precalculus

[summer ruin]

both operations are allowed if you're not multiplying by zero

otherwise no

[brittle shore]

How would you know if you are multiplying/dividing by zero without solving it through factoring?

[summer ruin]

because I can solve a much simpler equation tan(x)=0 or sin(x) = 0 (or I already know the answer to those because they're so simple)

first equation clearly holds for tan(x)=0 because both sides are zero, you just need to show that sec(x) is defined for all x for which tan(x)=0

so by considering tan(x) != 0 you can divide both sides after you've found all the solutions to the case tan(x)=0 or shown that there aren't any

[brittle shore]

What about in the case when you are not specifically solving for something

Like in a trig proof. Is it certain that sin(x) != 0 when you multiply by sinx/sinx ?

[summer ruin]

you need to restrict x to all such points that sin(x) isn't zero

which may make the original domain for x smaller, which can cause problems with finding solutions

but you then just revisit the case where sin(x) = 0

well I'm talking about solutions still, but it's all the same in proofs - you don't want to multiply or divide by zero almost ever since it just could make the proof invalid or some silly thing like 1 = 2

[brittle shore]

So if you are proving equality of two different functions? Where f(x) is manipulated to obtain g(x).
Is
f(x)?=g(x)
(sinx*f(x))/sinx) ?= g(x)
A valid step in the proof?

[summer ruin]

if you specify that sin(x) != 0, sure

but again that kind of proof would then be only valid for x such that sin(x) isn't zero

so functions in theory could be different at these points

[brittle shore]

So if f(x) definitely is equal to g(x), (f(x)sinx)/sinx is no longer equal to g(x) for all real numbers.

[summer ruin]

yes it has an infinite number of points where it's undefined now

[brittle shore]

Yea that seems like common sense, I guess it’s just weird to think about because when I was taught trig proofs and tested on trig identity proofs I feel like I did things like multiplying by sinx/sinx seemed to hold no consequence, and so did the teacher.

Perhaps it’s implied sinx != 0 ?

[summer ruin]

perhaps it didn't have consequences or you could've shown that the case sin(x)=0 is trivial or that the original function isn't defined at these points anyway so it doesn't matter

like csc(x), which is undefined when sin(x) turns to zero so multiplying and dividing by sine doesn't change anything about it

[brittle shore]

Yea I also suppose that the approach was to manipulate one side only without touching the other side. So if as long as that sinx makes it’s way out of the denominator which it must to make the sides appear exactly the same, than the domain should not have been effected.

Like the domain was only effected temporarily. Maybe that’s technically wrong still not sure, but I guess that’s why it never caused a problem in proving the two equal to each other

[brisk pasture]

Is 5 the only factor?

[burnt hare]

hm i think + or - sqrt -6 would work

[brisk pasture]

Why?

[burnt hare]

Well if you factor it you get (x-5)(x^2-6)=0

sorry

[brisk pasture]

(x-5)(x^2+6)?

[burnt hare]

(x-5)(x^2+6)=0 **

yes

[brisk pasture]

=0

[burnt hare]

then you just set each equal to zero etc

x^2=-6

x= plus or minus rad. -6

[brisk pasture]

@burnt hare you wouldn't have to use the quadratic formula?

[burnt hare]

you couldn’t use the quadratic

that’s just for ax^2+bx+c=0 no ?

i’m not exactly sure how to use the method yall using

[brisk pasture]

If for example 3x^2+8=0

a=3, b=8, c=0

@burnt hare but b would be 0, so a=3, b=0, c=8

[summer ruin]

why use quadratic formula for an elementary quadratic anyway

quadratics of the form ax^2+bx and ax^2+c are much quicker solved without quadratic formula

[opaque mango]

guys i don't understand why it graph the the down the x -axis.

is it because v.a has -1?

[brisk pasture]

I got the answer

[viscid thistle]

Do you Guys know what grade you are Supposed to do Pre - Calc In?

Because I am in 5th Grade..

And Doing it…

Hello? Anyone?

Can Someone Please Help Me?

[lime frost]

usually people do it in 6th…

[rugged light]

log (x+1) - log x = 1

what is

[scenic sable]

use the properties of the logarithm; $\log_a b - \log_b c = \log_a (\frac{b}{c})$. Apply this to the problem you’re having: $\log (\frac{x+1}{x})$ = 1
and now get rid of the log.

[obsidian monolith]

Quarktastic

[gentle sedge]

not related but I've always wanted to ask what is this input language and where to learn it?

[supple owl]

You know that log(10)=1 so the inside should be equal to 10 and then you get some algebra and you will get x=1/9

[vapid plaza]

#resources message

[scenic sable]

It’s LaTeX. Just Google LaTeX pdf and you’ll have a ton of resources.

[scenic sable]

http://static.latexstudio.net/wp-content/uploads/2015/03/LaTeX_Beginners_Guide.pdf https://wch.github.io/latexsheet/latexsheet.pdf

[gentle sedge]

thanks

[supple owl]

Man who pinged

[viscid thistle]

Why is x^y continous for all x > 0? Can't x = -1 and y = 0

[eternal star]

How can I find arctangent for any given real number. Without using a calc?

[viscid thistle]

Oh, Okay, Then…

@lime frost Because I am in 5th Grade (The Original Message had a Typo), So at Least I am 1 Grade Ahead…

[fiery ravine]

Hey if anyone can help me, im pretty rusty with math, and bad at the format they want on computer. If anyone can help me through this one, greatly be appreciated

this might more be algebra 2, not sure

[glossy onyx]

@fiery ravine Since the bases are the same you can set the exponents equal to each other

then solve for x

[marsh cave]

still need help with this, its rational root theorem

[fiery ravine]

thanks derpedee!

it was -2

when there is a e on both sides hit it with a log lol

[grave sorrel]

6 exponent -6

[slim steppe]

Take the limit as x approaches to 1 and negative 1, and you'll see that the function approaches to positive infinity and negative infinity as x approaches to 1 and negative 1

[obsidian monolith]

Jagteshwar(JPS838898)

[frank vine]

Anyone??

[dim cloud]

Cortés decided to change his plans and to circumvent Cholula before proceeding directly to Tenochtitlan.

[willow bear]

no, there is not.

[willow bear]

that looks a lot harder than typical precalc fare...

