x = rcos(t)
y = rsin(t)
#help-10
knotty crow
but it's given in that form? I don't see equation here
like you mean x = 4cos(t) and y = 5sin(t) ?
timid saddle
ya, I tried to convert by solving for t
knotty crow
so u see cos(t) = x/4 and sin(t) = y/5
timid saddle
and then plugging it in
timid saddle
can i get another hint?
knotty crow
plug values of cos(t) (it's x/4) and sin(t) (it's y/5) into this equation
then you'll get rid of t
timid saddle
oh right its just the equation of the circle. right
knotty crow
close to the circle but not, it's an ellipse
timid saddle
right not a perfect circle cuz radius isn't the same throughout
so equation is just (x/4)^2 + (y/5)^2 = 1
thanks again.
.close
raw spade
How tf do you allign a split to the left of the page?
obtuse pebbleBOT
How tf do you allign a split to the left of the page?
raw spade
Like I want to be abled to align everything to the left, however the split wants to chill in the center
wooden cipher
Maybe try \newline?
pseudo swift
How tf do you allign a split to the left of the page?
could you post the code here instead of using a picture ?
raw spade
Maybe try \newline?
I just tried it
pseudo swift
the purple also doesn't help tbh
raw spade
could you post the code here instead of using a picture ?
Yeah sure hold on
wooden cipher
I didnt work?
raw spade
I didnt work?
Nope
pseudo swift
you can enclose the code inside ```latex ```
wooden cipher
Aw shucks
raw spade
\documentclass[a4paper,12pt]{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{array}
\usepackage{fullpage}
\usepackage[utf8]{inputenc}
\setlength\parindent{0pt}
\begin{document}
\large
\textbf{Assingment 3: Math 3202}
\normalsize
\flushleft{
- \begin{equation*}
\nabla \times F=\left|\begin{array}{ccc}e_{1} & e_{2} & e_{3} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ \sin y & x \cos y+\cos z & -y \sin z\end{array}\right|=\langle-\sin z+\sin z, 0, \cos y-\cos y\rangle=\langle 0,0,0\rangle
\end{equation*}
Thus, $F$ is conservative
\begin{equation*}
\frac{\partial \phi}{\partial y}=x \cos y+\cos z \Rightarrow \phi=x \sin y+y \cos z+f(x, z) \Rightarrow \frac{\partial \phi}{\partial x}=\sin y+\frac{\partial f}{\partial x}, \frac{\partial \phi}{\partial z}=-y \sin z+\frac{\partial f}{\partial z}
\end{equation*}
Thus, $\phi=x \sin y+y \cos z$}
\begin{equation*}
\begin{split}
&r(0)=\langle 0,0,0\rangle \text {, and } r(1)=\langle\ 1, \frac{\pi}{2},\pi \rangle \
&\int_{c} F \cdot d r=\phi(r(1))-\phi(r(0))= 1-\frac{\pi}{2}
\end{split}
\end{equation*}
\end{document}
warm shaleBOT
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pseudo swift
i mean i don't want to see it compiled
just put your code inside `` and ``
it will be less of a mess
raw spade
‘ ‘ \documentclass[a4paper,12pt]{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{array}
\usepackage{fullpage}
\usepackage[utf8]{inputenc}
\setlength\parindent{0pt}
\begin{document}
\large
\textbf{Assingment 3: Math 3202}
\normalsize
\flushleft{
- \begin{equation*}
\nabla \times F=\left|\begin{array}{ccc}e_{1} & e_{2} & e_{3} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ \sin y & x \cos y+\cos z & -y \sin z\end{array}\right|=\langle-\sin z+\sin z, 0, \cos y-\cos y\rangle=\langle 0,0,0\rangle
\end{equation*}
Thus, $F$ is conservative
\begin{equation*}
\frac{\partial \phi}{\partial y}=x \cos y+\cos z \Rightarrow \phi=x \sin y+y \cos z+f(x, z) \Rightarrow \frac{\partial \phi}{\partial x}=\sin y+\frac{\partial f}{\partial x}, \frac{\partial \phi}{\partial z}=-y \sin z+\frac{\partial f}{\partial z}
\end{equation*}
Thus, $\phi=x \sin y+y \cos z$}
\begin{equation*}
\begin{split}
&r(0)=\langle 0,0,0\rangle \text {, and } r(1)=\langle\ 1, \frac{\pi}{2},\pi \rangle \
&\int_{c} F \cdot d r=\phi(r(1))-\phi(r(0))= 1-\frac{\pi}{2}
\end{split}
\end{equation*}
\end{document} ‘ ‘
warm shaleBOT
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pseudo swift
just post your file maybe
that's easy at least
and discord colours it by default
and no weird discord formatting shenanigans
raw spade
just post your file maybe
The latex editing software I use doesn’t allow me to rn
pseudo swift
ah
ok well just use my backticks instead of the weird backticks you found
that'll do
pseudo swift
ok well just use my backticks instead of the weird backticks you found
as in copy the backticks I posted earlier
or here ``
raw spade
`` \documentclass[a4paper,12pt]{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{array}
\usepackage{fullpage}
\usepackage[utf8]{inputenc}
\setlength\parindent{0pt}
\begin{document}
\large
\textbf{Assingment 3: Math 3202}
\normalsize
\flushleft{
- \begin{equation*}
\nabla \times F=\left|\begin{array}{ccc}e_{1} & e_{2} & e_{3} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ \sin y & x \cos y+\cos z & -y \sin z\end{array}\right|=\langle-\sin z+\sin z, 0, \cos y-\cos y\rangle=\langle 0,0,0\rangle
\end{equation*}
Thus, $F$ is conservative
\begin{equation*}
\frac{\partial \phi}{\partial y}=x \cos y+\cos z \Rightarrow \phi=x \sin y+y \cos z+f(x, z) \Rightarrow \frac{\partial \phi}{\partial x}=\sin y+\frac{\partial f}{\partial x}, \frac{\partial \phi}{\partial z}=-y \sin z+\frac{\partial f}{\partial z}
\end{equation*}
Thus, $\phi=x \sin y+y \cos z$}
\begin{equation*}
\begin{split}
&r(0)=\langle 0,0,0\rangle \text {, and } r(1)=\langle\ 1, \frac{\pi}{2},\pi \rangle \
&\int_{c} F \cdot d r=\phi(r(1))-\phi(r(0))= 1-\frac{\pi}{2}
\end{split}
\end{equation*}
\end{document}``
warm shaleBOT
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pseudo swift
uh wat
raw spade
I’ll try again
raw spade
`` \documentclass[a4paper,12pt]{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{array}
\usepackage{fullpage}
\usepackage[utf8]{inputenc}
\setlength\parindent{0pt}
\begin{document}
\large
\textbf{Assingment 3: Math 3202}
\normalsize
\flushleft{
- \begin{equation*}
\nabla \times F=\left|\begin{array}{ccc}e_{1} & e_{2} & e_{3} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ \sin y & x \cos y+\cos z & -y \sin z\end{array}\right|=\langle-\sin z+\sin z, 0, \cos y-\cos y\rangle=\langle 0,0,0\rangle
\end{equation*}
Thus, $F$ is conservative
\begin{equation*}
\frac{\partial \phi}{\partial y}=x \cos y+\cos z \Rightarrow \phi=x \sin y+y \cos z+f(x, z) \Rightarrow \frac{\partial \phi}{\partial x}=\sin y+\frac{\partial f}{\partial x}, \frac{\partial \phi}{\partial z}=-y \sin z+\frac{\partial f}{\partial z}
\end{equation*}
Thus, $\phi=x \sin y+y \cos z$}
\begin{equation*}
\begin{split}
&r(0)=\langle 0,0,0\rangle \text {, and } r(1)=\langle\ 1, \frac{\pi}{2},\pi \rangle \
&\int_{c} F \cdot d r=\phi(r(1))-\phi(r(0))= 1-\frac{\pi}{2}
\end{split}
\end{equation*}
\end{document}``
warm shaleBOT
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pseudo swift
ok cool
uh so what do you want aligned again?
