precalculus

[maiden pebble]

w/e?

[remote veldt]

whatever

[maiden pebble]

Thanks for the enlightening prediction Ann

[willow bear]

it's not a prediction, i'm just being passive-aggressive and salty

and you could've been even more explicit in your sarcasm

[maiden pebble]

https://i.imgur.com/vBk12pa.png

WHen checking this problem, I have been getting 0 on top, but infinite in the denominator

Does this still fit l'hopital's rule?

[viscid thistle]

Yes take derivative of top and bottom

[maiden pebble]

Seperatly or together?

[viscid thistle]

create a new fraction with the derivative of 3x on top and the derivative of ln(x+1) on the bottom

[maiden pebble]

I got 3/1/0

Wait hold up

no I didnt

[viscid thistle]

what is the derivative of ln(x+1)

[maiden pebble]

yea im getting that

1/x+1

[remote veldt]

@maiden pebble 0/infinity is not l’hopitals, it’s just zero. this case is however not 0/infinity, it’s 0/0

[maiden pebble]

hohere?

[remote veldt]

**

[maiden pebble]

oh ok

[remote veldt]

Corrected

[viscid thistle]

So you get 3 / (1 / (x + 1))

[maiden pebble]

Ooh thanks Nicholas, I get what ur saying now

3/1/1 = 3

[viscid thistle]

yeah

make sure that

when you create the new fraction

you still write the limit notation

until you plug in a value that makes it determinate

[maiden pebble]

oh, I havent been doing that

Good to know

[willow bear]

@viscid thistle you do realize that 0/infty is just 0 right

[viscid thistle]

but here it's not infinity on the bottom ?

[maiden pebble]

https://i.imgur.com/EJAe51v.png

Would i be able to use l'hopitals rule here?

[craggy dune]

yes if you want to

[maiden pebble]

I solved for the limit of the numberator and got 1/infinte

[viscid thistle]

what are the cases when you can use l'hopital's rule

do you know

[maiden pebble]

When the top and the bottom are either 0 or +-infinity

and in this case I got neither

[patent beacon]

You have inf/inf in this case

[craggy dune]

what's log x for x to inf?

[maiden pebble]

Kaynex how did you get inf on the top?

oh, I took the derivative of the top

Thanks Kaynex

Deekan I didnt understad what u said last

[craggy dune]

im curious why you think you don't think there is inf on top

what do you get if you do lim x to inf ln(x)

[maiden pebble]

I took the derivative then plugged in inf

so i did 1/inf

but I see now that I am not meant to take the derivative when checking the problem

[viscid thistle]

exact

first you plug in

[elfin crescent]

ive never done limits, or derivatives in precalc b4

[patent beacon]

Only the cool kids do calculus in pre-calculus

[elfin crescent]

l hop in precalc?

[maiden pebble]

Yea Ive gotten a lot of comments about my class doing a lot of derivatives and stuff in precalc

People tell me to go to #calculus but then when I talk there they expect me to know everything already

[viscid thistle]

also bridge, you asked why ln(infinity) is infinity

[maiden pebble]

I might have but I know why

[viscid thistle]

okok

cool

[maiden pebble]

thx for helpin

[elfin crescent]

you guys have any tips/resources for self studying precalc?

[viscid thistle]

prof leonard

[maiden pebble]

I just got a Calc for DUmmies book

[craggy dune]

khan academy

[elfin crescent]

hm okay

[maiden pebble]

it was like 25$

[elfin crescent]

precalc is basically alg 2+trig right?

[maiden pebble]

350 ish pages

Yea, I call it alg 3

[craggy dune]

pre calc is a bit of everything really

[elfin crescent]

alright

im gonna start using this server more, i always had it xd

[maiden pebble]

Does i have a derivative>?

[languid crane]

i has a derivative

its 0

[elfin crescent]

@languid crane the imaginary number i right?

@viscid thistle as long as i follow through with prof leonards video i should be able to learn everything for precalc right?

[viscid thistle]

fo sho

[maiden pebble]

prof leonard?

[elfin crescent]

alrighty ty

@maiden pebble he's a great math youtuber, if you don't know him you should check him out. he's really great i used him for calc and stat!

[maiden pebble]

no way hes better than my boy sal

Oh he does stat too?

That would have been helpful this year since I prob got a 2 on my AP Stat test

[viscid thistle]

i legit bought this t shirt for my sibling lol

he's lucky tho

he gets to explain what a derivative is for 3 hours

my school is like ok we have 10 minutes to explain what a derivative is

here's the equation now figure it out

[maiden pebble]

holy crap

his videos are like 3 hours long

[viscid thistle]

yeah but the theory is like 30 minutes to an hour

the rest he does examples

[arctic mantle]

I tried plugging rcos(t) into x but didn't get anywhere, im assuming i need to get rid of t but not sure how to it

I also have no clue how to do the rest of the question so all help is appreciated 🙂

[rich flint]

@arctic mantle it seems they want you to plug-in for x and solve for r and y

So your expected to have variables in your final solutions

Basically don’t get rid of t.

[rigid siren]

Have you studied the standard form of the parametric equation for an ellipse?

If you haven't here's a hint: it has something to do with $sin^2(\theta) + cos^2(\theta) = 1$

[obsidian monolith]

Mains of the Universe:

[reef jasper]

hello how do u factor this out

#69

[past meadow]

if its not clear, i would suggest letting $\sin(x)=u$ so that you have $2u^2+5u-3$

[obsidian monolith]

Sneaky:

[past meadow]

the expression in u is a quadratic, which i think you should know how to factor

[reef jasper]

oh yeah tnx

[past meadow]

no worries

[tawny nacelle]

#69 nice

[sterile notch]

So apparenty: x^4 + 2 x^2 - 4 x + 8 = (x^2 - 2 x + 2) (x^2 + 2 x + 4)

How would have I figured that out?