...

what level of education are you at

[lime frost]

(x-5)(x-(2-i))(x-(2+i))

[frank vine]

College precalc 1

[coarse kernel]

then it must be a sketching/function analysis question

like you find the maximum number of solutions and then just guess them via either direct guessing or sketching

[coarse kernel]

here you can have two solutions at maximum (try to figure out why)

but wait... both solutions are irrational. something must be wrong here

[coarse kernel]

and both are not expressible in elementary functions

[willow bear]

that question was addressed to someone who's since gotten themselves banned

[frank vine]

ah ok

[frank vine]

yeah that why i had a problem trying to find analytically and was ablle to do it graphically

aparently have to go into real analysis to find ans anaytically

since the solutions use the Lambert W Function

and use the complex and the real branch of it

[minor forge]

Why are people doing limits on pre calc

[coarse kernel]

yes, and I don't think that precalc curriculum includes it

it's rather calculus

[viscid thistle]

?

precalc in 6th grade? tf?

genius or typo?

[coarse kernel]

that's why you should specify the country you study in :)

[olive jewel]

I think you mean years in college?

Or just the system you study in is not similar to most curriculums?

[mellow vigil]

just saying why do u have discord in 5th grade

dont u have to be 13 years old ig

[vapid plaza]

1. They are actually under 13 and is unaware of the ToS
2. They said that to flex how young they learnt precalc
3. They are still in 5th grade after repeating for many years

[viscid thistle]

I need help with logarithms I'm not home rn but I will send problem dm me

[viscid thistle]

Wait is there pre calc in college

[slim steppe]

Perhaps in some community college

[viscid thistle]

Lmfao

I didn't know you could take pre calc in college

I take pre calc in school so I assumed by college you wouldn't be able to take it anymore

[raven lily]

If I have r = 3 sin theta, and I want to convert it into the rectangular form, how would you go about it?

My first move was to divide by sin theta

but I'm like, that won't do

[near cove]

Yeah, I need to take precalc 1 and 2 before taking calc 1

Im in a cc

[uncut latch]

Honestly, you don't typically do precalculus in year 5; at least, not in my country. That said, if you are actually doing precalculus (and properly) in year 5, then you are definitely ahead. But do not stop there; there are just so much more to mathematics than just calculus. I want to encourage you to stay humble and keep up with your good work.

[pastel hull]

ah i need help in pre cal any one active

circle problem

[rare fiber]

Just post your question

Preferably in the help section

[gentle sedge]

which question?

[summer elbow]

is this correct ?

[coarse kernel]

yes, but I don't think that improper integrals are studied in precalc curriculum

[summer elbow]

cool tyty

beautiful notation fr

[summer elbow]

do you consider the limit of a/a as a -> 0 infinity in calculus ?

[zinc fulcrum]

what do you mean by "do you"?

[summer elbow]

anyone

[zinc fulcrum]

anyone what? haha

[summer elbow]

what do you think 0/0 is

[zinc fulcrum]

0/0 is undefined, limit of a/a as a -> 0 is 1

[summer elbow]

ok so

actually wait I gtg

I'll complete later

today

[zinc fulcrum]

lol ok

[summer elbow]

ok so lets say you're working on a problem

you find a method and you get to 0/0 in the end

now there are 3 ways to solve

by choosing which rule you apply you can get different results

first one is the 0 * anything is 0

second is anything / 0 is infinity

and last is anything / itself is 1

0/0 can either be 0, 1, or infinity

now looking at the problem

lets say you can visually and analytically see that it can't be 0 or 1

but it can be infinity

so is it fair to pick one answer depending on the question at hand ?

[gentle sedge]

is undefined

[vapid plaza]

The answer is “undefined”.

[summer elbow]

yes but we do define the limit

that's what I'm talking about

taking the limit of a/a as a -> 0 is indeterminate

but a/0 and 0/a do have a limit

and any number else divided by itself is 1

so taking a different approach to the problem and using a limit of one problem to get there can yield the 3 answers

just depends on which one you pick

but if the problem results in 0/0 and it visually cannot be 0 or 1, then it has to be infinity which is the 3rd result ?

[vapid plaza]

It could also be 2, or 3, or -1, or literally other number you can think of.

[obsidian monolith]

NotMyself

[mellow vigil]

And every n satidfies that ig

[mellow vigil]

^

[slim steppe]

The terminology for that is, indeterminate. So, if you have 1/x (which is called a hyperbola), where x approaches to zero, meaning that we're approaching the function from the left and right side. So if you take -.000001 and .0000001 and values much smaller than these and plug that into x, you'll see that the function approaches to negative and positive infinite, there go it has an indeterminate limit

[rugged light]

log 4x-1 - log x-2 = log 5
help me plis

[hazy hollow]

since there is no base shown for any log we assume the base is base of 10. (common logarithm). so
A= 10
B= 4x-1
C= x-2.

[rugged light]

thank you very much

[zinc fulcrum]

to add on to what was already said...

1. 0/0 isn't something to "solve" it's just undefined, it's part of the definition of division and that doesn't have anything to do with limits
2. i don't think you're looking at rules like "anything / itself is 1" properly. a better way to say that would be "anything not zero / itself is 1"
   rules like that have conditions on the variables, they aren't all just "this holds for every real number"
   aaaaand anything / 0 is infinity is questionable haha

[summer elbow]

alright, so what if you were solving a problem and got 0/0 as the answer

what do you do then

[zinc fulcrum]

example?

[summer elbow]

I ran into one yesterday but It's too long to talk about rn

but isn't it fair to say It's possible to run into that somewhere in mathematics

also

[zinc fulcrum]

oh yea i remember you said this

[summer elbow]

visually speaking, the answer can't be 0 or 1 but can be inf

[zinc fulcrum]

the answer to what??

[summer elbow]

so I picked infinity as the answer and it was correct

[summer elbow]

well, In short I was trying to think of a formula of calculating the bounds of the domain of x^2

then I thought that if we were to theoretically drop the parabola down the x-axis and let It's offset from the x-axis be the limit as it goes to infinity, the roots r1 and r2 will approach the bounds of the domain

lim{a -> inf}[f(x) - a]

[zinc fulcrum]

wdym bounds of the domain of x^2?

