glacial umbra
or around 21
#help-42
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velvet osprey
time for round 2 of this shit. im hitting a snag again.
all given info is written and marked on the diagram (except the 45° angles, which i derived). goal is to find the area of ABC.
the area of AOB is obviously 10 but i can't seem to get my hands on any other measurement.
nimble harbor
looks like O is a center of a circle of radius 2
frozen rampart
greetings
nimble harbor
but why is AB = 10
frozen rampart
O is the incenter of ABC
velvet osprey
but why is AB = 10
given
frozen rampart
aha yes this is easy
velvet osprey
O is the incenter of ABC
hm.
velvet osprey
yeah sure i do. but do i know either leg?
frozen rampart
now since O is incenter
velvet osprey
yeah hold on i got it
frozen rampart
O ok
velvet osprey
lemme scribble
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rough ferry
find a parametric and vector equation for the plane, x-2y+3z-6=0
rough ferry
help 🙏
we know the normal is (1, -2, 3)
and the x intercept is (6, 0, 0)
then what do I do next
glad parrot
x = 2y -3z and so you can have two vectors
(x,y,z) = (2y-3z, y, z) = y(...) + z(...)
Just add the point you found and you have the vector form
And so you can deduce the parametrisation
calm coralBOT
@rough ferry Has your question been resolved?
rough ferry
Just add the point you found and you have the vector form
i dont get this, can u show me how?
rough ferry
x = 2y -3z and so you can have two vectors
what are the 2 vectors
glad parrot
(x,y,z) = (2y-3z, y, z) = y(...) + z(...)
See here
rough ferry
i see i don't understand
rough ferry
You have 2y in the first component
the first component is 2y-3z?
@glad parrot can I just set y = 1, z= 0 then from x = 2y - 3z to get a point Q, then from Q and P. we get (8, 1, 0). do the same with y = 0, z = 1, and get a new vector (-3, 0,1)
is that right
<@&286206848099549185>
glad parrot
Getting it complicated and not really methodologic
glad parrot
the first component is 2y-3z?
Yes
But for now we take care of the amount of y
We'll do the z amount after
rough ferry
Getting it complicated and not really methodologic
is it right though
I don't understand your approoach
glad parrot
I think yes
glad parrot
I don't understand your approoach
Getting the directionnal vectors
glad parrot
See the definition of vectors equation
glad parrot
so there is a point (2y-3z, y, z)
Not quite
Its more like
Every point on the plane will verify this
rough ferry
can you use the dot product to find the two vectors?
I think I've seen someone do that before
glad parrot
Its using bazooka to kill a mosquito
And i don't know how you will use it tho
You need two vectors to do so
rough ferry
You need two vectors to do so
alright ic, how can I be sure my answer is right
with an online calc or somoething
glad parrot
Geogebra maybe
solar dust
Does anyone know if this is correct?
calm coralBOT
Does anyone know if this is correct?
solar dust
If not help would be appreciated
Proof by contrapositive or contraposition are allowed
eternal shard
Does anyone know if this is correct?
Why both even? Shouldn't m and n be both odd?
You would like to show if both m and n are odd, then m(n+2) is even.
The negation of m even or n even is m not even and n not even or m odd and n odd.
solar dust
We want to do a proof by contraposition.
1 - We assume m and n are odd.
2 - m = 2k + 1
3 - n = 2j + 1
4 - Then m (n+2) = (2k+1) ((2j +1) +2) = (2k+1) (2j+2)
Would this be correct?
But i dont know how to continue this
eternal shard
m(n+2) = (2k+1)(2(2l+1)+2) = (2k+1)(4l+4) = 4(2k+1)(l+1)
eternal shard
Would this be correct?
I don't know how you got 21 suddenly.
solar dust
~~m(n+2) = (2k+1)(2(2l+1)+2) = (2k+1)(4l+4) = 4(2k+1)(l+1)~~
Where did you get the first two from here "(2(2l+1)+2)" ? Why are we multiplying it with 2?
eternal shard
Oh typo.
eternal shard
We want to do a proof by contraposition.
1 - We assume m and n are odd.
2 - m =...
