#help-28
1 messages · Page 257 of 1
Yes, if a point lies on a curve it means that its coordinates must satisfy the equation of the curve.
Wdym?
Yh is it acceptable to do that or do I need to eliminate m
this room is occupied by @lunar grove
And then rearrange to form the equation
So how to make my own channel?
You just have to show that this equality holds.
The markscheme derived the equation ig?
By eliminating m
Ig that's just a way to go about it eh?
Yes, it's possible, because we're supposed to prove it if m varies.
So it doesn't depend on m.
Also got a qucik question and ts kinda random and basic
But i want to ask
We got a locus
In parametric form
So like a circle
Simple example
X=costheta y=sin theta
And then dy/dx can be found by doing dy/dtheta x dtheta/dx
Ik the algebra works completely
What's like the concept tho like intuitively
Ofc doing it directly makes sense to me where u differintiate eq of a circle
And sub in those values
It's more useful dealing with implicit functions
I mean, when you can't directly solve for y in terms of x (or whatever the variable is)
And what you said is called chain rule
Yh yh ik it's chain rule haha but idk was kinda weird to think abt
But sure ig the algebra works fine
Yeah, notice that dy/dtheta x dtheta/dx is just dy/dx in fact
Yh yh ik ik
There's a 3 blue one brown video explaining the intuition behind the chain rule
You could probably apply a modified version of that intuition to your specific case
A visual explanation of what the chain rule and product rule are, and why they are true.
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The video is introductory level but I loved watching it personally anyway
9 minutes in is chain rule
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is 61.5 correct
im hvin doubts
Yeah
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help 💔 🥀
ping me when someone's here
No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/
OHH okay thanks, lemme post it rq , it is a RLLY dumb problem btw so just don't judge 😔
√200y^4
lemme send my steps rq
$$\sqrt{200y^4}$$
@ember solstice
Is that the question?
√10^3 * y^4
√10^2 * 10 * y^2 * y^2
√10 * 10 * y * y
i finally got:
10y√10y
and ik its wrong since chatgpt said so but i CAN'T figure out what is wrong
How did you change $200 \to 10^3$?
@ember solstice
Also, can you write your work on a paper? It's hard to understand it in text.
Also, your simplification of $\sqrt{y^2 \times y^2}$ seems off.
@ember solstice
ohh
what's wrong about it?
so that i can fix it
You changed $\sqrt{y^2 \cdot y^2} \implies \sqrt{y} \cdot y$. \
But both $y$s should have come out of the square root.
@ember solstice
oh yeah, since square root of y^2 is y right?
This is not proper simplification of 200.
chat gpt told me that i had to change anything squared to the base thing
so i started oding that w equations
wai
10 x 10 x 2
is that better
That happens when take the base thing out of the square root. But you didn't do that here. Does that make sense?
so if i take the y outside the square root, THEN i have to just remove the exponent?
wai but for previous problems i did, it were different
let me look for one of them really quick
So far for this problem
It’s like that
And usually the numbers outside the square root are squared
Or cubed
Or quadrupled I think that’s what we call smth with ^4
Here, some of the ys should have come out of the square root.
so in here, wai wha do i do next
teh problem, so far it's √10 * 10 * 2
$$\sqrt{10 \times 10 \times 2}$$
Can you rewrite $10\times10$ to have an exponent?
@ember solstice
Now I'm sure you can simplify $\sqrt{10^2\times2}$.
@ember solstice
Smth like this 😭
waitt, how do we know whether or not we're going to take something out of the square root or not
You no longer need to factor this because you can directly use the exponent to help bring it out of the square root.
It must have ...^2
OHH OKAYY
So can't we bring the 10 square, and the two y squares out of hte square root?
that leaves two by itself then
is that the answer
Once you take the ...^2 out of the square root, you should drop the exponent.
You should only drop the ^2 exponents.
In general, if you take something out of the square root, its exponent should be divided by 2.
