#help-0
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ay
Ima wait until someone asks me what I want
okay
What I currently need is double checking my answers to algebraic questions
I'm gonna just download lightshot to show them here
The first three need to be simplified, 4th must have the root evaluated.
okie where’s ur work
1 sec
ello
how do you take so crisp quality img
I used my pc camera
quality is super crisp
lmfaooooo
i’m literally die hahahahha
is ur ans to 2 $(256)^{\frac{1}{8}}$
swaggofishballs
it's 256 powered to 1/8
simplify
It says I need to answer with a single exponent
oh
256 is a power of 2
ya u can still simplify
Write
Write with a single exponent
swedish?
its Swedish
oooo
yes I can answer it with 2
I can also do that, one moment
done
Now let me show my 4th answer and solution, caught in 4k
finding the x
okie
I like being precise when it comes to solving
I need to get an undergrad-level problem solving skills in order to attend undergrad courses next year
It didn't take me long to do it
Because I'm simplifying the equations
presentation error
why is my name green
but ya ure dealing with equations
don’t need to have that = sign
when ure simplifying expressions
then ya
If you look at the point where I eliminate the 3's, I don't put the equal signs because it would be incorrect to do that
^^
an example
im just saying this to be safe HAHAH bec I’ve been penalised for it before 😭
In these cases, I'd definitely not write with equals signs
yepp! that’s gd
though, if I'm simplifying only one side of the equation, it should be correct when it comes to the first steps that I did
Tomorrow I have my first math exam, I don't know what I can do further to improve
ya the steps u did are fine (:
just the = sign
u seem to have the knowledge so ya
all the best ya

thanks
@wanton tusk GREEN
GREEN
@neon berry Has your question been resolved?
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I know that with the power rule, the derivative of this function should be df/dx = -1x^-2
But how do we get there algebraically?
limit definition
first principals? f'(x) = lim h -> 0, [f(x + h) - f(x)]/h
use tex
Alright but then you are left with 1/x - 1/x
No
wut?
He says it over here
if you directly substitute h as 0 you get something divided by 0
Well you can't let it run to 0 immediately. You would divide by 0
you have to "remove" the denominator before applying your limit
so you have to do other work first
simplify
Alright well if I try to then I end up with (x/x(x+dx) - x+dx/x(x+dx))/dx
Is this the moment to let dx -> 0?
Then it would be x/x^2 - x/x^2
Still not possible I suppose
You would still divide by 0
Right so multiply the numerator by dx first?
But then still I would run into the same problem
Oh wait you can just combine the fractions
so -dx / x(x + dx)?
How do I cancel the dxs?
You forgot a dx at the bottom
Well you have a dx in the numerator and in the denominator. Those cancel
also I thought you could only leave out dx^2 because those are negligible not dx
right but in which numerator and denominator exactly? because I got 2 of both xD
I have the full process written in latex
do you want me to send it or you want to continue yourself?
@quartz cave
$$\frac{\frac{1}{x + h} - \frac{1}{x}}{h} = \frac{x}{xh(x + h)} - \frac{x + h}{xh(x + h)} = \frac{-h}{xh(x + h)} = \frac{-1}{x(x + h)}$$
$$\lim_{h \to 0} \frac{\frac{1}{x + h} - \frac{1}{x}}{h} = \lim_{h \to 0} \frac{-1}{x(x + h)} = \frac{-1}{x(x + 0)} = \frac{-1}{x^2}$$
Hmm let me see
RedstonePlayz09
Alright
So how does -h turn into -1?
Oh I see
it cancels out
So this was not correct
It should be
(x/xdx(x+dx) - x+dx/xdx(x+dx))/dx instead of (x/x(x+dx) - x+dx/x(x+dx))/dx correct?
Oh no I see
In my step I leave the over dx
But you add it to the denominator in one go do you not?
In the first step, I take the h in the denominator and distribute it to the denominators of both fractions, while also making a common denominator.
Right exactly so you do 2 steps at once correct?