Ok but it should be (2k+1)(2l+3)
solar dust
oh yep
Then m (n+2) = (2k+1) ((2j +1) +2) = (2k+1) (2j+3)
(changed the l to an j, looks better)
but what do i do then?
eternal shard
Indeed, this looks troublesome.
solar dust
Yep
Im not 100% sure if this is the right approach either. I know that m and n must be odd since we do a proof by contraposition. And 2k+1 and 2j+1 should be okay too (bc 2k and 2j alone would be even) but idk
eternal shard
Actually, this can't be.
solar dust
hm
eternal shard
The implication seems wrong. Let m = 3 and n = 1. Then m(n+2) = 3(1+2) = 9 is odd, but neither m nor n are even.
Are you sure you have the correct statement?
solar dust
let me check
aah
i think its
∃𝑚, 𝑛 ∈ ℕ (𝑚 (𝑛 + 1) 𝑜𝑑𝑑 ⟹ (𝑚 odd ∨ 𝑛 odd)
does this make more sense?
this is an old exam apparently
eternal shard
Indeed, because you end up getting a factor 2, when you do the contraposition.
(2k+1)(2j+1+1) = (2k+1)(2j+2) = 2(2k+1)(j+1)
solar dust
So we assume the samething that m and n are 2k + 1 and 2j +1? And then we get this output right
eternal shard
Yes.
solar dust
(2k+1)(2j+1+1) = (2k+1)(2j+2) = 2(2k+1)(j+1)
okay and how does this prove that m or n is odd though? in the case of contrapositive that it is even?
eternal shard
Previously, you had (2k+1)(2j+1) which is impossible to be even.
solar dust
yep
eternal shard
So you actually showed there doesn't exist one m or n to make the statement true.
eternal shard
okay and how does this prove that m or n is odd though? in the case of contrapos...
You assume m and n are odd. Then you consider m(n+2)
and end up showing that this leads to an even number, due to the factor 2.
That was the proof for the contraposition.
solar dust
okay thank you very much ill note it
potent lotusBOT
eternal shard
You may want to get familiar with De Morgan's laws, so to effectively write the contraposition of statements.
Often, the contraposition is way easier to prove and leads to the same outcome - proving the original statement anyway.
calm coralBOT
@solar dust Has your question been resolved?
blazing coyote
I don't get 3.114
calm coralBOT
I don't get 3.114
alpine stone
Rewrite v as a linear combination of v_i’s and apply the definition of phi_i’s
blazing coyote
Rewrite v as a linear combination of v_i’s and apply the definition of phi_i’s
yea, but how do we know phi is a LT
alpine stone
They are defined as linear functionals
iron ore
Could somebody tell me why for the first one the minimum is -1 and not 1. And check the other two to see if they are correct
tall moon
Could somebody tell me why for the first one the minimum is -1 and not 1. And ch...
the first one its written that "Minimum=-1" no?
iron ore
the first one its written that "Minimum=-1" no?
Yea but I don’t understand why it is -1 instead of 1
devout burrow
Could somebody tell me why for the first one the minimum is -1 and not 1. And ch...
Your reasoniing from the wheel diagram was pretty correct
tall moon
oh wait nvm
iron ore
Wdym?
tall moon
its cause the waterwheel goes under the water
devout burrow
its cause the waterwheel goes under the water
iron ore
So even tho it doesn’t really make sense without the story since it is in the story it is -1?
tall moon
Could somebody tell me why for the first one the minimum is -1 and not 1. And ch...
for 5 shouldnt it start at the maximum?
iron ore
for 5 shouldnt it start at the maximum?
Oh yea I already fixed that I forgot to take a pic of the new version
tall moon
So even tho it doesn’t really make sense without the story since it is in the st...
it makes sense in the story, the water wheel goes 1 below the water, here the water is 0 so if you go under 1 it will be -1
iron ore
it makes sense in the story, the water wheel goes 1 below the water, here the wa...