Ohh okay
yes
okay so overall, from what i did in the whole question wai i'll write it all down in order
Accidentally added square to 10
At the end
is it right
what is this
$\sqrt{200y^4} \
\sqrt{2 \cdot 10^2 \cdot y^2 \cdot y^2} \
10 \cdot y \cdot y \sqrt{2} \
10y^2\sqrt{2}$
Astar777
finally
Damn
this is wrong
lol
$\sqrt{y^2} = y$
Astar777
you are taking y^2 out
its done like this
wdym
if i take the y outside the square root, THEN i have to just remove the exponent, right?
yeah
wai are u saying that's wrong or, the way i did it is wrong and that i should've done it a diff way?
Astar777
the way u did it is wrong
ohh
what is the absoloute value signs for
beacuse y can be both positive and negative
$\sqrt{(-2)^2}$ and $\sqrt{2^2}$ both equal to 2
Astar777
ohh nvm that makes sense
so if i fix this, would it all become correct?
or is there more
this is correct too for your case
i just said this to be precise

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Don't know where to begin with. ( any hint_)
Convert all the log base 4,16 to log base 2
And the log base 9 to log base 3
hey we met again
Indeed
it was very long
Okay
There is an error in your conversion
log_4(z) should be 2log_2(z)
uh wait what
$\log_{a^b} x = \frac{1}{b} \log_a$
RUSTY_DUSTBIN
let me try
ok
@green merlin
oh
I took 4 as base in third equation because i do not think it will give 4root in any case
Indeed
@green merlin
Hold up
should i squre both side?
oh how
The original equation
i do not understand what u said
the last line here
You have $0.5\log_2{z} + 0.25\log_2{x} + 0.25\log_2{y} = 2$
Carbonite
oh
Multiply both sides by 4 might make it easier
Multiply both sides by 4, use log(a) +log(b) = log(ab), then change the 8 on RHS into log(?)
@minor swift Has your question been resolved?
Yo, I solved the problem, can I just send the uhh process or should I just give hints instead?
Helpers aren't allowed to give answers
As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.
It's a system where u got 3 unknown values and 3 equations, what means you can solve it. You're almost there
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@minor swift Has your question been resolved?
sorry but the electricity went off
well let me leave
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Let there be $T:\mathbb{R}{3}[x] \to \mathbb{R}^{3}, and S:\mathbb{R}^{3} \to \mathbb{R}{3}[x]$ linear transformations.
prove that if dim(Im(TS)) = 2, then $1 \leq dim(Im(ST)) \leq 2$.
Horsi135
@leaden mica Has your question been resolved?
I didn't really found a way to approach it, The only thing i thought of was that 1. there isn't a difference between R^3 and R_3[x] they are the same just with a different writing, and i thought about converting to matrices but I don't think it would work.
Think about it
Could the rank of ST be 3?
Could it be 0?
if it is 0 then either dim(Im(S)) = 0 which is impossible because that would mean that dim(Im(TS)) = 0 or dim(T(Im(S)) = 0 and for that I'm not sure
I have a question, isn't dim(R_3[x]) = 4?
no, R_3[x] are all of the polynomials such that the highest degree of x is 2, p(x)=ax^2+bx+c
But a base of polynomials of degree 2 is {x^2; x ; 1} which means dim = 3
ah okay
yes, dim=3
suppose dim Im ST = 0. since dim Im TS = 2, it means that dim im T ≥ 2 and dim Im S ≥ 2 which means dim ker S , dim ker T ≤ 1 by rank nullity theorem but from dim Im ST = 0, we have Im T ⊆ ker S which contradicts dim im T ≥ 2
And a base of polynomials of degree 3 is {x^3; x^2; x; 1}, so dim(R_3[x]) = 4, no?
yes, but R_3[x] is polynomials of atmost degree 2 here
Oh, I see!
but what about dim ST equals 3?
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can somone explain the underlined and circled ones
@unreal wren Has your question been resolved?