Given only: x^4 + 2 x^2 - 4 x + 8

[radiant fiber]

Um trinomial multiplication? Ig

[rigid siren]

You know that a quartic is the product of two quadratics (ax^2 + bx + c)(dx^2+ex+f). Expand and compare coefficients to get a linear system of equations.

You could also do it as a product of a linear term and cubic.

You might run into complex coefficients this way though.

[rigid siren]

Also, this form is known as the depressed quartic, and there are a few general ways to solve it, usually involving some clever rearrangement and auxiliary parameters. They aren't covered in the school curriculum usually, but look up Ferrari's Method for one way to do it.

[willow bear]

quartic*

"quadric" means something different

[rigid siren]

Whoops, yeah.

[rich flint]

@reef jasper did you figure it out? If not you can also use the quadratic equation for this

Be mindful of the fact that 2 and 3 are prime numbers so it shouldn’t take that long to figure out the solution

[fervent girder]

how to find b?

[frozen needle]

use a ii

[rigid siren]

Find values where $(g \circ f)(x) - f(x) = 0$

[obsidian monolith]

Mains of the Universe:

[willow bear]

have you already done (a)(ii)?

[tepid basin]

Can anyone help me with this ?

[lethal spoke]

f(x)=4, g(x)=1

[viscid thistle]

that is 200 iq

[broken echo]

Wow

[viscid thistle]

i don't think i handle such smartness

[alpine basin]

is [-infinity, 6) U (6, infinity) the proper way to write x != 6 in interval notation?

[pale bison]

x in (-infinity, 6) U (6, infinity)

[viscid thistle]

you can also say R - {6}

[alpine basin]

wait @pale bison why is the left parentheses thicker then the others

[pale bison]

to highlight your mistake

you used [ instead of (

[alpine basin]

oh mb ye

infinity can never be [ right?

[pale bison]

right

[alpine basin]

thanks

[pale bison]

np

[reef jasper]

which identities would i use

how would i do #12

[viscid thistle]

what do u have to do

find theta?

[reef jasper]

no just find the exact value

[viscid thistle]

well u can re-write cot (theta) in terms of tan

then rationalize the denominator

[reef jasper]

oh mb #17

[viscid thistle]

lmaoooooooooooo

[reef jasper]

lol sorry

[viscid thistle]

no worries man

all g

✊

looks at a unit circle tbh

[reef jasper]

pi/12 isn’t on it tho

[viscid thistle]

pi/12 is half of pi/6

plus, 2sin(theta/12)cos(theta/12) = sin2(pi/12)

which equals sin(pi/6)

which equals 1/2

boom

next

[reef jasper]

where did cos(theta12)go

[viscid thistle]

idk

its part of the identity tho

https://opencurriculum.org/5261/proofs-of-trigonometric-identities-i/#:~:text=That's all it takes.,It's a simple proof%2C really.

[reef jasper]

ok tnx

[viscid thistle]

np

[jade heron]

@viscid thistle sin(pi/6) = 1/2, not sin(pi/12)

sin(pi/12) should be

[obsidian monolith]

L'Âne 🍐:

[viscid thistle]

k

[reef jasper]

i’m suppose to use an half angle identities in #41

but when i plug it in it gives a weird #

[viscid thistle]

https://socratic.org/questions/how-do-you-find-the-exact-functional-value-cos-7pi-12-using-the-cosine-sum-or-di

[reef jasper]

oh tnx

wait y is the textbook answer diff

[harsh cipher]

Hi

question

question c

when I find the common ratio of 1/3

when I subtract 1/3 from 7/9 i get 4/9

can someone show me how that is same as 7/3

[uncut mulch]

why are you subtracting 1/3 from 7/9?

[harsh cipher]

to see if the common ration make sense

[uncut mulch]

why are you subtracting

[harsh cipher]

ratio*

[uncut mulch]

this is a geometric sequence

how did you get 1/3?

[harsh cipher]

r = t2/t1

t2= 7/3

t1= 7

r = common ratio

[uncut mulch]

and to check you should evaluate t3/t2

[harsh cipher]

then i need to multiply 1/3 to t2 to get t3?

yes

I need to start practicing trial and error and develop discipline to become good at math. When I start asking for answers immediately, it's unlikely that I will become a problem solver or a "thinker"

sigh

[uncut mulch]

@reef jasper the method shown, uses compound angle instead of half angles (which is better in this case)
if you must use half angles you could simplify it to -sin(pi/12) = -sin( (pi/6)/2) and continue from there

the final values are equivalent, but the half angle method results in something less simple and takes some effort to simplify it further.

[reef jasper]

how do i even start with #75

[jade heron]

expand out cos(2t)

as a function of sines

so cos(2t) = cos^2 (t)- sin^2(t)
= cos^2(t) + sin^2(t) - 2sin^2(t)
= 1 - 2sin^2(t)

then just let t = sin^-1 (4/5)

@reef jasper

[viscid thistle]

Wth

[jade heron]

? @viscid thistle

[viscid thistle]

lmao

cos(2t)=1-2sin^2(t) doesn't seem right to me

tru

Neither wolfram agrees

So copswing

https://socratic.org/questions/how-do-you-simplify-cos-2sin-1-4-5

@reef jasper

[uncut mulch]

,w cos(2t)=1-2sin^2(t)

[obsidian monolith]

Display results online and refine query

Requested by ramonov#0921

[viscid thistle]

Wut

lmaoooooooooooooooo

Wait a sec

Actually

Im blind

I read sin instead of cos

Nothing happened

[jade heron]

👍

[viscid thistle]

agreed

nothing happened

[elfin crescent]

besides textbook problems, do you guys have any practice problems you would recommend for since i am self studying precalc course?

