precalculus

[viscid thistle]

that can be considered "basic-pre-calc"

[bright jetty]

lol

that's basic

[viscid thistle]

NO!

Its not as bad as you think!

Its really not

[quartz garnet]

Terrible parentheses skills

x')

[viscid thistle]

You canlearn this and be pretty good

[bright jetty]

teach me

[quartz garnet]

As long as you remember the things from before, like algebra and stuffs, precal is really easy

[viscid thistle]

well, I might chip in time to time but I got my own math studies, maybe when I get my sleep problem under controll, I'll help you

[quartz garnet]

(for the record there doesn't exist precal in my country, we just skip from algebra to calculus right away)

[viscid thistle]

Its weird

Is trig a seperate class/

?

[quartz garnet]

mm?

Basic trigs are learnt in 9th grade (middle school)
harder ones are learnt in Calculus I, usually in 11th grade

[viscid thistle]

🤔

huh

** I mean, I guess **

[quartz garnet]

Yeah

When I first heard of precal, I was confizzled as to for what purpose it exists

[viscid thistle]

lol

[willow bear]

@bright jetty were you trying to get sin(x) raised to the 99th power?

=tex \sin^{99}(x)

[bright jetty]

@willow bear honestly I'm just messing with the bot seeing how it works just joined yesterday

figuring out its kinks

[willow bear]

#bot-testing's for that

[bright jetty]

mb

[dusk solstice]

yo

I can a test tommorow

can anyone halp me out D;

its pretty late also

[willow bear]

help you out during the test? no

we can help you review though

[dusk solstice]

Yeah im taking the test tmrw

and need help now >_<

[willow bear]

so review

[dusk solstice]

yis :D

[willow bear]

ok what is the test on?

[dusk solstice]

rational functions

[willow bear]

what about them

[dusk solstice]

Unit 3: Radical and Rational Functions

what is the difference idek

lul

between a radical and rational function :/

[willow bear]

...radical functions are allowed to contain square roots

[dusk solstice]

i just know how to do the steps

[willow bear]

what are you asked to do with these functions, generally?

[dusk solstice]

o

[willow bear]

graph?

[dusk solstice]

sum of solutions

simplify

[willow bear]

can you show an example of a problem you're having trouble with?

[dusk solstice]

identify asymptotes

discontinuities

so annoiying this content

ya ok

[willow bear]

right, that's multiple choice

what are the options?

[dusk solstice]

tis the answer sheet

i should be able to solve the problem without looking at the options right xd

[willow bear]

"Which of the following..." implies the question is multiple choice

[dusk solstice]

i dun wanna look at the answer

ya so I can still solve it then check right

[willow bear]

there might be different things you may be able to simplify your thing to

anyway, what have you tried so far for doing that?

[dusk solstice]

UM

simplifieng

basically

but im terrible at math logic

[willow bear]

well what have you done? can you show me what you've got so far?

[dusk solstice]

i basically just multiplied the whole demoninator out

[willow bear]

?

[dusk solstice]

so i multiplied it by itself

on both sides

= 0 (the stuff)

[willow bear]

uh

[dusk solstice]

i think ive done it wrong :/

[willow bear]

can you just show what you've written down?

because i can't seem to understand what exactly you did and even then it seems wrong

[dusk solstice]

so this is the first step

i dont get this part

_<

how am i supposed to know

that

[willow bear]

i mean that's not the only way to do that

you could do something like...

=tex \frac{x - \frac{1}{x}}{1 - \frac{1}{x^2}} = \frac{x^2(x - \frac{1}{x})}{x^2(1 - \frac{1}{x^2})}

[willow bear]

that

to get rid of the nested fractions

since x^2 will take care of them both

so that becomes x(x^2 - 1)/(x^2 - 1), or just x, when x != ±1

[dusk solstice]

das a bit complex

harder

LOL

[willow bear]

is it?

[dusk solstice]

idk looks

[willow bear]

it's just multiplying the num and denom by the same thing

a/b = (ac)/(bc)

[dusk solstice]

oOO

so what happens on the numerator

after they're multiplied

wait

bottom becomes 0 right

so it doesnt work

[willow bear]

no it doesn't

[dusk solstice]

oh nvm

bottom becomes x^2

[willow bear]

x^2 - 1

that's what the denominator becomes

[dusk solstice]

but

x^2 * 1/x^2

= 1 right

[willow bear]

yes

[dusk solstice]

ok

then

on the numerator u get

x^3-x?

