x+y+z+g=50
#help-10
hushed moat
this looks right
hushed moat
yea first question looks right also
rigid gust
Can anyone help me with this ?
obtuse pebbleBOT
Can anyone help me with this ?
forest sinew
wait why cant you just integrate
rigid gust
What do you mean ?
forest sinew
like just integrate it using the power rule
this is $\int_1^2\frac{1}{x^2} \dd x$ right
warm shaleBOT
jan Niku
rigid gust
Yes
rigid gust
But I haven't yet learned the FTC
It's 1 chapter away ?
Can this be solved without FTC ?
plain stag
it looks like they want a riemann sum
obtuse pebbleBOT
rigid gust
.reopen
obtuse pebbleBOT
ā
rigid gust
.close
hollow sail
Can someone help me with 9i I csnt do it
trail cloak
high lily
have you tried anything yet?
hollow sail
Yep
high lily
can you show what you tried
hollow sail
I
high lily
i'd recommend starting from the left side
and try applying the double angle identities for sine and cosine
hollow sail
Okok
gilded needle
if all else fails, you could probably brute-force it by cross multiplying by the denominators and simplifying using the product formulas until you get an equation that is obviously true
hollow sail
Okok thx
cursive trellis
obtuse pebbleBOT
cursive trellis
For domain i got
that x is R except for 1
but i dont know how to find the range
kind hawk
Check for which y you can solve f(x)=y
cursive trellis
you want me to just insert random numbers?
high lily
you could consider long division / decomposition
cursive trellis
after dividing i get
1+(1/x-1)
oh ok so the range is Real numbers except 1?
.close
obtuse pebbleBOT
cursive trellis
thanks
obtuse pebbleBOT
Available help channel!
toxic hollow
obtuse pebbleBOT
toxic hollow
(t1,t2) is the point on the circle and I used the fact theyāre equal distance
However Iāve been stuck on the next part for quite some time
timber fox
ok so radius of the circle is 1 , its centred at origin, so we can say (1,0) is there, and we are given (2,0) to be double the distance, sooooo (3/2,0) has to exist on the curve
now to show it
:_:
@toxic hollow
you could just plug x=3/2 into your equation
you should get y = 0
also you could put your equation into general form of a hyperbola to get the intersection point
but that seems pretty painful
obtuse pebbleBOT
@toxic hollow Has your question been resolved?
simple viper
open
timid silo
Yes you can
sage geode
$\sin{x}+\cos{x}=\sin{x}\cos{x} \$
warm shaleBOT
A Lonely Bean
sage geode
Didn't mean to, wait
simple viper
Yea i mean its right
This is the problem
i just wanted to know if we can find x cus this is not the full question actually
sage geode
$\sin{x}+\cos{x}=\sin{x}\cos{x} $
$\sin^2{x} + 2\sin{x}\cos{x}+\cos^2{x}=(\sin{x}\cos{x})^2 \$
$1+2\sin{x}\cos{x}=(\sin{x}\cos{x})^2 \$
$1=(\sin{x}\cos{x})^2-2\sin{x}\cos{x} \$
$2=(\sin{x}\cos{x})^2-2\sin{x}\cos{x}+1=(\sin{x}\cos{x}-1)^2 \$
$\sin{x}\cos{x}-1=-\sqrt{2} \$ (apparently it can't be the positive root because the maximum value of $\sin{x}\cos{x}-1$ is lower than that)
$\ \sin{x}\cos{x} = 1-\sqrt{2}$
warm shaleBOT
A Lonely Bean
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
simple viper
No no
So i said
its not the full problem right?
This info helps
cus we had to find
Cos^3 + sin^3
Wait wait
simple viper
That's not the end though
just one last thing
u took sqrt both sides right
after that
how we neglected the other value
sage geode
**A Lonely Bean**
Compile Error! Click the <:errors:504266556054962196> reaction...
