#help-36
1 messages · Page 11 of 1
But doesn’t that mean u go from R^2 to something else
well to go to something else we would have to make our matrix from 2 x n to (not 2 x n), so something like 3 x 2
that would be R^3
for example, say we had a function
$f : \mathbb{R}^2 \rightarrow \mathbb{R}^3$ such that $f(<x, y>) = <x, y, x + y>$
MellowDramaLlama
then that would take 2d Vectors and push them onto the 3d plane
How does that make sense logically tho
what do you mean?
How am I allowed to make a line I draw on a page into a line that exists in 3D
Man that sounds terribly phrased
Idk how to ask it lol
well keep in mind that the 3D plane is nothing more than us look at the x, y plane from straight above
hold on let me draw some visual aids
Thanks
Yes
Sure
I think I understand this more clearly thinking about it now
The nature of the vector being <2,3> doesn’t change
It’s just popping out of the page now
yep!
well almost lol
it's still on a flat plane
oh wait
for our function, it would be <2, 3, 5>
yes that would be popping out of the page
but <2, 3, 0> would be flat on the xy plane
Ah i c
see? not so bad 🙂
now when you get to dimensions >= 4, then it's really hard to visualize
but the math checks out
yea, thats true. its hard to just understand from the textbook cuz unfortunately my college only offered the online course and its asychronous with no lectures, just textbook, webassigns, and tests, so its a bit tough to learn
oh boy
isnt 4d time
thanks so much man, i really appreciate the explanations
of course, happy to help 🙂
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Can i get tips or advice to solve c
How can I learn Graph Theory step by step?
?
Didn't you understand?
Its my question
No, I just asked something to people to help me.
Yah but ure taking my space
its like the question you had earlier
split the shaded region into smaller regions
parts a,b) will be helpful in calculating areas of those smaller regions
How about the small semis
you shouldnt look at it like that i think
seperate it into two pieces, DOA and AOB
since you know the angles of each of the triangles, you can calculate their area and take away the shaded parts
by treating them as smaller triangles
wdym by small semis
what unshaded quad on the left
if quad was short for quadrilateral, the unshaded part isn't a quadrilateral
@final tangle how to find that
applying the appropriate formulae for areas of sectors, triangles,segments
the radius / relevant sides are known
you've pretty much already determined the angles you need as well in parts a,b
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Normally wouldn't we say 400$ on a win, because we paid 100 and won 500, hence 400
and thus the expected value would be 0
@feral pivot Has your question been resolved?
@feral pivot Has your question been resolved?
The wording is kind of vague, but I agree with you.
@feral pivot
Maybe they meant that if she wins, she profits $500
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Hey
Today we talked about sets but i didn't understand this:
let A be a set, then what is supA and infA?
If A is a subset of some ordered set, for example the real numbers
Then sup(A) is the smallest element of the larger set, which is greater than every element of A
For example if A is the set of all real numbers less than 2,
$A = {x \in \R: x < 2}$, then $\sup(A) = 2$
tatpoj
inf is the same, but on the lower end instead
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can anyone explain to me how to determine if a relation is a function
I still dont understand it
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Heracles needs to fight the Lernean Hydra for one of his labors. A Hydra has many heads, and to defeat it, you must remove them all.
Heracles has two swords, and he can only use one of them at a time. The smaller sword will sever 21 of the Hydra’s heads with a single stroke. The larger one will sever 100 heads with each stroke, but then the Hydra (unless it has just been killed) will grow 2023 new heads! If the Hydra initially has 101 heads, is it possible for Heracles to defeat it with his present swords or does he need to procure another sword?
NOTE: A sword can only remove a particular number of heads, no fewer and no more. If, for example, the Hydra has just 5 heads, they cannot be cut off with either of the swords.
is this an invariant question?
if so, whats the invariant
also, i dont think its possible, because every time he severs some heads, they increase by 2023
with is much larger than the largest amount taken away
so it will always be larger
unless im missing smth
The heads only grow if you use the larger sword.