[willow bear]

x^3 - x, which factors to x(x^2-1)

[dusk solstice]

ah

can u do the math thing again

and represent that result

mathbot do ur magic

[willow bear]

=tex \frac{x - \frac{1}{x}}{1 - \frac{1}{x^2}} = \frac{x^2(x - \frac{1}{x})}{x^2(1 - \frac{1}{x^2})} = \frac{x(x^2-1)}{x^2-1}

[willow bear]

= x

with holes at -1, 0 and 1

[dusk solstice]

0 and 1?

how come

[willow bear]

bc you had x^2-1 in the denominator, which becomes zero at 1 and -1

and you had x in the subfractions' denominators, which becomes zero at 0

i'm surprised -1 wasn't a surprise to you, but 0 and 1 were!

[dusk solstice]

OHwaat

no cuz i see it like

theres a hole if there is the same number on the top and bottom

same root or watever

so confused ><

i thought when the denominator = 0 thats number is an asymptope

[willow bear]

p(x)q(x)/q(x) is equal to p(x) with holes wherever q(x) was 0

[dusk solstice]

so then

how do u identify

the vertical

asymptote

s

[willow bear]

when the denominator is zero, but the numerator is not

[dusk solstice]

yeah so

[willow bear]

you have no asymptotes

only holes

[dusk solstice]

OH i see

makes sense

awesome

whew ok

need 2 min break and ill be back

[patent beacon]

You can also tell that your function has no asymptotes, because x has no asymptotes.

[dusk solstice]

wat u mean

ok im back

where did everyone go

:(

[patent beacon]

Heyo.
So your rational function will simplify to x.
x has no asymptotes, so your function has none either

[dense trellis]

Double angles has been pretty difficult for me

How would you do Sin 22.5?

ik it becomes Sin 45

but isnt sin 45 1/Sqrt2

[willow bear]

=tex \sin^2(x/2) = \frac{1-\cos(x)}{2}

[willow bear]

=tex \sin^2(45^\circ/2) = \frac{1 - \frac{\sqrt{2}}{2}}{2} = \frac{2-\sqrt{2}}{4}

[willow bear]

=tex \sin(22.5^\circ) = \sqrt{\frac{2-\sqrt{2}}{4}} = \frac{\sqrt{2-\sqrt{2}}}{2}

[willow bear]

@dense trellis

[dense trellis]

ohhh ok i understand now thanks!

[dusk solstice]

@willow bear u there

[willow bear]

now i am

[dusk solstice]

O

can we do a few more problems :D

[willow bear]

sure

[dusk solstice]

alright heres one

https://gyazo.com/94c798d4699d90e3cca1f9b8f2a6e281

so first off

im thinking

LCD?

[willow bear]

mm yup

[dusk solstice]

which is

[willow bear]

you'll arrive at a quadratic

(x+1)(x-1) is your common denom

[dusk solstice]

y

doesnt that give u

[willow bear]

well you've only got two denominators there really

x-1 and x+1

...no?!

that'd be (x+1)^2

and x^2 + x + 1 can't be factored over the reals anyway lul

yeah anyway you'll arrive at a quadratic, which you can clean up and solve

[dusk solstice]

o but hold on

(x-1)(x+1) isnt a factor of 2

so how come

u can use it

[willow bear]

so what

it's 2

2's a constant

don't care about it

[dusk solstice]

erm

alright

so then

[willow bear]

independently of what we're doing right now, you can double both sides if you wanna avoid fractions

[dusk solstice]

i get

yea

can u put this in the mathbot

what im about to type

2(x^2-2x+1)+8x+8=5(x^2-1)

[willow bear]

uhh yup looks correct

[dusk solstice]

then that becomes

-(3x+5)(x-3)

[willow bear]

does it? gimme a moment

2x^2 + 4x + 10 = 5x^2 - 5

3x^2 - 4x - 15 = 0

yyyeah your factorization seems to be correct

[dusk solstice]

soo then

-5/3 + 3

[willow bear]

yup

so that'll be 4/3

... as expected, fwiw

[dusk solstice]

AH

makes sense

ok next

https://gyazo.com/d716cb676e53e155c4b76881914f7992

immediately i know

H.A. : 2/3

[willow bear]

yup

well

y = 2/3

it's a line

[dusk solstice]

is that it

only one

hA

[willow bear]

well a function can only have two horizontal asymptotes at most; one at +∞ and another at -∞

but for rational functions, these always coincide

[dusk solstice]

So

-2/3 is another

?

[willow bear]

no

[dusk solstice]

den

how about for a radical function

how do i know if its radical function

or rational function

how to differentiaate

lul

of

sqrt

u said

?

[willow bear]

rational functions contain radicals yes

well

yknow

square roots

cube roots

n'th roots

:p

rational functions are always polynomial/polynomial

[dusk solstice]

o gawd i hate radicals

[willow bear]

they can be somewhat nasty to deal with, true

[dusk solstice]

https://gyazo.com/d716cb676e53e155c4b76881914f7992

anyways

HA: 2/3

VA:

[willow bear]

yeah, the one horizontal asymptote is y = 2/3

[dusk solstice]

So any rational function will only have one HA to confirm?