Read my comment in parenthesis
sage geode
It's fine
drifting loom
can someone help me with this
obtuse pebbleBOT
can someone help me with this
drifting loom
?
lost tree
When theta is a small angle sine theta is approximately equal to theta
lost tree
What is the force exerted on one leaf by the other
lost tree
And E = ?
lost tree
Right u can express r in terms of l and theta
lost tree
I suppose u must consider there to be like a tension force along each leaf
drifting loom
ok
lost tree
Set its vertical component = mg, and horizontal component equal to electrostatic force
lost tree
?
drifting loom
then i found out tantheta/2
lemme send
see
i missed r^2in last equation
srry for that
I fixed it
lost tree
tantheta is approx equal to theta when theta is smll
drifting loom
now what to do
ok
then
i am getting this
but its not in options
what to do
where i did wrong
lost tree
RĀ² is not lĀ²
drifting loom
ohh
drifting loom
is electric force is acting on horizontal direction
then i have to take l in terms of sintheta right?
lost tree
Yes
lost tree
R=2lsinĀ½theta
drifting loom
how
r=lsintheta1/2theta
this is ot
?
can you explain me plss
am i considering only one point thats why ??
@lost tree
help
drifting loom
.reopen
obtuse pebbleBOT
ā
drifting loom
helps
obtuse pebbleBOT
@drifting loom Has your question been resolved?
drifting loom
<@&286206848099549185>
obtuse pebbleBOT
@drifting loom Has your question been resolved?
lost tree
hey
drifting loom
yes
lost tree
do u get it
warped flume
obtuse pebbleBOT
warped flume
i was thinking that the goat can graze in a half circle
the mark scheme gives me whatever the hell this is
can someone please explain
main rose
Remember the the goat attached with a rope, so it can wrap around corners
warped flume
Remember the the goat attached with a rope, so it can wrap around corners
how's angle FAH alpha + pi/3
main rose
how's angle FAH alpha + pi/3
Try drawing a line parallel tot BC through A and apply Z and F angles
obtuse pebbleBOT
@warped flume Has your question been resolved?
cobalt valve
obtuse pebbleBOT
cobalt valve
Can someone help me solve this problem?
I do not have any idea how to approach this one
upbeat gazelle
You may use the Quick exponentiation of matrices
obtuse pebbleBOT
@cobalt valve Has your question been resolved?
strong atlas
if p(x) = ax+b
g(x) = bx+a
and p(g(1)) = g(p(1)) ,then find the value of p(1)
strong atlas
options:
A)1
B)0
C)-1
fierce lagoon
What have you tried
sand hinge
What is the main question?
sand hinge
What is the main question?
?????
fierce lagoon
Scrol up
I think you did p(g(1)) incorrectly
g(1) is a + b
p(a+b) = a^2 + ab + b
Oh but the ab's cancel out
Eventually
strong atlas
ya
warm shaleBOT
Umbraleviathan
fierce lagoon
It suddenly becomes clear
Use difference of squares
@strong atlas you know how I got that?
strong atlas
Ya,after this we have that b+a =1,
OHHHH
Damn
Iām so dumb
Thnx a lot
@fierce lagoon
.close
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
plain grove
Hi guys Iām having trouble simplifying this
obtuse pebbleBOT
Hi guys Iām having trouble simplifying this
slim lake
Sure send
plain grove
,rotate
warm shaleBOT
plain grove
Is it 7 log 3 x^2?
plain grove
How come x isnāt squared
timid silo
or log3(x^7)
plain grove
Oh
fierce lagoon
Why would x be squared
warm shaleBOT
plain grove
That first law
timid silo
Thats if they have the same leading coefficient
fierce lagoon
Yrah
plain grove
,rotate
warm shaleBOT
timid silo
try taking both sides to the power of e
plain grove
What like how
obtuse pebbleBOT
@plain grove Has your question been resolved?
upper fog
move log e to the right
combine the two
1=loge(7x)
7x=e^1
x = e/7
idk why they dont just write ln
for log e
@plain grove
obtuse pebbleBOT
upper fog
idk how to use it
but you just move log e (x) to the right
and use the addition property to combine the two
obtuse pebbleBOT
Available help channel!
cursive tundra
Hey guys
cursive tundra
I have a question regarding statistical terms for linear trends
I forgot the word to describe the high growth rate in 1.
Whereas in 2. I can think of words like ābecoming more flat, slowing down, etcā
But for 1. What words would be appropriate to describe it? Like in contrast to 2.
For 3. I think can say maybe it started off very ___ and then it slowed down and became more flat etc?
warm canopy
Slope, gradient
cursive tundra
Yeah yeah but i wanna describe the slope
And cannot find the right words to describe a slope like in 1.
As compared to what i said about 2
tardy epoch
And cannot find the right words to describe a slope like in 1.