the phi = stuff with the rest?
at the bottom I mean
raw spade
uh so what do you want aligned again?
I just want the split section which is the part with r(0) to be shifted left instead of the middle of the page
For some reason /flushleft did not work
pseudo swift
ah I found something
@raw spade
googling for "math display mode left aligned"
there's an flalign environment in amsmath which flushes equations to the left
tldr
but you should probably read to get a code example
or here's the one you want
\begin{flalign*}
r(0)&=\langle 0,0,0\rangle \text {, and } r(1)=\langle\ 1, \frac{\pi}{2},\pi \rangle &\\
\int_{c} F \cdot d r&=\phi(r(1))-\phi(r(0))= 1-\frac{\pi}{2}&
\end{flalign*}
the & allow you to specify which parts should be at the same level vertically
ie all the parts at the level of the 1st & on different lines should be aligned
same if there are multiple &
the ending & (at the end of the line) seems to be really important for flalign (according to the post)
in general I prefer align vs equation tbh
timid silo
not sure where to go after i equal it to 0
obtuse pebbleBOT
not sure where to go after i equal it to 0
timid silo
im at cos(3z) = 1/2
timid silo
yes i do
willow ravine
So setup the 30-60-90 triangle, assign the appropriate side lengths, and pick which angle has cos(x) = 1/2
Set whatever you decide x is = 3z to solve for z
timid silo
could u draw it out for me
im having a hard time understanding
this is how they did it
fierce lagoon
im at cos(3z) = 1/2
take the arccosine of both sides
timid silo
ohh
willow ravine
No the function is discontinuous at some points
knotty crow
how we can talk about discontinuities at the point which are not in the domain
fierce lagoon
That function isn't continuous
willow ravine
Cos(3z) = 1/2
Is in the domain
Cosine exists between [-1, 1]
1/2 is inside of that
knotty crow
not really, for these values h(z) isn't defined, we can't divide by zero, so?
knotty crow
also you'd say 1/x isn't continuous?
fierce lagoon
That is discontinuous
willow ravine
That’s what we are trying to find, where it is discontinuous at.
fierce lagoon
also you'd say 1/x isn't continuous?
It isn't
1/x isn't continuous since it has a vertical asymptote
willow ravine
Modus, the question doesn’t state it is continuous. He is trying to find where it is discontinuous.
fierce lagoon
It's literally discontinuous: look at the graph
Note that the domain of where we're looking for discontinuities isn't restricted
It's gonna contain every discontinuity
knotty crow
My book says:
Function is continuous if it's continuous at each point of its domain.
timid silo
is there an online graphing calc
tardy epoch
agrees with modus:
https://en.wikipedia.org/wiki/Classification_of_discontinuities
fierce lagoon
My book says:
Function is continuous if it's continuous at each point of its dom...
Yeah but the domain is -infinity to infinity for his problem
fierce lagoon
It never specified any restrictions
It's gonna default to -infinity by infinity
Within that set of all real numebrs, there will be some that will result in discontinuities
knotty crow
if you solve equation: (x^2 - x)/x = 0 you treat 0 as the solution because domain isn't given?
fierce lagoon
0 is gonna be a discontinuity lol
0 is within the set of all real numbers
I don't know what you're trying to argue here
And it is discontinuous within the all-reals domain
zenith raft
I think the problem here is that under the definitions mathematicians use, things like 1/x and 1/(2-4cos(2x)) are continuous, but the problem does not want you to use those definitions (unfortunately???)
knotty crow
My maths book, Wikipedia and other sites on the internet says that theorem is true, we analyze only domain, I'm trying to decide what's the right way
zenith raft
that is the "right" way
fierce lagoon
That is the right way
But look at the problem
It doesn't specify any domain
So it's a default assumption to be all reals
-infinity to infinity
zenith raft
well I'd say h is not discontinuous anywhere, but I know that's not what they want
fierce lagoon
No (in agreement lol)
They wanna find places where h(z) is discontinuous for $z\in\mathbb{R}$
warm shaleBOT
Umbraleviathan
zenith raft
there is no meaning of continuous or discontinuous where a function isn't defined
or at least...
no one really cares about that lol
fierce lagoon
I mean at this point it's going to a discussion beyond David's level I think
zenith raft
yea I agree
zenith raft
but I just wanted to say, I think that's the source of the argument
willow ravine
Modus, no one is trying to say this function is continuous. The problem literally says “find where the DISCONTINUITY is”
zenith raft
well I agree with the point that it's not discontinuous anywhere under formal definitions
willow ravine
Look the function is discontinuous at (pi/3) + 2pi n
timid silo
isn't it just discounious when 2-4cos(3z)=0 ?
so solve for z, and then generate a set using ±a where a is the period of the function.
fierce lagoon
Yeah
willow ravine
yes
fierce lagoon
But like
knotty crow
Modus, no one is trying to say this function is continuous. The problem literall...