Yea but I feel like without the story it would be wrong if that makes sense
Where the other problems correct?
topaz raft
@upper sparrow 
calm coralBOT
@iron ore Has your question been resolved?
dense thistle
26
fiery basalt
f(x - 1) translates the graph from f(x) right by one unit
what will happen to the distance after that?
dense thistle
what will happen to the distance after that?
just move to right?
maybe nothing changes
fiery basalt
yes, the distance doesn't change
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lavish wind
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lavish wind
I don't understand where this v came from, or why we need it when applying the chain rule
This is my chain rule
oh are you just throwing in any random constant so you can use the previous property
stoic oyster
oh are you just throwing in any random constant so you can use the previous prop...
yeah, (f°g)'(t0) is a linear transformation anyway, evaluating it at a vector v makes sense
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serene talon
here f in mathbb C^mathbb R
is phi(c)=0 exactly saying f(x) is continuous at x=c, or is it strictly stronger, like some sort of uniform continuity at a point?
serene talon
which doesnt make sense, uniform continuity at a point, so, im not sure
leaden thunder
if f(x) = x for x !=0, and f(0) = 1, what's phi(0) ?
serene talon
if f(x) = x for x !=0, and f(0) = 1, what's phi(0) ?
let's see, thats just the diameter of the ball, so inf delta>0 of 2delta, so 0?
leaden thunder
oh i got that wrong
serene talon
no wait its 1
leaden thunder
phi(0) = 1 since the supremum is 1 for any interval around x=0.
leaden thunder
what's rca
serene talon
prove f is continuous at a point c iff phi(c)=0, and deduce that the set of points of continuity of such an f is G-delta
real and complex analysis by Rudin
so I know that this is continuity, but it looks stronger than continuity
i did prove phi is upper semicontinuous if that's relevant
calm coralBOT
@serene talon Has your question been resolved?
calm coralBOT
@serene talon Has your question been resolved?
remote mural
Try to picture different discontinuities at x=c
Namely jump, removable, infinite and wild/oscillatory
serene talon
okay so if if is continuous at c, then for any eps/2 > 0 , there is a delta>0,
|x-c|< delta ⇒ |f(x)-f(c)|<eps/2
Now for any s, t in B(c,delta)
|f(s)-f(t)|=|f(s)-f(c)+f(c)-f(t)|≤|f(s)-f(c)|+|f(t)-f(c)|< eps/2 + eps/2 = eps
hence
sup{s,t in B(c,delta} |f(s)-f(t)|< eps
this holds for every eps > 0 , so phi(c)=0, this is one direction
that is, continuity at c implies phi(c)=0
suppose phi(c)=0, we want to show continuity at c
to say:
inf{delta > 0}sup{s,t in B(c,delta)|f(s)-f(t)|=0
by the definition of inf, implies that for every eps >0, there is a delta, call it delta0,
sup{s,t in B(c,delta0)}|f(s)-f(t)| < epsilon
note that c in B(c,delta0)
|f(s)-f(c)|≤sup{s,t in B(c,delta0)}|f(s)-f(t)| because c is one of the values of t which one is taking the sup over
is there a cleaner way to say that?
remote mural
Hmm
serene talon
thus |f(s)-f(c)|≤sup{s,t, in B(c,delta0)}|f(s)-f(t)|<epsilon and by transitivity blah blah
this is supposed to imply that the set of points of continuity of f is G-delta, let's see why.
well the set of continuity points of f is ker[phi]=phi^{-1}{0}
remote mural
is there a cleaner way to say that?
I don't think so
serene talon
which is also phi^{-1}((-infty, 0]),so ker[phi] is closed? is that true
yes I think so
wait maybe not im thinking
no it's not true
$$\ker[\varphi] = {x: \varphi(x)=0}$$
$$=\bigcap_{n \in \mathbb N \setminus{0}}\left{x: 0 \le \varphi(x) \le \frac 1 n}\right}$$
potent lotusBOT
gfauxpas
$$\ker[\varphi] = \{x: \varphi(x)=0\}$$
$$=\bigcap_{n \in \mathbb N \setminus\{0\}}\left\{x: 0 \le \varphi(x) \le \frac 1 n}\right\}$$
```Compilation error:```! Extra }, or forgotten \right.
l.57 ...}\left\{x: 0 \le \varphi(x) \le \frac 1 n}
\right\}$$
I've deleted a group-closing symbol because it seems to be
spurious, as in `$x}$'. But perhaps the } is legitimate and
you forgot something else, as in `\hbox{$x}'. In such cases
the way to recover is to insert both the forgotten and the
deleted material, e.g., by typing `I$}'.