[summer elbow]

well

the domain of x^2 is +/- infinity

aka (-inf, +inf)

these are the bounds of the interval

I wanted to find a way to calculate these bounds myself with a formula

[summer elbow]

so I made this

after doing that you'll end up with a weird quadratic basically

I then take the quadratic formula of that quadratic to find the 2 intersections / roots / bounds on the x-axis

my calculations reduced to +/- (0/0)

[zinc fulcrum]

😕

[summer elbow]

btw

that's the first time I answered it

second time I used a different calculation technique and straight up goes +/- inf

which is correct

but if you look at the parabola on desmos or something

the domain is clearly not +/- 0 since it goes out of it pretty quickly

and same goes for +/- 1

which leaves +/- infinity to be the only available candidate

I picked it then checked online and it was correct

[zinc fulcrum]

sorry but i have no idea what you're saying rn

[summer elbow]

if you basically drop x^2 down the x-axis It's 2 intersection points will become the bounds of It's domain at infinity
that's the whole point of me formula
so what I did is saying

now Q{} is a notation I made, basically it means taking the quadratic formula and getting the 2 roots r1 and r2

now we'll take the limit of the dropping function thing
It's gonna give is the new weird ass quadratic formula that has It's roots as the domain bounds
now we want the roots of that new quadratic
so we'll take the quadratic formula of that new function
It'll give us r1 and r2
and the domain will be the interval [r1, r2]
where all values of x live inside that interval
but not outside

at the end we reach the conclusion the domain is from -0/0 to +0/0

or that r1,2 = +/- 0/0

one cool thing I noticed is

the - that came before the plus or minus flips it and makes it minus or plus

which makes sense when you consider the order

r1 = -inf, r2 = inf

I know it doesn't matter since It's on a zero but multiplying the whole thing by -1 preserves it

[zinc fulcrum]

i'm at a loss for words

[summer elbow]

is that a good or bad thing 😅

[zinc fulcrum]

good for me, bad for you

[summer elbow]

lol so is the formula bad

[zinc fulcrum]

what you just said makes 0 sense to me .-.

like not even a little

i cannot understand anything in those images

what is this and what does it have to do with the domain of a function (and what is the domain of a function to you)?

[summer elbow]

ok talk later today I gtg again

[zinc fulcrum]

ok before you go there is one thing i really need to say

it's

??????????????????????????????????????????????????????????????????

that's all, enjoy your day 🙂

[brave sorrel]

[split steppe]

trying to review for a test and this has been boggling my mind

in the numerator: cos x + cos x sin x - (1 - sin² x)
it became: cos x (1 + sin x) - (1 - sin x) (1 + sin x)

how did "cos x + cos x sin x" become "cos x (1 + sin x)"?

[summer ruin]

factoring

[trim solar]

Hey bitches

(x/(y+z)) + (y/(x+z)) + (z/(x+y)) = 4

x y and z are positive integers

What are the numbers

[peak pasture]

this is a cool problem

[trim solar]

Yea but what's the solution

[peak pasture]

https://math.stackexchange.com/questions/2192461/find-answer-of-fracxyz-fracyxz-fraczyx-4

its not as simple as you think lol

[zinc fulcrum]

https://ami.uni-eszterhazy.hu/uploads/papers/finalpdf/AMI_43_from29to41.pdf

[zinc fulcrum]

probably a troll question lol

[peak pasture]

probably lol

[trim solar]

Bruh so I have to use Google to know it?

[zinc fulcrum]

you don't have to do anything

[formal crow]

massa

[light crescent]

im pretty sure thats a double angle theorem

[viscid thistle]

is that to do with the chain rule by any chance?

[split prairie]

no

the chain rule is used for subtition and slope

[light crescent]

it doesnt deal with chain rule

its just an identity u have to know

that and the half angle theorems

although your pre calculus class will most likely give u a formula sheet with it if you are doing standardized testing

[heavy sierra]

How is this converted?

[summer ruin]

4 = 2^2

[hazy hollow]

cube root just means divide by three, base of 4 is 2^2. Divide exponent by 3.

check it on a calculator. they both = 1.587etc

[slim steppe]

By factoring out "cos x"

[slim steppe]

That's right

[paper sand]

Helppp

[coarse kernel]

use Partial Fraction Decomposition

[gilded karma]

if you take cosx common, you get cosx(1+sinx)

[gilded karma]

Just divide the numerator by the denominator using long division

[viscid thistle]

That doesn’t look like precalc to me

[summer ruin]

write h(x) in terms of f(x) and g(x)

[fallow brook]

PLS HELP ME

[ashen skiff]

sin(-x) = -sin(x)

csc(-x) = -1/sin(x)

so the first step would be to rewrite it as -1/sin(x) + sin(x)

does that help?

[vagrant wing]

Why + sin(x) ?

[keen anchor]

I don't know what you mean by that

[gilded karma]

h(x) = kx^6 + BC^6x + C^6x
h(x) = g(x)(f(x)+1)
h(3) = g(3)(f(3)+1)
h(3) = 4x(8+1)
h(3) = 4*9
h(3) = 36

[keen anchor]

hmm, i over-thunk that

Alright make sense

[jolly niche]

I'm stuck on one of these questions and it would be really helpful if anyone can help explain how to do these. I tried watching many different YouTube videos and I still can't get the grasp of it. These are Improper Integrals.

[summer ruin]

you're supposed to already know how to integrate both of these functions in indefinite integrals

[peak sinew]

Isn't log(10) = 1 for log base 10 and not natural log?

[supple owl]

Yea for base 10

And for natural log log(e)=1 where e is Euler constant rufly 2.71

[peak sinew]

I see

[summer kiln]

Can anyone help me with proving this identity?

[viscid thistle]

looks more like integral calculus

[viscid thistle]

after 15 min, you can tag helpers

[midnight cloak]

if we take sinΘ + cosΘ = a,
then the equation will turn into :
2sinΘcosΘ/ a-1 =a+1
= 2sinΘcosΘ= (a+1)(a-1)
= 2sinΘcosΘ= a^2 -1
now, a^2= (sinΘ+cosΘ)^2
= sin^2Θ +cos^2Θ + 2sinΘcosΘ because (a+b)^2= a^2 +2ab +b^2
= 2sinΘcosΘ = (sin^2Θ +cos^2Θ + 2sinΘcosΘ) -1
= 1- 2sinΘcosΘ -1 because sin^2Θ +cos^2Θ=1
2sinΘcosΘ =2sinΘcosΘ

oh wait

was i not supposed to answer?