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no
sorry
you are not
ooh
@outer storm get to your channel pls
thanks
we have
4,12,36,...
do you see what sequence this is?
so my way of thinking was: 4 x 3 = 12, 12 x 3 = 16, so i thought it was like multiples of 3 so i divided the 708 number by 3 and got that n = but its wrong
yes i think the sequence is times 3
not sure
right
so 4*3^2=36
in general we see that
4 * 3^0 = 4
4 * 3^1 = 12
4 * 3^2= 36
so in general
right
4*3^(n-1)
the n-1 is because our first element of the sequence has exponent of 0 in this notation
but we want to start n from 1
yes that makes sense
it says that n isnt defined
n=1 means we look at the first element
it is given that 708588 is an element of the sequence
so we can set them equal
4*3^(n-1)=708588
3^(n-1)=177147
log_3(177147)=n-1
n=log_3(177147)+1
,w log_3(177147)+1
looks good
does this step make sense to you?
great
$a^{b}=c \ log_{a}(c)=b$
~Martin
i always read the log as
a equals c if taken to the power of b
base equals () if taken to power of result
oh no
n=log_3(177147)+1
this is what we have
however
we can calculate this
,w log_3(177147)+1
you spooked me for a second haha
spooky season
fr
idk if i can continue using this channel but heres the next (if i can)
sure thing
what do they mean with common ratio?
im not used to english terms here
but a_n are easy
$a_{1}=\frac{9^{1+3}}{6}=\frac{9^{4}}{6}=\frac{6561}{6}=1093.5$
~Martin
i think its 3
because they are all divisible by 3
ok 3 is not correct
lol
i have unlimited tries btw so
i think with common ratio they mean a_n/a_n-1
or something like that
oh great
$\frac{9^{n+3}}{6}:\frac{9^{n+4}}{6}=1:9$
~Martin
so it is either 9 or 1/9
9 is correct
,w (9^(1+3))/6
,w (9^(2+3))/6
,w 9*1093.5
got that correct!! thank you!!
this is what i did
lol
great stuff
so 5 more questions left!! almost there
we start with 2.5
multiply by 7 each time
multiply
2.5 n x 7
why would u multiply twice?
so 2.5*7^(n-1) should work
btw
we could avoid this (n-1)
this comes up often
we can avoid this if we say the sequence starts with the 0th term
but in math that is uncommon
that is a programming thing though
there this is usual
lets start with the first sequence
we can split it up into 2 sequences
the sequence of nominators and the sequeence of denominators
3,4,5,6,...
3
3+1
3+1+1=3+2
and so on
so we get
3+n
16,25,36,49,...
do you notice something with these numbers?
25+9=34 which is not 36
like +1 for numerator and +9 for derminator
so nope
is starts with 16
perfect squares
try to get a formula
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hiya
lets call this rule f
f(12)=81
f(6)=9
f(8)=25
f(10)=49
and so on
notice that 81,9,25,49 and so on are all perfect squares
f(12)=9^2
f(6)=3^2
f(8)=5^2
f(10)=7^2
notice something else here?
@crude lagoon Has your question been resolved?
not quite
12^9 and 9^2
6^2 and 3^2
8^2 and 5^2
there is a pattern
f(x)=(x-3)^2
so i jus leave it like this?
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Is there an easier way to do this?
Basically, I was given this stupid problem:
For Triangle ABC, $\frac{\sin(0.5(A-B))}{\cos(0.5C)} = \frac{12}{15}$. Sides $BC$ and $AB$ are 40 and 50, respectively. Find the exact length of AC
Umbraleviathan
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ℝamonov
that's not what the question says
it says to find the asymptote, but it says to factor it out
and in the video he says 2x - 2 factors into 2x * (x - 1)
note that the x^2+x was factored too
I'm pretty sure they don't say that
because it doesn't make sense to me why that is true
factoring is the reverse of expanding
it's still the application of the distributive property
are you able to expand
a(b+c)?
yes
cause I am distributing a
missing ()
yea i'm writing it fast
write slow and properly
but ok that helps so much because in the video he just says oh yea that factors into that
without explaining what it means
otherwise the default assumption is misunderstanding
2x-2 = 2(x-1)
x^2+x = x(x+1)
overall (2x-2)(x^2+x) = 2 * (x-1) * x * (x+1)
= 2x(x-1)(x+1)
Here he cancels out a -x and gets 1/2x (x + 1), and he says replace the denominator, which in this case is 2x (x + 1) to 0
so that would just make it 1/0? how does he arrive that both x are either 0 or -1
by solving
2x(x+1) = 0
oh wait
he set to 0
ok not replacing the whole equation to 0
I got confused
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No clue on how to start
@wicked skiff Has your question been resolved?
@wicked skiff Has your question been resolved?
so i would use the second formula?
no the first one and the fourth one: sin(6x) + sin(4x) = 2 sin(6x + 4x / 2) cos(6x-4x / 2)
do the same of the denominator
how would i do it for the denominator if its 2
no I mean the cos(6x) + cos(4x)
so im using 2 different formulas
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✅
ok 2cos(6x+4x/2)cos(6x-4x/2)
U need to simplify this?