[craggy dune]

khan mastery

[elfin crescent]

alright

if a graph has an inverse function, that graph is a function right?

[viscid thistle]

yes

[elfin crescent]

alright

[viscid thistle]

being a function is a necessary condition for a relation to be an inverse function

[viscid thistle]

What notation would this be called

defining a function by writing out the ordered pairs that it consists of?

don't know if that has a notational name

Me neither

[viscid thistle]

set of ordered pairs i think

i found it

yes that is a set of ordered pairs

ok thanks

im stupid sorry

but I don't think it has a simpler name than that

yeah

you could probably say "ordered pair notation" and people would know what you mean

but it's not an official name

Would this be correct?

hm well

your commas look like they are separating and statements in some places and or statements in others

which is pretty nitpicky

but otherwise nothing wrong

the domain could read like
x such that -5 < x <= -1 AND 1 < = x < 5 AND x in R

while you want
x such that
-5 < x <= -1 OR 1 < = x < 5
AND
x in R

could you write that out

on paper

if you can

I can't decide if $\vee$ is appropriate for this context or not

[obsidian monolith]

Botn:

Botn:

[viscid thistle]

I guess it is, but be careful using it as the word "or" anywhere else

is this for a high school class?

no

just writing notes for next year

so i dont need to remember everything from this year

[remote veldt]

0 bytes

[viscid thistle]

ok

you can use the vee symbol here instead of a comma but learn more about logic notation before using it elsewhere as a replacement for the word "or"

since it should really only be used in a few specific situations

if writing as properly as possible, at least

what is the vee symbol?

$\vee$

[obsidian monolith]

Botn:

[viscid thistle]

oh

I never used this before

Is the way I wrote it fine?

because I dont know what this is

with commas?

technically not really

it can be inferred what you mean

so instead of commas

there would be a vee symbol

and thats all?

in the one place you wanted an "or" not an "and" yes

because if that comma meant "and" like commas usually do for conditions

what does vee mean?

then your domain would be the empty set

it's shorthand for the word "or" in some logical contexts

by comma I meant and

like set builder notation

the domain is __ and _-

oh, well you don't want and, do you?

there is a break in between

x such that x is inside two disjoint intervals?

that's the empty set

which is not what your domain actually is

x such that -5 < x <= -1 AND 1 < = x < 5 AND x in R

wait why

there are no such x

so v

x would need to be both less than -1 and greater than 1

where there is and

x such that 2 conditions are met:

1. -5 < x <= -1 OR 1 < = x < 5
   AND
2. x in R

is what you want

well

actually, I'd probably rewrite a bit

after reading that

${x\in\bR \mid -5<x\leq -1 \vee 1\leq x<5}$

[obsidian monolith]

Botn:

[viscid thistle]

ok

${x\in\bR \mid -5<x\leq -1 \wedge 1\leq x<5}$ (where $\wedge$ means "and") is the empty set

and that's what commas would typically read as

"and"

ok thanks

[obsidian monolith]

Botn:

[viscid thistle]

oh also, I might add, it's intuitively similar to $\cap$ and $\cup$

[obsidian monolith]

Botn:

[viscid thistle]

those mean "and" and "or" (in a sense) when working with sets

same thing in logic, the symbols are very similar looking

but with statements instead of sets

so when you were doing the interval notation you wrote the interval as a union of two intervals

you would need to use the same thinking here

either of the two statements (equivalent to intervals)
-5 < x <= -1
or
1 < = x < 5
need to be met in order for x to qualify for the set, not both of the statements

[sterile notch]

what does it mean when theres a line on top of a complex number

liek z but upside down

[willow bear]

$\overline{z}$ denotes the complex conjugate of $z$

[obsidian monolith]

Ann:

[willow bear]

if $z = x + yi$ then $\overline{z} = x - yi$

[obsidian monolith]

Ann:

[sterile notch]

Oh right, thanks heaps!

[wide ocean]

can someone explain how to isolate x in these?

[keen sigil]

I'm not sure how to finish this. Could someone let me know if I did something wrong. If not I just don't know where to go after this

thanks :)

[viscid thistle]

correct 🙂

is there are relation between cos^2 and sin^2?

[keen sigil]

oh cos^2 is 1 - sin^2

[viscid thistle]

you got it

[keen sigil]

so it would be sin^2x(1-sin^2x)

[viscid thistle]

whats the answer then

[keen sigil]

and sin^2x-sin^3x

i mean

[viscid thistle]

without ^2

[keen sigil]

wait if it's cos^2 why is it not sin^2?

[viscid thistle]

so it would be sin^2x(1-sin^2x)
the first is just sinx, look at what you wrote in the picture

[keen sigil]

oh my bad

that seems right

[viscid thistle]

which is answer D

[keen sigil]

yea

thanks :)

[viscid thistle]

np

[desert pendant]

@wide ocean Multiply everything by the product of the denominators, for Q1, multiply everything by (x^2-x)(x-1), and stuff will cancel with the denominators, and you won't have fractions anymore, for Q2, multiply everything by x(x-5), and same thing

[wide ocean]

@desert pendant thanks so much!

[viscid thistle]

can I get some help with this?

[frozen needle]

integration by parts gives good things

[viscid thistle]

ok lemme see

?

i’ll try like this

[frozen needle]

what's the reasoning behing this ?

[viscid thistle]

well i put n inside the integral

and i integrate by parts?