[summer kiln]

Idk

But thanks bro

[midnight cloak]

welcome 😊

[gilded karma]

Multiply and divide by sintheta + cos theta + 1

= 2sinΘcosΘ * (sinΘ + cosΘ + 1) / (sin^2 Θ + cos^2 Θ + 2sinΘcosΘ -1)
= 2sinΘcosΘ * (sinΘ + cosΘ + 1) / (1 - 1 + 2sinΘcosΘ)
= 2sinΘcosΘ * (sinΘ + cosΘ + 1) / (2sinΘcosΘ)
= (sinΘ + cosΘ + 1)

[summer kiln]

🙏🙏thanks

[warm garnet]

Any video recommandations on how to solve this problem (A,B,C) are matrices

[willow bear]

,rccw

[willow bear]

AX - C = 2X - 2B

this can be written as (A-2I)X = C - 2B

[quasi mesa]

what is this?

im super confused

[violet garnet]

ln is a natural logarithm

you can re write it as Log with the base e

and solve it as you would solve any other log (assuming you know how to)

[viscid thistle]

divide both sides by 2

you then get ln (5x) = 2

now 5x = e^2

now divide both sides by 5

x = e^2/5

there you go

now if you want to find the exact answer

then just put that in you calculator and you will get

x = 1.477811

[thick thunder]

How do you integrate both sides of the equation using indefinite integrals
2y(dy)=x

[viscid thistle]

try asking that in the calculus section #calculus

[thick thunder]

Thanks

[ionic hemlock]

can someone help

[lofty sage]

the figure is too small

[ionic hemlock]

sorry

@lofty sage

[heady fable]

Right triangles

What are you able to do here

Like

If you're given $\theta=\pi$, then can you just calculate $g\left(\frac{\theta}{2}\right)$ or not

[obsidian monolith]

Lachlan

[heady fable]

Anyways

You're obviously going to want to find what theta is

It'll be in terms of $a$

[obsidian monolith]

Lachlan

[heady fable]

Note that the circle you're using has radius $\sqrt{20}$

[obsidian monolith]

Lachlan

[heady fable]

Which is 2\sqrt(5)

So as a hint, you already know that the angle is greater than $\frac{\pi}{2}$ because it passes the $y$ axs

[obsidian monolith]

Lachlan

[heady fable]

Now, just form a right triangle using the points $(0,0)$, $(a,4)$, and $(0,\sqrt{20})$

[obsidian monolith]

Lachlan

[heady fable]

Use trig to figure out the angle

You know all the side lengths

Oh mb

Sorry that's wrong

You don't use the point $(0,\sqrt{20})$ but you can get by without it

[obsidian monolith]

Lachlan

[heady fable]

It's $(0,0)$, $(a,4)$, $(0,4)$

[obsidian monolith]

Lachlan

[viscid thistle]

that is a little bit wrong

[heady fable]

Yeah you think

Anyways

[heady fable]

Triangle using these points,

[viscid thistle]

you cant say if the point (0, sqrt(20)) divides the angle thetha

[heady fable]

?

Sorry I'm tired lmao

[heady fable]

^^

The angle theta is pi/2 + whatever the angle in that triangle is

And you know opposite and adjacent so $\tan(\phi)=\frac{a}{4}$

[obsidian monolith]

Lachlan

[heady fable]

$\phi=\arctan\left(\frac{a}{4}\right)$

[obsidian monolith]

Lachlan

[heady fable]

$\theta=\phi+\frac{\pi}{2}$

[obsidian monolith]

Lachlan

[ionic hemlock]

hey i tried something a little diferfent

i ended up with this

[heady fable]

What did you try?

[ionic hemlock]

well i solved for cos theta

then i plugged it into the half angle

not sure if my math was right

[heady fable]

Interesting

[ionic hemlock]

i think im good on that problem

[heady fable]

$\cos\left(\frac{\theta}{2}\right)=\cos\left(\frac{1}{2}\arctan\left(\frac{a}{4}\right)+\frac{\pi}{4}\right)$

[ionic hemlock]

i have something else im struggling on

could you help?

[heady fable]

Sure

[obsidian monolith]

Lachlan

[heady fable]

Why

These types of questions smh

There's literally no point in even learning how to do this

[ionic hemlock]

agreed

[heady fable]

Calculators exist 🙄

Anyways, you'll want to first evaluate the inside

[viscid thistle]

trigo is all about to learn how to not use calculator

[heady fable]

Who says trigo

[heady fable]

Yeah and it's dumb

A waste of time if you ask me

[viscid thistle]

if you see trigonometry is just manipulations

[viscid thistle]

partially true

[heady fable]

There's no reason to even learn how to evaluate something like that

A calculator can do it in less than a second

And every high-level math course uses calculators because it's an utter waste of time to do it by hand

[viscid thistle]

btw @ionic hemlock how did you got this answer?

[heady fable]

It should depend on a...

[viscid thistle]

you know what

thats true

[ionic hemlock]

well

[viscid thistle]

but doing long calculations improve your retention powerr

[ionic hemlock]

i solved for theta

cos theta

from the triange

[heady fable]

Which triangle

[ionic hemlock]

then i used that cos theta for the half angle formula

the given one

[viscid thistle]

how did you get rid of a

[ionic hemlock]

pythag theorem

it gives the radious

[heady fable]

If you think about it there is no way to get rid of a

[viscid thistle]

radius is known already

[ionic hemlock]

radius is ur c

[viscid thistle]

that is sqrt(20)

[heady fable]

The point is that, here, every other field would be satisfied with this lol

[viscid thistle]

lol

[ionic hemlock]

so can you help with this

[heady fable]

No reason to simplify it

Honestly

I'd rather not

www.wolframalpha.com

[ionic hemlock]

i dont think i could plug that in there

[viscid thistle]

Sorry but my hands are numb on inverse trigonometric functions currently

[heady fable]

,w calculate sin^2(1/2 * arccos(-12/13))

[lofty sage]

i found this

[ionic hemlock]

how di u get +?

also u need to rationalize

[lofty sage]

cos(theta) is negative

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by Lachlan#2738

[heady fable]

Lmao how are you guys getting these answers

[ionic hemlock]

wym

[ionic hemlock]

wym

its not giving soplution for the wolframapha

[lofty sage]

what?