2cos(8x)cos(4x)
yeah thats it
2 sin(6x + 4x / 2) cos(6x-4x / 2)= 2sin(8x)cos(4x)
Need to write it in terms of tangents
now plug it into the original equation and reduce you see the numerator and denominator have both cos(4x)
mmk
If you expand this all the way you should end up@with a simple term tan(__)
how do you get sin8x + sin4x?
and no it is not one
write it down and show it to me your paper
the denominator too
okay
now take 2sin(5x)cos(x) / 2 cos(5x) cos(x)
reduce it what do you get
tan(5x)
there is your answer
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Hello
I’m doing a question on related rates and it’s asking me to find how the fast the angle of a triangle is changing
I tried to do it on my own and this is what I got
Does this seem correct?
Also dx/dt and dy/dt are given
X is 6 and y is 8 too
I thought it would be easier to just leave them as variables so that’s what I did
<@&286206848099549185>
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f - f(x+z0) + C1 + C2 = sum of residues
thats what i got but idk what theorem to apply to do this
like idk how we can find C1, C2 or show they vanish
@void crest Has your question been resolved?
yeshh
the first choice is wrong lol
it was a random
ok
nice
whats it tho
should be last one i think
should be 3rd not last

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ty

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So it's like a upper and lower limit of a set?
And in the example provided, the infA should be 0 right?
no, the infimum of {x ∈ R : x < 2} is not 0. it either doesn't have an infimum at all or its infimum is -∞ depending on who you ask
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@normal hill Has your question been resolved?
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How do I get better at math
I'm in college algebra, yet I'm behind by years because I was lazy and my high school was shit
ok
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can anyone help me
how old are you?
what grade are you in ?
7th
use this rule
i did
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Anyone tell me about, how can I PDE by solve neural network?
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We toss coin until either the sequence of HHT appears, or HTH. WHat is probability that HHT appears first. How about unfair coin with probability p
first why it's 2/3
did you use the abracadbra theorem?
never heard of it
sounds mildly interesting
i win if hht comes first
skip the Ts at the start however many
then
HHT: i win
HTH: i lose
HHH: i win
HTT: we start over
there's 2 ways for me to win and 1 to lose, the fourth option doesn;t change my odds
it probably works the same for arbitrary p
$\frac{p²(1-p) + p³}{p²(1-p) + p²(1-p) + p³}$
hm
we start all over at HTT since we didn't get either options we're looking for?
that's too short it scares me
like it's the same procedure to determine who wins from that point
because it ends with T we have the same chances as in the beginning from that point
so itll be a waste of a turn?
thats why we start all over at that turn?
@sturdy cypress
im just tryna understand why we start all over when we get HTT
<@&286206848099549185>
like why did you say that?
the probability doesn't change from the initial
htt helps neither of us to win later
it doesn't start with H
thats why we start all over?
i don't understand what you mean
since HTT helps neither of us win, thats why we would start all over if any of us flip that sequence HTT right?
we read the text from the beginning
why would you lose if you flip HTH?
what
it's in the text of the problem
HTH is the sequence that keeps HHT from appearing first if it appears first
i'm sorry i couldn't help, srsly
its okay at least you tried, ty
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whats 2 plus 2 and 1 divided by 2
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do any of you know how to find an optimal curve based on certain parameters?
Well, what parameters are you trying to have it best fit?
one sec
its hard to explain
essentially i am doing a model, where i have 675.46cm^2 of surface area (material)
and i need to make a curve that gives me the highest maximum, with the volume of revolution from 0-that maximum being 750cm^3
heres a picture
ah i get it now
and im changing the curve accordingly
if u want the bigger picture, I have to model a wine jug, where 750ml of wine has the biggest surface area touching air, where i have 675.46cm^2 of glass for the bottom part
so do you know the equation for the surface area of a solid of revolution?
yes i do, I have the equations all down
one sec
first one is SA of a solid of revolution, the second one is volume of revolution and the third is an expression for how "tall" (y component) the jug is
so really what you want to do here is make a system of equations to solve for a and b, right? that's what it looks like at first glance
yes
but would that output the widest possible r? or just any r?
my teacher said i could use optimization as in using differentiation but i dont see how
i wanna find the values of a and b that make r the largest
hm that's a tough question, but you should probably try solving the equations for a and b to start off
If a and b are variable, then maybe you can model an equation to show how the r changes with respect to a and b and then differentiate that to find the maxima?
that's not right
how did you get this?
and yeah a and b are variable so maybe i can think of something there?
i equalled the two eqn
.