[willow bear]

for vertical asymptotes, you'll wanna find when the denominator equals zero, and check if either of those values make the numerator zero

and no, a rational function can have as many vertical asymptotes as you want

[dusk solstice]

cant i just factor top and bottom

and check

[willow bear]

you can

sure

[dusk solstice]

hmm

depends on the equation, sometimes u can just substitute

right

[willow bear]

define substitute?

[dusk solstice]

put in 0 for x

bcuz its quicker

[willow bear]

i mean

[dusk solstice]

or sry not 0

[willow bear]

why would you do that lol

[dusk solstice]

any number

that would lead to 0

coz

LIKE

SAAy

you had

x^8 - 1/ x^3 - 2

if u use ur brain

NO WAIt

[willow bear]

=tex \frac{x^8 - 1}{x^3 - 2}

[willow bear]

this?

[dusk solstice]

nunu

this

x^8-1/x^2-1

u instantly can put in 1 for x

and its a VA

[willow bear]

uhhhh no it's not

[dusk solstice]

O)_O

[willow bear]

if the numerator equals zero you'll just get an x intercept lul

[dusk solstice]

OH

fuck

the top is

x^8-3

:)

:^)

what happens if u get 0/0

when u plug x in

hole

?

[willow bear]

honestly, 0/0 can be whatever

you'll have to factor and see if any instances of the "offending" factor (i.e. the one that becomes zero at that point) remain in the denominator

if they do you get a VA

if they don't you get a hole

[dusk solstice]

ok fuk dis

back to teh problem

https://gyazo.com/d716cb676e53e155c4b76881914f7992

3x(x

uhh

brain fart

[willow bear]

the discriminant of that denom is 169 if i didn't screw up my arithmetic so you should get rational roots there

== 5^2 + 4 * 12 * 3

[granite stirrup]

169

[willow bear]

yup

k

so (5 ± 13)/6 are your two roots

and thus two potential VA points

they are... uh

3 and -4/3

yeah

== 2 * (-4/3)^2 - 7 * (-4/3) + 3

[granite stirrup]

15.88888889

[willow bear]

== 2 * 3^2 - 7 * 3 + 3

[granite stirrup]

0

[willow bear]

yeah k so

[dusk solstice]

hold your horses

[willow bear]

mh?

[dusk solstice]

im still on page 1

xD

trying to factor dis

[willow bear]

yeurgh

you'll get (x-3) factors in the numerator and denominator, one of each

which you can cancel out, so none of that remains in denom

hence there's a hole at x=3 but no VA

[dusk solstice]

oo

no

there is a va though

[willow bear]

there is a VA but it's not at x = 3

[dusk solstice]

yah

[willow bear]

you've still got that (x + 4/3) factor there

well up to a constant

[dusk solstice]

so -4/3

is the va

done

next

[willow bear]

x = -4/3 yes

[dusk solstice]

u god

[willow bear]

yeah that's it really, you've got your horizontal asymptote taken case of and we've just done the vertical ones

[dusk solstice]

can u math bot the factored forms

wait nvm

i got it

(2x-1)(x-3)/(3x+4)(x-3)? correct?

[willow bear]

hhh looks correct yes

[dusk solstice]

im seriously contemplating how long its gonna take me to factor these on the actual test

kms

i wanna get extra time

FUUk it takes me 2 long to get the factored form

why is my brain so guess and checky

i have 0 math logic

XD

[willow bear]

tbh

these are non-monic quadratics

they're super hard to factor mentally even for me

[dusk solstice]

🤖

i hope they give us easier factorization on the test

https://gyazo.com/bc523f70f5237ca2fe6f623f2f396779

so

i guess we have to do limites

or

actually hm

[willow bear]

no

you're asked for the asymptotic degree

which is just deg(num) - deg(denom)

[dusk solstice]

it doesnt say use limits

to represent end behavior

[willow bear]

it doesn't and you don

't

need to

[dusk solstice]

for our rubric

y

h

what is teh end behavior asymptote

[willow bear]

have you ever done polynomial division?

[dusk solstice]

yeah

lol

im rusty

i need to work on my long division

its rusty

LOL

[willow bear]

yeah ok so you know the concept great

[dusk solstice]

so i basically gotta

divide it

[willow bear]

ya don't

[dusk solstice]

then take the degree

[willow bear]

and i'm about to tell you why

[dusk solstice]

o-O

[willow bear]

so you know how you can rewrite a rational function as the sum of a polynomial and a "proper" rational function?
proper here meaning deg(num) < deg(denom), and that numerator is the remainder you'll get after doing the polynomial long division

[dusk solstice]

i guess

[willow bear]

yeah so

A/B = Q + R/B, with deg(R) < deg(B)

(all capital letters here refer to polynomials)

in other words A = BQ + R

and you know that the degree of a product of two polynomials is the sum of the factors' degrees right?