[ionic hemlock]

like teh steps

smth else

um p1cee

why is it negative

i mean positive

[heady fable]

What are you even doing to get that

There should be no reason why a isn't included in the answer

[viscid thistle]

yeah

please show the steps

[ionic hemlock]

pyhat theorme guys

for the triange

will solev for A

[lofty sage]

oh, i saw the sin formula that have - cos

[ionic hemlock]

A^2 +B^2=C^2

[viscid thistle]

on which triangle, umm coordinates?

[viscid thistle]

Heh?

[heady fable]

On the triangle formed by connecting the blue segment to the y axis?

[ionic hemlock]

yes

[lofty sage]

so is that @ionic hemlock

[ionic hemlock]

then rationlize correct

[heady fable]

And you're trying to evaluate cos(that angle)?

[ionic hemlock]

which should give u mine?

[heady fable]

Because this is obviously $\frac{a}{4}$

[ionic hemlock]

wait how did u get 2?

[heady fable]

I give up 💀 💀

[lofty sage]

Yeah

i forget this

[ionic hemlock]

how did u get 2root5

[lofty sage]

so just find a

sub in equation

[heady fable]

But you don't know what that angle is lol

[lofty sage]

(a,4) is in x² + y² = 20

u know a

[heady fable]

Oh

Did not see the equation lmao

Nvm 💀

[viscid thistle]

lol yes

i am a dumbass

[ionic hemlock]

pr1ce

how did u get 2root5

[lofty sage]

i made a mistake

actually ur answer is right

this 2 dont exist

because i found cos(theta) = -2/sqrt(5)

but is -1/sqrt(5)

@ionic hemlock

[ionic hemlock]

kk ty

[viscid thistle]

For some reason

[viscid thistle]

I am getting this 😅

Help @ionic hemlock @lofty sage @heady fable

[lofty sage]

how?

show everything u did

[lofty sage]

u didnt square the 2 that is dividing

[viscid thistle]

Oh yed

Yes

But

Would it make s difference?

[lofty sage]

sure

[viscid thistle]

Yeah actually wait

I think I got it now

Silly mistakes sucks by life

[lofty sage]

u use a good way to do this

[viscid thistle]

Is it alright now?

[viscid thistle]

Thanks bro

[lofty sage]

hm wait, is not 44 is 40

[viscid thistle]

I should actually do suicide

Yeah

[lofty sage]

and this 20 that is multiplying

[lofty sage]

i saw other mistake

u switch 20 to 1/20

[viscid thistle]

Today I am gonna cry

I am making all of them today

[lofty sage]

calm down bro mistakes is normal

[viscid thistle]

Yeah

Ofc

I am joking

I just do maths for maths

And not for calculations

[lofty sage]

😄

[viscid thistle]

Bro

Finally I'm here

Is it alright

[lofty sage]

i think yes, u can simplify more writing sqrt(20) = 4sqrt(5)

[viscid thistle]

Yeah ofc

[lofty sage]

and simplify with the denominator etc...

[viscid thistle]

Alright finally?

And thank you for carrying me through my calculation disasters 😅

[lofty sage]

HAHAHSHSH

no way bro

i saw this now

u use a = 2 but the point (a,4) is in the second quadrant so the 'x' is negative

[viscid thistle]

🤡

I'm done

And you're so good at observing bruh

[viscid thistle]

And that is because of my disastrous diagram

[lofty sage]

then u will get this but with - but this 2 dont exist

i dont know what is wrong to u find with this 2

[viscid thistle]

So basically this method is wrong

Right bro?

I thought I could use the angular bisectors here

But that doesn't seem to work

Oh I got it

My mistake

It is not necessary for the Angular bisector to be perpendicular

[lofty sage]

no, its right

[viscid thistle]

But surely it will pas through the mid point

[lofty sage]

i see

u didnt simplify the sqrt(20) with 20

[viscid thistle]

With -2?

[lofty sage]

(10 -2sqrt(5))/20
simplifying everything by 2
(5 -sqrt(5))/10

[lofty sage]

yeah

[viscid thistle]

Maybe yes

[lofty sage]

i think u put 4before the root of 20 but is 2

[viscid thistle]

Oh yeah actually

https://tenor.com/view/its-extremely-bad-ash-c4etech-this-is-so-bad-this-is-the-worst-gif-20669322

My calculations

And yeah

Maybe my method works

I should have attempted this question on the notebook

I always mess up on the board

[lofty sage]

at this time i cant judge u, my mind turn off after 11:59 PM

[viscid thistle]

Ah I see

You can

What's wrong with judging

Calculation mistakes are childish

Only kids do that

[ionic hemlock]

can someone explaion this

[obsidian hinge]

damn that is a weirdly tough problem

you'll most likely want to use the fact that arccos(x) + arcsin(x) = pi/2

and then use the angle addition formula for sin

which should all evaluate out nicely

[hexed wraith]

@ionic hemlock
You can also scroll down to MATH HELP channels

[viscid thistle]

You can use the identity $\cos 2a=1-2\sin^2a$ you have $\sin^2$ convert into cos double the angle that will get rid of the half then you will have cos (arccos)

[obsidian monolith]

QuantumMath

[cosmic herald]

A terminology question. What do you call a function `$f : A \rightarrow B \cup { \perp }$ such that for every $x$, if $f(x) \neq \perp$, then $f$ is continuous at $x$

[obsidian monolith]

BlokirKominfo

[viscid thistle]

are you ready for IB HL Math Placement Test?

[peak sinew]

Just apply sin²x=(1-cos2x)/2 bet

Either its correct or I'm too dumb to be left alive idk

Wait "using radicals" What does that mean 💀

Man idk ig my solution is incorrect smh💀

[zinc fulcrum]

never heard of that so idk if it has a name lol but you could say $f_{{x\in A: f(x) \neq \perp}}$ (meaning $f$ restricted to ${x\in A: f(x) \neq \perp}$) is continuous

[obsidian monolith]

💜𝓁𝒶𝓎𝓁𝒶💜

[summer kiln]

Yes

[viscid thistle]

https://cdn.discordapp.com/attachments/1039585772141039637/1040703729852239972/C69BC9BC-BC39-4B90-8AAC-E1FFBBA43EF9.jpg

help

[worldly glacier]

anyone know how to show sin(x)/x is decreasing on [0,1] algebraiclly?

without using derivatives

[zinc fulcrum]

what qualifies as algebraically?