You're gonna need to draw a better picture
1 and 2 both look like lines
1 has slope > 1 and 2 has slope between 0 and 1
Assuming the axes are the same scale
3 could be sqrt or log
cursive tundra
Yeah 3 is similar to logs but for 1 and 2 they are just linear no?
tardy epoch
1 and 2 both look like lines
That's what I said
cursive tundra
Yeah and i would like to know how i can describe a slope that results in something like 1
Like very upward
tardy epoch
Steep slope
cursive tundra
In contrast to the flatter line in 2
Ohhhhh
STEEP
YES THANK U!
And for 2, is flat appropriate?
Any other terms?
rich tapir
Anyone got a good tool for expanding polynomials, no sites seems to like mine
$(1+x+x^4+x^9+x^{16}+x^{25}+x^{36})*(x^{-144}+2x^{-120}+x^{-100}+2x^{-96}+2x^{-80}+2x^{-72}+x^{-64}+2x^{-60}+4x^{-48}+2x^{-40}+x^{-36}+2x^{-32}+4x^{-24}+2x^{-20}+3x^{-16}+2x^{-12}+2x^{-8}+x^{-4}+13)$ 
š¤§
obtuse pebbleBOT
Anyone got a good tool for expanding polynomials, no sites seems to like mine
$(...
warm shaleBOT
alihsaas
trail cliff
oh my
,ask (1+x+x^4+x^9+x^{16}+x^{25}+x^{36})*(x^{-144}+2x^{-120}+x^{-100}+2x^{-96}+2x^{-80}+2x^{-72}+x^{-64}+2x^{-60}+4x^{-48}+2x^{-40}+x^{-36}+2x^{-32}+4x^{-24}+2x^{-20}+3x^{-16}+2x^{-12}+2x^{-8}+x^{-4}+13)
trail cliff
š¤
timid silo
,w expand (1+x+x^4+x^9+x^{16}+x^{25}+x^{36})*(x^{-144}+2x^{-120}+x^{-100}+2x^{-96}+2x^{-80}+2x^{-72}+x^{-64}+2x^{-60}+4x^{-48}+2x^{-40}+x^{-36}+2x^{-32 }+4x^{-24}+2x^{-20}+3x^{-16}+2x^{-12}+2x^{-8}+x^{-4}+13)
warm shaleBOT
trail cliff
š
rich tapir
trail cliff
I would not allow wolfram to hit me with that
timid silo
you could also do it by hand if you have the time
rich tapir
My favorite hobby, expanding product of polynomials 
well, I dont think there is escaping that fate
thanks
.close
karmic hedge
obtuse pebbleBOT
https://gyazo.com/d31779f7e757262de189f189fa30ef94.png
tranquil quiver
I know that IIvII is sqrt(3) and IIwII is sqrt(13)
do you
tranquil quiver
how do you calculate it
karmic hedge
sqrt(3^2 + 2^2)
little warren
karmic hedge
sqrt((-1) + 2)^2 like this?
little warren
no no
but you didn't put the "-" in the paren
which will eventually give -1 instead of 1
karmic hedge
wow really stupid mistake haha
tranquil quiver
thats unfortunate
little warren
obtuse pebbleBOT
little warren
cya š
oak knot
Hi, I have some short answer questions about general format, etc.
The unit I am looking at is Functions and quadratic equations
-
When completing the square, do you have to set the ax^2 + bx + c equation equal to zero?
-
When partial factoring, is it correct to set f(x) to be equal to c (in ax^2 + bx + c) , thus having ax^2 + bx to be equal to 0 before starting?
-
What should my conclusion be if the number under the radical is a negative (and thus there are no x-intercepts)
Thanks !
obtuse pebbleBOT
Hi, I have some short answer questions about general format, etc.
The unit I am ...
solemn trout
- Yes, you do set the quadratic equal to zero.
timid silo
you only have to set the term equal to 0 if youre looking for roots/zeros
wild swallow
1. Yes, you do set the quadratic equal to zero.
no you don't 
solemn trout
I presume the question asker is solving the roots of quadratics.
wild swallow
you still don't need to set anything to 0 to complete the square
solemn trout
Since that is what "completing the square" generally eludes to.
wild swallow
oak knot
<:uhhhh:826921081490636851>
uhhhhhh
wild swallow
that's just not right
completing the square is a separate process to solving for the roots of a quadratic
solemn trout
If you are solving for roots, then yes, you set the quadratic to equal to 0.
oak knot
wild swallow
but you do that every time you solve for roots
timid silo
like, solve
xĀ²+4x+3=-1
would you set this equal to 0?
wild swallow
you can complete the square in contexts other than solving for roots
a basic example is trying to find the vertex of the parabola
solemn trout
- You'll have no real solutions. Hence no x-intercepts. That being said, you should not say the quadratic has no roots at all.
solemn trout
a basic example is trying to find the vertex of the parabola
That is fair.
solemn trout
Then you don't need to set it to zero.