Multiplying by the LCD and distributing the h
Because then I follow allong what is happening
$$\frac{\frac{1}{x + h} - \frac{1}{x}}{h} = \frac{1}{h(x + h)} - \frac{1}{xh} = \frac{x}{xh(x + h)} - \frac{x + h}{xh(x + h)}$$
RedstonePlayz09
@quartz cave Has your question been resolved?
Right, perfect thank you
Now I am onto the derivative of sqrt(x)
Good luck!
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hi, so i wanted to ask that what works interms of solving quadratic equations, for middle term do you use the lcm or another method that is other than taking lcm
im currently learning the lcm one and ive entirely forgotten about another one that was without taking the lcm one
and i find taking lcm a little harder
for solving quadratic equations specificalli
thanks
lcm u mean prime factorization?
if u find factoring tedious
just use quadratic formula
or just complete the square
thats if u dont like the guessing part
this one
i wanna know which one is more preferred
methods of factorizing a quadratic?
it just depends person to person
sometimes its kinda hard to find two numbers which add up to something
lcm one
and there product is something
oh
so if u cannot do it in case
u can always use quadratic formula
there's no LCM method for factorizing a quadratic
thats the last thing u wanna use
i think he means middle term split
she haha
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help :')) ive been trying to solve this but i keep erasing my answers because they might be wrong
@alpine sable Has your question been resolved?
show everything you have tried
i deleted them- but im rlly only struggling with finding the ratio
@alpine sable Has your question been resolved?
.close
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How do you solve these kind of problems?
Divide by numerator and denominator by 4^x (for the first limit)
Then you should be able to apply limit arithmetic.
@wispy urchin Has your question been resolved?
Thank you!
.close
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Hey, I need to prove that 1-1/2+1/3-1/4+.... = ln(2) but I'm not sure where I am wrong in this
yea its true
but if you want to switch sum and integral, you need to prove why its possible

Unless you are an engineer or physicist
lol
@lilac silo Has your question been resolved?
<@&286206848099549185>
i mean
your integral looks like its wrong
g(0) = 1
and now you're missing both the constant term and the x term
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how can I tell how many roots this equastion has?
uhh 
well 8x^2 is dominating term so for very large values it will just shoot off to infinity, at least thats what i believe
have you checked on desmos?
Well, f(0) = -1
and this function is continuous
so you have a positive root
and since this is an even function
you also have a negative root
I'm not sure how to prove these are all of the roots though
Any inequality or equation with this will be very hard
Well, if there are anymore roots, there must be at least 4.
if we can show its derivative is positive for x> 0 and even function is then ig we should be good no?
Yeah
you mean odd function?
Proving that f''(x) is positive will also work
,w plot 8x^2 -ln(1+x^2) -cos(2x)
i mean f(x)
ah
f(x) is even right
for some reason i saw the ln and was like nah that's not even
well even is easy to show
this >0 for x>0 maybe hard
i think second derivative would be useful here
If we can show that f''(x) > 0 it would go like this:
Assume that there are more than 2 real roots, meaning there are at least 4 (since f is even).
Between the 2 positive roots and the 2 negative roots, we have a point where f'(x) = 0.
So in between those 2 points, there is a point where f''(x) = 0.
But f''(x) > 0, contradiction.
hmm
$\frac{2(8x^4+17x^2+7)}{(1+x^2)^2}+4cos(2x)$
monikanicity
We need to show that this left fraction is bigger than 4.
Then this whole thing will be positive.
@broken beacon Has your question been resolved?
the guy better not go afk
I think he already did
yeh I'm here guys
use rouche's theorem 
8x^2 dominates -ln(1+x^2) - cos(2x) on the boundary of the unit disc
so they have the same number of zeroes counting multiplicity

It's way too complicated for the test this question from HAHA
Maybe you are right about solving this problem with the theorem but it was not thought in the course this question from
monikanicity
Isn't this obviously >4
ok what the heck
my next stop after linear algebra is complex analysis
or maybe I'll finish calculus first
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use sin and cos formulas
ok thanks ✅
What to do here??
idk but my teachers explaining it now
x=y
idk what to do after i find out my cos and sin
@meager lotus Has your question been resolved?
