#help-23
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So I'm encountering a ton of issues trying to get euler angles into a format that is easy to implement in control theory. Right now I only have the euler angles that a 6-DOF quaternion block outputs. These angles being interdependent are making it really difficult to figure out what the output format is, and convert it to something like angle axis representation.
Basically I am getting very lost trying to convert my euler angles the block gives me into something a little easier to use like angle axis representation (example image of them).
@devout glade Has your question been resolved?
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hi i’m having trouble figuring out the differences between B and C
read the question again. "neither is a blue" and "not a blue pair" are different things right?
yes
i believe the answer i have for C is the one i have right
<@&286206848099549185>
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how do i do this
i have gotten to the precalculation that delta is equal to epsilon/9
but idk what to do from here
oh all it is asking for is that precalculation
lol
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what’s good
i need help with this
The parabolas have to hit the labeled dots on the graph
and i have 4 equations
@lime light Has your question been resolved?
no lol
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how would i go about answering the first and last question?
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does this make sense or is it even correct 💀
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Can someone help me with 27-32
Can you walk me through it
,rotate
can sum 1 help
what have you tried so far
I put it in its form idk what to do now
show your work
$\left(\frac{x-2}{x^2 - 9}\right)\left(\frac{x-3}{2-x}\right)$
texaspb
yea
ok... do you think u can simplify that any further?
no
why not?
idk how
I suggest factoring the first denominator
hint: use the difference of squares formula
what is that
$a^{2} - b^{2} = (a + b)(a - b)$
texaspb
When an expression can be viewed as the difference of two perfect squares, i.e. a²-b², then we can factor it as (a+b)(a-b). For example, x²-25 can be factored as (x+5)(x-5). This method is based on the pattern (a+b)(a-b)=a²-b², which can be verified by expanding the parentheses in (a+b)(a-b).
Watch the next lesson: https://www.khanacademy.org/m...
I suggest watching this
ok im watching
do you notice a simplification that could help you in this formula here
(x+3)(x-3)?
yes
so now you can just substitute, ending up with
$\left(\frac{x-2}{(x +3)(x - 3)}\right)\left(\frac{x-3}{2-x}\right)$
texaspb
what else can you do here
x-2 and 2-x
well firstly you could simplify the two (x - 3)s terms right
yea
so we're just left with $\frac{x-2}{(x + 3)(2-x)}$
texaspb
idk
switch 2-x to x-2
how so?
idk
you can factor out a minus sign
how does that work
like this $\frac{(-1)(2 - x)}{(x + 3)(2-x)}$
texaspb
distribute the (-1) over the term (2 -x) and you'll get the same thing we had the last step
now you can simplify the (2 - x)s
we're left with $-\frac{1}{(x + 3)}$
texaspb
ok
what's the domain of this new function
-3
I mean 3
why 3?
uh cause -3+3 is 0
that doesn't make any sense
for the denominator
so our domain is $\bR \setminus {-3}$
texaspb
do you see the reason why
isnt the domain from the original function
this
no but you have to simplify the expression first
it's what we've been doing this whole time
in my book it gives me the domain for this
$(s \cdot t)(x) = \left(\frac{x-2}{x^2 - 9}\right)\left(\frac{x-3}{2-x}\right) = \left(\frac{x-2}{(x +3)(x - 3)}\right)\left(\frac{x-3}{2-x}\right) = \frac{(-1)(2 - x)}{(x + 3)(2-x)} = -\frac{1}{(x + 3)}$
texaspb
can you send me a picture of that
ok can we do the next one
ok ill try the next one
ok ty
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(23^19 + 19^19) & (23^23+19^23) what are the common factors ?
How to approach this
@full fox Has your question been resolved?
<@&286206848099549185>
Ok so I found something,
If n is odd then hcf of (x^n + a^n ) will be (x+a) ?
$(23^{19}+19^{19})\ &\ (23^{23}+19^{23})$
ItzAine
Im not sure about this one, it doesnt seem factorable to me.