(I guess I predicted wrong)
oak knot
Would setting it to 0 be wrong
wild swallow
it would not be right
solemn trout
You do that if you are solving for the roots.
oak knot
uhhh
wild swallow
you would be better off writing f(x) = ...
oak knot
i am so sorry what
wild swallow
instead of 0 = ...
wild swallow
then you would be solving for x in the equation or something
solemn trout
- From what I can parse, I'm inclined to say no. Can you give more context behind the question?
wild swallow
not finding a vertex
lone berry
Critical points are where the derivative = 0, right?
Just checking cuz I haven't done this in a while and I honestly forget
oak knot
then you would be solving for x in the equation or something
MM OH I see
Yeah okay that makes sense
wild swallow
Critical points are where the derivative = 0, right?
sometimes
oak knot
2. From what I can parse, I'm inclined to say no. Can you give more context behi...
Let me see
solemn trout
@oak knot What class is this by the way? Is it Calculus? Or Algebra?
oak knot
solemn trout
Well, that is a way of finding the vertex.
oak knot
The other im assuming is completing the square? and then (-h, k)
solemn trout
- It seems the process you were taught is simply for finding the vertex. The idea being that the vertex is equidistant from
x=0and the pointx =/= 0such thatf(x)=c.
So in the context of this process, then yes.
Just understand that what you are doing is solving for f(x)=c, which is generally easy to do.
solemn trout
- No in general. Only if you are solving for roots.
- Yes for your question in particular.
- Write no real solutions.
oak knot
.close
nimble harbor
i have no idea on how to do this, am i supposed to find the sum of the quadratic equation or else?
obtuse pebbleBOT
i have no idea on how to do this, am i supposed to find the sum of the quadratic...
clever temple
just a simple question, why is it called diamond shape when it's not close to the shape of real diamond at all
obtuse pebbleBOT
just a simple question, why is it called diamond shape when it's not close to th...
clever temple
i feel like rhombus should be used more instead of diamond
how is real diamond close to the diamond shape at all
im really irritated
can someone explain
clever temple
this one has 5 sides
trail musk
there's also baseball diamonds
warm canopy
you can find geometric diamonds on a diamond
clever temple
you can find geometric diamonds on a diamond
oh those little shapes?
i think i can see them
trail musk
warm canopy
but really diamond is a vague mathematical term and youre better off using rhombus
wanton dagger
I know i need to make a quadratic equation here
obtuse pebbleBOT
I know i need to make a quadratic equation here
sharp pecan
yesw
sharp pecan
yes
timid silo
Total area is 49
sharp pecan
uwu
timid silo
Itās a square so you can sqrt it and get the side lengths of the veranda
Since it surrounds the hut
timid silo
I figured it out for the quadratic equation
wanton dagger
I see
nimble harbor
the equation is 0.2t^2-5.6t+50.2=30 isn't it? just want to clarify since the solution is odd
obtuse pebbleBOT
the equation is 0.2t^2-5.6t+50.2=30 isn't it? just want to clarify since the sol...
timid silo
Time is continuous so it can have decimal points
If thatās what youāre asking
You canāt have negative time though
nimble harbor
hmm i just didn't arrive at a whole number, my solution is $\frac{5.6 +or- \sqrt{15.2}}{0.4}$
warm shaleBOT
sus
timid silo
hmm i just didn't arrive at a whole number, my solution is $\frac{5.6 +or- \sqrt...
Yes this is fine
You can have a fraction of a month (1 week)
You can check the table of values to verify it yourself
nimble harbor
i wouldn't be able to answer the question though with the radical, any ideas?
timid silo
hmm i just didn't arrive at a whole number, my solution is $\frac{5.6 +or- \sqrt...