[frozen needle]

clearly the n is annoying you, it would be natural to try to make it go away by making a 1/(n+1) appear

[viscid thistle]

ohh wait

i think

1/n+1 is the integral of x^n

from 0 to 1

[harsh smelt]

(nx)' is derivative ?

wait from what it comes in second one

[viscid thistle]

i stopped doing that

i’m trying smth else

what Tuong said

[harsh smelt]

,w x^2+4x+5

[obsidian monolith]

Display results online and refine query

Requested by Commander Vimes#7532

[harsh smelt]

nah does not have roots (real) to use partial fractions

[viscid thistle]

Hopital doesnt work right

[frozen needle]

you've got something looking like
$$n\int_0^1x^nf(x)\dd x$$
with $f$ a differentiable function, and you'd like to make a $\frac 1{n+1}$ appear... there's a quite natural integration by parts to do here

[obsidian monolith]

Tuong:

[harsh smelt]

integration by parts + completting the square should do

[viscid thistle]

i did

[harsh smelt]

,w integral 1/(x^2+4x+5)

[obsidian monolith]

Display results online and refine query

Requested by Commander Vimes#7532

[harsh smelt]

why you ve got x^(n+1)

[frozen needle]

idk what you've done

[viscid thistle]

wait a sec

,rotate

[frozen needle]

ah you forgot a write a minus sign

[harsh smelt]

,w integral arctan(x+2)

[viscid thistle]

,rotate

jesus

[harsh smelt]

wtf

[viscid thistle]

how do i rotate mine

[frozen needle]

stop everything

[harsh smelt]

,w integral arctan(x+2)x^(n-1)

[viscid thistle]

i used that what i did in my pic

and i integrated by parts

[obsidian monolith]

Display results online and refine query

Requested by Commander Vimes#7532

[frozen needle]

you've got something wrong

let's go back in the past a little bit

[viscid thistle]

i said that the derivative of x^n+1/n+1 is x^n

[harsh smelt]

meh, my way is too complicated lol

[frozen needle]

https://cdn.discordapp.com/attachments/363224154469826562/722785006669529088/94730921676115968.png

[viscid thistle]

i made an 1/n+1 appear

lemme write it

[harsh smelt]

OH

ye i see

look

[frozen needle]

for all $n\in\bbN$ you have $\displaystyle n\int_0^1x^nf(x)\dd x=\frac{n}{n+1}f(1)-\frac n{n+1}\int_0^1x^{n+1}f'(x)\dd x$, and this is the natural integration by parts to do

[obsidian monolith]

Tuong:

[harsh smelt]

$\int x^n f(x) dx = f(x) \frac {x^n}{n+1} - \int f(x)x^n dx$

[viscid thistle]

,rotate

[viscid thistle]

i think i did the same

the first thing is 1/10(n+1)

[frozen needle]

your parenthesing is a bit meh

[viscid thistle]

oh

[obsidian monolith]

Commander Vimes:

[viscid thistle]

now f’(x) is arctan(x+2)

[frozen needle]

err not quite

[harsh smelt]

now f’(x) is arctan(x+2)
no forget about that i did not profitable sub

[frozen needle]

more like -(2x+4)/(x²+4x+5)²

[viscid thistle]

😮

i dont get it

[harsh smelt]

hence

[frozen needle]

$$n\int_0^1x^nf(x)\dd x=\frac{n}{n+1}f(1)-\frac n{n+1}\int_0^1x^{n+1}f'(x)\dd x$$

[obsidian monolith]

Tuong:

[viscid thistle]

yes

[harsh smelt]

why

[frozen needle]

and f(x) is 1/(x²+4x+5)

[viscid thistle]

,rotate

[frozen needle]

so f'(x) is -(2x+4)/(x²+4x+5)²

[viscid thistle]

in this case is this

[harsh smelt]

$n\int_{0}^{1} x^n f(x) dx = n(f(1) \frac {1^n}{n+1} - \int_{0}^{1} f(x)x^n dx$)

[obsidian monolith]

Commander Vimes:

[frozen needle]

so

[viscid thistle]

ok now

[frozen needle]

n/(n+1) tends to 1, no mystery

[viscid thistle]

yes

n/10(n+1) tends to 1/10

[frozen needle]

would you be able to show that $\displaystyle\lim_{n\to+\infty}\int_0^1x^{n+1}f'(x)\dd x=0$ ?

[obsidian monolith]

Tuong:

[viscid thistle]

hmm

lemme see

i think so

i should try a squeeze

[frozen needle]

Well try

[viscid thistle]

,rotate

[viscid thistle]

mb this?

[frozen needle]

that ineq seems hard to show if it's even true

[viscid thistle]

well

hmm

[harsh smelt]

should be f(1)/2 as limit, no?

[frozen needle]

you could go for something simpler

[viscid thistle]

welp the lower bound is 0

[harsh smelt]

like i've got
$n\int_{0}^{1} x^n f(x)dx = \frac{nf(1)}{2(n+1)}$

[obsidian monolith]

Commander Vimes:

[willow bear]

[frozen needle]

$$0\leq\frac{2x+4}{(x^2+4x+5)^2} \leq\frac 6{25}$$

[viscid thistle]

i am confusion

[obsidian monolith]

Tuong:

[viscid thistle]

😮

[frozen needle]

when x is in [0,1]

[viscid thistle]

oh

but

arent the bounds supposed to have the same limit?