What I meant is if we have
(x^n + a^n) & (x^m+a^m)
Both m,n being odd, then (x+a) will be hcf ....I read it somewhere but I'm not sure if it's true
try plugging it into a calculator and see if the results are the same
Yeah let me try with smaller numbers
Yeah I tried a few numbers, it is working, the hcf is (x+a)
nice
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Differentiate pi^pi
This is 0 because pi^pi is a constant and differentiating a constant always results in 0
Just want to confirm if this is a case
if this is the case*
yeah that is weird
Im differentiating with respect to x
yea so ur right with it being 0
like d/dx pi^x?
yes
yes thats right
yes
okay thank you
np
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i have question √-18 * √-12 so result of each of them is
√-18 = 3i√2 - √-12 = 2i√3
so √-18 * √-12 = (3i√2)(2i√3)
ok but how can i solve this (3i√2)(2i√3) ???
Use the fact that i * i equals -1
bro explain more
There are 2 i’s in your final expression. When u multiply them they will be -1. And then u can multiply the rest of the numbers as usual
So the i and i become -1, 3 and 2 become 6, and sqrt 2 and sqrt3 become sqrt 6
$\ -6\sqrt(6)$
Pauli Excluder
ok there another thing
now if i did (3√2)(2√3) in calculator will be 6√6 why - 6√6?
Because there was also two i’s tucked away in your (3√2)(2√3) that together equal -1.
They multiply to be -1, you multiply that with (3√2)(2√3), you get -(3√2)(2√3), which is - 6sqrt(6)
Yup!
aha we get -1 outside and multiply all numbers by -1?
Exactumundo
Commutative property of multiplication baby
As long all the numbers are being multiplied together, it doesnt matter what order you multiply them
They can commute and move all around into any order you want
$\ -1(3\sqrt(2))$
$\ -1(3\sqrt(2))$
Pauli Excluder
?
Mohamed.
Compile Error! Click the
reaction for more information.
(You may edit your message to recompile.)
$\ -1\ (3\sqrt(2)((2\sqrt(3)$
You shouldn’t be putting backslash before the parenthesis
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I am on my last attempt and need help, I don’t understand what is wrong about the answer here.
Before it gave me a new question, this is the answer to the similar one I got wrong.
I corrected that mistake in the newer question, just different numbers yet it states my inputted answer is incorrect
Old & New questions work
This is pre calc*
This is the current question that I got wrong, trying to figure out what is wrong about it to fix and get correct
<@&286206848099549185>
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hi deep
oh no my fairy again
oh
goddess help room 2

he is dying
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fairylog 

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what does lambda mean/whats does it represent
the symbol that looks like an upside down y
yes that one
you can use it for anything
ah
no context really
its just like using x then
yes

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Would I round up in this situation?
Question says: Reported to the correct number of significant figures, ((0.091 + 4.78)) * (14)
Ah
I see what the question is really about
4.871 is the addition
Since the least number of significant figures is on 4.78, we round it down to two decimal places, giving us 4.87
Multiply buy 14
68.18
Brain no work
68.2?
since three significant figures
that's wrong too
what
oh
oohhhh
***14 ***
TWO SIGNIFICANT NUMBERS
you hoe HIDING FROM ME
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The question is: How many square meters, m^2, are in 465 square kilometers, km^2?
So, most people know 1km is 1000m, but hypothetically speaking, would it be possible to solve this not knowing that?
no
So, essentially I should memorize all of the milli, micro, nano, etc exponents
wha-
that's memorisation lol
if its some uncommon unit the conversion should be given
you should know milli and kilo at least
kilo is 10^3 I believe
mega is 10^6 I got that from a previous problem
Not sure what 4 or 5 is
4 or 5?
10^4 or 10^5
ya
Huh, interesting. Why's that?
why do you think
That's the problem, I don't 
its called myria but completely obsolete
having too many prefixes is redundant
what do you want to represent every measurement or result as in the ones
with incrementing every three you can use, ones, tens, hundreds
suitable
and over complicating it
it would just make things harder
do you really want to change prefixes every power of 10?