[dusk solstice]

maybe

:p

[willow bear]

so you can write deg(A) = deg(BQ + R) = deg(BQ) (since deg(R) < deg(B) < deg(BQ)) = deg(B) + deg(Q)

so deg(A) = deg(B) + deg(Q)

deg(Q) = deg(A) - deg(B)

[dusk solstice]

whaaaa

my brain exploded 2 comments ago

if its x^8+1/x^4+1

wouldn't it become

x^4+1?

[willow bear]

it would not

[dusk solstice]

??

1/1 = 1

[willow bear]

(a+b)/(c+d) is not a/c + b/d

[dusk solstice]

WHAT

BUT Why math

why u do dis to me

so then

they

're

individual

attributes

sotherefore

[willow bear]

k so

let's back off a bit shall we

[dusk solstice]

._.

k

[willow bear]

ok

so you know what the degree of a polynomial is

right?

[dusk solstice]

yea

[willow bear]

yeah

so

so i'll have to make sure you need to know two things:

[dusk solstice]

if a polynomial had a degree of 8

it would be

p(x) = ax^8 + ax^7 + ax^6 + ax^5 +ax^4 + ax^3 + ax^2 + ax + a?

[willow bear]

those coefficients need not be all the same lel

i mean if you insist you could write it in general as

=tex p(x) = a_8 x^8 + a_7 x^7 + a_6 x^6 + a_5 x^5 + a_4 x^4 + a_3 x^3 + a_2 x^2 + a_1 x + a_0

[willow bear]

but anyway

i'll have to make sure you know two things

1. if you add two polynomials of different degrees, then the degree of the sum is the higher degree of the two addends

so e.g. the sum of a 10th degree polynomial and a 7th degree polynomial has degree 10

are we clear on that?

[dusk solstice]

sure

[willow bear]

k

and the other thing

the degree of a product of two polynomials is the sum of the degrees of the factors

so e.g. the product of an 11th degree polynomial and a 6th degree polynomial has degree 17

are we clear on that?

[dusk solstice]

yh

[willow bear]

k good

so

polynomial division

[dusk solstice]

._.

[willow bear]

we'll get to that in a moment!

let me finish!

patience fam

[dusk solstice]

pls cont :D

[willow bear]

k

so there's a theorem that says that for any two polynomials A and B, you can perform division with remainder on them

so in other words you can find two polynomials Q and R such that A = BQ + R and deg(R) < deg(B)

and the theorem also states that these are unique

to make it clearer why division is involved we can switch to rational functions for a moment and divide through by B, giving A/B = Q + R/B

are we good on that?

[dusk solstice]

wat are

a = bq+r

r is remainder?

[willow bear]

R is the remainder and Q is the quotient yes

[dusk solstice]

ok

i get it sort of

[willow bear]

k alright so

[dusk solstice]

yea

thats like a logical mathematical way of thikning of it

[willow bear]

deg(A) = deg(BQ + R)

(bc they're the same & shiz)

[dusk solstice]

OK

[willow bear]

deg(R) < deg(B) < deg(BQ), so deg(BQ+R) = deg(BQ)

[dusk solstice]

what is the deg of r

why

[willow bear]

we don't know what exactly it is

[dusk solstice]

and why is it less

than b

,-,

[willow bear]

well i mean

it's the polynomial equivalent of the remainder being less than the divisor when you're working with numbers

😛

but with polynomials, they're compared by their degrees

patience

we'll get there in a moment

ok so

deg(BQ) = deg(B) + deg(Q)

so deg(A) = deg(B) + deg(Q)

and so deg(Q) = deg(A) - deg(B)

[dusk solstice]

k

[willow bear]

right, so!

=tex \frac{A(x)}{B(x)} = Q(x) + \frac{R(x)}{B(x)}

[dusk solstice]

xD

[willow bear]

i'm about to make said point

so as x grows larger and larger

that rightmost fraction, R(x)/B(x), is gonna get very small

because deg(R) < deg(B), as we said earlier

and so A(x)/B(x) will be approximately equal to Q(x)

[dusk solstice]

we werent taught this in school

but alright

:D

i like it

[willow bear]

so Q(x), which is a polynomial, is called the end behavior asymptote of A/B

and with that massive detour now over!

you're only asked for its degree

and we know deg(Q) = deg(A) - deg(B)

[dusk solstice]

so ur telling

me

that

all

this

time

it

was

just

the

degree

of

the

end

result

final

product

q

thingy

xd

[willow bear]

so all you need to do is take the degree of the numerator, the degree of the denominator, and subtract them

what we've just gone through is the why

[dusk solstice]

https://gyazo.com/bc523f70f5237ca2fe6f623f2f396779

so

the asnwer

is

4

[willow bear]

👏💯👌😄

[dusk solstice]

fucking

hell

you're

a

maths

genious

[willow bear]

thanks lol

[dusk solstice]

so lemme try the limits thing

i guess

when x -> infinity

lim f(x)= infinity

when x -> - infinity

lim f(x) = inifniy

?