How did you get this
nimble harbor
quadratic formula 
timid silo
You should have done the quadratic formula for 0.2x^2-5.6x+20.2
nocturne minnow
You should have done the quadratic formula for 0.2x^2-5.6x+20.2
They did
$(-5.6)^2 - 4 \cdot 0.2 \cdot 20.2 = 15.2$
warm shaleBOT
dldh06
timid silo
Oh my bad
nimble harbor
what do i do with the results now then, can't i simplify it anymore?
i think i'll just.. answer it with the radical lol
that's probably what they expect
alright thanks
.close
grim mirage
Let ${X_n}$ be i.i.d. with $\mathbb{E}[X_1] = 0,\ \mathbb{E}[X^2_1] = 1$. Set $S_n = \sum_{i=1}^n X_i$. Show that $S_n/(n^{1/2}\log(n)) \rightarrow_{a.s.} 0$.
obtuse pebbleBOT
Let $\{X_n\}$ be i.i.d. with $\mathbb{E}[X_1] = 0,\ \mathbb{E}[X^2_1] = 1$. Set ...
warm shaleBOT
Quaerendo_Invenietis
grim mirage
To show almost sure convergence I would think to use the strong law of large numbers.
However, there seems to be something funky about this.
Actually, maybe not.
.close
mild igloo
Need help with dis
obtuse pebbleBOT
Need help with dis
mild igloo
ASAP A$AP
timid silo
What
mild igloo
Just solve my math problem
timid silo
That wouldnāt be helping you though
Giving you the answer is not helping you solve it
tardy epoch
Just solve my math problem
It's against the server rules to give away answers. Work on it yourself and ask when you get stuck
fierce lagoon
We ain't giving answers lol
timid silo
I ain't know math
Do you know what youāre dealing with in the problem
mild igloo
Nope
timid silo
Do you know what an exponential function is
An exponential function is a function to the x power
What youāre dealing with here is ab^x
timid silo
What youāre dealing with here is ab^x
- c*
mild igloo
No
timid silo
How
mild igloo
Idk
tired skiff
you need a calculator for that problem
teal turret
timid silo
desmos.com
Are there tables of values in desmos ?
teal turret
it's regression
Oops didnāt read that far up sorry
tired skiff
ur good
timid silo
He can pick two points and find it with elimination
tired skiff
it has to be super exact though
timid silo
Yeah but he can use a normal calculator for that
mild igloo
Man y'all fr can't do the math for me
timid silo
bro
mild igloo
Screw this server I'm out I got a day for this I said I needed this ASAP A$AP
Suck my balls
timid silo
š
lost tree
š
tired skiff
ok then
teal turret
He doesnāt have that dawg in him I guess
tired skiff
He doesnāt have that dawg in him I guess
lol
timid silo
.close
tired skiff
he left so.. i guess we just wait
obtuse pebbleBOT
@mild igloo Has your question been resolved?
tribal star
could someone help me solve this matrix row equation
obtuse pebbleBOT
could someone help me solve this matrix row equation
tribal star
i donāt know what to do next but i have to change it to its reduced echelon form
looks like this
nocturne minnow
Do you know what reduced echelon form is?
tribal star
isnt it supposed to be where the whole table is zeros but theres a diagonal path of ones going across it
the last column on the right side isnt a part of the table i forgot to say that
nocturne minnow
isnt it supposed to be where the whole table is zeros but theres a diagonal path...
Yes
tribal star
so then what would i do to change this table to that, cause im stuck and i dont know what to do to change it
nocturne minnow
Did you attempt something?
nocturne minnow
You have 3 different row operations
swapping two rows, multiplying a row by a number, and adding a multiple of one row to the contents of another row
You just need to apply those three repeatedly until you get it into RREF
tribal star
heres what i got to
im planning on swapping row 1 or 3 with row 6
i dont know what else yet though
nocturne minnow
You have 3 different row operations
> swapping two rows, multiplying a row by a ...
.
tribal star
is there like a guide or something on when you know which operation to do
nocturne minnow
You want the first row, first column to be the one with the 1, meaning all the other column elements should be 0
So you need to apply row operations to make the rest of the column 0s
tribal star
so you see how in this, i have 2 rows in the bottom table that sum up to 84 and 87 right
i would be able to get one of those rows down to around the same numbers as the other rows but how would i get the other row
nocturne minnow
You know that you need 1s along the diagonals so you need to making the rest of the column 0s, so use that knowledge and the three row operations to do that
oak bear
Yeah as long as it is separated by the line, youv gotta use pemdas or whatever your school calls it
The fraction line, is that what you're asking?