[worldly glacier]

like using trig identities onyl

[zinc fulcrum]

welp

i have no idea then :c

[viscid thistle]

can someone recommend trigonometry text books

that is mainly focused on calculus

Plane Trigonometry By SL Loney

Ig that would work

is it heavy on calculus? because i need that desperately

Not on calculus

It is only on trigonometry

But the concepts are well divided

And I don't think there's any books in the market teaching calculus based trigonometry

All Trigonometry is used in calculus smh

[viscid thistle]

In the first question if you put the values of graph in the function you could form equations to solve for a and b

Like

A+B = 3 when you put (1,0)

B = 4 when you put (0,4)

[viscid thistle]

thats a problem.. the problem is that the precalculus course i took with all the trigonometry didn't prepare me for calculus. i need a book that can teach me more advanced problems with calculus terminology..

thx for the suggestion

Ah I see

[viscid thistle]

Scam

[viscid thistle]

Welcome

[viscid thistle]

but... im taking calculus with analytic geometry, thats why the trigonometry is so problematic

[viscid thistle]

Actually you know what you don't need to get the trigonometry book then, just get a trivial calculus book and practice the questions they will teach you trigonometry eventually

[viscid thistle]

Geometry is interesting

[viscid thistle]

i tried.. im 1 month away from the semester ending and i still dont get trigonometry

[wraith parcel]

hai

[viscid thistle]

no way. im only excited for integrals

[viscid thistle]

That's very strange

Have u practised accordingly

[viscid thistle]

Me too

[viscid thistle]

yuh. i have 100% on the homework through the semester, never skipped a single class, take notes, participate. but the trigonometry is absolute

[wraith parcel]

help conic section

[viscid thistle]

Yeah?

[viscid thistle]

Ah man that's sad

Trigonometry is eating you

of course its sad, i'm trying daily

I think you should watch some YouTube lectures on trigonometry

i just pray they remove it from the curiculum in the future

Wait I have a youtube playlist for you

[viscid thistle]

help

[viscid thistle]

https://youtube.com/playlist?list=PL24P3-cdGKKtIc6XaGYF0zjOXJ-bpnVUX

Try to complete it

Maybe it will help you out

[viscid thistle]

All the three questions?

I told you the first one

[viscid thistle]

this looks good, just from the title. i will take a look thanks 🙂

does he discuss calculus?

Nope

But there's a lot of trigonometry for you

That will make you better fr

[viscid thistle]

Answer to second is B

[viscid thistle]

And for third question, read this https://owlcation.com/stem/Descartes-Rule-of-Signs

It's easy read it up

[wraith parcel]

Plot the object on a cartesian plane and take its measurements (for example: radius and length). Lastly, write down the equation of the conic section in standard form.

tabang

[viscid thistle]

can you show the question please

[calm locust]

Whats Df from this?

[peak sinew]

Df = differentiation?

[calm locust]

yes

|x| / {0} right?

[viscid thistle]

Try using Arihant Trigonometry Book. Search for the full pdf and download it.

[viscid thistle]

$
\begin{aligned}
\frac{\text{d}f}{\text{d}x} &= \frac{\text{d}\sqrt{\frac{1-4x}{-x}}}{\text{d}x} \
&= \frac{\text{d}\sqrt{-\frac1{x}+\frac{4\cancel{x}}{\cancel{x}}}}{\text{d}x} \
&= \frac12\cdot\frac{\text{d}(4-\frac1{x})}{\text{d}x} \
&= \frac12\cdot(0-(-1)\cdot\frac1{x^2}) \
&= \frac1{2x^2} \
\end{aligned}
\ \
\boxed{\frac{\text{d}f}{\text{d}x} = \frac1{2x^2}}
$

[obsidian monolith]

Gi

[calm locust]

No no thats not what I am asking

Thats the question

The question was what values can x have and what values cant x have

Like x!= 0

[viscid thistle]

well the function √x is defined for all x>=0
and it looks like you already got the fraction part.
so just solve for which values of (1-4x)/(-x) are greater than 0

[calm locust]

are greater or equal to zero I think

[viscid thistle]

ya true but you already know it can't be 0 because of the division by 0 in there, so I just left it out

[calm locust]

wait

so it can be lower or greater than 0

only no 0

any guide on this

am I right

[viscid thistle]

what about the region where one one is positive and the other is negative

[calm locust]

wdym?

[viscid thistle]

one sec

hm is it okay if I show you a graph of it?
in my bed and don't feel like working through the algebra on my phone lol

[calm locust]

yeah sure

[viscid thistle]

see how for this region all your -x values are negative, and 1-4x is positive. so when you do 1-4x / -x you'll get a negative value your sqrt function will be undefined

[calm locust]

so it should be from 0 to + infinity?

or?

[viscid thistle]

nah the starting value for like the right side is going to be the value where both are going to be negative

over (0,0.25) blue is positive

(and green is negative)

but afterwards, from (0.25, infinity) blue is negative and green is negative so what's inside the square root is positive

||(-infinity,0) U [0.25, Infinity)||

for harder problems like this, I think what we did in my class was draw it out on a number line. which parts were negative and which parts were positive

[calm locust]

I think x shouldnt be 0

[viscid thistle]

ya

the ) means I'm not including it in my region

also for if you get more problems like this, this is how I learned to solve em. drawing out a little number line then drawing points where a function crosse the x axis

the work looks kinda sloppy my bad lol but on phone

[calm locust]

yeah no problem thanks for helping me

[viscid thistle]

bet no worries

lemme know if there's anything I should explain more on

[viscid thistle]

Lol alright

yeah what did you mean by Df If you didn't want the derivative lol

[calm locust]

its short cut for domain

I mean thats what we use

[viscid thistle]

ohhhh

[calm locust]

yes

Like this

@viscid thistle @viscid thistle

[viscid thistle]

interesting I never seen this notation before

[peak sinew]

Fr

[viscid thistle]

But then you said yes to this too lol

But don't worry all good

[calm locust]

well that was my bad

thanks

[peak sinew]

Help I'm stuck

Nvm that sounded sus 💀

Just help smh

[calm locust]

this is calculus right?

no precalculus

or I am wrong?

[peak sinew]

What's the difference? 💀

[calm locust]

well precalculus is for functions I think

and calculus is for more complex stuff like yours

[peak sinew]

And calculus is just a God level channel

[calm locust]

yes

[peak sinew]

I don't get what da hell is even happening there

Nvm lemme shift that question there then

[calm locust]

How do I draw this graph?