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Need help with line integrals. tried different yt vids. not one gives proper explanation about parameterisation or etc etc
@upper grail Has your question been resolved?
@upper grail Has your question been resolved?
@upper grail Has your question been resolved?
@upper grail Has your question been resolved?
which ones have you tried so far?
I need help with my tenth grade homework on Linear equations
@upper grail Has your question been resolved?
Mostly uni questions and some examples found on internet. Still stuck on how to do parameterization
@upper grail Has your question been resolved?
What exactly do you mean with "how to do parametrization" ?
It's not always obvious how to parametrize a given curve if that's what you are asking
(Depends on how you are given that curve of course)
how would u parameterise a curve?
what type of curve
if its just a straight line u can just use
given it starts at x1 and ends at x2
x1(1-t)+x2t
for t in 0,1
no its not a straight line, also am not understanding what u did there
Well it depends on the curve
...
You can parametrize a line by x1+t(x2-x1), where x1 is your startpoint, x2 is the endpoint and as t runs from 0 to 1 you run between them. What they wrote is just this but rearranged
Ahh got it now
Thanks alot both of u @ornate condor @mortal trellis
hm
i still find my way of writing more intuitive
np harri

imma just close dis too
.close
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Yeah it's interesting what people see as more intuitive. I personally like my way a lot more
Don't just close a channel lol
✅
wot oki
hello, this is preoccupied channel, you can read on how to make ur own channel in #❓how-to-get-help
So I need to Ask to this people? is it? I thought Help-0 need to start
currently help-0 is occupied, you can choose one from available help section
the ones you’ve circled
ok, Thanks👍

.close
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Initially I thought the question hadn't been answered since the last message wasn't sent by op, then I read above and closed it
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.open
This is what im trying to solve, i knew how to solve it a year ago but now i have forgotten everything. I know that there's a short way. I don't want to just calculate the values of everything.
Factor out powers of 3
factor 3^2
like cross them out?
No, put them out of brackets

3^2 (3^3 + 1) on top and 3^2 (3^2 + 1) on bottom
and then i cross out 3² right?
Yes
and then i just calculate the rest?
Yeah
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in a chessboard of nXn, we define that p(i,j) is the placement of a square in that chessboard where i = 1,2,3,4 ... n and j=1,2,3,4 ... n
why in the second picture it says that two squares given that the first square is in p(i,j) and second square is in p(i',j') are diagonal, i+i' = j+j'
any value of p(i,j) all it diagonals can be found by adding the same number to i and j
it should be like this p(i+n, j+n) where n is an integer
I get it from logic and proofs book
specifically the application of satisfiability
when we add them we find that
2i+n = 2j+n
meaning for this to be equal i and j must have the same value
and that's not the case for different values of i and j
Could someone explain please
@hard remnant Has your question been resolved?
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Hi everyone
hi
I'll send the question just gimme a sec
First of all let A be a subset of N
If A is finite then a part of the theorem is already proved
Now if A is infinite then I have to prove |A|=|N|
Now first of all if I consider the function g:A→N where g(x)=x+1
Assume that g(a)=g(b) for some a,b included in A
Then a+1=b+1 which means a=b
So this function is injective
why +1? you can just consider the inclusion map from A into N w/o that complication
well it's not fatal
just a bit odd
||i assume now you will want an injection the other way around to apply csb, which is where the least-element principle comes in||
Now let x=y-1 where x belongs to A and y belongs to N then g(y-1)=y-1+1=y then this function is also surjective
Then it is bijective and thus |A|=|N|
wait can you state more clearly your function N -> A
your g is not surjective unless A happens to be all of N !
That's what someone told me but the theorem states that for every infinite set A in N |A|=|N|
How should I do this
Hey guys am I doing this right?
try to think about the least element principle
Excuse me but I didn't understand what you mean here
@earnest topaz channel busy please move
g is not surjective even if A = N because of the +1 
don't worry about it for now.