[frozen needle]

that's just 1 ineq, the whole work isn't done yet

[viscid thistle]

oh wait

we also have the x^n+1

i forgot

[frozen needle]

yes

[viscid thistle]

makes sense then

[frozen needle]

and with the integral, it all goes nice

[viscid thistle]

yup

thank you

[frozen needle]

You're welcome 🍻

idk what you did Commander Vimes

[tawny nacelle]

no one gets what he does

[vague aurora]

idk what i gained but i proved sec is the reciprocal of cos

[viscid thistle]

wot

[vague aurora]

yees

[viscid thistle]

U can't imo

[uncut mulch]

hmmm you could try applying
sqrt(x^2) = x for x>=0

which gets you something very nice

[pale bison]

that's clever

[carmine birch]

we used this formula in school all the time

[serene heath]

ew

just square it tbh

[uncut mulch]

you could also consider
$$ 7 \pm 4\sqrt{3} = 2^2 \pm 2\cdot2\sqrt{3} + (\sqrt{3})^2$$

[obsidian monolith]

ramonov:

[uncut mulch]

7 = 4+3 = 2^2 + sqrt(3)^2

that approach is if can recognise certain patterns and are very comfortable with binomial expansion

applying sqrt(x^2) = x for x>=0
here will be sufficient for simplifying

wdym?

since your expression is positive, you can square it to get something easier to manipulate and then take the square root

$= \sqrt{\br{\sqrt{ 7 - 4\sqrt{3}} + \sqrt{ 7 + 4\sqrt{3}}}^2}$

[obsidian monolith]

ramonov:

[uncut mulch]

expand and simplify the stuff inside

[viscid thistle]

how do i do this

[remote veldt]

,rotate

[viscid thistle]

number 1

[willow bear]

ok what's giving you trouble here

[viscid thistle]

how to approach it

i just did difference of cubes but idrk where to go from there

[willow bear]

hm.

[viscid thistle]

so how do i do it

[alpine basin]

can anyone help me find the average rate of change for this problem?

[viscid thistle]

9x^2

is the derivative of your function

[alpine basin]

What you mean?

3h^2 + 9h + 9 is the answer but im not sure how to get to that point

[uncut mulch]

$\frac{\Delta j(x)}{\Delta x}$

[obsidian monolith]

ramonov:

[muted summit]

need big help: Determine the slope of the tangent line 2x^3 +2xy^2 -y^3 =21 at (2, 1)

[uncut mulch]

any reason why part of the equation is on a separate line?

[muted summit]

no that's a mistake

[uncut mulch]

implicit differentiation
sub in coordinates of your point
isolate dy/dx

[muted summit]

yeah but the issue is the point isnt on the equation

[uncut mulch]

hmm

might be a typo then

maybe (2, -1) xor +y^3

[muted summit]

so it wouldnt be possible to find the slope if the point isnt on the equation?

[uncut mulch]

no

unless you're misrepresenting the problem

[muted summit]

don't think so copy and pasted straight from the pdf

il inform my teach bout it thanks

[uncut mulch]

theres another type of question asking for info about tangents passing through points that aren't on the curve

[muted summit]

oh i see but this question is most likley wrong since it's multiple choice giving slope

[viscid thistle]

what r the options

[muted summit]

a) -5/26 b) 26/11 c) -11/26 b) 11/5

ramonov is most likley right the (2,1) should bne (2,-1) i tried it and got 26/11 which is one of the answers in the MC

when one negative wastes hours

[viscid thistle]

there is a simple form to this but i dont know how to arrive to it

long division leads nowhere

[loud sierra]

If two numbers divide to give 0

The top one must be zero

@viscid thistle

Set your top equation to be equal to 0

[viscid thistle]

yeah i know but i need the form after division

so i can find the other solutions in terms of a

here's the solution but again i dont understand how this jump was made

how one would do this algebraically without messing around with the cubic formula

[patent beacon]

K hol up lol

[viscid thistle]

problem 2 of spivak chapter 9 :(

why does he make them start out annoying

[loud sierra]

I’m sure there’s a better method to do this, but one method would thinking about it like this

If the x^3 coefficient is 2,

Then the only possible combinations of x is (x+coefficient) (x+coefficient) and 2x+ coefficient

And since a3 is the constant a must be the constant in the factorisation

So you have (x+a)(x+a)(2x+a)

And now you can try out the different signs to see which one it is

Obv it’s a bit of trial and error

[viscid thistle]

oh that is a very interesting way of doing it

[loud sierra]

I’m sure there’s a more better way to do this

Like an actual mathematical way instead of a logical way

[loud sierra]

@viscid thistle you could further go to say since the a^3 is positive, then two of the there should be two -a and one +a since only that works to give a^3. Three positive a would lead you to having more terms,

Then it becomes a question of placing the 2 minus and 1 plus

Then the 2 minis must be with the x-a since there is into a -3ax^2 remaining

Sorry for not sending this earlier I was eating dinner lol

[viscid thistle]

no worries

i guess you could intuitively start off by writing x-a twice, because then its a difference of squares and you're expecting a cancellation

[loud sierra]

Did u figure out an algebra method

[viscid thistle]

i dont think there's an analytical expression for it except the cubic formula

but that thing is monstrous so

[loud sierra]

Ye exactly

Like thats what I was thinking since it’d too complicated to put into a cubic formula

What class is this anyway lol

Like what grade is this for

[viscid thistle]

im not sure, im just doing this for fun

its a problem from spivak's calculus, chapter 9 problem 2

usually the book doesn't have computationally heavy problems like this

so i was a bit stumped when i first saw it

[loud sierra]

Ah nice haha

[viscid thistle]

but if you're asking what grade im in then i think its grade 11? (year 12 in the uk)

thanks for the help

[loud sierra]

I thought this was part of school curriculum or something

Np

Glad to hear ur doing maths for fun lmao

[viscid thistle]

at 2am no less haha

[loud sierra]

The best time 😉

[willow bear]

the image you posted is called an inequality

[tawny nacelle]

are you sure you know what bitch means ?

yeah right ofc

you should say thanks queen

nice

[willow bear]

uhh yeah i would appreciate not being called a bitch

it's kinda rude

what's your native language @stark vale

ты меня сукой назвал(а), так-то

в смысле??

нахуй пошел

<@&268886789983436800> this person called me a bitch and then said "you are a slut and so's your mom".