[viscid thistle]

Thanks

[pulsar harness]

How would you solve something that has a log(x) term on one side = to x^2 term
e.g. -x^2 -2x +4 = log2(x+1) without plotting a graph, purely algebraically, please @pulsar harness If you know how.

[viscid thistle]

this is what i did for it

[calm locust]

what are you doing in this part?

[viscid thistle]

finding the oblique asymptote

[hushed sphinx]

Unless you're lucky and there's an integer solution you can guess, the best you can do with such equations in general is approximate a solution numerically.

In this case x=1 is a solution, but I don't think there's a systematic way to discover that other than "try some small integers first".

(A bit of ad-hoc thought shows you need only try integers that make x+1 a power of 2, but can can hardly be called systematic).

[granite ice]

can someone show me how to do number 27

[viscid thistle]

what do you think of this?

like let me know if theres any step here you dont get

[granite ice]

can u explain how u got the 2nd line

and thank you btw

[shell cipher]

Does anyone know how to do this

[static canopy]

do linear function questions belong here or in prealg-and-algebra

[viscid thistle]

Using the fact that sin(2A) = 2 sin(A)cos(A)

That shouldn't be difficult

You already know sin(A)

And cos(A) = √(1-(sinA)^2)

Get the answer

That's it

[viscid thistle]

You could ask anywhere

But it is better to ask it here

[static canopy]

@viscid thistle, can you try #prealg-and-algebra message
i'm stuck on knowing weather my assumptions are correct

**
how to use y-y1=m(x-x1)
to find the linear function of a if we know two points for eg: (-1,1), (1,-1) intercept the slope
ps: thus far, I've gotten only the slope m=-(1/2)

[viscid thistle]

Do you wanna find the equation of the straight line?

By these points (-1,1) , (1,-1)

[static canopy]

Yes

[viscid thistle]

Then what's difficult

You know y-y1 = m(x-x1)

Take y1 = 1, x1 = -1

And let me check the slope

Which is obviously y2-y1/x2-x1

[viscid thistle]

You know what, the slope is incorrect

It should be m = -1

number 7 not sure why theres an intersection at the polw

i did wrong math for first one and got right answer

other two I understand

[static canopy]

re-did it, you're correct it's -1

so that part required two points to find the slope, then to find linear function equation
it requires only one point

[viscid thistle]

Yeah exactly

[latent gate]

does anyone have any good textbooks/worksheets for practicing laws of logarithms and exponents? All the stuff that I have found seems to be very simple, so if anyone has any challenging questions please ping me @latent gate

[viscid thistle]

@latent gate https://cat.wordpandit.com/cat-logarithms-questions/amp/

Maybe try these

[latent gate]

Alright, Ill give it a try. Thanks!

[viscid thistle]

And it's not worth spending a lot of time on these

Exponents and Logarithms

You grab the concept and keep moving ahead

[static canopy]

to find equation of function pendicular to y=3x-4 that intercepts (1,1)
using Ax+Bx=C and rewriting y=3x-4 to -3x+y=-4 I got: m=3

was the approach for slope correct, and now use y-y1=m(x-x1) with the intercepted point?

[shell cipher]

Thanx

[static canopy]

question asked for linear equation, i got y=1 as final

description: pendicular to y=3x-4, intercepts (1,1)

unrelated question (perndicular to type), from y-2x-4=0 can we assume A=1, B=2 and C=4

[shell cipher]

Ik that I have to use sin(2A)=2sin(A)cos(A) but idk what to use next. Can anyone help.

[static canopy]

and replace Ax+By=C accordingly?

to find slope

1.) Pendicular to y-2x-4=0, intercepts (1,1)
answer: y=2x-1

[viscid thistle]

Just remember (sinA)^2 + (cosA)^2 = 1

And if you can get sin and cos as they are interelated through this equation

[viscid thistle]

Correct

And now get the slope of the perpendicular line by using m1.m2 = -1

That will be -1/3

And you know a point that is (1,1)

[viscid thistle]

Now just use this

[static canopy]

if instead of pendicular it's normal

does slope become 3?

[viscid thistle]

That's the different word

But same meaning

No yes m = 3 ofc

The slope of the perpendicular line is -1/3 then

[static canopy]

if it intercepts (2,0) it results in y=3x-6

[viscid thistle]

I don't understand what your question is

[viscid thistle]

If you have calculated it maybe it's right

Or maybe not

[static canopy]

I think i got the concept down 🙂

horizontal lines have slope 0, do vertical ones have a similar rule?

i found on google undefined, which sounded right but not useful

i think it's 0 also?

[viscid thistle]

Actually slope is equal to tan(angle between the positive X axis and the line)

And for the perpendicular line (vertical) it's 90 degrees with the Axis

And if you know tan(90°) is undefined

And once you learn derivatives you will understand it better

Leave it for now

[static canopy]

https://media.tenor.com/eRn4LABoUlwAAAAM/back-to-the-future-taking-notes.gif

I got a question that says: passes through (1,-1) and y-point intercepts on 3
should I assume (0,3)?

yes omg

dumb question

[hazy hollow]

log(8k), expand using the product rule

[viscid thistle]

can you factor out something from: -x(2x-1)^-3/2 + (2x-1)^-1/2 ?

[static canopy]

can 2x(sq)-4x+2=0 be refactored further than
2(x(sq)-2x+2)=0

sq = squared

[opal tree]

log(ab) = log(a) + log(b)

[opal tree]

You can factor out (2x-1)^(-1/2)

,w graph 2x^2 - 4x + 2

Oh shit I'm dumb

[opal tree]

Yeah

x^2 - 2x + 2 is gonna be one of the most common perfect square trinomials

[shell cipher]

Does anyone know how to solve this question.

[sour abyss]

u need to use a double angle formula

[sour abyss]

i got -7/25 but i might be wrong

[shell cipher]

That’s right I think

[sour abyss]

i dont have a lot of practice with usng double angle formulas

im only in year 9

[shell cipher]

I’m way older and rly bad at this tbh.