Preview: Tightpage -1310720 -1310720 1310720 1310720
[1{/usr/local/texlive/2023/texmf-var/fonts/map/pdftex/updmap/pdftex.map}]```
serene talon
ah then this proves it's gdelta
wait i did the wrong inquality signs
but I got it, thanks everyone
.close
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true cave
Hey, I am looking for a what I think is more rigorous proof of the 2x2 Jacobian than with the determinant.\
So what I'm trying to prove is that \$\iint_Rf(x(u,v),y(u,v))dxdy=\iint_{R'}f(x(u,v),y(u,v))(\pdv{x}{u}\pdv{y}{v}-\pdv{x}{v}\pdv{y}{u})dudv$\
This is basically just a reverse u-sub in 2 dimensions, so here is my proof for 1-dimensional u-sub that i tried to replicate:\
Let $f$, $u$ and $w$ be functions such that $\dv{w}{u}=f(u(x))$\
Then, $w=\int\dv{w}{u}du=\int\dv{w}{x}dx$\
$\int f(u)du=\int\dv{w}{u}\dv{u}{x}dx$\
$\int f(u)du=\int f(u(x))u'(x)dx$\
So for the two-dimensional case I tried a similar proof, hoping to find back the original function somewhere.\
Let $f, x, y$ and $w$ be functions such that $\frac{\partial^2w}{\partial x\partial y}=f(x(u,v),y(u,v))$\
$w=\iint\frac{\partial^2w}{\partial x\partial y}dxdy=\iint\frac{\partial^2w}{\partial u\partial v}dudv$\
Now using the multivariable chain rule: $\pdv{w}{v}=\pdv{w}{x}\pdv{x}{v}+\pdv{w}{y}\pdv{y}{v}$\
$\iint f(x(u,v),y(u,v))dxdy=\iint\pdv{u}(\pdv{w}{x}\pdv{x}{v}+\pdv{w}{y}\pdv{y}{v})dudv$\
Now we use the product rule\
$=\iint(\pdv{x}{v}\pdv{u}(\pdv{w}{x})+\frac{\partial^2x}{\partial u\partial v}\pdv{w}{x}+\pdv{y}{v}\pdv{u}(\pdv{w}{y})+\pdv{w}{y}\frac{\partial^2y}{\partial u\partial v})dudv$\
We use the multivariable chain rule on $\pdv{w}{x}$ and $\pdv{w}{y}$:\
$=\iint(\pdv{x}{v}(\pdv[2]{w}{x}\pdv{x}{u}+\frac{\partial^2w}{\partial x\partial y}\pdv{y}{u})+\frac{\partial^2x}{\partial u\partial v}\pdv{w}{x}+\pdv{y}{v}(\frac{\partial^2w}{\partial x\partial y}\pdv{x}{u}+\pdv[2]{w}{y}\pdv{y}{u})+\pdv{w}{y}\frac{\partial^2y}{\partial u\partial v})dudv$\
NOT COMPLETE PROOF SEE LATER
true cave
oh accidentally pressed enter my proof is not done yet wait a sec
potent lotusBOT
pixel
calm coralBOT
@true cave Has your question been resolved?