.
and started simplifying
but idk it doesnt seem to be very nice
or doable
maybe a better option would be to solve the second (easier) integral for a or b by integrating it, and then plug that value into the first integral so you only have one of them
hm
i solved them both, and yeah the first one is simpler
but i dont know if i can solve for either a or b
one second
it looks like you can solve it for a
$-\frac{b^3}{24a^2}+\frac{b^3}{8a^2}-\frac{2.61155b}{a}$
lawl
yeah i got that too, so you can just multiply both sides by a^2 and then use the quadratic formula?
lawl
$a=\frac{-\left(12\left(5.2231\right)b\right)\pm\sqrt{\left(12\left(5.2231\right)b\right)^{2}-4\left(\frac{18000}{\pi}\right)\left(2b^{3}\right)}}{2\left(\frac{18000}{\pi}\right)}$
demaw
here you're basically there you can solve it from here
but apply the quadratic formula to what
yeah but theres an a on the oter side too...
move them all to one side
i just did that to preserve the 5.2231, if you simplify my expression you should get the same thing that you would get by doing the quadratic formula on the expression you got
i would assume after solving for a like this, you can probably plug it in to the equation for the radius? and then just find the maxima by differentiating with respect to change in b?
yeah i suppose so
well here goes nothing
ill try rq so ill leave this open
for now tho tysm u saved my adhd ass
hah i hope it works man. i have no idea if it will but it should i think
i hope it will
if not ill turn this in and figure it out with my teacher
its only a first draft so
wait for this, why isnt ur (2b^3) negative
hm, you might be right about that, if you think it's supposed to be negative just make it negative ill double check my work
ill redo mine too xD
sorry
@dull dirge Has your question been resolved?
wait but @smoky shore this is still a in terms of b
yeah, my thought was that you can replace all substitute all instances of a so you can find the maxima of b and then solve backwards for a
confusion
oh
i get it
oh nooo
thats gonna be pain
dont i need to integrate the other eqn then too
oh noooo
$\frac{\pi \left(5.2231\left(-0.5b\sqrt{b^2+1}-0.5\ln \left|b+\sqrt{b^2+1}\right|\right)+\frac{b\sqrt{b^2+1}+\ln \left|b+\sqrt{b^2+1}\right|-2b\left(b^2+1\right)^{\frac{3}{2}}-4b^2\left(-b\sqrt{b^2+1}-\ln \left|b+\sqrt{b^2+1}\right|\right)}{32\cdot \frac{-12\cdot :5.2231b\pm \sqrt{\left(12\cdot :5.2231b\right)^2-4\cdot \frac{18000}{\pi }\left(-2b^3\right)}}{2\cdot \frac{18000}{\pi }}}\right)}{\frac{-12\cdot :5.2231b\pm \sqrt{\left(12\cdot :5.2231b\right)^2-4\cdot \frac{18000}{\pi }\left(-2b^3\right)}}{2\cdot \frac{18000}{\pi }}}$
lawl
hmm
delicious
<@&286206848099549185> anyone know a site that can find b for me from this soup?
.w $\frac{\pi \left(5.2231\left(-0.5b\sqrt{b^2+1}-0.5\ln \left|b+\sqrt{b^2+1}\right|\right)+\frac{b\sqrt{b^2+1}+\ln \left|b+\sqrt{b^2+1}\right|-2b\left(b^2+1\right)^{\frac{3}{2}}-4b^2\left(-b\sqrt{b^2+1}-\ln \left|b+\sqrt{b^2+1}\right|\right)}{32\cdot \frac{-12\cdot :5.2231b\pm \sqrt{\left(12\cdot :5.2231b\right)^2-4\cdot \frac{18000}{\pi }\left(-2b^3\right)}}{2\cdot \frac{18000}{\pi }}}\right)}{\frac{-12\cdot :5.2231b\pm \sqrt{\left(12\cdot :5.2231b\right)^2-4\cdot \frac{18000}{\pi }\left(-2b^3\right)}}{2\cdot \frac{18000}{\pi }}}$
internet
yikes
hmm
wolfram alpha lol
does wolfram take latex
oh wait its past the max chars
i havent been able to
no... fffffffffffffffff
can i DL a wolfram alpha with more max chars
yeah symbolab is just deding on this
idk what the problem is
its just b with a bunch of things around it
solve!