yes
is that what you really want to do to yourself when you're writing down measurements
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need help writing the equation of curve L vs θ and creating a vertical asymptote graph
ig you could start with theta = |1/L|
then adjust to fit the data
unless u want something that goes through all points
how do i do that
There are fitting tools which exist that will try to find a function which will attempt to go through all the points. Like syrenate said it seems to follow a 1/x curve with some modifications needed. I dont use desmos much but id assume they have some regression tool. I would recommend plugging in all the positive points and see what it comes up with. Dont plug in all the negative points tho
i have plotted my points
Lemme get on my pc and ill see if i can find the regression stuff
if u have the data on a spreadsheet you can just copy paste it in to get a table like this. then you should write something along the line of
$$y_1~\frac{a}{x_1}+c$$ and see what it comes up with. the 2nd screenshot is an example of a regression
seems ~ does not like showing up in latex
Duh Hello
then write
y_1~a/x_1+c
below
so now that is a basic regression of the points that you have. u should see that the negative points are just the same as this one but negative. hence u have some okish approximations with the given function
so the program decided that
$$y=\frac{655.233}{x}-5.65523$$is the best fitting curve. im sure there is something that will be more accurate but not sure exactly how accurate you need it
Duh Hello
ye thats what i would do, there is def some more sophisticated way to get this more accurate, not sure exactly how accurate u need to be tho
is that the equation of curve?
yea
what about the negative points
assuming symmetry about the x axis it will just be the same but negative
however feel free to try and do the same for the negative points and see how similar they are
but i see your points are the exact same so it will be the exact same
so $$y_2=-y=-\frac{655.233}{x}+5.65523$$
Duh Hello
yep
but if i was doing this manually i would probably try to define $\frac{a}{x}=g(x)$ and then have $c=-g(90)$ as its important for u to have it be 0 as x goes to 90
Duh Hello
oh
but i dont really see a clean way of doing that
after playing around with it a bit i found that $$y=\frac{265.685}{x^{0.56}}-\frac{265.685}{90^{0.56}}$$ fits it really nicely
Duh Hello
negative points will just be the negative of this
and then you should have nice curves following the points
obviously feel free to fine tune it, i just did it by eye. https://www.desmos.com/calculator/824fhsbowp
just changing the r is what i did till it looked nice
thanks
no problem
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I'm a bit confused on how to construct a bijection between $[0, 1)$ and $(0, 1)$.
DerpZ
I tried a piecewise function at first but that didn't work obviously
I've also tried $f(x) = \frac{2x-1}{(2x-1)^2-1}$
DerpZ
I feel like I'm missing something here, am I supposed to do something with a set of open intervals?
@zinc token Has your question been resolved?
@zinc token you are trying to proof that f is bijetive from 0,1 to 0,1?
Proving that f is bijective doesn't means there is a bijection ?@zinc token
DerpZ
It's not
I'm saying that my choice of f is wrong yeah
Cuz f (0) isn't 0
^
It's*
Yeah
how about you find a bijection between [0,1) and R and a bijection between (0,1) and R
DerpZ
Yeah
Let me check
or tan
Wait I have $f: (-1, 1) \to \mathbb{R}$
DerpZ
but hmm finding one for [0,1) seems tricky
DerpZ
I saw on stack they constructed a bunch of open intervals or something?
I didn't understand it at all tho
tan(πx-π/2) works for (0,1) too
(This question is from understanding analysis)
oh ok seems this is impossible with continuous functions, we need something more clever
Yeah I thought so too hmm
The annoying bit is the 0 ugh
What if I "pull out" the rationals
you could pull out all elements in the form of 1-1/n and do something different to them, while mapping all other reals to itself
Yeah yeah
HMM
Yeah I think that's it
I'll play around with it myself for a bit, thanks!
i think its for topological reasons that the preimage of an open set of a continuous function has to be open too
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@zinc token what if say
you send 0 to 1/2
then 1/2 to 1/4
and 1/4 to 1/8
etc
and keep everything else where they are
Hmm
.reopen
✅
send 1-1/n to 1-1/(n+1)
you could do this a number of ways
the one im suggesting is
send 0 to 1/2
and 1/2^n to 1/2^(n+1)
and you can undo this map
just by sending 1/2^n to 1/2^(n-1)
and sending 1/2 back to 0
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lol i didnt come up with it
one of my profs taught me this
im just studying this for fun so I didn't have anyone to teach me this oof
this is interesting yeah
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For f(x)=x^2+2x and g(x)=5x+1 find (fog)(x)
you mean $f \circ g$?
illuminator3
yes
show work
you have to put g into f
You need to input g into f.
what should it look like after input
,, (f \circ g)(x) = f(g(x))
illuminator3
yes
chris what's your goal here
@agile sable Has your question been resolved?
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I’m interested in the placement of the i
Does it matter if i is in the numerator, denominator, or in front or behind the fraction altogether?