It was other problem, read this:
(I recommend to stop arguing tho, it won't help david)
fierce lagoon
THE DOMAIN IS ALL RWALS DKSSJJDOS
You assume all reals
This. Is. What. I've. Been. Saying bdjajhdkslekejdjf
willow ravine
The discontinuity happens when cos(3z) = 1/2
fierce lagoon
Which is in the all reals set
willow ravine
1/2 is inside of the domain of cosine
knotty crow
OK GUYS YOU'RE RIGHT, JUST STOP IT DOESN'T HELP ANYONE PLEASE
zenith raft
if the domain is all reals, then h(5pi/9) must be something you can evaluate
willow ravine
btw it's weird question since that function is continuous I think?
You think it is continuous ^
zenith raft
**OK GUYS YOU'RE RIGHT, JUST STOP IT DOESN'T HELP ANYONE PLEASE**
no you are right 😭
timid silo
what do i do after i equal it to zero and solve for z
math help chat drama? ill get my popcorn
willow ravine
You just add 2pi*n
timid silo
z = .349
willow ravine
To z
timid silo
why do i add? and whats n?
willow ravine
n is an integer
willow ravine
It repeats every 2pi
willow ravine
You know, I’m just gonna pop out of this chat. I hope Modus can convince you this equation is continuous and the problem is wrong by trying to make you find its discontinuity.👍🏼
timid silo
ight, david you okay with me to take over?
yeah sure
so just a quick review
this 1/x type equation is discontinous when the bottom of the fraction is equal to 0
(because your deviding by zero)
yes
so the point that the function will be discounious is when 2-4cos(3z) = 0
yes
i got z = .349
@timid silo
yeah one sec im verifing
we get cos(3x)=1/2
yes
we take the arccos of both sides
3x = cos^-1(1/2)
yep
now add 2pi * n?
can u explain why 2pi * n?
3x = pi/3
x = pi/9
huh
hold on wait im missing something
we should get 2
2 points?
yes
how come?
take a look at desmos
cause cos(3x)=1/2 has 2 answers
(well 2 answers in 1 period)
knotty crow
Generally solution to the equation:
$$\cos x = a$$
where $-1 \leq a \leq 1$, is:
$$x =\arccos a + 2\pi n \vee x = -\arccos a + 2\pi n$$
$$n \in \mathbb{Z}$$
this is because cosine is even function (cos(x) = cos(-x))
timid silo
oh yeah forgot about that lol
knotty crow
and 2 pi n is because period of cosine is 2pi
timid silo
yes
warm shaleBOT
Modus
timid silo
i thought 2pi was the whole circle
yep, that general for cos(x) not cos(3x)
then again if we define x = 3z and repalce x
Generally solution to the equation:
$$\cos 3z = a$$
where $-1 \leq a \leq 1$, is:
$$3z =\arccos a + 2\pi n \vee x = -\arccos a + 2\pi n$$
$$n \in \mathbb{Z}$$
warm shaleBOT
aspwil
knotty crow
i thought 2pi was the whole circle
you can think that way, it's also true if you use unit circle to find solution, fine
timid silo
so when using the unit circle to solve
do i always use 2pi * n?
yes, because
cos(x) = cos(x+2pi*1) = cos(x+2pi*2) = cos(x+2pi*3) = cos(x+2pi*4) ...
ooo ok
so cos(x) = cos(x+2pi*n) for any n in integers
$\cos(x)=\cos(x+2\pi n)$
warm shaleBOT
aspwil
warm shaleBOT
aspwil
timid silo
im writing notes rn so what is the reason i should put for adding 2pi * n
like a quick explanation
something like
"when we add 2pi to the cos we get the same answer as before [cos(x) = cos(x+2pi)]
therefor if we say cos(x) = 2 that means that cos(x+2pi) = 2,
and for any integer n we choose, cos(x+2pin) = 2.
therefor if we take the cos^-1 of both sides we get x+2pin = cos^-1(2),
so when solving for x we need to take into account this 2pin, thus leaving us with x = cos^-1(2)-2pin"```so do we only do that when we take the cos^-1 of both sides?
and to push it a little further -2pin can be changed to +2pin because n is all integers, (... 3, 2, 1, 0, -1, -2, -3...) so if we make it negative we still get all the integers.
meaning that n = -n
and thus
-2*pi*n = -2*pi*-n = 2*pi*n
ooo ok thank u
timid silo
so do we only do that when we take the cos^-1 of both sides?
well yeah kinda, it depends on the situation
so what we should get is
$3z = {\frac{\pi}{3},\frac{5\pi}{3}}+2\pi n\z = {\frac{\pi}{9},\frac{5\pi}{9}}+\frac{2\pi n}{3}$
warm shaleBOT
aspwil
timid silo
because
how does 0.349 turn into pi/9
$\cos(3z)=\frac{1}{2}\ \cos(3z+2\pi n)=\frac{1}{2}\ 3z+2\pi n=\cos^{-1}(\frac{1}{2})\3z = {\frac{\pi}{3},\frac{5\pi}{3}}+2\pi n\z = {\frac{\pi}{9},\frac{5\pi}{9}}+\frac{2\pi n}{3} $
$\cos(3z)=\frac{1}{2}\ \cos(3z+2\pi n)=\frac{1}{2}\ 3z+2\pi n=\cos^{-1}(\frac{1}{2})\3z = {\frac{\pi}{3},\frac{5\pi}{3}}+2\pi n\z = {\frac{\pi}{9},\frac{5\pi}{9}}+\frac{2\pi n}{3}$
warm shaleBOT
aspwil
timid silo
how does 0.349 turn into pi/9
0.349 is just pi/9 rounded
does that explain everything you need?
or is there anything your still unclear about?
(if you end up needing more help dm me, and ill come back)
obtuse pebbleBOT
@timid silo Has your question been resolved?
timid silo
@timid silo thank u i will
crisp dune
What does finding the avg value of a function do ?
obtuse pebbleBOT
What does finding the avg value of a function do ?