What's the least element principle
It says that for every non empty set there is a least element that set right
But what does that have to do with this
Hey yea, just saw the format of this sever has changed, been away for a while
Thanks mate imma catch on
least element principle will help you create the function from N -> A
Let me think about it for a moment
Is the function f:N→A where f(x)=x+c such that x for x in N and a constant c such that c starts from 0 and for c=0, f(x) is the least element in A a surjection ?
no you're kind of overdoing it...
let A be your subset of N.
it has a least element, call it a_1.
A \ {a_1} is also a subset of N, and also has a least element. call it a_2.
there is a way to describe in words the relationship of a_2 to the original set A.
if you want to think on your own from here that's fine, but do still answer my question here.
the least element principle has everything to do with this, by the way.
So your question here is to find a relation between the element a_2 and the original set A ?
of course it belongs to A, but that isn't what i'm looking for
I don't think it's the description you want bc it's obvious that it belongs to A
in the set {4, 20, 69, 533, 895475, ...}
4 is the smallest element
20 is the ______
2nd smallest one
I thought of that but I didn't think that this Is what you want
you can similarly construct the third smallest, and the fourth smallest, and so on.
Yea I can do this continuously since A is infinite
in fact this alone already gives you a bijection between A and N, since every element of A will be numbered in this way eventually!
and injectivity is obvious as each n'th smallest element is obviously different from any below it, by construction.
And each element in N won't have more than 1 a that gives it as an output since each a is greater than the one before it
So it is also surjective
Then it's bijective and thus |A|=|N|
yes
But I just still have to write the expression of the function
nyeh
Which is f(a)=a_n
i think writing f: N -> A, f(n) = n'th smallest element of A
will be just fine
and for the inverse you could send a in A to the number of elements in A that are less than or equal to a, if you so desired
But with the condition that n= c+1 for some c in N is this correct
Ok but how to write nth smallest element in A symbolically (instead of verbally )
are you penalized for non symbolic expressions or just allergic to them personally?
No I don't have problems with them but its good to know how to use the symbols
i don't think it's worthwhile to worry about a symbolic expression for f, honestly.
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Hello i dont know how to start on answering this question
Show that a^4 + a^2 + 1 = (a^2 - a + 1) (a^2 + a + 1)
Expand the RHS
II think you can turn (a^2 - a + 1) = (a+1)^2
First figure out the prime factors of 111
No
Still wrong
still no
that is wrong never mind
Alright thank you i think i got this now :)
.close
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Hi can anyone give a hint for my problem?
we have a sequence of $n>=6$ numbers, first 3 numbers arbitrary, others follow $a_{i+3}=2a_{i+2}-a_{i+1}+a_{i}$, we know $a_2=-2, a_3=6, a_{n-2}=2022,a_n=6226$, find the sum of the sequence
AlterNet
@versed island Has your question been resolved?
$a_{i+3}+a_{i+4}=2a_{i+3}+a_{i+2}+a_i$, so {$\sum_{i=1}^m a_i=-a_{m-2}-2a_2-a_1+2\sum_{i=1}^{m-1}a_i$}
Toby
(that should be correct now)
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Okay here's a problem I've been trying to solve:
A cucumber weighs 500g. It's 96% water. It dries. Now it's 92% water. How much weighs the cucumber?
tries?
oh it must be dries*
Maybe “dries”
because of the lack of information of what it's 92% off I assume it just lacks water after drying
yes.
let the dried amount be x
initial water=480, final water=480-x, final weight=500-x, final water is also=0.92(500-x)=480-x
on solving- 0.92(500-x)=480-x
u will get ur ans
wdym how much weighs the cucumber? isnt it already given
How much weighs it after drying I meant
ok
x=250, so final weight=250
Okay where was my reasoning wrong then
Cause I assumed 0.04 * 500 of non-water was kept
And that the WATER decreased so we get 0.92 * 500
Now it's 92% water
the word now
yea cuz some of its weight dried
@proud igloo Has your question been resolved?
Okay, I didn't look at your calculation, but with a slow thought process and your clue you gave me right there I managed to get 250.