[willow bear]

yup

[dusk solstice]

ez

ok

we spent

way 2 long

on dat

guess what time it is

for me

[willow bear]

22:41 for me, so i'm going to guess it's between 11:41 and 15:41 for you, based on the potentially incorrect assumption that you're from the US?

[dusk solstice]

LOL

u will be

very delighted

to know

that

its fucking

3:45

right now

and i havent slept yet :D

bcuz im studying

for the test

today

at 13:50

💩

but i have to get up

at 6

should i sleep now

or keep studying

that is the question

[willow bear]

yes

sleep

sleep

deffo sleep

[dusk solstice]

👶🏿

ok

IM about to pass out anyways

add me

o u already did

nice

thanks for all the help today

ur my saviour

im ded

gnight man

[timid plank]

How would you solve sin(3x) = -1 ?

[deft shoal]

3x = arcsin(-1)

3x = 3pi/2

@timid plank

You can solve from there. 😉

[timid plank]

ASWAIT

the heck?

how did I over complicate it?

execute me rn

[patent beacon]

@timid plank
Are you supposed to solve over a certain interval?

[timid plank]

yes but I can do that myself

I over looked it

and was doing it using sum and difference formulas

instead of simply using arcsin

[patent beacon]

Note that
arcsin(-1) = 3pi/2, 7pi/2, ect

[viscid thistle]

^^

also there's only one angle in the unit circle for which sin x = -1

[timid plank]

yes I know

[patent beacon]

Cool cool. Didn't want you to get the wrong answer

[viscid thistle]

but for other values such as arcsin(1/2) you'd have to evaluate both angles that allow for that possibility

i got partially correct answers many times for forgetting to note the second angle that satisfied the condition

[timid plank]

@patent beacon I already knew the answer I just didn't how to get to it, realized I was over complicating it

[split valley]

hey everyone, question:

how do I get type inverse trig functions into some like Desmos?

[willow bear]

arcsin

arccos

arctan

[split valley]

I can't seem to get the right answers (the angle)

[willow bear]

what are you trying to do?

[split valley]

what I mean is:

well I was trying to copy/paste an image but it's not working

a problem gives the answer of the arctan of 5/9 = 29.1 degrees

I get keep getting .507?

what am I missing

[willow bear]

radians?

== atan(5/9)

[granite stirrup]

0.5070985

[willow bear]

== atand(5/9)

[granite stirrup]

29.0546041

[willow bear]

0.507 radians = 29.05 degrees

[split valley]

okay, thanks. the text didn't make it clear that I would be dealing with radians I guess that's just implied whenever you're dealing with trig

I'm kinda going through a self-teaching phase regarding this stuff

[ebon spoke]

http://prntscr.com/hjlcuz

<@&286206848099549185> Can someone please show me how I'd find the midline here

The radian is kind of confusing

[viscid thistle]

what do you mean by midline

[ebon spoke]

the average

from the highest to lowest point

[viscid thistle]

oh

[ebon spoke]

i think thats what its asking

[viscid thistle]

i'd say find the function value of the minimum

then take the max value + min value and divide that result by 2

to get the average in between them

i mean you could use calc but since you posted it in precalc i assumed you don't want to use calculus stuff

[ebon spoke]

how would i find it

[viscid thistle]

do you remember how to find extrema

[ebon spoke]

it looks like 1 but

no

dont know what that is

[viscid thistle]

um

maximum / minimum = extrema

in other words

[ebon spoke]

oh

[viscid thistle]

can you see how the maximum is what you were given

[ebon spoke]

yeah

the coordinate ?