@crisp dune Has your question been resolved?
willow ravine
Do you have a picture of the question problem?
sage lintel
How can I eliminate the parameter in these parametric equations
x = 3-cos2t
y = 1 + sin2t
I solved for t in the first equation and ended up with t = arccos(3-x)/2
I then substituted that in the second equation and ended up with 1 + sqrt(1 - (3 - x)^2)
However my professor said the equation should be an ellipse, and mine looks like a circle
obtuse pebbleBOT
How can I eliminate the parameter in these parametric equations
x = 3-cos2t
y = ...
sage lintel
ok i'll try that
however is there a reason my method doesn't work?
idk if I made some erorr in my simplification
short zodiac
Yours is a more complicated way of expressing it, especially the square root can lead to loss of some of the curve values
Better to do it simply
sage lintel
gotcha, just to check does the answer turn out to be 1 = (y-1)^2 + (3-x)^2
thank you!
.close
obtuse pebbleBOT
sage lintel
Better to do it simply
sorry one followup question, how do you use the identity when you have sin and cos with different things inside (2t and t )
.reopen
obtuse pebbleBOT
✅
short zodiac
In that case you have no alternative but to express sin or cos in terms of the other
This identity can't be used for different arguments
It'll depend on the qn
sage lintel
.close
timid silo
Very simple but hard for me to understand. X/x=0 because they cancel (basic algebra). If x=7 then x/x=1. Why do we count it as 0 and not 0 or 1?
obtuse pebbleBOT
Very simple but hard for me to understand. X/x=0 because they cancel (basic alge...
cursive jolt
X/x=0 because they cancel
?
willow ravine
x/x =1
valid crown
x-x = 0
willow ravine
X/X does not equal X-X
willow ravine
xy/xy does not equal 0
zenith raft
for any real number x other than 0, x/x = 1 because 1/x is defined to be the number that when multiplied by x, is equal to 1
and x/x is another way of writing x*(1/x)
timid silo
correct
timid silo
not sure where to go after this
obtuse pebbleBOT
not sure where to go after this
timid silo
2x = cos^-1(0)
timid silo
radians
willow ravine
pi/2 and 3pi/2
timid silo
how?
willow ravine
Are the two points
timid silo
how do u get that
willow ravine
Look at radians column and cosine column
Also looking at the graph of cosine from 0 to 2pi
It hits the x-axis (equals 0) twice
willow ravine
You may need to lessen your period on this one.
So let cos^-1(0) be equal to pi/2
And add just pi*n
timid silo
i thought cos(0) = pi/2
willow ravine
No
Cos(pi/2) = 0
And cos (3pi/2)= 0
Where before you only had one value from [0, 2pi] for x. In this case you have 2 values.
timid silo
ooo
willow ravine
So, since it only takes pi radians to get from pi/2 to 3pi/2
timid silo
what quadrant would the second solution be in
willow ravine
The solutions lie on the y-axis
Not a specific quadrant
Well hang on, it’s cos 2x
And you want 2x = pi/2
So x=pi/4
Is the first quadrant
And 2x = 3pi/2, x= 3pi/4
Is in the second quadrant
obtuse pebbleBOT
@timid silo Has your question been resolved?
timid silo
ohhh okay thank u sm
willow ravine
obtuse pebbleBOT
willow ravine
Complex variables: this is what I got so far:
Is this correct so far?
Also, is the easiest way to o go about this is convert them all into cos(t) + isin(t)?
wooden cipher
Oh my god ok
I think i can handle this one
If my internet works >:(
Split the integral to its separate terms
For example the first terms becomes
4/i [e^4it] evaluated from 0 to 2pi
willow ravine
Okay. I’ll try to do all of them
wooden cipher
So 4/i[e^8pi(i)-1] which is
4/i[1-1]=0
brave bramble
You were correct Poisoned, that the first two terms are just zero because Cauchy.
The last term is the one that throws a fit
willow ravine
brave bramble
Avoid expanding it. I'm pretty sure that, with that sub, it just becomes cos³θ
willow ravine
I saw somewhere that any z^x is 0 unless x=-1 hence the 3/z term is the only one but I can’t use that if I don’t understand that enough to explain it to my professor.
Apparently 1/x integrated is 2pi*i
Which would be exactly 6pii/2pii
brave bramble
You know Cauchy-Goursat, that any closed integral around an analytic region is 0?
willow ravine
Yes I knew that for the first two terms
But the 1/z^3
I couldn’t use that for it
brave bramble
That's enough to say, for positive n, that zⁿ integrates to 0
brave bramble
We also have that any function with an anti-derivative integrates to 0
That's enough to kill 3/z³
brave bramble
It does not!
brave bramble
For any contour that doesn't include the negative x-axis, I would agree with that
Actually, no I don't think that's even true. I need to remind myself why no version of ln is an anti-derivative for 1/z
willow ravine
Also did I do the second one on the right side of my original post correctly?
brave bramble
The big deal is that every polynomial term and inverse polynomial term integrates to 0, leaving 3/z
willow ravine
brave bramble
That's essentially the residue theorem, that it's enough to find the constant on 1/z
brave bramble
I admit I'm bad with Cauchy's formula
willow ravine
Maybe Riemann or Ahoron can help
willow ravine
a pii? Not 2pii
high snow
How can I solve this?
obtuse pebbleBOT
How can I solve this?
high snow
You can use prime factorization on both numbers and multiply the common factors to find the gcf
ruby shoal
How bout that LCM
high snow
Yea
grizzled shore
huh
oh
$\frac{1}{12x} + \frac{1}{12y}=\frac{1}{12}\left(\frac{1}{x}+\frac{1}{y}\right)=\frac{x+y}{12xy}, 12xy = 144 xy = 12$ find $xy = 12$ where x and y has no common factors
wooden cipher
Heres a little tidbit GCF×LCM is the product of your two numbers
Still do have to satisfy the gcf and lcm separately tho
grizzled shore
the only things with x y is that they cant have any common factors except 1
warm shaleBOT
Frosst
obtuse pebbleBOT
@high snow Has your question been resolved?
ruby shoal
the only things with x y is that they cant have any common factors except 1
Says who
grizzled shore
if they had another common factor
then HCF wouldn't be 12
it'd be 12*whatever their common factor was
ie. if x = 2, y = 6, a = 24, b = 72, their HCF would be 12 * 2 = 24, and LCM = 144
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
compact gate
When integrating by parts, we define u and dv. Then we differentiate u and integrate dv to find v. This is all well and good. The part that confuses me is why we leave out the constant when integrating dv. Could someone help me understand why we're allowed to do this?
obtuse pebbleBOT
When integrating by parts, we define u and dv. Then we differentiate u and integ...
chilly veldt
Why would this be wrong
obtuse pebbleBOT
Why would this be wrong
tardy epoch
take a derivative and see if you get the integrand
obtuse pebbleBOT
@chilly veldt Has your question been resolved?
chilly veldt
take a derivative and see if you get the integrand
Ahh I get it
You can only pull constants out
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
timid silo
Sos
timid silo
Ans for (c)
Why can't I
- Translate curve 1 unit in positive x direction
- Reflect curve about y axis
3.scale factor of 3 parallel to y axis
Is this wrong? Whats the right way?