Yeah, thank you
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this is continius when x=pi/3 ((tan(x)-tan(((π)/(3))))/(x-((π)/(3)))
right?
dude the number of brackets
that's a lot of brackets...
this right
hm
yea
np
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shank's is a villan?
maybe
potential
like 80%
yh
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Are you allowed to use the word “greater” when describing distribution?
@gilded vale Has your question been resolved?
So how can I describe distribution?
Like there is an equal/unequal distribution?
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How do you solve this?
@untold panther Has your question been resolved?
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i dont get which one would be uncalculatable
since each of them are rewrites of each other
btw this is a practice test
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help
I am confused because I am trying to solve this problem: 3x-4=12
How do I solve it?
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I'm not sure where to progress on this
well you can start by moving the -2 to the outside
and remembering that the derivative of cot(x) is -csc^2(x)
and derivative of cscx is -cot(x)*csc(x)
then you should just be able to use the chain rule like normal
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thank you
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Hello
Currently struggling to solve the separable equation for y
My work is posted above
Thank you in advance.
Im stuck on which one to choose can someone help the answer is 61
@sudden herald Hey, I am currently occupying this channel. Do you mind reposting in a free help channel?
sorry i dont know how to use this
Fix this:
Do you mind elaborating for me?
That is actually just a typo.
That's what I mean. To fix the typo.
Let me correct.
Will do, but my question still stands on the final isolation of y.
A integral calculator returns this for the integral of e^y - y e^(-y).
Right, but that is the Maxima calculation
the "hand" method provides what I have
through integration by parts
To be fair, you did never say what you performed integration by parts on.
A lot has been left to the reader to try and understand what you've done on each line.
$\int (e^y - y e^{-y})\dd{y} = 2\int \sin x \dd{x}$
stabulo
From using a integration calculator, it would seem you should end up with:
$e^y + (y + 1)e^{-y} = -2\cos x + c$.
stabulo
Alternatively, $e^{-y}(e^{2y} + y + 1) = -2\cos x + c$.
stabulo
Hmm, a question similar to this is going to be on my test tomorrow and I don't see how I can derive -2cosx + c from the integral
Sine integrates to a negative. That's why it's -2cos(x).
Correct, rather, I don't understand how $2\int \sin x \dd{x} = \int (e^y - y e^{-y})\dd{y}$
Thasis
Okay I'm sorry I understand now
I just disagree with your results for the integrations.
Thasis
Implicit functions can be harder to solve for y than solving the original equation.
Right, but given the initial condition y(0) = 2; does one not have to find y to then find the value of c in order to get the exact solution to the DE?
No. You can just replace y with 2 and x with 0 in the implicit solution to determine c.
Yeah I suppose that makes sense. I don't particularly know why I thought otherwise
🙂
Therefore, $y(0)=2 \longrightarrow e^{-2}(e^{4}+3)=2cos(0)+c \Rightarrow e^{2}+3e^{-2}=c$
?
Thasis
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Need help with this
Its dimensional analysis
I can turn Klygenes into Widgets no problem but idk how to turn Klygenes/Seilas into Widgets/Dueseldorfs
These are just made up units i need to convert
so it'd be like
the answer for klygenes -> widgets divided by seilas -> dueseldorfs?
so for example if kly -> widgets is 10 and seilas -> duese is 5
i can just do 10/5?
im confused on that
okay ill try it out and show you
yea im stumped
what is the name of what you just did?
when i look for dimensional analysis videos i just get the regular one
not the one where its something/something
okay, i'll watch that then try it out and if im still confused ill show you what i got
oh shit
this shit is actually easy
thanks for the resource ill try out the problem now
I got this
so just assuming that the conversions I put there are correct did I do it right?