Does TSTST stand for team selection test selection test lol
Yes
You can get to the red one after submitting a=b=c=t^3
And the second one after submitting a=b=c=t
Well if I get to t^3 after applying the function t^9-t^3 times to t^9, and I get to t after applying the function t^3-t times to t^3; that would mean I get to t after applying the function t^9-t times to t^9
Because I just do those 2 things back to back
If this doesn’t make sense I can explain with a diagram
I know these alr
I don't get it sorry
You don’t see how those 2 equations gets us to the equation you underlined
I mean if I apply f to 5 a times and get to 10, and I apply f to 10 b times and get to 30 then applying f to 5 a+b times would make you get to 30 because you just do those 2 things back to back
I get it already
Oh okay
How about the circled paragraph?
Ill look when i have paper and pencil
@unreal wren Has your question been resolved?
There is no problem in the proof of the lemma that says f^t^2-t(t^2)=t right
No, i just don't understand the proof
if t = abc, shouldnt it be f^(t-a) (t) instead of f^(t-a) (a)?
Yes first line in the circled part seems to be mistyped
how did they get from first line to second line??
I was like maybe it isn’t mistyped and somehow provable but it is just a wrong equation when you consider the function we are proving it is supposed to be (which clearly works)
Second line we will prove, i dont know why they just gave it and didnt bother like it is obvious
oh, so you mean there are a lot of steps between first and second line, but they didnt show it?
Yeah well prove the second line using some things we have so far
Consider the first term in the left side of the equation (in the second line)
yeah
So I used that
Yeah I know
Instead of applying the function to t^2 t^2-ab times im applying t^2-t times and then t-ab times, we get to t after the first step like you pointed out
yes it makes so much sense
No need to be sorry lol
I mean proof isn’t done because they still dont turn into the exact things in the RHS
We will prove that the equation in the second line is correct, I don’t know how exactly they got to it tho there might be an easier way im not seeing
Maybe it is actually a simple observation
Oh wait the second line is so obvious actually
Forget about this well get to the second line equation in a way simpler way
yay
This is unnecessary apparently
It is just the equation that is given in the question
Just notice ab , t and c multiply to t^2
oh yeah
wait wait i will process
but why would it equal RHS?
i still dont see it
It is just the equation given in the question
I’ll look into it
ok
I am taking this back it was obvious turns out
Okay third line is provable using first two lines but well do some arranging and stuff
The transformation I thought was necessary to prove the second line is actually useful here
i kind of see it
Oh great
all we need to do is prove that f^(t^2 - ab) (t^2) + f^(t^2 - c) (t^2) = f^(t-ab) (t)
How did we get to this
It is probably correct i just need it to be explained
uh + f^(t-c) (c)
forgot to type that
oh i know
You’re saying we just need to prove this?
f^(t-c) (c) = f^(t^2-c) (t^2) so they cancel out
That is great because we will prove it with just one step
yeah
Yes they are equal but why are they equal?
Yeah
We are using the fact that applying f t^2-c times is the same thing as …
Fill the blank
Just making sure it is all understood before moving on
Wow great
Okay so after arranging and using this trick are we convinced third line is true?
yes
Nice
This is equivalent to the third line we proved earlier
Just look at the definition of the function g_t()
wait
lemme process
oh yeah lol
didnt see that
that means only 1 underlined one leftt
Btw the reason they noted ab|t for that equation is because we are working with positive integers and the original equation given in the problem only works with abc being a multiple of a b and c
And they noted ab<t because they defined g_t() to work only with inputs smaller than t
Okay
I need to read the parts between the underlined we just went through and the next underlined and comprehend them too rn
I’m reading the solution for the first time
No not at all dont worry
I don’t understand why the line right before the line you marked is correct
But I see how it is used to prove the line you underlined
which part of the line?
theres many equalities there i understand all of them
Why does pg_t(a)+qg_t(b)= g_t(a^p.b^q)
Ohh yeah
And getting to g_t(s) was useful because t is s^2 and we know how f^(k^2-k)(k^2) behaves
They made a typo in that line btw
It needs to be f^(s^2-s)(s^2)-s=0
They wrote f^(s^2-s)(s)-s=0
Look at definition of g_t it needs to have t in there which is s^2
And we know why it is correct when it is s^2 there, because of the first lemma we proved
Oh okay here we go
The line you didn’t underline that you just explained to me says this right
yes
Well p divides 0, it also divides the first term on the left hand side (because there is a p in it)
What does this say
a+b=c , a b and c are all integers. If p divides a and c what does this say?
p divides q g_t(b)
Yeah
oooh
These are all integers so we can come to conclusions like this
yes
Well there are lots of numbers that are divisible by p, not just 0? How do we know g_t(b) is 0?
