[willow bear]

technically, you missed the case a = b

[ocean plover]

Shoot. So in the case of a = b, how should I go about explaining that?

[willow bear]

well, what does max(a,b) return when a=b? and what about the formula?

[ocean plover]

max(a, b) returns a = b, but I don't know how to write it in words. Suppose b = a, then max (a, b) = max (a, a) = (a + a + a - a) / 2 = 2a / 2 = a = b?

[willow bear]

when a = b, max(a,b) = a (you could also say max(a,b) = b; here it doesn't matter)

then (a + a + |a-a|)/2 = (2a + 0)/2 = a

essentially the same thing you wrote

[ocean plover]

Oh cool. Thanks!

Trying to learn some Precalculus before 11th grade.

[unique basin]

Im trying to find the area of the triangle with calculus. What am I doing wrong. I get 4 but the answer should be 6

[willow bear]

where did that integral come from

[unique basin]

Found it here

[willow bear]

that applies to right triangles only.

[cyan flame]

Is the triganomic Identity Sin^2x+Cos^2x=1

The same as sinx+cosx=1

Just sqrt everything?

[serene heath]

no

[willow bear]

$\sqrt{x + y} \neq \sqrt{x} + \sqrt{y}$

[obsidian monolith]

Ann:

[willow bear]

also, "triganomic" isn't a word

you meant "trigonometric"

[cyan flame]

Oh oof

Okay thanks

[elfin night]

x = 0

[echo plaza]

y = 0

[wet hare]

I am asked to find a positive and negative coterminal for 750°. The way I got the previous ones were by getting -360° and adding it with the given degree and then using 360° and adding it. Can someone explain to me as to why this is incorrect?

By adding like I did with the previous exercises I got this.

[viscid thistle]

i dont get how it has to be greater than 1

there is no square root in the inverse

[rigid beacon]

domain of inverse can only be as big as the range of the original function (or smaller)

it can't be larger

[wet hare]

[earnest wedge]

hi, anyone have examples I can find about
Convert a polar equation to rectangular form.

[ruby otter]

@echo plaza you free?

[echo plaza]

uh

yeah

[ruby otter]

if there's modulus in wavy curve the graph always look symmetric ??

[serene heath]

wot

[ruby otter]

no i actually have this doubt for a long time

if we have function having modulus

the wavy curve of that function looks symmetric ??

[serene heath]

can u give an example

if f(-x)=f(x) then its symmetric about the y axis

for eg: y=|x|

[willow bear]

"wavy curve"?

[ruby otter]

$ \frac{ (2|x| -2) ( ( |x| -1 )^2 +2 ) (|x|+1)(|x|-3)}{x^2-2|x|}$

[obsidian monolith]

Radical Ninja:

[ruby otter]

solve the ineq for >0

[serene heath]

[echo plaza]

split it up

[ruby otter]

i.e. putting + - + - in graph

[echo plaza]

x>0 and x<0

[ruby otter]

to find where it is +ve and -ve

[echo plaza]

2 cases

[serene heath]

isnt that always positive tho

[ruby otter]

you can't do like that

nah

[serene heath]

hmm

so many cases

[echo plaza]

consider when x is positive and negative tbh

[ruby otter]

its very big

and finally you have to put wavy curve anyways

instead you can directly do that from this step

[echo plaza]

fine

consider when each individual part is positive and negative

[ruby otter]

no you can do like this right??

equate single term to zero and find the point

[echo plaza]

4 total

because the other is a square + a positive number

so always positive

[ruby otter]

ya so x= +-1 , +-3 , 0 , +-2

yaa i have done till this step

after this while plotting im confused

[echo plaza]

you need to graph it?

[ruby otter]

to find where it is <0 or >0 like that

[echo plaza]

you don't need to graph to do that

[ruby otter]

then how will you??

[echo plaza]

just check whether the number of negative factors is odd or even

if it's even and none of them are 0, it's positive

[ruby otter]

you have to substitute ??

[echo plaza]

no

for each of these 4 factors

find when they're positive and when they're negative

[ruby otter]

how to find that without substituting some value for x??

[echo plaza]

oh it's even just 3

the blue one is also clearly always positive

[ruby otter]

ya next

[echo plaza]

just solve $|x|-3>0$ for example

[obsidian monolith]

CaptainLightning:

[echo plaza]

add 3 to both sides if you aren't comfortable yet

[ruby otter]

im not a newbie

wait lemme show you what im saying

See

[echo plaza]

yes

[ruby otter]

its not a prob when its the function is normal

but when it has mod in it, it gets weird

it seems we need to check for each term

[buoyant oxide]

who can help with precalc?

[viscid thistle]

nice

[tall wraith]

Hey, im having problems with my precalc for double angle and half angle identities. The question is for a value on sin(∆/2) when i have the value of tan(∆)= 5/12. Im probably supposed to use the formula of sin (theta)/2 which is ±√(1+cos(theta))/2 but i cant for the life of me figure out the steps to get to sin(∆/2) with that bit of formula. Can anyone help with the steps to get to the value of it?