In front or behind, no
hunter hunter
Hmmmm
numerator and denominator, yes
hunterX HUnter
Why wouldn’t in front or behind work?
i is just a number tho
exactly
doesnt matter where it is
Multiply by √3/√3 to remove radical on denominator
But the i started in front to begin with
yea, why not ?
Keeping in front is not valid?
why ?
Oh because it would say entire fraction is negative discriminant?
If having i in numerator only it states only numerator is negative discriminant
Do you know what the term discriminant refers to in math?
discriminant = b^2 - 4ac
with a polynomial ax^2 + bx +c, a=/=0
Discriminant refers to whatever value is underneath of the radicand symbol
yea and ?
i dont understand your problem with i
I’m asking why the i cannot be in front or behind
^
Oh, I read this as numerator only
wherever
doesnt matter
well
dont transfer something from the numerator to the denominator
but put it anywhere you want
“In front or behind, no” not sure what this is referring to
Oh I see
snow is quite righty
Got it now thanks
$i\sqrt{3} = \sqrt{3} i$
Herels
Why you blasting anime memes when I’m asking a question? Kind of distracting
“Hunter hunter” ?
Why?
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I am confused, Are elements of the set of natural numbers ordinals? and cardinal numbers are the cardinality of such sets
How should one think about this
I know ordinals are; first, second third fourth and cardinals are; 1,2,3,4.. (The former referring to an ordering within a set and the latter refering to the cardinality of a set, hence the name)
But in ZFC we construct the Von Neuman Ordinals and declare this set to be N
Implying the elements of N are ordinals
is this correct?
@cyan pecan Has your question been resolved?
<@&286206848099549185>
@cyan pecan Has your question been resolved?
So well
Ordinals and Cardinals can both be thought of as sets
And in that case, every cardinal is an ordinal
Every natural number, ω, ω1, ... are the cardinals
so yeah every natural number is an ordinal and a cardinal
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usually cardinals are thought of as equivalence classes on set size
not by their set definition
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I don’t know what to do for #17
Tan(26)
Oh why is it tan cuz I got cos
You have the adjacent and you are trying to find the opposite of the angle
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i know this is false but can someone show me how you would solve for it
like full steps
$-2 \ln 2^2 =\ln((2^2)^{-2})=\ln(4^{-2})=\ln(1/16)$ \ \ $e^{\ln(1/16)}=1/16$
Civil Service Exam is good
,w e^{-2 \ln 2^2}
.close
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hi
$\frac{x^2}{xy-y^2} - \frac{x^2+y^2}{x^2-y^2} - \frac{y^2}{xy+y^2}$
Raid
I factorized them in the following way
$\frac{x^2}{y(x-y)} - \frac{x^2+y^2}{(x-y)(x+y)} - \frac{y^2}{y(x+y)}$
what could be a common demoninator?
Raid
Is it (x-y)(x+y) * y ?
👍
so why the hell
my result is not right ahah
sending the pic
$\frac{x^2}{y(x-y)} - \frac{x^2+y^2}{(x-y)(x+y)} - \frac{y^2}{y(x+y)}$
sorry for the quality
are my divisions right? I don't think so
Raid
calculator
this looks correct
and then it cancels really nicely
they dont have the same letteral part
what common term can you take out?
x
ok
x(x^2 - y^2)
easiest way to solve problem is to use photomath
?
hacks irl
yea but I am practicing ahah
ahhh
do you know difference of squares?
yea
x((x-2)(x+2)) ?
not quite
is the ^ supposed to represent power?
yea
x(x+y)(x-y)
then just divide
but I do I get x/y
Raid
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is f(x+y)=f(x)+f(y)?
always?
no lol
apparently it works only for matrixes then
in other words linear transformations of vectors
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(light reading https://en.wikipedia.org/wiki/Cauchy's_functional_equation)
oh baby cauchy equation
so it's true for f(x)=ax
but if you invoke axiom of choice
then you can get some wacky discontinuous functions
👀 that's an answer
it's a very interesting functional equation
that makes a lot of sense
matrices are in the form ax=f(x)
explains it perfectly
a matrix x vector
deamn
but also there are some other weird solutions
and you can't prove if those other solutions exist or not
unless you assume axiom of choice
I can't define a pathological solution explicitly
no clue what that means i gotta continue studying now tho
.close
oh already did that
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The awnser is -39 right?
are you sure?
Or is it 39 becayse
what does |x| mean
Theres no negatives at all
Im nit sure but ik how to solve them in like problems
(hint: |x|>=0 always)
@west acorn Has your question been resolved?