true cave
on the bot isnt updating anymore
ill send it again
Hey, I am looking for a what I think is more rigorous proof of the 2x2 Jacobian than with the determinant.\
So what I'm trying to prove is that \$\iint_Rf(x(u,v),y(u,v))dxdy=\iint_{R'}f(x(u,v),y(u,v))(\pdv{x}{u}\pdv{y}{v}-\pdv{x}{v}\pdv{y}{u})dudv$\
This is basically just a reverse u-sub in 2 dimensions, so here is my proof for 1-dimensional u-sub that i tried to replicate:\
Let $f$, $u$ and $w$ be functions such that $\dv{w}{u}=f(u(x))$\
Then, $w=\int\dv{w}{u}du=\int\dv{w}{x}dx$\
$\int f(u)du=\int\dv{w}{u}\dv{u}{x}dx$\
$\int f(u)du=\int f(u(x))u'(x)dx$\
So for the two-dimensional case I tried a similar proof, hoping to find back the original function somewhere.\
Let $f, x, y$ and $w$ be functions such that $\frac{\partial^2w}{\partial x\partial y}=f(x(u,v),y(u,v))$\
$w=\iint\frac{\partial^2w}{\partial x\partial y}dxdy=\iint\frac{\partial^2w}{\partial u\partial v}dudv$\
Now using the multivariable chain rule: $\pdv{w}{v}=\pdv{w}{x}\pdv{x}{v}+\pdv{w}{y}\pdv{y}{v}$\
$\iint f(x(u,v),y(u,v))dxdy=\iint\pdv{u}(\pdv{w}{x}\pdv{x}{v}+\pdv{w}{y}\pdv{y}{v})dudv$\
Now we use the product rule\
$=\iint(\pdv{x}{v}\pdv{u}(\pdv{w}{x})+\frac{\partial^2x}{\partial u\partial v}\pdv{w}{x}+\pdv{y}{v}\pdv{u}(\pdv{w}{y})+\pdv{w}{y}\frac{\partial^2y}{\partial u\partial v})dudv$\
We use the multivariable chain rule on $\pdv{w}{x}$ and $\pdv{w}{y}$:\
see later for next part
uhh
it's not rendering everything
potent lotusBOT
pixel
true cave
$=\iint(\pdv{x}{v}(\pdv[2]{w}{x}\pdv{x}{u}+\frac{\partial^2w}{\partial x\partial y}\pdv{y}{u})+\frac{\partial^2x}{\partial u\partial v}\pdv{w}{x}+\pdv{y}{v}(\frac{\partial^2w}{\partial x\partial y}\pdv{x}{u}+\pdv[2]{w}{y}\pdv{y}{u})+\pdv{w}{y}\frac{\partial^2y}{\partial u\partial v})dudv$\
potent lotusBOT
pixel
true cave
why isn't it rendering everything
i'll write it out on paper then i guess
but i hope you see where i'm going, yet i don't see how i am going to get to the jacobian if i have that second partial derivative of x and stuff
final form of my attempt
my partial signs might look weird sometimes but they always indicate a partial derivative not full with d/dx
calm coralBOT
@true cave Has your question been resolved?
warm warren
Have you seen this video? https://youtu.be/wCZ1VEmVjVo?si=XkXISxpwX5IuMPGO
near crag
I believe the person tried to work without the notion of a determinant
I skimmed through the video and it seems like it kinda misses the mark
true cave
yes, exactly what you are saying
warm warren
In the second line how did you get that $\iint\pdv{w}{x}{y},dxdy = \iint\pdv{w}{u}{v},dudv$
potent lotusBOT
frosst
true cave
i thought because with single integrals $w=\int\dv{w}{x}dx=\int\dv{w}{u}du$, for two integrals it must be that it works like that for partial derivatives
potent lotusBOT
pixel
true cave
like the partial derivatives and integrals just cancel
warm warren
Is this w(x(u))?
true cave
**pixel**
have you looked at the top part of this?
there i described how i would do it for single integrals
so this is w(u(x))
i see my choice of first doing u(x) and then x(u,v) is a bit weird now
warm warren
**pixel**
Maybe I’m just stupid because I fail to see how this holds for every choice of u
I feel like you need to put some conditions on u
Substitution does not always work for integration
true cave
oh, i didn't know that. do you have an example where substitution wouldn't work?