My way^
Please help idk how anymore
obtuse pebbleBOT
@timid silo Has your question been resolved?
obtuse pebbleBOT
hexed agate
Why in the positive direction of x?
obtuse pebbleBOT
Available help channel!
pulsar raft
I have problems understanding this proof. How come that the inequality (1) holds? Is it because we have defined ${x_n}\rightarrow d$ and therefore the two terms in the sequence $x_n+x_m$ will both be very close to $d$, giving us something like $(\frac{d+d}{2})^2\geq d^2$
And for (2), how can we see that this is a Cauchy sequence? Do we have to take the square root of the expression and see that $0\leq x_n-x_m\leq\epsilon$?
obtuse pebbleBOT
I have problems understanding this proof. How come that the inequality (1) holds...
warm shaleBOT
simonpetri
obtuse pebbleBOT
@pulsar raft Has your question been resolved?
kind hawk
y=1/2 (x_n+x_m) is in K because K is convex, so norm(y)>= inf{all norms of K}=d
for the second we know that 2(norm(x_n)^2+norm(x_m)^2)-4d^2->2(d^2+d^2)-4d^2=0
so by the squeeze theorem norm(x_n-x_m) also has to go to 0
pulsar raft
So for (1), By convexity, y=1/2(x_n+x_m) is in K and because d=inf{all norms of K} we must have that norm(y) must either be d or some norm greater than d.
kind hawk
yes
pulsar raft
And if i understand correctly, for (2) we have that as x_n and a_m will both eventually be equal to d somewhere in the sequence. Then norm(x_n-x_m)^2 will be squeezed in between zero and zero eventually in the sequence.
pulsar raft
very close to d then 🙂
pulsar raft
Yes I see it now. Thank you very much. I really got lost somewhere and could not figure out what was happening.
.close
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
solar elm
can anyone help me with this
solar elm
also this
2. During a month with 30 days, a baseball team plays at least one game a
day, but no more than 45 games. Show that there must be a period of some
number of consecutive days during which the team must play exactly 14
games.
Solution: Let aj be the number of games played on or before the j-th day of
the month. Then
a1 , a2 ,..., a30 is an increasing sequence of distinct positive integers, with 1 ≤ aj
≤ 45. Moreover, a1 + 14, a2 + 14,..., a30 + 14 is also an increasing sequence of
distinct positive integers, with 15 ≤ aj + 14 ≤ 59.
The 60 positive integers a 1 , a2 ,..., a30 , a1 + 14, a2 + 14,..., a30 + 14 are all less
than or
equal to 59. Hence, by the pigeonhole principle two of these integers are
equal. Because the
integers aj , j = 1, 2,..., 30 are all distinct and the integers a j + 14, j = 1, 2,...,
30 are all distinct, there must be indices i and j with ai = aj + 14. This means
that exactly 14 games were played from day j + 1 to day i.
royal basin
that's... the solution to the same problem
obtuse pebbleBOT
@solar elm Has your question been resolved?
subtle adder
The digits 2, 3, 4, 7 and 8 will be put in random order to make a positive five-digit integer. What is the probability that the resulting integer will be divisible by 11? Express your answer as a common fraction.
obtuse pebbleBOT
The digits 2, 3, 4, 7 and 8 will be put in random order to make a positive five-...
pseudo swift
The digits 2, 3, 4, 7 and 8 will be put in random order to make a positive five-...
well first do you know a systematic way of checking divisibility by 11 ?
that would probably help
subtle adder
the only divisiblity rule of 11 i know is adding the alternate digits and finding the difference between them
i got the total possibles numbers
$5! = 120$
warm shaleBOT
☙ Kiwwi ❧
subtle adder
possibilities
pseudo swift
the only divisiblity rule of 11 i know is adding the alternate digits and findin...
ok you know this one nice
yeah you got 120 possible orderings of you digits
now you know you'll have pretty much 2 types of digits when considering divisbility by 11
you'll have the ones counted positively and the ones counted negatively
let's say I fix the two digits which are counted negatively
how can you check such an arrangement could work ?
pseudo swift
do you mean checking if the number is divisible or not?
I mean yeah that's the whole point
we want to see which choices give us numbers divisble by 11
subtle adder
i am aware of how to do that, but i don't now how to count the number of possibilities that gives a number divisble by 11
pseudo swift
so yeah if I give you the two "negative" numbers, the three "positive" numbers are already chosen
so you can check divisibility this way
subtle adder
i don't understand
pseudo swift
at what point
subtle adder
fixing numbers
i'm sorry, but could you do the problem in its entirety? I'm not able to understand the idea of fixing numbers
pseudo swift
I'm getting to it yeah
the point is, the choice of the two negative digits forces what the 3 positive digits have to be
now, how many ways are there to choose 2 negative digits (among the 5 we got) ?
subtle adder
5C2
pseudo swift
yeah so 10
subtle adder
10
pseudo swift
the number of ways you can choose the two negative digits, gives you the number of ways of splitting the 5 digits among the two classes positive and negative
it doesn't give you all the orderings possible
(you still have to account for them inside the positive and negative classes)
but divisibility by 11 doesn't care about those internal orderings
so that's why I only looked at splitting positive/negative
pseudo swift
10
so now you got yourself 10 cases to check
take all the choices of 2 negative digits possible
and check if you got divisibility by 11 for each choice
subtle adder
how?
pseudo swift
how, like the big picture ?
or how to check divisibility by 11 ?
I could do one example to show you
subtle adder
please do so
pseudo swift
our set of digits we take from is 2, 3, 4, 7 and 8
let's say the negative digits are 2 and 3
then the positive digits are 4, 7, 8 right ?