Ah shit
Thanks a lot
Physics
My teacher didn’t teach us DA
He taught us sigfigs for like 1 day and then 1 day they did an assignment without a lesson on DA but I wasn’t here and then boom had to take a quiz
So I’m tryna learn it rn before I retake it
he could’ve just put normal conversion units but he wants to make his class fail lmao
I already have 3 F’s worth 70% of my grade because I was in a group and they didn’t do work
So imma good it gets better
I hope so
Thanks a lot for the help
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can someone help with #1?
i got the identity matrix and the row operations,
not sure what to do nex
t
You need to do the same operations to A to find the inverse of A if you're looking for it
after i find the invers
e
do i do those same operations
on the 3x3 identiy matrix
@alpine sable
to get the elementary matrices
or are they obtained another way
Idk, ask helpers after 15m
alright thank you
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is it possible for me to get your help on 2.1?
Yeah sure thing!
.reopen
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if we assume A to be noninvertible
that would mean that the column vectors are independent
then the solutions are not all identically 0?
which could contradict
@flint tree whenever you get time please let me know, thank you sm
okay, dont need help
FTIM proves for me
thank you guys
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for 10a I would think that it would just be $$ \frac{1}{7!} = \frac{1}{5040} $$
sytez
you have to remember that there are 52 cards in a deck
why would that matter tho is my question?
the order of the 7 cards dealt shouldn't be affected by the number of total cards
there's 7! permutations of the cards and only 1 is in ascending order no?
one sec
alr
it should be 52P7 though
because you are picking 7 cards from 52
textbook answer
oh I guess that is equal to 1/7! lol
,wolf 1/7!
i have no idea what the actual answer is
the textbook doesn't match up with either of what i thought
it wouldnt be 1/52P7
it is 1/7!
oh i just realized
answer key thru me off because it started with a 4
i see
thanks
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forgot to ask how to do b
Hiii can I say what I need help with now?
sytez
52^7 being the total number of possiblities
it wouldnt be 52^7, as each time you hand out a card, the total stack decreases by 1
im thinking it might be 13P7 because there's 13 denominations and they can only be used once
I figured out one way
how?
dont always have to use ncrs and pcrs, could always just find the probability of each single draw
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Hello!
@lunar girder Has your question been resolved?
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Cool question, thanks for sharing
I need help
theres 2 of each side for the box
calculate the given 3 sides
it should be twice of that
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How do you solve this ? I know how to do it if there's only one || on the left but not if there's 2
do a sign chart analysis for separate cases to remove the absolute values.
What's a sign chart analysis?
on the real line, find where each of the individual absolute values change signs
they're different for each so you'll have 3 combinations
for example for x > 2, you know both terms are positive
so you just solve for (x-2) + (2x+3) = 6
do the same for x<-1/2 and -1/2 < x < 2
by combinations i mean (x-2) - (2x+3) or -(x-2) + (2x+3) or -(x-2) - (2x+3)
should only be 3 unless i made a mistake
I still don't understand, can you please draw it for me?
no
draw it yourself and share it here
something like that
but instead of - + -, you replace it with (x-2) or -(x-2)
similarly for the 2x+3 term
Ah so that's what you meant by real line
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Hey guys
I think?
yea ig, idk if i don’t see what you’ve done
is this right
yes
well you’re skipping some terms but ig yea
and now write numerator in form of (1-sin^2)
helloo
oh hi
are you trying to show the left hand side?
yes
$\frac{sinC - sin(C) cos^{2}(C)}{cos^{2}C}$
you ate some terms
duckiescute!
also 
I think it should be that
oh
then you factor out sinC
oh yeah I saw my mistake

I turned 1-sin(C) into cos^2(C)
oh how did you get 1-sin(C)
I think I ended up adding it instead 💀
ohh hahaha
wait where'd cos^2(C) come from
making the denominator the same
okayy
okay so left hand side
changed into sin and cos
should I make that -sin(C)/1 actually
or
hm
nah
okay
yes
what denominator does it have
right
uhh
the one with -
oh 1
we want it to have the same denominator as the one on the left
what’s the denominator
for the left
cos(C)
cos^2(C)
oh huh
cosC x cosC = cos^2 C
do I multiply it already
yes
why multiplied?
duckiescute!
duckiescute!
yeah
so to make common denominator
oh
we multiply it’s numerator and denominator by 4
so we multiply both the numerator and denominator
what do we multiply the numerator and denominator by
by cos^2(C)
YES

so this?