[gloomy spade]

@tall wraith use pythagorean theorem to find the hypotenuse of the triangle, then you’ll have sin(∆)

and you can use sin(∆) to solve for sin(∆/2)

[marble crown]

wait guys wtf

i got root2 -1

but the answer is 2 times root2

my integral was

pls help

[willow bear]

#calculus

[marble crown]

wait this isnt precalc?

i dont live in america lol not familiar with syllabus sorry

[short sorrel]

thats fair, but generally, precalc is what comes before calculus

it usually ends at about limits

maybe introduces the derivative but it doesnt go too into depth

hence the "pre"

integration is definitely straight-up calculus

[tawny nacelle]

they didnt even teach us limits in pre calc

P AT H E T I C

[short sorrel]

yah, it does vary from place to place

my precalc class introduced ln but didnt introduce e in any other context

lmao

it was basically

"here's ln, it has base e. what's base e? its important trust me but thats beyond the scope of this class ™ "

"but yeah, you'll see it around."

[sand pewter]

Same!

Just goes "Sometimes we use base e. For what? Find out in the next episode!"

[viscid thistle]

lmao

[rocky bison]

@sand pewter

,$ \log_e\left(x\right)\equiv\ln\left(x\right)

Issa all

[obsidian monolith]

Pseudo:

[sand pewter]

Well, I know that. I was just chiding how it was being taught initially. Don't mind my stupidity.

[zenith locust]

Tbh whats considered precalc varies a lot generally

We didnt do complex numbers or logs until after i did integration

[sand pewter]

We did logarithms, but not complex numbers.

[true rose]

If there are 2 independent events a and b

P(a)=0.2

P(b)=0.6

What’s the chance that a or b occurs

It’s just adding right? Or do you have to subtract something as well

[limber bone]

yes

[serene heath]

$P(A \cup B)=P(A)+P(B)-P(A \cap B)$

[obsidian monolith]

the one n only:

[serene heath]

@true rose

[ruby otter]

$P(A \cup B)=P(A)+P(B)$

[obsidian monolith]

Radical Ninja:

[patent beacon]

Only if A and B are independent

[ruby otter]

it is given they're independent

[serene heath]

thats if theyre mutually exclusive

[ruby otter]

nah

[serene heath]

well yes

[ruby otter]

what is P(A|B)

[serene heath]

because if theyre not mutually exclusive youre counting their intersection twice

?

[ruby otter]

mutually exclusive doesn't matter here

as long as their probabilities are independent of each other

it doesn't matter they're mutually exclusive or not

[willow bear]

Only if A and B are independent

wrong

[serene heath]

i think you might be mixing up independent and mutually exclusive?

[willow bear]

how is the meaning of "mutually exclusive" not obvious from the term itself, tbh

[ruby otter]

well
$ if $$P(A \cap B) = 0 $

[obsidian monolith]

Radical Ninja:

[serene heath]

yea thats mutually exclusive

[ruby otter]

wait i think im confusing

[serene heath]

[ruby otter]

ya the thing is

its mutually exclusive

i made a small mistake

[viscid thistle]

anybody know how this happened

https://i.imgur.com/84zrpUx.png

[patent beacon]

Multiply top and bottom by eˣ

[viscid thistle]

thank you

i worked it out you are right

[faint mason]

At what point should I know algebra, trigonometry and logarithms to be able to understand calculus fluently?

[gloomy spade]

@faint mason well.. calculus isnt just algebra, trig, and logs

but you should be fluent in these topics to get a better grasp at calc

[faint mason]

I mean, so far, just from vaguely hearing things and watching people solve complicated problems, I have a really really rough idea of the stuff calculus does. (I think)

I think I understand what a limit is, just not how to compute it.
but with derivatives and integrals i get completely lost as to how people got ln(x) as the derivative of 1/x for example

[gloomy spade]

tbh, i think you should be asking in #calculus

#precalculus is just a preparation for calculus, so if you want an idea of what actual calc is like, ask there

[faint mason]

Sure, then.

[gloomy spade]

... if no one answers, you could try #math-discussion @faint mason

[faint mason]

thank you athlete

[gilded leaf]

https://cdn.discordapp.com/attachments/269573202018041856/567177741347192877/Math_q.PNG does anyone know how to solve thise

25 a

[autumn violet]

@gilded leaf What is P

[gaunt gate]

can someone tell me how to multiply a 3x2 and 2x3 matrix ?

[willow bear]

the same way you would multiply any two compatibly sized matrices?

what is it about doing a 3x2 by 2x3 multiplication that's throwing you off?

@gaunt gate

[gaunt gate]

tysm for the response but a friend managed to help me out

I was just confused on like what was supposed to be multiplied by what in order to give it the proper amount of rows and collums

columns *

[viscid thistle]

anyone

online?

[languid crane]

1985 people online

actually more

just ask

[tawny nacelle]

lies

1986 people online

git gud

[languid crane]

lmao

[viscid thistle]

1984

[ruby otter]

sec^2(x)-tan^2(x)

Has period pi/2

How ??

Even it's discontinuous and the holes in the graph repeat with period pi/2

It is a constant function right??