(and then the idea is to repeat that argument for every possible choice of negative digits)
fluid snow
$$\frac{1}{\sqrt[3]{1} + \sqrt[3]{2} + \sqrt[3]{4}} + \frac{1}{\sqrt[3]{4} + \sqrt[3]{6} + \sqrt[3]{9}} + \frac{1}{\sqrt[3]{9} + \sqrt[3]{12} + \sqrt[3]{16}}$$
\
Find the sum.
obtuse pebbleBOT
$$\frac{1}{\sqrt[3]{1} + \sqrt[3]{2} + \sqrt[3]{4}} + \frac{1}{\sqrt[3]{4} + \sq...
high lily
but they're not
warm shaleBOT
Kepe
high lily
no
fluid snow
But the problem is the 3rd root
high lily
those equations you wrote are all false
fluid snow
uh, what am i doing lol
high lily
consider factorisation for a difference of two cubes
the denominators are in the form a^2 + ab + b^2
fluid snow
the denominators are in the form a^2 + ab + b^2
yeah
fluid snow
consider factorisation for a difference of two cubes
why difference, there are only sums in the denominators?
or does it turn into a difference bc of reciprocal
high lily
i'd recommend looking it up so you can see it more clearly
regal coral
maybe simplifying them would give another perspective?
fluid snow
maybe simplifying them would give another perspective?
wdym with simplifying them
fluid snow
consider factorisation for a difference of two cubes
$$a^3 - b^3 = (a - b)(a^2 + ab + b^3)$$
warm shaleBOT
Kepe
regal coral
the denominator is clearly your problem, you could try removing the radical
fluid snow
oh, so the latter bracket
is what's in the denominators
so I should multiply by (a-b)/(a - b) each frac
high lily
yes, multiply numerator and denominator by (a-b)
regal coral
you can also look up for how to evaluate radicals thingy
worthy cargo
for some reason i can't find the factorizatjon of two perfect squares what is it?
obtuse pebbleBOT
for some reason i can't find the factorizatjon of two perfect squares what is it...
worthy cargo
a^2 + b^2
nocturne minnow
nocturne sun
I mean you can say (a + b)^2 - 2ab but that's not really factoring I guess
worthy cargo
oh i see i was thinking of ^
but i also get what you mean
alright thanks
.close
hasty linden
quick question
obtuse pebbleBOT
quick question
hasty linden
on the second line
did we just divide the demoniator by n
dont worry about the following steps
pale mortar
dont worry about the following steps
From what i see yes
pale mortar
Well
What happens there is that the limit is multiply by (1/n)/(1/n)
So the numerator is going outside and the denominator to the denominator
obtuse pebbleBOT
@hasty linden Has your question been resolved?
proven monolith
imma need some help on this equation
obtuse pebbleBOT
imma need some help on this equation
high lily
what have you tried
proven monolith
I tried to look for something to factorise with
velvet sundial
dismantle the x2 - 4
proven monolith
thats what i did but i don't know how to continue
high lily
were you successful in factorising x^2-4
velvet sundial
I dismantled x² - 4
so what've you got after you dismantled it?
high lily
dismantled?
proven monolith
x.x - 2.2 + (2x + 5) (x + 2) = 0
proven monolith
mhm
high lily
there's a factorisation identity for that
simple cradle
$$(a+b)^2 = a^2 + b^2$$
warm shaleBOT
yüce tengri
high lily
no
velvet sundial
WHAT NO
proven monolith
i'm kind of confused rn
high lily
ignore yuce
proven monolith
alright
high lily
x^2 - 4 is a difference of two squares
there's a factorisation identity for that
if you don't know it, look up difference of two squares
proven monolith
I might know it but I don't study in english
gimme a sec
oh that the 3rd identity
Yeah can I just use that?
Literally just solve the equation
high lily
I don't know how the list of identities you're using is numbered
high lily
calling it 3rd means nothing to me
proven monolith
fair enough I thought you guys would call it the same thing
high lily
so yeh, you have a difference of
two squares and you can use the identity for that
proven monolith
yeah I'll do it rn, I'll show you my answer in a bit
(x +2) (x -2) + (2x + 5) (x + 2) = 0
factorise with (x+2)
i got
velvet sundial
yeah that's right
proven monolith
(x +2) (x - 2 +2x + 5) = 0
(x + 2) (x + 3) = 0
So now I just use the zero factor thingy
right
either x + 2 = 0 or x + 3 = 0
high lily
messed up simplification
proven monolith
i think so
velvet sundial
where do you get the extra 3
high lily
3x+3 is fine
velvet sundial
oh yeah
proven monolith
2x + x
- 2 - 5
wait no
5 - 2
alright i think i carry on thanks for the help!
.close
hazy vine
3+9/5+27/25+81/125+243/625+...
obtuse pebbleBOT
3+9/5+27/25+81/125+243/625+...
nocturne sun
3+9/5+27/25+81/125+243/625+...
$\sum_{n=1}^{\infty} \frac{3^n}{5^{n - 1}}$
this is what you have right?
warm shaleBOT
Doggo
hazy vine
nop
nocturne sun
hm?
hazy vine
my question is literally what I typed
but I guess I could find the sequence and write it like that
nocturne sun
alright so it's a geometric series
$\sum_{n=1}^{\infty} \frac{3^n}{5^{n - 1}} = \sum_{n=1}^{\infty} 5 (\frac35)^n$
hm wait I may have done it wrong
gimme a sec
yeah let me edit
warm shaleBOT
Doggo
nocturne sun
alright
nocturne sun
yeah should be correct
lucid delta
Hey I'm stuck on a logic problem and I can't find any sources on how to convert from CNF to CCNF
obtuse pebbleBOT
Hey I'm stuck on a logic problem and I can't find any sources on how to convert ...
@lucid delta Has your question been resolved?
obtuse pebbleBOT
@lucid delta Has your question been resolved?
obtuse pebbleBOT
@lucid delta Has your question been resolved?
obtuse pebbleBOT
@lucid delta Has your question been resolved?
willow ravine
Well I take that back. The false part that is.