[viscid thistle]

no i mean

[ebon spoke]

http://prntscr.com/hjlcuz

[viscid thistle]

are you in calculus or not

xd

[ebon spoke]

no

trigh

[viscid thistle]

oh

[ebon spoke]

trig*

[viscid thistle]

i mean

there's no such thing as a trig class afaik

[ebon spoke]

then pre calc

[viscid thistle]

alright

do you know what a derivative is

[ebon spoke]

heard of it

so no xd

[viscid thistle]

well thats not good enough so lets just try this

you know your max is 6.7

and you're told the amplitude of this sin wave is 4.1

so that means that the "center" of the wave is at a value of 6.7 - 4.1

==6.7-4.1

[granite stirrup]

2.6

[viscid thistle]

there

if you were to draw a straight line y = 2.6

that would be the "center" of your wave

[ebon spoke]

thats it..?

smh

[viscid thistle]

no hold on

[ebon spoke]

oh

[viscid thistle]

oh wait

actually it is

but

the minimum

is therefore

2.6 - 4.1

==2.6 - 4.1

[granite stirrup]

-1.5

[viscid thistle]

-1.5

[ebon spoke]

how does that work ^

[viscid thistle]

and it seems to make sense with the graph too

ok so

have the graph in front of you

[ebon spoke]

https://images-ext-1.discordapp.net/external/jNyz1T_8uegPoqeNhQ8Pz2PB3mZkcfRTkL2vRTGFC-0/https/image.prntscr.com/image/xq_RUk9NQciCfec_OPEMGw.png?width=621&height=438

[viscid thistle]

do you have it in front ?

[ebon spoke]

yeah

[viscid thistle]

ok so

do you remember what "amplitude" meant

[ebon spoke]

highest point to lowest right?

[viscid thistle]

oh

there

its the distance from the "middle" of the wave to the crest

that in itself tells you that if you're given a maximum, you can subtract it by the amplitude to get the "center" of the wave

if you're given a minimum, you can add it to the amplitude to get the "center" of the wave

does that make more sense now ?

[ebon spoke]

yeah it does

[viscid thistle]

👍

[still yew]

I got stuck on this as well, so I'll listen in

[ebon spoke]

thx

[viscid thistle]

its just a question to test for you understanding what amplitude means

to hammer the point home, if i told you you started at the following point

and i tell you that point is a minimum

(the black one that says min)

how would you get to the middle of the wave ?

[still yew]

True, so amplitude - mininum, or maximum - amplitude?

[viscid thistle]

(the green line)

what do you do in this scenario to get from the min to the center of the wave

@still yew

[still yew]

Find the amplitude from the min?

[viscid thistle]

ok let me put it in more simple terms

you start at the black dot

[still yew]

K

[viscid thistle]

figure you have a tan line like this

[still yew]

Yes

[viscid thistle]

where f(min) is the y value of your minimum

what is separating this tangent line y = f(min) from the center of the wave

what is the distance separating it

[still yew]

The period of the part of the wave ur at?

[viscid thistle]

yep

but dont be confused

just cause our minimum is below the center

amplitude is not negative

like if i told you that amplitude was 2

then it is NOT -2 for the red section

it's just 2

so in either case

you know that the distance separating the 2 lines is the amplitude

(which is always positive, as we just saw !)

[still yew]

Since the amplitude is going both directions up and down?

[viscid thistle]

yes

its absolute, there's no negative associated with it

[still yew]

Thx for the help, I gtg now, see ya

[viscid thistle]

@still yew that's how we do it, for the maximum its just y = f(max) - Amp

if you have questions later, ask

[still yew]

K thx

[fossil cove]

Could somebody explain the basics of precalculus? It sounds really interesting

[nimble oxide]

Sure,

Hell

🙃

[halcyon gate]

Can someone explain to me why these are two similar questions, but the way to solve them is different?

And if I use one way for the other I get a different answer.

[fossil cove]

ah, I understand @nimble oxide

[calm whale]

@halcyon gate whats your question??

they pretty much do the same thing 😄

are you confused about something or?

can you show me your working

[halcyon gate]

Sorry I was in class, still unable to show you some work

I'll be back soon, thank you! I am very confused about this

[calm whale]

just tag me when you get back

[hexed ermine]

nooooooooooooo

I got a 99 on my last precal test .-.

[deft shoal]

Those hurt the most. 😕

[fossil cove]

What does it mean when you get 99?

[gilded hazel]

it means they only knew 1 % of the material

[azure vine]

It means he was in the 99th percentile of his class on the test

[fossil cove]

Oh, damn

That is the worst

[nimble oxide]

Oh, @hexed ermine ..... Try thinking you know the material, then getting a 66.6% on a test 😐 ...

[split valley]

Hey all, another basic trig question

The decimal answer a text gives for a cos110 is -0.342

No matter whether I convert the decimal value given in Desmos from radians > degrees or degrees > radians I can't get this answer

what could I be doing wrong?

[willow bear]

what are you trying to figure out?

cos(110°)?

== cosd(110)

[granite stirrup]

-0.34202014

[willow bear]

@split valley

[split valley]

okay, I was able to get that going

I was using google to convert -.999 radians to degrees and it was giving me -57.23

any idea why?