Bc I wrote this down but ik it’s definitely wrong

[sour abyss]

because of one of the double angle formulas cos2θ=1-2sin²θ

[shell cipher]

So did you just plus in the value of sin into cos2(A)=1-2sin^2(A)

*plug

[sour abyss]

damn, my writing is messy

[slate crane]

try solving for theta first

[sour abyss]

wouldnt it be easier to just go straight to cos2θ if there is a formula that basically lets u do that

[shell cipher]

No worries, thnx a lot

[slate crane]

or just square sin(theta) and plug into the identity formula.. it would get you -7/25

it’s what you have

[viscid thistle]

Did you know that in Canada precalculus is called Advanced Functions

[sour abyss]

til

[median hare]

use the standard triangle

u'd get it

[viscid thistle]

Ofc course, you could do this

I am just simplifying things so that he doesn't get confused

You could learn hunderd formulae in trigonometry

[sour abyss]

ic

i prefer memorising hundred of formulas, its easier for me

[viscid thistle]

But (sinA)^2 + (cosA)^2 = 1 is universal

[viscid thistle]

Wrong approach

You're blocking your mind to think

[sour abyss]

i did a test and i did the working out all correct but the sentence i wrote was wrong, i suck at anything that doesnt include just memorisation and practice

[viscid thistle]

I can understand

Some people are more better at memorizing rather than analytical skills

I kinda hate memorizing

It's better to just think every question with basics

[viscid thistle]

And you're still in 9 so ofc you will only encounter formula based questions

[gusty fable]

[analog meadow]

hey guys quick question, how do I know the principal axis of a hyperbola? is it just the bigger number?

[sage moon]

How can this equation be vizualized

[summer ruin]

which one

[static canopy]

-x2+2x+2

[sage moon]

I solved it by root of 3 approximately to 1.732

[viscid thistle]

can you find the roots using the rational root theorem without trying and failing

like i want it to all be right

im sure there is one out there since apparently math can explain the nature

so why cant it be like that with roots

[viscid thistle]

I have a logs quiz on precalc on monday

we use this book: http://patemath.weebly.com/uploads/5/2/5/8/52589185/precalculusbook.pdf

some sullivan 9th edition

[signal torrent]

Could someone help walk me through the process this person used to solve this question?

i dont really understand how they got pi/2 - theta = 4pi/9

Or why they needed to show that 1/cos4pi/9 is equal to a

[summer ruin]

that doesn't seem at all related to the question

[summer ruin]

because that's the transformation of the original equation

you could do most of these steps in your head, but does it really matter anyway

[signal torrent]

ah i see thanks

[sour abyss]

Get rid of the square root by turning it into a surd

U can factorise the bottom

Acctually, i probably git it wrong

[peak sinew]

Lol its logarithms

[peak sinew]

Use properties of logs

Got this pic in this server itself

Use them

[sour abyss]

I can see

The answer i got was like log e( (2x +1)/(x-3)(x+3)

But that is definitely wrong

Oh wait i know what i did weong

[sour abyss]

U cant

After some thinking

Because u double the angle

After h find it

So u still end up doing another double angle formula

[vapid plaza]

cos^2 theta = 0 does not lead to solutions

[viscid thistle]

Is the Square root of a/b the same as rad a / rad b

[slim steppe]

Yes

[viscid thistle]

thank you

[hazy hollow]

It says this is wrong
idk why
common base since none is given is 10 right?
and its formated where inside the parenthesis is the subscript thats how the teacher said to input them
i tried k inside parenthesis also

[cunning parcel]

Try log(8) + log(k)?

[hazy hollow]

didnt work

[cunning parcel]

Idk then, sorry

[hazy hollow]

thx anyway

[summer ruin]

common base since none is given is 10 right?
no, it depends on the notation, some people use log to exclusively mean base 10 logairthm, some use log to mean natural log and in some (more advanced) cases it can be any base log

[summer ruin]

yes or you could just set the calculator to radians

[acoustic gale]

I got undefined when I did that to get my answer in radians.

[summer ruin]

you input the 2.82 into arccos?

either you're doing something wrong or your calculator doesn't understand it

[peak sinew]

Ig the ans is

$$
\left[\frac{1}{2} \ln (x-1)+\ln (2 x+1)\right]-[\ln (x+3)+\ln (x-3)]
$$

[obsidian monolith]

anand

[peak sinew]

Try log base e

[sour abyss]

woudlnt log e x-3 be on the bottom, not tp?

[peak sinew]

Wydm?

[sour abyss]

oh wait nvm, there are brackets

i see it now

i couldnt figure it out before ebcause i didnt see that bracket, which caused me some issues

[peak sinew]

Lol

[static canopy]

lost at these type of questions
"nese" = "if"
"gjeni" = "find"

solving when degree is known is a breeze

[gentle sedge]

albanian 👀

[gentle sedge]

ctg is cot?

[summer ruin]

you know that alpha is between 0 and pi/2

at what angle is sine(alpha) = 1/2 ?

solving this you'll figure out alpha

then just plug it in into other functions

[gusty fable]

maybe

[static canopy]

yes 😊

[static canopy]

cotangent

[static canopy]

so between 0 degrees to 90 degree?

[summer ruin]

yes

[static canopy]

sine of 90 degree is equal to 1?

cos is 0

and then continue based of those for other functions?

[summer ruin]

that's not really relevant to the question

sin(alpha) = 1/2

then what's alpha if alpha is between 0 and pi/2?

[static canopy]

?

or sin(30*)=1/2

and cos(30*)=sq(3)/2

[summer ruin]

sine is on the vertical axis

so the answer is alpha = pi/6

[strong plover]

does differentiation count as precalc? sorry im not from the us

[static canopy]

ahh, so it's pi/6, and sin(pi/6) equals 1/2

and it's on 30 degrees

cos(30 degree) equals sq(3)/2

[peak sinew]

Yup

[static canopy]

is this correct?

[peak sinew]

What's cty ?

[static canopy]

ctg

[peak sinew]

Or ctg

[static canopy]

cotangent

[peak sinew]

Yeah what's that?

DA HELL

[static canopy]

yes

[peak sinew]

OH

Cot

You guys call cot ctg?

💀

Nvm I'm no one to judge

But how did you get ctg?

[static canopy]

1/tangent

[peak sinew]

Yeah so why do I see an addition sign there?

[static canopy]

it's an asterisk

we use that for multiplication

it looks like asterisk, it needs to be a dot

[peak sinew]

Understood

But how are u multiplying 1/1 with 1/root3 to get root3?

Ah wait nvm

[static canopy]

1/2*

yes

[peak sinew]

Just write it directly next time

💀

[peak sinew]

This is part 1 ryt

Do the other ones as well

[static canopy]

so its correct?