Maybe just give me an example
Yeah
You gonna use pemdas here
oak bear
Yes this is right
dim yacht
(4/2) - (2/6) + (5/11)
oak bear
I mean this is
Its just in order
No
The rules were applied
Because its addition and subtraction only
If subtraction went first, then it goes first
Oooooh
That's what you're confused about
oak bear
You're way of thinking pemdas is wrong
dim yacht
you use that
oak bear
It should be PE-MD-AS
dim yacht
pe(m or d, whichever comes first in the equation)(a or s, whichever comes first in the equation)
oak bear
It should be PE-MD-AS
A and S goes last,
dim yacht
it could also be written as pedmsa
oak bear
So they're the last priority
Yeah its still the same
Just another name for it
Yeah it is right
But you got it all wrong
BI-DM-AS
oak bear
If A goes first, then it's firsy
dim yacht
left to right
oak bear
If S goes first, then it goes first
dim yacht
so then move on to multiplication/division
if there are no brackets
which would be the fractions
oak bear
You should probably watch a yt video first about bidmas
You got all the basics for bidmas wrong
dim yacht
the answer should be 2.1212 etc or 70/33
oak bear
I brb gotta do something
dim yacht
70/33 is the most simplified it can be
yes it is
yes
can you call so I can explain it to you?
what does your calculator say it is
BI - MD - AS
if there is no BI, move onto MD
MD
fraction is division
4/2 is 2
2/6 is .3333333
5/11 is .45454545
2 - .333333 + .45454545 is equal to 2.1212121212
do you understand at all?
also (4/2 - 2/6) + 5/11 should still give you 2.12121212 or 70/33
can you tell me what you put in your calculator
no
thats not true
it's whichever comes first in the equation
ok but thats how it is
thats just an acronym
the correct way to learn it should be
BI(M/D)(A/S)
they do
yea
yea
5/3 + 5/11 now
55/33 + 15/33
= 70/33
no it doesnt
it's saying whichever comes first in the equation
not addition first
so 5 - 2 + 3
what would that be
dude
I don't know how to explain it to you
dim yacht
so 5 - 2 + 3
this is equal to 6
5 - 2 is 3
3+3 is 6
its left to right
to show order between brackets, exponents, multiplication/division, and addition/subtraction
the point of bidmas is to show that you always do brackets first
then exponents
then multiplication or division
whichever comes first in the equation
same with subtraction and addition
because thats what bidmas says to do
bidmas can be rewritten as bimdsa
and it's the exact same
you do whatever comes first in the expression or equation
yes
solve it
step by step
you do it left to right
not addition before subtraction
in your expression subtraction is first
so you'd do subtraction
thats just how it works
it's the conventional method somebody came up with that everybody uses
it doesnt
Im telling you how it works and you refuse to listen to me
whole dock
The bracket comes first
dim yacht
solve your equation step by step for me please
whole dock
So you solve the numbers in bracket first
dim yacht
(9+4) - 4 + 5
whole dock
$(9 + 4) - 4 + 5 = 13 - 4 + 5 = 14$
warm shaleBOT
ColdTee
dim yacht
for the second step it's not 13 + 5 -4
it's 13 - 4 + 5
in that order
because in your expression that you created subtraction comes first
thats not true
put it into a calculator
I've explained why
yes but you do subtraction first
since subtraction is first in the expression
going from left to right
as I've said before, it is to show the order of operations
it should be B - I - (M or D, whichever comes first in the expression) - (A or S, whichever comes first in the expression)
and as I said before it can be rewritten as bimdsa instead of bidmas
because it's simply telling you to do either multiplication or division in the order it appears in the expression
and addition or subtraction in the order it appears in the expression
PEMDAS is the same as BIDMAS
yes
same with multiplication and division
yes
multiplication and division both do
so solve this
5/7 - 2/3 + 4/3
close
you have order of operations correct
but multiplying and simplifying the fractions was wrong
at 1/21 + 4/3
it should be 1/21 + 28/21
not 3/84 + 63/84
yep
good job
Im proud of you
nice
how old are you?
if you don't mind me asking
Math is fascinating once you learn to enjoy it and not worry about tests and such
nice
dm me anytime you have a question about math
Ill be on periodically everyday pretty much
16
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
lone vault
how do i simplify cos(x)/tan(x)(1 - sinx)?