It is always true if r is true though.
obtuse pebbleBOT
@lucid delta Has your question been resolved?
tardy epoch
i see 4 vectors
two blue ones look like they could lie on the same line so one's a scalar multiple of the other
royal basin
ok right so
From what I was told, two vectors are only linearly dependent if they are parallel,
this works only if you're dealing with a vector space consisting of like... geometrically drawn vectors, as in those little arrow thingies
i.e. that visual image of linear independence holds only for R^2 and R^3
for this, they give you two functions and ask if the restrictions of these functions to [0,1] are linearly independent as elements of C[0,1].
by "the restrictions of these functions to [0,1]", i mean this:
obtuse pebbleBOT
tardy epoch
no
obtuse pebbleBOT
Available help channel!
pulsar coral
can someone help me with this
obtuse pebbleBOT
can someone help me with this
pulsar coral
timid silo
take two points
(x1,y1),(x2,y2)
pulsar coral
i got it
timber osprey
Hi, my struggle continues with probability
Two 6sided dice are thrown continuously at the same time, until the multiplication of the resulting numbers from the 2 dice are odd
What's the probability of this ^ happening in the third try?
obtuse pebbleBOT
Hi, my struggle continues with probability
Two 6sided dice are thrown continuou...
tardy epoch
Hi, my struggle continues with probability
Two 6sided dice are thrown continuou...
Use conditional probability
Probably want to find the probability mass function for the products of the two dice
timber osprey
Use conditional probability
P(A|B)?
timber osprey
Probably want to find the probability mass function for the products of the two ...
I don't think I've learned this
From conditional probability, I only know the Bayes' theorem
tardy epoch
You probably have. Just called it something else
Unless you prove something about the symmetry or something and show that X has equal probability of being odd and even
timber osprey
Unless you prove something about the symmetry or something and show that X has e...
I think my solution may be right
First I specified the probability of the product being odd
3/6 * 3/6 = 1/4
So, the probably of NOT being odd, aka even, is 3/4
We want this to be odd on the third try
So, the first try must be even, the second try must be even, the third try must be odd
=> 3/4 * 3/4 * 1/4 = 9/64
short zodiac
Yes, I think this is correct
stone rain
hello
blissful seal
how can i solve this
obtuse pebbleBOT
how can i solve this
cloud dove
you know that when x=0 that y=2
fickle turret
What have you tried so far?
cloud dove
try plugging in 0 for the exponents and see what happens
blissful seal
What have you tried so far?
dont know what the x power means and whats the number under it indicates
blissful seal
try plugging in 0 for the exponents and see what happens
and which answer am i looking for
cloud dove
well, when you plug in 0 for x, you try to solve for f(0)
fickle turret
dont know what the x power means and whats the number under it indicates
the x means how often the number below is multiplied with itself
cloud dove
and see what matches up w/ the graph
from there, you should have a good idea where to go from that point
blissful seal
the x means how often the number below is multiplied with itself
i know that but like what does it indicate on the graph
blissful seal
im really lost tbh
cloud dove
for starters, id do like
$f(0) = -2 * 3^0$
warm shaleBOT
meek
blissful seal
that would equal to -2
blissful seal
aha ok
cloud dove
then figure out which out of them is the correct one
you can try plugging x=1 to determine that then
blissful seal
thank you dude
cloud dove
nwnw
blissful seal
.close
obtuse pebbleBOT
fickle turret
There are still two of them
cloud dove
you can try plugging x=1 to determine that then
yeah then do this
to make sure you dont pick the wrong one out of the two
blissful seal
.reopen
obtuse pebbleBOT
✅
blissful seal
to make sure you dont pick the wrong one out of the two
yeah i just did that and found 2 so plug x=1 and do the same thing?
cloud dove
yeah looking at the graph, it looks itd be (1,6)
cloud dove
oh no i just looked at the graph lool
blissful seal
and how did 1,6 come up xD
cloud dove
but for b) if you do $f(1) = 2*(1/3)^1\rightarrow 2/3$
warm shaleBOT
meek
cloud dove
which means that the graph should be going from 2-> 2/3
blissful seal
you lost me here
cloud dove
but in d) if you do $f(1) = 2*3^1\rightarrow 6$
warm shaleBOT
meek
cloud dove
which means the graph would be increasing
cloud dove
well it doesnt really matter
if it's (b) then it's going to lower values after 0
but if its (d) then its increasing after 0
and the graph given is increasing
are u following so far?
blissful seal
yes
blissful seal
it is thank you
Tbh
like every other question seem to be hard for me even though i studied
cloud dove
i wish you much luck
blissful seal
i wish you much luck
Thank you.
fickle turret
Maybe something to mention:
(1/3)^x = 3^(-x) which is just 3^x reflected on Y-axis
cloud dove
oh smart
fickle turret
You're welcome l. Meek did the most
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
icy yacht
So I have a question. If dec 21 at 2020 at 6am was the last update of my app, what day would it be 5 updates evenly scattered from that day to the current day at 6am?
obtuse pebbleBOT
So I have a question. If dec 21 at 2020 at 6am was the last update of my app, wh...
@icy yacht Has your question been resolved?
somber mural
your question doesn't make sense
icy yacht
Like I have an app on my phone that was installed on dec 21 2020 at 7pm and after that update, there’s 5 more after it. The last update of the 5 was yesterday at 7pm. My question is was for the 4 updates before yesterday’s one, if I evenly distributed it from yesterday to the other day (dec 21 2020) what would each month and day be for all 4?
obtuse pebbleBOT
@icy yacht Has your question been resolved?
obtuse pebbleBOT
@icy yacht Has your question been resolved?
obtuse pebbleBOT
@icy yacht Has your question been resolved?
icy yacht
What’s that?
kindred silo
not sure how to do this at all
obtuse pebbleBOT
not sure how to do this at all
royal basin
are you familiar with things such as double-angle identities?
namely the one for sin(2x)
royal basin
$\sin(2x) = 2\sin(x)\cos(x)$
warm shaleBOT
Ann
royal basin
does this ring any bells? or have you only heard wind of it
royal basin
well notice that what you have here can be written as [2sin(pi/8)cos(pi/8)]^2
low halo
Natural numbers m, n, k when divided by 7 give the remainders 3, 5 and 1 respectively, then the number mk + n gives the remainder when divided by 7 ?
obtuse pebbleBOT
Natural numbers m, n, k when divided by 7 give the remainders 3, 5 and 1 respect...
low halo
Sorry for posting so many times my internet breaks due to a storm
And the solution is 1 i mean the remainder
proven zephyr
so... what's the question?
low halo
Idk i have written all i was given
pallid canyon
Are you looking for the remainder of mk + n when divided by 7?
low halo
Yea i cant figure out hot to get 1 basicaly
proven zephyr
oh ok
valid crown
(3×1 + 5)mod7
kind hawk
if you want to, write m=7x+3, n=7y+5 and k=7z+1, where x,y,z are some integers
then multiply out (7x+3)(7z+1)+(7y+5) and take the remainder when you divide it by 7
valid crown
altough good tactic
kind hawk
thanks
low halo
Yea its fine i get what he meant
kind hawk
but the great thing about remainders is that it doesn't matter if you first take remainders and then add/multiply or first add/multiply and after that take remainders
maybe if you are interested try proving that for yourself
valid crown
last thing to observe is any non 7×variable part adds up to 8
timid silo
then multiply out (7x+3)(7z+1)+(7y+5) and take the remainder when you divide it ...
you could just do modular arithmetic, without doing what denascite has written out.
kind hawk
well if they knew what modular arithmetic is then they wouldn't have asked I think
timid silo
then maybe we should explain modular arithmetic.
obtuse pebbleBOT
@low halo Has your question been resolved?