You’re looking at q|g_t(a) line we showed p|g_t(b)
oh sorry lol
what g_t(b) is 0?
g_t(b) > -b because just look at the definition of g_t() and remember that f gives positive outputs (N->N)
We will prove that yeah
wait let me process
oh yeah i see it
But I am not going to use the fact that g_t(b)>-b so forget that for a second
ok
We have shown p divides g_t(b). Well there are lots of numbers that are divisible by p, not just 0. How do we know g_t(b) is 0?
is it because g_t(a) is divisible by q?
No forget that too
Sure I’m waiting I’ll tell you how when you want
i dont have any idea
It is because it didnt matter which prime we choose as p. Like lets say q=37 a=5 b=3 and we chose 41 to be p. Well we could choose 43 to be p too couldn’t we? And p still would divide g_t(b)
So g_t(b) wouldn’t be only divisible by 41 but also by 43 and also by any of the infinite amount of primes we could choose p to be
oh wow
And I can only think of 1 integer that would satisfy that
wait wait
i need to process this
Sure
how about the restirction s = a^p x b^q?
Doesn’t really matter
why not?
If we chose p to be 59 s would be a^59.b^q (whatever we chose for those 3). If we chose p to be 61 s would be a^61.b^q
But we could always carry out the proof for p|g_t(a) the same way we just did
err let me think
p isnt infinitely many primes simultaneously, it is just that p can be any of them it wants and always divides g_t(a)
okk i think i get it
Continue this thought
For any positive integer you’d choose a prime bigger than it and for any negative integer you’d choose a prime bigger than its absolute value and they wouldn’t divide
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I am supposed to find the center mass of the space defined by y^2 = x and x^2=y but im finding some wrong i think integral of dE is wrong and is supposed to be 1/3 but idk why
i am blind srry 
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How do i show that 214x215-214 is a square of something i know this must be easy but how
214 common
214x215 - 214 = 214(215 - 1) = 214^2
Damn thanks alot
!occupied
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Oh sorry
Np
consider: 214 * 215 - 214 * 1 = 214 * ?
oh ffs
i honestly wanna call !nosols on this
i can delete it, if you want (though its already too late), i guess i could have not included the last equality
Ty
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Could someone please help me with the second shift? I know x moves to 0 but I can't get y
yes
I mean, my x shift is correct, yes?
4 down vertically, and flip over x axis
flip over first
then 4 down
otherwise that -4 becomes +4 after u multiply the entire thing by -1
now lets go about our thingy
we start from (0,1), now move right once we're at (1,1)
we got flipped so we're at (1,-1)
then we brought down 4 more units
so
(1, -5)
ya?
Yes, that's a. Now b starts from 1, e. moves right so x value becomes 2, correct?
yeah, then it goes to -e and down 4 more
ya
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So i have a question when should we use bounding to find a max and when u cannot use it ?
Cuz i was doing an exercise from MAT 2023 paper and in tge first questions i used like the y coordinate of the max by completing the sequare method but then i used it for the last question and it didnt work so im confused to when to use each method
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Build a 2nd degree equation with coefficients as whole numbers that have as roots:
my bad
Nah lol I’m joking
Gl with ur question
Build a 2nd degree equation with coefficients as whole numbers that have as roots:
@haughty ember Has your question been resolved?
Use a(x-x1)(x-x2)
x1, x2 are roots
Then choose a so that all the coefficients are whole numbers
Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).
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Can someone help me with these graphing questions? Im not sure how they work
You have a picture?
alright, so does that mean its going from increasing to decreasing?
No
They want you to find on the graph the intervals where the function is decreasing
i know
Alright
I don't know 😅
Which answer did you find?