i get that not in all functions helps you with integrating
but it's not that the function makes it incorrect
warm warren
For example, solve $\int_{-1}^1 x^2 , dx$ with $u=x^2$
potent lotusBOT
frosst
warm warren
Not every substitution will work for solving integrals
In the sense that solvable ≠ is the right solution
warm warren
It happens when u^-1 isnt a function on the domain of the integral
Something like that
true cave
yeah
so i think we need to put the restriction that the function is injective (or non-injective i don't really know the right term in this case) on the domain
warm warren
The idea of the jacobian is to locally approximate the change of coords in terms of a linear transformation of a suitable dimension
First off you need the change of cords to actually be differentiable even
true cave
so you're saying that there isn't an algebraic proof like for u-sub for the jacobian
warm warren
So you can at least know that dw/dx exists
I’m not saying there isn’t (I don’t know) but you need to place a lot more restrictions before some of those statements are true
calm coralBOT
warm warren
okay
I would argue that if you first understand the jacobian proof then you can just replace everything with the partials because that’s what the jacobian is equal to anyway
It may form a better guide to show you what to do next
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delicate sable
Hi just need some help
calm coralBOT
Hi just need some help
delicate sable
I have no idea how they got this answer
This is part a
Just don't know how they got the answer
for part b not a
real olive
for part b not a
Please show part B then, what you have provided is only part A
delicate sable
Hi just need some help
...
in the mark scheme it said 12 cubed
but i have no idea how and why that gets you the answer
real olive
Oh I apologize, I did not scroll up
torpid canopy
anyone? pls
notice the radius is 12 times greater right
So the volume is 12^3 times greater
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tall moon
Given polynomial $p(x)$ such that $p^2(x)+p(x^2)=2x^2$ for all real $x$. If $p(x) \neq 1$ then find the sum of all possible $p(10)$
potent lotusBOT
skissue.in.a.teacup
tall moon
if p(x) is linear then you get
$$a^2x^2+2abx+b^2+ax^2+b=2x^2$$
$$a^2+a=2 \quad b=0$$
potent lotusBOT
skissue.in.a.teacup
tall moon
a=-2 or a=1 so -20 and 10 are some of the possibilities
idk about higher power cause technichally its possible but im not sure about the condition/requirements for it to work
honestly i dont get the p(x)=/=1 cause it doesent even hold for the functional equation
calm coralBOT
@tall moon Has your question been resolved?
calm coralBOT
@tall moon Has your question been resolved?
calm coralBOT
@tall moon Has your question been resolved?
remote mural
idk about higher power cause technichally its possible but im not sure about the...
The RHS has degree 2 so p²(x) + p(x²) must also have degree 2
tall moon
The RHS has degree 2 so p²(x) + p(x²) must also have degree 2
what if p(x)^2 has a higher degree but p^2(x) and p(x^2) cancel out the higher terms
remote mural
what if p(x)^2 has a higher degree but p^2(x) and p(x^2) cancel out the higher t...
Let p(x) have degree n
Then p²(x) + p(x²) has maximum possible degree max{deg(p²(x)), deg(p(x²))} = 2n
RHS has degree 2, so the only solution is n=1, that p(x) is linear
tall moon
Let p(x) have degree n
Then p²(x) + p(x²) has maximum possible degree max{deg(p...
just because deg(p^2(x)) and deg(p(x^2)) is 2n doesent mean that deg(p^2(x)+p(x^2)) is also 2n
remote mural
Ah okay wait yeah
If it's 2(n-1) then n=2
Yes
Okay so say
$p(x) = \sum_{i=0}^n a_ix^i$
potent lotusBOT
à뜜
remote mural
$p(x^2) = \sum_{i=0}^n a_ix^{2i}$
potent lotusBOT
à뜜
remote mural
$p^2(x) = \left(\sum_{i=0}^n a_ix^i\right) = \sum_{k_0 + k_1 + \dots + k_n = 2}\frac{2!}{k_0!k_1!\dots k_n!}x_0^{k_0}x_1^{k_1}\dots x_n^{k_n}$
With $k_i \geq 0$ and $x_i = x^i$
potent lotusBOT
à뜜
remote mural
Natural solutions to $k_0 + k_1 + \dots + k_n = 2$ are clearly $k_i = 2$ and $k_j = 0$ where $i \neq j$
potent lotusBOT
à뜜
remote mural
Or $k_i = k_j = 1$ and $k_m = 0$ where $i \neq j$ and $m \neq i$ and $m \neq j$
potent lotusBOT
à뜜
fickle steeple
Let p(x) have degree n
Then p²(x) + p(x²) has maximum possible degree max{deg(p...