@solar swift can u think of another way to put this
into that
id have to see the original expression
this is where the quadratic eqn came from
yeah but where did that come from
this
@solar swift
i did the integration, rearranged it and did a quadratic formula
now i need to put it into that other integral
but symbolab and wolfram just die if i try
you can literally plug this into wolfram alpha to find what a and b are
<@&286206848099549185>
wait what is this?
the 2nd integral you sent?
heres the wolfram page
no but the way ur writing in wolfram
the int stuff is latex but the rest is just words
well shit
this absolutely doesnt give me the right answer
could u help me out from the start? i am hopeless at this point
depends on what it is tbh
essentially i am doing a model, where i have 675.46cm^2 of surface area (material), and i need to make a curve that gives me the highest maximum, with the volume of revolution from 0-that maximum being 750cm^3. The y intercept is fixed at 5.2231.
so what i did is make 3 equations
and im adjusting a and b to get the highest r, but they need to follow the two equations so the parameters are followed
this is a drawing i made of it
if u want the bigger picture, I have to model a wine jug, where 750ml of wine has the biggest surface area touching air, where i have 675.46cm^2 of glass for the bottom part
yeah ngl i dont really know revolution stuff, it doesnt get taught until 3rd year uni where i live
the revolution stuff is chill, idk how to do the optimization
its just 2 equations
<@&286206848099549185>
you didnt square f
yeah i mean i check wolframalpha for the first one and the a,b values never make the SA equation equal to what you want
i think it might either be the curve of the interval points
yeah
im fucked
theres no way im gonna solve this
i dont even know how i could possibly create an equation for r out of this
cos thats what i need really
a way to express r that incorporates the volume and SA
also the r equation is wrong
wym how
nvm no it isnt
<@&286206848099549185>
What can I do for you?
hey
I'm working on an optimization question, where I am trying to find the optimal quadratic maximizing the y component of the maximum, with 2 parameters
The surface area of the solid of revolution = 675.46cm^2, and the volume of revolution from 0-maximum =750cm^3
The y intercept must = 5.2231
i did a fancy illustration
i can show u what ive done already
i somehow need to combine these i think, though im not really sure
@coarse yacht u got any idea?
I would advise to compute this integral
In function of a and b
And plug it here
You can express $ax^2 + bx + c $ using $r$ in the canonical form
black_couscous
Yes
You know you can express it that way $y = (x- \frac{b}{2a})^2 + smthg$
black_couscous
really?
black_couscous
wait but how would i find that lol
$(b/2a)^{2}= r +5,2231$
black_couscous
okay, now that i have that how can i incorporate the other 2 eqns?
Yup
lawl
You can express b/2a as a function of r
Here
lawl
@coarse yacht can u explain pls
Yes
Yes
ok i will try
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I got the first part
But not sure how to do the second one without multiplying and making the factors of f(x)
Is there some other approach?
@tranquil pine Has your question been resolved?
@tranquil pine possibly symmetry?
you know f is symmetric in x=6
so the roots should add to that maybe
,w (x-1)(x-3)(x-5)(x-7)+15=0

twerk
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Need help with b
Anything you've tried
@gleaming mantle ^
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ak is ksin(1/k)
so how is the limt of ak =1 as k goes to inf?
cuz it woul be inf *0
where's the question
are you just trying to show each term converges?
instead of the series?
that's indeterminate form. you should have learned something about l'hopital or something else
oh right i totaly forgot about lhopital xd
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I need help with my math. So the topic is, Finding the equation of a quadratic function. How to find the quotation of a quadratic function given: (a) table of value, (b) graph, (c) zeros; y= ax² + bx + c, is determined by 3 points. Point is named by its ordered pair (x,y). Find the value of a,b and c. To get the answer substitute 3 ordered pairs (x,y) in y=ax²+ bx + c.
I've already got the a and c now I can't get the value of b
show work
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Can you ask more clearly
@outer storm Has your question been resolved?
oh hey its you again!
these would be θ/2 as well
Use trig only for right triangles
you have to find the base and height values for the area right
so you use the shaded right triangle and find those values using trigonometry
you have to have a right triangle to write those things you know
yeh sorry but i was just trying to say you have to use the same right triangle for the second case as well
so it would be θ/2 only
can u tell me the length of red line?
are you sure
ah yeh
no no
first
before using trigonometry u need to have a right triangle
can you outline the right triangle on it?
right
now do you see what x is
yep
similarly the perpendicular would be rsin(θ/2)
now is it fine?
i get that but
that second triangle you wont have cosθ and sinθ right
those would be cosθ/2 and sinθ/2 no
ah its fine then i will stop
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im bacl
i got a problem
where this turns into 0...
which idk i think it should be impossible
or better said i have to do simultaneous eqn of these 3 things
<@&286206848099549185>
<@&286206848099549185>
@dull dirge Has your question been resolved?
<@&286206848099549185>
@dull dirge Has your question been resolved?
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,rccw