(-inf, -4) U (0, inf) sorry i dont have an infinity key
Don't worry
Mmh this is correct though
I don't know 🤷♂️
Let me try something
Let's see if done other helpers have a different opinion
Oh weird
But okay lol
Tests with pen and paper still are the best in my opinion
Because you never know if the websites are bugged 😬
Hey if you're trying to learn calculus
I am a shill for 3 blue 1 brown's intuitive explanations of the more entry-level concepts
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In this first video of the series, we see how unraveling the nuances of a simple geometry que...
It won't in anyway replace working on exercises and it's not without its faults, but it could help
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How is cauchy schwarz used in this step? (what are alpha and beta?)
I don't understand how this can be true for all i..
for example I can write the rhs as
$$\norm{\Psi}\norm{\psi\otimes\phi}=1$$
but this upper bounds it with one, I cannot use that the two states are proportional
schadowpop
o sorry maybe this provides more context and is clearer
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I need some practice materials and refernce materials for congruence
Like it should have lots of examples and some good theory
Here have 10 random questions 
Thanks a lot now some theory?
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I have these double integral volume questions and I have no clue how to set up the integrals, Im having a hard time what I need to set as my bounds, but other then that I can do partial differentiaion and integration pretty well I just need help understanding how to set the integrals up
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can anyone help me with my math its really hard and i dont get it
just send the question
heres the thing
i dont understand how to do it
I dont want the awnser
i need help understanding it
any question in particular or the whole thing
the concept you're looking for is called collecting like terms
terms?
variables
yeah, terms. that's the word in algebra for things being added or subtracted
no, variables are individual letters. don't confuse.
oop mb
we were told terms and variables are the same
yay
what's 5 apples plus 3 apples?
8
8 what
8 apples
right
yay
5 apples + 3 apples = 8 apples
ok
let's try another one
what's 18 pencils minus 7 pencils
11 pencils
right
... im not taking any clients at the minute sorry
its ok i was kidding
anyway
well yeah it was a stand in for letters
like i literally replaced the variable a (scary) with apples (intuitive)
but the logic is still the same
i was then going to ask you for something like "6 billion + 15 billion" or something along these lines and then segue into things like "6x + 8x"
but it looks like you got it already
so 14x
just as
5 apples + 3 apples = 8 apples
so too
5a + 3a becomes 8a
yep you're getting it
ok so why don't you try doing the first six questions from here
don't worry about the rest just yet, we'll get to them
you're technically but not morally right
I can explain further
OP wants to learn how to do exactly this
no need, ive already gotten started
Would you like me to draw a diagram
:D
I wanna become a math teacher some day ngl
calvin i appreciate your enthusiasm but i can handle this guy myself
on which one
Okay
I think he needs a diagram
i can handle it, thank you.
D:
5
right
ok right well i did ask you to do the first six
so now you see how some of the terms contain the letter a while others contain c, right
yea
yeah so the idea is
ones with different letters don't interact so to speak
in the lingo, they're called unlike terms
lingo?
math-speak
ok
but "lingo" generally means language or words specific to some discipline or subject
ok
so for example
so will it be ac
if you had a question that asked to simplify
15a + 8b
there would be nothing to do, as the two terms here are unlike and so don't interact with each other
on the other hand, with your question:
3a + 5c - a + 2c
you have two "a" terms:
- 3a
- -a
and two "c" terms: - 5c
- 2c
you collect these separately in the same way you've been doing thus far
you can think of this as:
3 apples, plus 5 carrots, minus an apple, plus another two carrots
you count the apples and carrots separately
idk
(with a for apples and c for carrots)
so 2a 6c
8
nope
8 ac?
2a+6c, that's your final answer
no it's a very different thing from 8ac
ok
anyway, this sort of logic should let you do up to and including q18
note also that raw numbers (ones without a variable) can only add/subtract with other raw numbers
wait
ok yeah you got me
5c+2c is 7c indeed
you said 6 carrots earlier and i didn't bother to double check
importantly though i hope you understood the logic
wdym stop
Probably means where he should do up to next
as i said earlier,
anyway, this sort of logic should let you do up to and including q18
so go until there
gonna repost the entire sheet for convenience
wdym
are you talking about the 3 and the -2 on the right
these are terms alright, they just don't have a letter
they are constants
or as i like to call them, raw numbers
note also that raw numbers (ones without a variable) can only add/subtract with other raw numbers
3-2 is 1. just 1. (and you should know what to do with 3x-x by this point)
This Algebra video tutorial explains how to combine like terms using the distributive property. It contains plenty of examples and practice problems.