obtuse pebbleBOT
how do i simplify cos(x)/tan(x)(1 - sinx)?
royal basin
is the (1-sin(x)) in the numerator or in the denominator?
fading musk
Use latex
royal basin
$\frac{\cos(x)}{\frac{\sin(x)}{\cos(x)}(1-\sin(x))}$
warm shaleBOT
Ann
lone vault
yea
fading musk
alr
obtuse pebbleBOT
@lone vault Has your question been resolved?
winter cargo
is it (1/sin)+1?
lone vault
no 1/sinx + 1
fading musk
(1+sin)/sin
you just committed a crime
rich cipher
integrate e ^ x * ln(x) dx can you help
fading musk
integrate e ^ x * ln(x) dx can you help
Type it in #help-1
wanton thistle
Hi
Can anyone help me solve this question?
obtuse pebbleBOT
Hi
Can anyone help me solve this question?
green epoch
Do you know the sin2x formula
wanton thistle
Yes
tall adder
wdym
tall adder
wait, do we have enough info to do this tho
i think yes
sinĀ²x + cosĀ²x = 1
royal basin
yeah, we need to know what sign cos(x/2) is
tall adder
yeah, we need to know what sign cos(x/2) is
through that law
royal basin
no
sin^2(x)+cos^2(x)=1 does not let you determine if cos(x) is positive or negative...
tardy epoch
Hi
Can anyone help me solve this question?
Are you given a quadrant for x
wanton thistle
Are you given a quadrant for x
No, whats in the picture is all thats given
tall adder
we dont need to know if it was pos or neg because its squared in the law
winter cargo
you need to find cos(x/2) first
tall adder
you need to find cos(x/2) first
true
next reef
Yeah
distant moth
x=2sin^{-1}(1/3)
Thats only if x/2 belongs from -pi/2 to pi/2?
tall adder
cos x/2 = 2ā2/3
winter cargo
you need to find cos(x/2) first
which is equal to $\sqrt{(1-(\frac{1}{3})^2)}$
wanton thistle
you need to find cos(x/2) first
Yeah its (2 * root 2)/3
warm shaleBOT
GameSwitch
tardy epoch
which is equal to $\sqrt{(1-(\frac{1}{3})^2)}$
How do you know to take the positive square root
winter cargo
yeah im coming to that
royal basin
x=2sin^{-1}(1/3)
2arcsin(1/3) is not the only possible value of x
there is also pi minus that
winter cargo
$=\pm\sqrt{1-\frac{1}{9}}$
$=\pm\sqrt{\frac{8}{9}}$
$=\pm\frac{2\sqrt{2}}{3}$
tardy epoch
Should be +- in all terms
warm shaleBOT
GameSwitch
winter cargo
like this?
wanton thistle
Yes
winter cargo
$sin(x)=2sin(\frac{x}{2})cos(\frac{x}{2})\$
$=\pm2(\frac{1}{3})(\frac{2\sqrt{2}}{3})$
$=\pm2(\frac{2\sqrt{2}}{9})\$
$=\pm\frac{4\sqrt{2}}{9}$
tardy epoch
wait, do we have enough info to do this tho
Yes we do
warm shaleBOT
GameSwitch
royal basin
@tardy epoch even sign?
tardy epoch
if the question were asking for x itself, we don't. But the sin(2x) and sin(pi-2x) are the same
Ah shit I didn't multiply by 2 enough hang on
royal basin
sin(4pi - 2x) = **-**sin(2x)
winter cargo
$cos(x) = \pm\sqrt{1-sin^2(x)}\$
$= \pm\sqrt{1-\frac{32}{81}}\$
$= \pm\sqrt{\frac{49}{81}}\$
$= \pm\frac{7}{9}$
warm shaleBOT
GameSwitch
winter cargo
we have cos(x) and sin(x) now
tardy epoch
sin(4pi - 2x) = **-**sin(2x)
Yea nevermind me I don't think so
tall adder
we have cos(x) and sin(x) now
finally, inner peace
tardy epoch
we have cos(x) and sin(x) now
You have two possibilities for each of them
tall adder
š§š§
tardy epoch
yeah im coming to that
Did you explain this yet
winter cargo
yeah thats why i put plus or minus wdym
wanton thistle
**GameSwitch**
Yeah that was what i was stuck on
winter cargo
$sin(2x) = 2sin(x)cos(x)\$
$= 2(\pm\frac{4\sqrt{2}}{9})(\pm\frac{7}{9})$
$= \pm2(\frac{28\sqrt{2}}{81})$
$= \pm\frac{56\sqrt{2}}{81}$
warm shaleBOT
GameSwitch
winter cargo
yeah thats what i got
warm shaleBOT
Result:
0.62853936105471
?