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Can someone help me understand
Please don't occupy multiple help channels.
what do you not understand, like what question means? or how to find what it means?
How to find what it means , I I know I want to substitute 7 for y in the equation
okay so substitute y=7 in I=3.44y+9
So 33.08 , or is that now how I’d do it
that will tell you percent of insta users in 7th year
3.44x7+9
,w 3.44(7) +9
Oh okay so that’s what the next part was asking
yes
Thank you would you be able to help me with one more it’s another word problem I don’t understand
I know the variable but the construct a series of expressions I don’t understand
,rccw
@lean otter Has your question been resolved?
@lean otter Has your question been resolved?
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read up on the algebra of polynomial functions
im not qualified enough to talk about this lmao tbh
but i think the answer is d
factorize and divide have fun
If you take x=1 f/g is only equal to the last answer i think
it does. f(x)/g(x) = f/g(x). write down f(x)/g(x). now to simplify factor them
$\frac {f}{g}(x) = \frac {f(x)}{g(x)}$
utrebsto
You should check a video may be
oh thats just a method to factor quadratics mate!
which gives us $\frac {2x^2-5x-3}{2x^2+x} = \frac {(x-3)(2x+1)}{x(2x+1)} = \frac {x-3}{x}$
utrebsto
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Does anyone know if I did anything wrong or is it right because I am very confused
,rotate
a can be negative, in which case |a|=-a=b>0 so it's wrong
|x| = x if x>0 and |x| = -x if x<0
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is this actually solvable
#❓how-to-get-help feynman
@topaz rain Has your question been resolved?
meme questions go into #chill
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how did they get x/2 as u here?
the bot just picked it
in order that the denominator be of the form sqrt(1-u^2)
then trig sub
$4u^{2} = x^{2}$
Kookiemon
ok yes i know that but like where did they pull x/2 from
4u^2 = x^2
2u = x
u = x/2
The goal with these types of integrals is to figure out a way to transform the expression into a form that you know how to solve.
sqrt(4 - 4u^2) = 2 * sqrt(1 - u^2)
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i need some help with bhaskara's formula, i havent quite understood it, and it just puzzles me while doing work
also please do realise that english isnt my main language, so my translation of math terms might be somewhat innacurate
you mean quadratic formula?
no like
yeah its this
b^2 - 4 . a . c
my teacher says that the 4 is negative
when multiplying
and i cant figure out why
you did the proof of this in class?
especially because i saw a video saying that the 4 is also negative
or did they just give you?.
but i saw somewhere else that it isnt??
it is negative
here's where im bugging
i'll just screenshoot
the sum of the cordinates of the vertexes
so
a = 2
b = 10
c = 12
100 - 4 . 2 . 12
so it'd be
(-4) . 2 . 12
which would be -96
how can i differentiate the positives from the negatives
if theres no parenthesis
i know how to do this, but negative numbers just screw up my calculations
yeah
so in our case its D = 10² - 4(2)(12)
mhm
well yes, its all fun and games until i have to put the - 4 into it
alright
dont forget
that
-4 = -1 * (4)
so if you have
-4 . 2 . 12
its like
-1 (4 . 2 . 12)
which is like
- (4 . 2 . 12)
oh
when you see a minus before a number
its like that number times -1
and since you can do multiplications in any order
you can do the minus one times whatever is left
at the end
so here
you just do
4 . 2 . 12 = 96
and put the minus before it
so D= 100 - 96
well
D = 4
heres what trips me up
isnt 96 just positive at the end of the day?
100 - 96
just leave it at that
forget what we did before
the 96 would just be positive in here
i'll just show my thought process through this multiplication
-4 . 2 . 12
-4 . 2
-8
-8 . 12
-96
what did i do wrong here to find it negative
like, my teacher wont mind if i put this in my written calculations to show how i got to a result
benjamin
this whole bimester
since the very start
i always did my calculations scared of doing any multiplication
now i understand this
just quickly if you have
benjamin
12 - a*b
yeah
ima edit messages
this part remains unchanged, even though the number is technically different
is this normal
no its not
do it with the whole expression from the beginning
like
12 - (-3)(2)
go on
12 - (-3) . 2
why did the - disapear?
didnt type
oh yeah
i edited
if you struggle dont forget about the hidden -1
well its still 12 - 6 though, isnt it?