Algebra - Free Formula Sheet: https://www.video-tutor.net/algebra-formula-sheet.html
_________________________________________________________________________________________...
mb watch this
ok
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Im a bit confused on how the following trig identity is derived:
, , tan(a-b) = \frac{tan(a)-tan(b)}{1+tan(a)tan(b)}
Rien
Rien
, , tan(a-b) = \frac{sin(a)cos(b)-cos(a)sin(b)}{cos(a)cos(b)+sin(a)sin(b)}
Rien
, , tan(a-b) = \frac{\frac{sin(a)}{cos(b)}-\frac{sin(b)}{cos(b)}}{1+sin(a)sin(b)}
Rien
, , tan(a-b) = \frac{tan(a)-tan(b)}{1+tan(a)tan(b)}
Rien
In this step you divide by Cos(a)Cos(b), but I don't quite understand how you can do that if cos(a)cos(b) could equal 0.
if cos(a) = 0 then tan(a) is undefined anyway
ditto for b
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Assignment: Find the eigenvalues and eigenvectors of a given matrix
what part r u struggling with
the getting the answer correct and finding where I did a mistake part
maybe some computation, have u tried doing it again slowly?
first lets see if u understand the concept, how did u find ur characterisitc polynomial?
I semi understand it but the tasks are very similar so I dont care so much about the theory part currently, just want to pass the practical part of the exam so I'm focusing on that
I'll try again
but I'm intrigued as to why this computation didnt work
have u tried using cofactor expansion along a row or column
no, but still if I use this method that I used I should get the correct answer
idk what method ur using but if its correct then maybe ur algebra is messed up
sarrus rule, and yea it probably is
That's for determinant though
Not for finding char poly
whats a char poly?
Characteristic polynomial
yea but we find the characteristic polynomial by passing A-lambda*I through a determinant
polynomials* rather
also I double checked and have every value correct
It should be (3-lambda)[(4-lambda)(-1-lambda) +4]
Hm idk what that is then
Tried again buy same result...
<@&286206848099549185> anyone know where I messed up with the Sarrus rule? because if not, I've just proven Sarrus wrong
did it two sepperate times and checked at least 4 times at this point but I still cant find the mistake
nevermind, it was actually correct lol
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did i get this right?
my answer was f(2) = 4, therefore x = 2
Are you sure there are no other solutions?
oh we need to solve for two cases?
yes
wdym
we should get negative x
do we sub numbers into x?
solve this eq
to solve f(x)
we consider which x s.t. f(x) = 4 for x < 0 and which x .s.t f(x) = 4 for x >= 0
u found that f(2) = 4 works for x >= 0
whats .s.t?
What is the question asking?
Two intersections
so how many x should there be satisfying y=4?
two
Hmm
consider -x = 4
what can x be
hint: multiply both sides by a number to get rid of the negative sign
💀
Most realest hint ever
appreciate it
are the solutions
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hi guys, I'm preparing calculus 1 and while I was doing the old exercises I got this limit that I don't understand. In the first part highlighted in red my teacher uses asymptoticity, and replaces e^x -1 with x, which is 0 for x->0. Then he replaces cot with cos/sen and I don't understand why it tends to 1/x. In the blue part, use the hierarchy of infinities to solve it. Can you explain to me why cos/sen is replaced with 1/x?
Well, $\cos(x)=1+O(x^2)$ as $x\to 0$ and $\sin(x)=x+O(x^3)$ as $x\to 0$, so (naturally?) $\cot(x)=\frac1x$ as $x\to 0$ (or I'm missing some use of $O$ notation, but you get the point)
;(
Really weird to not use $1-\frac12x^2$ though, for $\cos(x)$
;(
okay thanks i got it, it was pretty obvius my bad
well if 1 is enough...
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