[willow bear]

π, surely?

[ruby otter]

What??

[willow bear]

your holes occur at π/2 + kπ

[ruby otter]

Ya whatever

But how ?? It is a constant func

,w plot sec^2(x)-tan^2(x)

[obsidian monolith]

Display results online and refine query

Requested by Radical Ninja#7581

[serene heath]

tan^2=sec^2-1

[ruby otter]

I think theyrn't identical

That's why

[viscid thistle]

Nvm figured it out

[valid vector]

Is the Domain of this tangent graph correct?

[short sorrel]

why's that pi in the domain?

[valid vector]

Well I wrote that the domain was x cannot be = 1/2 + n(pi)

Is it wrong?

[short sorrel]

your asymptotes are at x = 1/2, x = -1/2, etc

theres no pi involved there

[valid vector]

So, what would my domain be? x cannot be = 1/2 + n(1)?

[short sorrel]

just smthn like

${x | x \neq \frac{1}{2} + n, n \in \mathbb{Z}}$

[obsidian monolith]

Namington:

[short sorrel]

if you wanna do proper set builder notation

but yeah, that'd just be

[valid vector]

Dang, that just confused me more.

[short sorrel]

"x cannot be 1/2 + n"

in plain english

[valid vector]

Oh

[short sorrel]

'for integer n'

[valid vector]

Ye

Thanks!

I think I got it

[short sorrel]

yea

normally there are gonna be pi values

in the restrictions on the domain of tan

but we already have that pi covered

since it's our b value

[valid vector]

Yeah, cause pi/b = pi/pi = 1

Right?

[short sorrel]

sure, yeah

[valid vector]

Oof, OK

[short sorrel]

basically

because its tan(pi*x)

we always have that pi there

[valid vector]

Yeah

[short sorrel]

for regular ol' tan(x), we get asymptotes when x = pi*n + pi/2

in tan(pi*x), the pi is already there

so are asymptotes are gonna be when x = n + 1/2

as that ends up getting multiplied by pi anyway

[valid vector]

OH!

Thx!

[viscid thistle]

https://cdn.discordapp.com/attachments/238956921581862913/567521663177064458/unknown.png

I am not sure if I am on the right path for solving this.
Would I find the angle now or did I set it up wrong?

[full garden]

please someone help me with this

how do I find the resultant displacment for this

it shows that its 13 km but idk where they got it form

even when I add all of those vectors i don't get 13

please help

[languid crane]

Why is this on precalculus

[full garden]

what?

please help

[gloomy spade]

@full garden because the person walked back 3m, the person's displacement is now 5m (in the y direction)

then you use Pythagorean theorem to find the resultant

[full garden]

ohh so thats where it came from

thank you bro

[viscid thistle]

https://media.discordapp.net/attachments/531367551201050644/567790112394575893/041619_incomplete.JPG?width=637&height=670

can anyone help me answer #2? i'm not sure if it's "always, sometimes or never"

[royal gull]

@viscid thistle what did you try?

[viscid thistle]

i'm still stumped. i wanna say the answer is always, because in this equation sin(x) = 1/2, but i don't think that's right

[royal gull]

if the answer was always, that means for every x sin(x)=1/2

which is not true - counterexample sin(0)=0

it the anwer was Never - that means there are no x's such taht sin(x)=1/2

If the answer was sometimes, that means that sin(x)=1/2 if there exists x that satisfies the condition, but its not for every x

[viscid thistle]

ah ok, that's a really good explanation my professor didn't give

thank you!

[royal gull]

you're welcome!

[glacial island]

someone told me precalc didnt help them for calc

[viscid thistle]

@glacial island someone lied to u tbh

[charred hull]

it called pre-calculus

it will help

you just dont notice

[slender river]

it helps i promise

[viscid thistle]

anyone on

anyone?

[willow bear]

jesus fuck just post your problem already

@viscid thistle

[viscid thistle]

@willow bear

yo what up dawg

[echo plaza]

[viscid thistle]

alright

it has to do with my precalc book

one sec

[rocky bison]

HE DOESNT EVEN HAVE THE QUESTION READY

[rocky bison]

what even is life

[willow bear]

,rotate

[viscid thistle]

here u go

[willow bear]

,rotate 180

[viscid thistle]

,rotate

[viscid thistle]

o

[willow bear]

so.... which problem do you need help with

[viscid thistle]

the circled one broo

boo : )

[willow bear]

ok, what's holding you up here?

[viscid thistle]

can you see it xD

idk how to do it

like what is the question asking for

[willow bear]

the question is giving you the graphs of two functions it names f and g

and asks you to find $(f \circ g)(0)$

[obsidian monolith]

Ann:

[viscid thistle]

ok

how do i find

dis

b0ss

[willow bear]

do you know what $f \circ g$ means?

[obsidian monolith]

Ann:

[viscid thistle]

ofc

[willow bear]

then what's the problem?

[viscid thistle]

im in quantum mathmatics

ofc

[willow bear]

what's holding you up, then?

[viscid thistle]

f(g((x)

is what it means

[willow bear]

yes

so