[willow bear]

-0.999 radians? come again?

[split valley]

In your first image in Desmos

the - 0.999 value is in Radians, is it not?

[willow bear]

what -0.999 value?

i only posted one image

and no, the output of cos or sin should not be interpreted as being in radians

[split valley]

In that image you posted in the Projector Mode tool box, if you click "radians" it gives you a value of - 0.999

[willow bear]

yes because then desmos interprets cos(110) as cos(110 radians)

rad vs deg is NOT about outputs

it's about inputs

[split valley]

hmm..., k thanks, I'll think on that a bit

[jolly turret]

Why is it that if you try to learn pre calc in khan academy the content is different than in some precalc book

in khan you get some usefull things for the next class which would be calc 1 but in books you get a bunch of stuff that doesnt seem like itll be any good

[willow bear]

what precalc is varies a TON

[severe verge]

i mean, they aren't made by the same people

of course they will differ

[hexed ermine]

Yeah most of the time I have no clue what is being said in here

but im in pre-cal lol

[gilded prairie]

hey there guys

im having an issue with this. i got this far but i feel must have done something wrong

[willow bear]

=tex (cos(x) + i\sin(x))(\cos(y) + i \sin(y)) = \cos(x+y) + i\sin(x+y)

[willow bear]

just sayin' :p

[gilded prairie]

what do I do about the coefficients in front of the parentheses?

[willow bear]

i mean they're just multipliers

and you're asked to multiply and divide anyway

[gilded prairie]

so do i distribute them or do i multiply the 3/4 and 4?

[willow bear]

no need to distribute anything

=tex z_1z_2 = 3(\cos(105^\circ) + i\sin(105^\circ))

[gilded prairie]

ok awesome thank you

now if i were to divide the z1/z2 would it still be adding cos (x+y) or would it be cos(x-y)

[willow bear]

x-y

[gilded prairie]

so would it turn into
=tex z_1/z_2 = 3(\cos(15^\circ) - i\sin(15^\circ))

?

=tex z_1/z_2 = 3(\cos(15^\circ) - i\sin(15^\circ))

[willow bear]

not 3 tho 😛

3/16

[gilded prairie]

ah finding lcd and dividing?

[willow bear]

"lcd"?

just divide those coefficients lel

[gilded prairie]

my bad yeah

thank you

[jolly turret]

this seems like an odd question because it gives you similar numbers with different exponents do you think theres a shortened way to do this or is just something you plug into you calc and brute force it?

[austere bramble]

would factor out a 2^(1/2)

then simplify the inside and redistribute

[jolly turret]

yup thats probably the way they wanted it done

[willow bear]

there will be nothing to distribute

=tex 2^{5/2} - 2^{3/2} = 2^{3/2} (2^1 - 2^0) = 2^{3/2}

[severe verge]

spooky latex

[jolly turret]

Yeah i guess because it was a simple subtraction problem i didnt think about taking out something

[viscid thistle]

i have a question on a word problem

the one with oil

how do you solve this 😂

[willow bear]

okay so

t is the number of years since 2000

and f(t) is oil production at time t, in millions of barrels

we're assuming f(t) = at + b for some constants a and b

you know f(0) = 4 and f(13) = 1.9

[viscid thistle]

hmm alright

[jolly turret]

can someone tell me how to solve this

i feel like you would get the inverse sign of 1/2 but i dont know what to do after

i know that gives and angle but that doesnt leave me with any sides or anything like that

[calm whale]

🤔

this is an interesting question nevertheless

[willow bear]

sine, not sign

also

[jolly turret]

right......

[willow bear]

sin^2(x) + cos^2(x) = 1

use that

[calm whale]

it challenges your current understanding of maths, the world and ultimately the universe

[jolly turret]

Ight ill try that

i had it for a precalc practice test it seems as if ill have to memorize these random identites if i want to go on to calc 1

[willow bear]

random?!

the one i just mentioned is the pythagorean theorem fam

[jolly turret]

is it?

[willow bear]

yes!

[jolly turret]

by the way how would i plug in what im givin to that? is t an angle?

[austere bramble]

you're given that sin(x)=1/2

so sin^2(x) =1/4

[jolly turret]

im still not exactly sure what to do with that because of the exponent

[austere bramble]

so sin^2(x)+cos^2(x)=1

but sin^2(x)=1/4

[jolly turret]

oh i see

[austere bramble]

so you get 1/4 +cos^2(x)=1

solve for cos(x)

[jolly turret]

its odd that sin^2(x) is the same as sin theta because if they both equal 1/2 then they equal each other

they seem to have similar notation just that one is squared and the other isnt

[willow bear]

what

okay first off, m8 is using x instead of theta here

second

sin(θ) = 1/2

[jolly turret]

wait no so to get the 1/4 did you do something to the 1/2

[willow bear]

sin^2(θ) = (1/2)^2 = ???