warm shaleBOT
Result:
0.62853936105471
same thing ig
wanton thistle
**GameSwitch**
Got it thanks
winter cargo
the numerical answer is
$\pm 0.97772789497$
warm shaleBOT
GameSwitch
winter cargo
@wanton thistle
obtuse pebbleBOT
@wanton thistle Has your question been resolved?
timid silo
obtuse pebbleBOT
timid silo
how do we do that
gilded needle
how do we do that
what is f?
alpine raven
^
timid silo
1/z
alpine raven
|z-3| = 1
\sqrt((z-3)^2) = 1
(z-3)^2 = 1
z-3 = 1 or z-3 = -1
z = 4 or z = 2
these are the only elements of K
gilded needle
presumably z is complex though
gilded needle
"the circle of radius 1" would be nonsense if it were not though
alpine raven
if its complex analysis, its out of my league sadly
gilded needle
if we let w = 1/z, then 1/w satisfies the equation |1/w - 3| = 1, try starting with that
timid silo
hmm?
gilded needle
the image f(K) is the set of all w in C such that |1/w - 3| = 1
timid silo
the image f(K) is the set of all w in C such that |1/w - 3| = 1
how did we get there
gilded needle
how did we get there
w = 1/z if and only if z = 1/w, right?
(assuming w and z are nonzero as is the case here)
so the set {z : |z-3| = 1} gets mapped to a corresponding set of points w, where each w is 1/z for one of the z's in that set
gilded needle
well the circle is the set {z : |z - 3| = 1}
timid silo
yea
gilded needle
each of those z's gets mapped to a corresponding w = 1/z
and w = 1/z if and only if z = 1/w
gilded needle
so the image is {w : |1/w - 3| = 1}
gilded needle
now it may be possible to simplify |1/w - 3| = 1 to make it more obvious what it is geometrically
but imo, it's a perfectly valid answer as is
warm shaleBOT
Bungo
gilded needle
so $|1/w - 3| = 1$ is the same as $|1-3w| = |w|$
warm shaleBOT
Bungo
gilded needle
you could square both sides
giving you $(1-3w)\overline{(1-3w)} = w\overline{w}$, or $1 - 3w - 3\overline{w} + 9|w|^2 = |w|^2$
warm shaleBOT
Bungo
gilded needle
maybe putting it in polar form would be helpful
timid silo
that doesnt seem simpler
gilded needle
nope, not really
timid silo
you could square both sides
why did u have the idea to square?
gilded needle
why did u have the idea to square?
somewhat standard thing to do with absolute values
btw, when in doubt you can try plotting the original and the image, i did so using the following matlab commands: ```
theta = 0:0.01:(2*pi);
K = exp(i.*theta) + 3;
image_of_K = 1./K;
figure;plot(K);
figure;plot(image_of_K);
figure;plot(K);hold on;plot(image_of_K,'r');
and got this:
which tells you that it's probably a circle, so that gives you something to shoot for
(blue = K, red = image of K)
gilded needle
i'm sure there are online tools you can use as well
not worth learning matlab just to do this kind of stuff, i happen to know it already and had it open on my desktop so i used it
little warren
plt all the way
timid silo
plt?
little warren
matplotlib.pyplot
gilded needle
yea python can do anything matlab can do and a lot more
gilded needle
import matplotlib and use that, that's one way
there are many other plotting packages as well
bokeh is a popular one
browser-based
timid silo
thanks
timid silo
so back to that
gilded needle
yeah basically take the equation i wrote above, rearrange it a bit to $8|w|^2 - 3w - 3\overline{w} = -1$, divide by 8, and complete the square on the LHS
warm shaleBOT
Bungo
gilded needle
that's what your instructor did
gilded needle
well not quite "complete the square", the complex version of that
timid silo
just wanna make sure i understand how we got to the |1/w - 3| = 1
gilded needle
f(K) is the set of f(z) for z in K
damn i'm typing idiotically, it's late here sorry
timid silo
dw