[viscid thistle]

[craggy dune]

🤔

[tawny nacelle]

[willow bear]

thank you whoever banned this idiot

[tawny nacelle]

google translate says they called u a snail

[willow bear]

no

first off don't trust gtran and second trust me they basically called me a prostitute

which i do not appreciate, as you can imagine

[tawny nacelle]

y did they call u a prostitute after u helped them

dis make no sens

[willow bear]

no clue, apparently in their incel mind the fact that i refuse to be called a bitch means i am to be degraded and denigrated as much as possible

[hearty vector]

Can anyone teach me Real zeroes of polynomials?

[willow bear]

too vague

is there a problem you're looking at rn

[hearty vector]

@willow bear

i remember doing something similar but i cant recall fully

[willow bear]

okay what is giving you trouble

[hearty vector]

on how to solve this

[willow bear]

look up "polynomial long division"

[hearty vector]

kk

tyty

[tame wedge]

Why isn't x*y<=1 the same as x<=1/y

[willow bear]

because y may be negative

[mint bobcat]

@hearty vector use synthetic division?

put the coefficients of the polynomial x2 +11x +31 into a bar

and divide it in a special way with -3

cuz for x +3
x = -3

disclaimer is that synthetic division only works when ure dividing a polynomial with a linear polynomial

i believe the quotient is x + 8

and the remainder is 7

to any expert reading this if im wrong plz correct me, i just started taking maths as a hobby so im still a novice in many parts

[jovial cairn]

There’s a generalization of synthetic division to work for arbitrary polynomials!

The Wikipedia page for it details it, it’s really cool @mint bobcat

[mint bobcat]

wait really?

[jovial cairn]

Yeah haha, take a look if you get a chance

[mint bobcat]

omg if i can use synthetic division for any polynomial itd be awesome

aight sure

[jovial cairn]

Polynomial long division is for suckahs

[mint bobcat]

im reading the page rn

and its like finding a treasure trove lmao

whoaaa

[viscid thistle]

Polynomial long division is for suckahs
@jovial cairn this, soooooooooooo much

[viscid thistle]

can I get some help with this?

thamks

i did this

and i’m kinda stuck

[willow bear]

how did you even make that happen

already looks hella suspicious if you ask me

[viscid thistle]

i put the 1/n inside the sum

and then inside the square root

[willow bear]

can you show your work

[viscid thistle]

lemme write it again cause it’s very chaotic

[willow bear]

are you sure you didn't mean $\lim_{n \to \infty} \sum_{k=1}^n \frac{1}{n} \frac{1}{\sqrt{1 + \frac{2k}{n} - \frac{1}{n}}}$

[obsidian monolith]

Ann:

[viscid thistle]

oh yes ffs

that one

[willow bear]

aight

[viscid thistle]

i forgot about the sqrt

now i should take the 1/n out of the sum

and me it so i have only k/n

i am stuck in a loop lol

@willow bear could u give me a hint?

[willow bear]

i'm trying to think of sth rn

[viscid thistle]

oh sorry

wait a minute

the 1/n tends to 0

so that means i can get rid of it, right?

then

im soo bad with limitss

[patent beacon]

Indeed you are looking at turning this into a reimann sum

[viscid thistle]

oh good

that 1/n scared me

for some reason

[frigid sapphire]

f(x) = {x + 3 for x < -1, x^2 for -1 <= x < 2}

@here

how can x be < -1

and also x >= -1 ?

I'm trying to find the domain of the function

[carmine birch]

x isnt < -1

x is a variable

that just states how the function behaves when x < -1

[viscid thistle]

@frigid sapphire so we are talking about this

$f(x)
\begin{cases} x+3 & x<1 \ x² & -1≤x<2 \end{cases}$

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

Right?

If so, note that the domain doesn't exist at x≥2

[frigid sapphire]

damn

oh well

I got it wrong on my homework

lol

but that's ok

thanks a lot guys

[keen sigil]

Could someone help me? I'm not sure where to start in proving this identity

[viscid thistle]

I think i'd start from left side

Hold on

So $\sin(A-B)=\sin(A)\cos(B)-\sin(B)\cos(A)$

[obsidian monolith]

Al𝟛dium:

[keen sigil]

yea

[viscid thistle]

------(free channel)-----

He already got it

[obsidian monolith]

nebula:

[viscid thistle]

Cube root of 8i = cube root of 8 x cube root of i.

[patent beacon]

Write it like 8(cos(iπ/2 + 2πk) + sin(iπ/2 + 2πk))

[viscid thistle]

Oh.

Lot easier to just do it mentally ngl.

[patent beacon]

Then Demoivre gives all three

[viscid thistle]

$i=\sqrt{-1}=(-1)^{\frac{1}{2}} \$
$i^{\frac{1}{3} = \sqrt{-1}^{\frac{1}{3}} = ?$

[obsidian monolith]

leviosa:
Compile Error! Click the reaction for details. (You may edit your message)

[viscid thistle]

Ugh.

Whatever.

Lmao

[obsidian monolith]

nebula:

[viscid thistle]

Lmaoooooo

[stuck lark]

@viscid thistle as kaynex said, rewrite 8i in polar form, though i'm not sure why he included i inside the trig functions

8i=8cis(pi/2+2pi k) where k is an integer. hitting that with de moivre gets you 3 distinct cube roots

https://www.khanacademy.org/math/precalculus/x9e81a4f98389efdf:complex/x9e81a4f98389efdf:complex-polar/v/polar-form-complex-number to start, KA probably has something on de moivre's theorem too

[drowsy karma]

How would you find this on a normal whole number graph? Would you just have to estimate?