[jolly turret]

a square root?

oh i see

Thanks for the help guys it still somewhat confuses me why you square it. I get that the other side is sin^2 but i dont see why you have to do it to the other side when you put it equal to it

One last problem i got i need some help with i normally dont ask much i try to search for the problem type and learn from there but i dont really know the problem type its a basic slope question but it makes no sense to me

i thought it was b but its actually a. can somebody explain? normaly a negative slope goes the other way from quad 2 to 4

[calm whale]

m is the slope 🤔

as in the gradient

as in rise over run

[jolly turret]

yes

[calm whale]

-r/s 🤔

or am i missing something

[jolly turret]

what?

[calm whale]

oh wyit

ah yes it should be b

[floral rose]

it's r/s

[jolly turret]

it said it was A lol

[floral rose]

should be the change in r / change in s, but whatever

it's wrong

[calm whale]

what hte fuck is s

[floral rose]

😆

[jolly turret]

dude s = x

[floral rose]

it's the x value

Δy/Δx
Δr/Δs

[calm whale]

see if its a

then it means that the slope is negative

and it clearly isnt

🤔

[jolly turret]

yep lol i thought i had it right but it said its wrong

[floral rose]

unless you look at the graph from right to left

🤔

[calm whale]

what like a left-right cordinate system

kindof thing

[jolly turret]

but either was it should be the same one

[floral rose]

yeah we're just meming

it should be r/s

[jolly turret]

if it was negative the line would be traveling on different quads right

hmm ill have to ask a friend lol

[floral rose]

the line would be going downwards from left to right

if the slope was negative

[calm whale]

negative slope = as x increases y decreases

positive slope = as x increases y increases

there u go

[jolly turret]

someone mind helping with this? i thought you would divide by two then add the 20 then just just multiply by 2 or am i getting it wrong i thought all you do is L*W is it different for squares?

[patent beacon]

@jolly turret
It's a square so the side lengths are equal. What must the side lengths be if the area is 1600?

[jolly turret]

well 1600/2

so 800 right

@patent beacon

[patent beacon]

So the area of a square is given by
A = s²

We can rearrange that to say
sqrt(A) = s

[jolly turret]

oh i see thank you i must have forgot that i guess that makes sense kinda i think the way i got my answer was messed up when i got the base by dividing

makes sense though thanks

[patent beacon]

If you need anymore help, feel free to ask!

[jolly turret]

Yeah thanks will do!

actually i do have a question say you had 4s^2 + 16k^2 all under a square root why is the answer 4 outside of the square root with whats inside being s^2 + 4k^2 i get that you take out a 4 from it but should you be able to take out a 2s + 4k instead

[patent beacon]

Whoa whoa whoa, you can't math like that. Every time you do, a puppy dies.

sqrt(a² + b²) ≠ a + b

[jolly turret]

lol

[patent beacon]

Instead, you CAN do this:
sqrt(ab) = sqrt(a)sqrt(b)

In fact, you can do that for any exponent

[jolly turret]

so then how does the 4 get out because things should be reversable so how would you put the 4 back in? by multiplication?

[patent beacon]

Let's take a look:
sqrt[4s² + 16k²]
= sqrt[4(s² + 4k²)]
= sqrt[4]sqrt[s² + 4k²]
= 2sqrt(s² + 4k²)

[jolly turret]

ohhh i see makes sense i think the thing in math that gets me so messed up is reading the problem incorrectly but thanks thatll help for when i take my placements i want to get into calc 1 because ive been in precalc for a bit now it seems like ive been marinating in the same math section for a long time. I get the begining parts of calc so i feel like i should be able to pass on to it

[patent beacon]

Sounds pretty cool! I wish you luck.

[halcyon gate]

Hey guys, can you explain this solution to me, I'm not sure how they got the shift 3 unit to the right.?

[jolly turret]

Study graph transformations

the line of a graph will move up or down or left or right depending on a number in your case (x-3) its the -3 and the + 1. Basically theyre saying y = x2 is the original equation and the rest are transformations you just got to answer in what way do they shift the line

[willow bear]

x2
x**^**2

[halcyon gate]

ohh ok

thank you

i got it

[sick whale]

Excuse me, I'm having trouble trying to find the leading Coefficient in this polynomial function

[calm whale]

have you tried

expanding them all out

😄

[sick whale]

Multiplying all of them together you mean?

[calm whale]

yes so youd get

y=ax^n + bx^n-2 ...+ something

they are asking you for a

=tex [coeffictient].r^4r^3r = [coefficient]r^8