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
ivory flare
Hellooo I’m back. I tried to solve this and I have no idea what’s happening. I tried a math app but it gave me part of the answer and showed no steps. The correct answer is -3,4
obtuse pebbleBOT
Hellooo I’m back. I tried to solve this and I have no idea what’s happening. I t...
ivory flare
ivory flare
I tried to remove the square
pine sail
There are no squares?
pine sail
Yes but your square root is all over 9w^2 - 108
Infact you didn't do anything to the RHS
swift marsh
Hellooo I’m back. I tried to solve this and I have no idea what’s happening. I t...
-3 is not a solution
And just solve the resulting quadratic
pine sail
There is NO NEED of square root.
And yes, (-3) is not a solution.
Because it is not in the domain.
ivory flare
This was the answer sheet the teacher provided
pine sail
They are wrong.
swift marsh
Teacher wasn’t paying attention
ivory flare
Infact you didn't do anything to the RHS
The what?
pine sail
9w^2 - 108 = 9w
ivory flare
Gosh if my teacher was actually good i wouldn't be here guys😭 but i cant believe the answer sheet is wrong
pine sail
9w^2 - 108 = 9w
Do you have any issue up until now?
ivory flare
Do you have any issue up until now?
Nope
swift marsh
It’s not that big of a deal they just forgot to check at the end
ivory flare
It’s not that big of a deal they just forgot to check at the end
Nah bro I promise you she didn't teach us how to solve complex logarithmic equations thats why im so lost
ivory flare
Can you solve a quadratic?
.... sort of...
pine sail
-3 is a correct solution
Really?
restive ridge
Yes
ivory flare
To the power of 2?
pine sail
RHS has no powers.
restive ridge
$\ln{-n} = \ln{-1} + \ln{n} = \ln{n} + (2k + 1)\pi i, k \in \mathbb{Z}$
warm shaleBOT
illuminator3
swift marsh
Lmao this is obviously over the real numbers
restive ridge
they used them in their attempted solution
ivory flare
I don't think beargur knows complex numbers yet.
My teacher was supposed to explain it but she made us do weird projects instead? And we barely understood so we couldn't teach each other
ivory flare
they used them in their attempted solution
Attempted😭😭
pine sail
But you are taking those in your solutions, interesting.
ivory flare
My teacher is nice she's just all over the place and scattered
pine sail
Then I'll have to agree, this isn't limited to real numbers my bad.
ivory flare
So my answers and methods are scattered
pine sail
So my answers and methods are scattered
Just solve the quadratic.
ivory flare
Then I'll have to agree, this isn't limited to real numbers my bad.
So -3 is a solution?
restive ridge
That's like a cool magic trick
yea it's due to the fact that $$\begin{align}e^{i\theta} = \cos{\theta} + i\sin{\theta} \ \implies e^{i(2k+1)\pi} = \cos{(2k+1)\pi} + i\sin{(2k+1)\pi} = -1 + i0 = -1 \ \implies \ln{-1} = i(2k + 1)\pi, k \in \mathbb{Z}\end{align}$$
swift marsh
Have you just started learning about logarithms
restive ridge
ok texit sux
pine sail
If a complex valued logarithm is acceptable then definitely (-3) is a solution.
ivory flare
Have you just started learning about logarithms
I'm taking my final tomorrow 
warm shaleBOT
illuminator3
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
pine sail
If you are just talking about real numbers, then it's just 4.
restive ridge
there
ivory flare
I thought i understood them but I apparently don't
swift marsh
I'm taking my final tomorrow <:melody:880877652574097489>
No I mean has whatever course you’ve been taking been your first exposure to them
ivory flare
No I mean has whatever course you’ve been taking been your first exposure to the...
Yeah my highschool classes. First exposure
ivory flare
Im in 11th grade taking the American curriculum
swift marsh
Yeah my highschool classes. First exposure
Then unless you have some weird educational order you’re probably not supposed to worry about complex-valued logarithms
ivory flare
Then unless you have some weird educational order you’re probably not supposed t...
Im not sure if we have an order at this point everyone is completely lost
@pine sail how do you solve that quadratic though?
pine sail
...
ivory flare
My guy said dot dot dot my bad
swift marsh
Usually you just reject the negative domains because once you first see logarithms and for quite a while we generally just consider them over the real numbers
pine sail
I think an Eleventh grader is expected to solve quadratics?
But I'll tell you.
You can factor it out.
Or use quadratic formula.
ivory flare
I think an Eleventh grader is expected to solve quadratics?
I think I would be able to if covid wasn't an issue and my teacher wasn't terrible
pine sail
Sure.
ivory flare
Or use quadratic formula.
Oh crap that thing
pine sail
Yes that thing.
ivory flare
Oh i tried factoring it out and i got stumped at w²-12
restive ridge
here you get taught solving quadratics in 7th/8th grade
pine sail
@restive ridge I still doubt there's complex numbers involved.
ivory flare
I uh took out the 9
ivory flare
here you get taught solving quadratics in 7th/8th grade
What curriculum are you taught i was taking linear equations
restive ridge
<@391270706186420224> I still doubt there's complex numbers involved.
wfa agrees with me tho
assuming a complex valued logarithm
ivory flare
Rather w^2 - w - 12
How did you get that
restive ridge
anyhow
swift marsh
Yeah I don’t think complex-valued logarithms are in a standard American curriculum, I’m not American but I did an SAT once and the only thing in it was just multiplying them, seems like a steep slope to climb
pine sail
Yeah I don’t think complex-valued logarithms are in a standard American curricul...
That's what I'm talking about.
ivory flare
Is that German