[keen sigil]

I'm not sure what it means when it asks me to express the function in terms of cosx

[fluid shore]

It means that there is a trigonometric identity that gets you from $\cos(\frac{1}{2}x)$ to $\cos(x)$. In particular, note that:

$$\cos(2\theta) = 2\cos^2(\theta)-1$$

[obsidian monolith]

Abhijeet Vats:

[keen sigil]

oh so it would just be cos2(1/2x)

so it's just cosx

[fluid shore]

Indeed.

[keen sigil]

alright, thanks

[fluid shore]

you're welcome

[muted marlin]

if Tan0=-5/2 and 0 is a quadrant 11 angle, what is cos0?

how will i be able to solve this?

[willow bear]

do not

use

zero

for

theta

and also there's no such thing as a quadrant 11, there's only four quadrants

[muted marlin]

II

sorry

[willow bear]

that's the roman numeral for 2, if anything

[muted marlin]

oh

ok

[willow bear]

so you have tan(theta) = -5/2 and theta is in quadrant 2.

[muted marlin]

yeah

so i am trying to figure out what is cos0

[willow bear]

cos(theta)

don't use the number zero for theta, or else someone else will come along and tell you cos(0) (i.e. the cosine of zero degrees) is 1, which is the wrong answer to your problem.

[muted marlin]

o

my bad sorry

[willow bear]

anyway.
you may first consider using that quadrant information to find out what sign cos(theta) is.

[muted marlin]

its called theta right?

[willow bear]

is it positive or is it negative?

i

yes

yes it's called theta

the letter θ is called theta

[muted marlin]

tan(theta)= -5/2

[willow bear]

this is not what i am asking you.

[muted marlin]

oh

its postive

[willow bear]

no.

[muted marlin]

wouldnt it be cause its in the second quadrant?

oh

im dumb

negative

[willow bear]

cos(theta) is NEGATIVE in the second quadrant.

yes.

[muted marlin]

sorry

i just remembered

[willow bear]

now.

consider the pythagorean identity

cos^2(θ) + sin^2(θ) = 1

or to put it in a form more relevant to your particular problem

1 + tan^2(θ) = 1/cos^2(θ)

[muted marlin]

wait I am confused

how did you get the 1+tan^2

[willow bear]

i divided through by cos^2(θ).

i divided both sides of cos^2(θ) + sin^2(θ) = 1 by cos^2(θ).

[muted marlin]

OH ok

wouldnt you use sohcahtoa

[willow bear]

your angle is not in the first quadrant, so it'd be tricky to construct a relevant right triangle without running up against complications that i'd rather not get into right now

[muted marlin]

ah ok

so if it was a in the first quadrant it be more easier

[willow bear]

not really

i mean

"sohcahtoa" is a mnemonic for the right-triangle definition of trig functions, which at this point is not applicable to most angles

[muted marlin]

sorry I really am bad with alol this

I been trying to write notes on it so this is really helpful

[willow bear]

i'm trying my best here

rest assured i've seen students be much less cooperative than you

[muted marlin]

yeah I am understanding a little bit

[willow bear]

anyway, once you have 1 + tan^2(θ) = 1/cos^2(θ) (and not forgetting that cos(θ) < 0), it's a matter of algebra to solve for cos(θ)

(remember that the value of tan(θ) is known.)

[muted marlin]

Oh so just solve the cos

for it in the end?

[willow bear]

you solve the equation

1 + tan^2(θ) = 1/cos^2(θ)
for cos(θ)

[muted marlin]

wait my calculator is weird

so I have to put that in my calculator right like what you just said

I get 0.99

[willow bear]

uh

no

since when did a simple algebra problem like this require the use of a calculator?

[muted marlin]

im really kind of bad with math so I use a calculator

[willow bear]

refrain from it as much as possible

[muted marlin]

oh ok

[willow bear]

like... you can just rewrite this thing as $\cos^2(\theta) = \frac{1}{1+\tan^2(\theta)}$... and then take the square root to give $\cos(\theta) = -\sqrt{\frac{1}{1 + \tan^2(\theta)}}$...

[obsidian monolith]

Ann:

[willow bear]

no calculator required, unless you're unable to simplify $\sqrt{\frac{1}{1 + (-\frac52)^2}}$ by hand i guess

[obsidian monolith]

Ann:

[muted marlin]

yeah im pretty bad with simplifying so I try to use a calculator so i dont mess up in the end

[willow bear]

no

[muted marlin]

but ill try to solve it without it

[willow bear]

calculators don't give you exact answers

[muted marlin]

so woulnt u get 2sqrt(29)/29

and then i can simplify it to 2/ sqrt29

[willow bear]

yeah, so your final answer is cos(theta) = -2/sqrt(29)

[muted marlin]

oh im dumb ok yeah my calc messed it up way to hard

I just kinda did it in my head

[willow bear]

again

oh

no thats even worse

if you do it in your head it's like a surefire way to fuck up

[muted marlin]

no not like in my head but i wrote it down

[willow bear]

do it on paper

[muted marlin]

and did it

and just simplified it

so try to do it all in paper and not on calculator right

[willow bear]

yes

the only time you should ever use a calculator is when you're asked for the value of something to some number of decimal places, and even then you should only use the calculator at the END once you've gone as far as you can on paper w/o rounding or approximations

[muted marlin]

ah ok tyy for your help

imma write in my notes

because i have a exam coming up soon so it definetly helped me

[frigid sapphire]

https://i.imgur.com/C3i0soi.png

how can I find the value of a?

[harsh cipher]

okay

why is the value of a = 2.92 in the general term?

ball started at 4m.

[willow bear]

presumably, the initial 4 meter height is not the height of any of its bounces

[harsh cipher]

okay I kind of got it

it has to be"the height of the bounce"

meaning after first bounce?

meh

[viscid thistle]

@harsh cipher It's 2.92 because 73% of 4 is 2.92

And the answer to B) is 4(.73⁸) ≈ 0.322

[harsh cipher]

hi

is (sqrt3)^5 (sqrt)243

oops no parentheses

wtf how did it get sqrt243

[viscid thistle]

3^5=243

[harsh cipher]

it has sqrt?

r = sqrt 3

you can't just multiply the 3 under the root

can you?

[viscid thistle]

I dont understand what is your question

[harsh cipher]

how is (root3)^5 = root 243

[carmine birch]

sqrt 3 = 3^1/2

so power rule

[harsh cipher]

and?

i know that

[viscid thistle]

that's allowed

[harsh cipher]

so you go back and forth

?

[carmine birch]

so (3^1/2)^5 is 3^5/2

god idk how to use texit

[viscid thistle]

$(\sqrt{3})^5 = 3^{5/2}=\sqrt{3^5}$

[carmine birch]

yes

[obsidian monolith]

HoboSas:

[harsh cipher]

yes

correct

thank you

[keen sigil]

if someone could help me figure out this problem that would be greatly appreciated

y = 2cos(2x) and y=sin(x)-1

trying to find x

[uncut mulch]

have you attempted anything yet?

[keen sigil]

idk where to start

i know i might need to use the identity for cos2x

which would equal cos^2x - sin^2x

[uncut mulch]

there are multiple forms for the double angle identity for cosine

[keen sigil]

oh right

my intuition tells me to use the (1-2sin^2x) one but idk if that's right

i could use the other one and somehow convert cos into 1 and sin^2x

[uncut mulch]

try it and see what happens

[drowsy karma]

For Method of Elimination, can I choose either x or y to eliminate? Or do I have to eliminate a specific one

[keen sigil]

I've gotten here and I'm not sure if it's right or where to go

[uncut mulch]

incorrect setup

those aren't the equations(s) you should be solving

you should be starting with:

$2\cos(2x) = \sin(x) - 1$

[obsidian monolith]

ramonov:

[uncut mulch]

applying the double angle identity: $\cos(2x) = 1 - 2\sin^2(x)$ will lead you to a quadratic in $\sin(x)$

[obsidian monolith]

ramonov:

[willow bear]

@drowsy karma it's up to you

[keen sigil]

oh i was trying to make the left match the right and doing systems of equations i was just overthinking it

this seems right

[dense creek]

Hi, I’m stuck at this question. Anyone can help me ?

[viscid thistle]

Not #precalculus at all.

[dense creek]

@viscid thistle I suppose to post to which channel ?

[viscid thistle]

Well

You have plenty of channels of adv maths

If you have no idea still, use #questions section

[wide ocean]

yeah that's not precal..

[heady jewel]

@dense creek thats not proper notation btw you cant make changing variable(k) and the bound the something in terms of k

so use a different changing variable

[dense creek]

So can I use N ?

[viscid thistle]

Not #precalculus not #precalculus not #precalculus

[dense creek]

@heady jewel

[heady jewel]

Anythi other than k

@dense creek

@viscid thistle relax notation is used everywhere

You dont want something like

$\sum_{k=1}^{k+1} k$

[obsidian monolith]

Lionel:

[viscid thistle]

@viscid thistle relax notation is used everywhere
@heady jewel and?

[heady jewel]

So hes clarifying notational problems

Im not discussing his problem

The one he posted

[viscid thistle]

Still not the place.

[heady jewel]

K

Its done

[warm crescent]

how is it obvious that difference of students height is less tham 3m?

heights of student *

[sour eagle]

Have you ever seen someone taller than 3m

[viscid thistle]

good argument, how do you prove that?

[sour eagle]

😔

[warm crescent]

NO

[serene heath]

lmao imagine being under 4m tall

[sour eagle]

yea the question is kinda dumb imo

[warm crescent]

thanks vroo

[sour eagle]

epic

😎

[warm crescent]

opening my maths notebook after almost 5 months

[sour eagle]

[warm crescent]

[viscid thistle]

...

[viscid thistle]

yah this is right @robust rover

[sour eagle]

what

[reef jasper]

how do i write a cosine function for this

[reef jasper]

hello?

[craggy dune]

🤔

what have you tried?

[reef jasper]

i got the amp and midline and i’m pretty sure i got the period

3cos(pix+1/2)-1

but i don’t get the same thing in desmos

[remote veldt]

put brackets around the (x + 1/2). If you want to transform f(ax) by a shift of b, it's f(a(x - b)) not f(ax - b)

[muted granite]

Anyone have a link to video or article how to do this on a TI-84? I can not find one. Thanks.

[viscid thistle]

u can do this using math by hand

matrix addition is simple

@muted granite

[muted granite]

yeah, but I am going to need to know the process when it gets more complicated.

[viscid thistle]

how can matrix addition get more complicated?

[reef jasper]

how do i start with this

i need an exact value

[craggy dune]

well you can use the stuff you learned at school or you can use wolfram

or just wait till someone answers your question

sometimes that works too

[uncut mulch]

express it in terms of a function you're more familiar with

should help a bit

[reef jasper]

i did 3pi/12 +14pi/12

but what i do after

[viscid thistle]

reduce the fractions

[reef jasper]

yeha pi/4 + 7pi/6

do i jsut do 1/cos for each of those and add them together?

[viscid thistle]

yah

[patent beacon]

Nu not at all lol. Know the sum identities? @reef jasper

[wide ocean]

Can someone pls explain why the reference angle has answers in ALL four quadrants?

I thought it would only be in Quadrant 1 and 4 since cos is positive

But since it's Cos squared, it can have negative values?

"CAST" rule?

[viscid thistle]

But since it's Cos squared, it can have negative values?
indeed

[wide ocean]

Ah okay, thanks

Is this always the case?

[viscid thistle]

btw, domain doesnt imply solutions

[wide ocean]

wym?