I got it from this, thanks
remote mural
I don't know if there is an easier way than brute forcing like this 
Oh wait
$p^2(x) = \sum_{i=0}^n (a_ix^i)^2 + 2\sum_{0<i+j < n, i \neq j}a_ia_jx^{i+j}$
potent lotusBOT
à뜜
remote mural
**à뜜**
Ignore this, this one is wrong
So
$p(x^2) + p^2(x)$
$= \sum_{i=0}^n(a_i^2 + a_i)x^{2i} + 2 \sum_{0<j+k<n, j \neq k}a_ja_kx^{j+k}$
potent lotusBOT
à뜜
remote mural
I'll brb in 5 minutes
tall moon
So
$p(x^2) + p^2(x)$
$= \sum_{i=0}^n(a_i^2 + a_i)x^{2i} + 2 \sum_{0<j+k<n, j ...
so deg of lhs can be less than 2n iff a_n^2+a_n=0
fickle steeple
Hey I am sending a pdf of the solution
Wait
@tall moon
Read it bro it's really simple explanation
aoeoe. did it the summation way, I just took his idea and avoided including summation to make it simple
remote mural
so deg of lhs can be less than 2n iff a_n^2+a_n=0
I think so
Also a slight correction
potent lotusBOT
à뜜
remote mural
Since I wrote 2 so j,k is counted only once
tall moon
oooh
remote mural
p(x) = -x²-1 is possible as well
p(x²) = -x⁴-1
p²(x) = x⁴ + 2x² + 1
Ah I messed up the indexing again
So
$p(x^2) + p^2(x)$
$= \sum_{i=0}^n(a_i^2 + a_i)x^{2i} + 2 \sum_{0\leq j<k \leq n}a_ja_kx^{j+k}$
fickle steeple
I am still trying to find where I did the mistake
potent lotusBOT
à뜜
remote mural
I am still trying to find where I did the mistake
Yeah, the question is tricky
tall moon
I am still trying to find where I did the mistake
i hink its the "but since the final expression has only an x^2 term, all other coefficients must be 0"
remote mural
**à뜜**
We can re-write the second sum here as:
$\sum_{i=0}^n(a_i^2 + a_i)x^{2i} + 2 \sum_{i=0}^n\sum_{j=0}^{i-1}a_ia_jx^{i+j}$
potent lotusBOT
à뜜
remote mural
I think we can play with this now. This is why I was not sure before if just setting a_i² + a_i = 0 would work or not in general to reduce the degree of LHS (considering p(x) is fixed)
tall moon
**à뜜**
since the x^1 coefficient is 0, then if a_1 is 0 then a_0 isnt 0, which means that a_3 is 0 as x^3 coefficient is 0, which should spiral to every odd number
remote mural
Because reducing 2n to 2n-2 by setting a_n² + a_n = 0 is correct, but not for general 2(n-k)
tall moon
uhh bad news guys its getting late and i have to get up early tomorrow, sorry im gonna have to close this, ill open up a new one tommorow though, much sorry 🙏
.close
calm coralBOT
fickle steeple
I ain't doing this 💀 @remote mural did you get any real answer
We still haven't used f(x)≠1 lol
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low roost
I need help identifying if the two following graphs are isomorphic
Let G1=<V, E1>, G2=<V, E2>, V={1,2,...,100}. E1 is the set of all pairs {a, b} where a and b are in V, so that |a-b|=10 or 90. E2 is the set of all pairs {a, b} where a and b are in V, so that |a-b|=89 or 11.
low roost
If they are isomorphic I'd like a hint as to how to construct a bijection F:V->V that proves the isomorphism
The closest I got was F takes x and returns (x+floor(x/10)) mod 100, but it doesn't seem to work for all possibilities of |a-b|
<@&286206848099549185>
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copper star
is it possible to
calm coralBOT
is it possible to
copper star
primitive (2x) divided by x squared + 1
glad parrot
Donc 2x/(x² +1) c'est ça ?
velvet osprey
primitive (2x) divided by x squared + 1
you should never write it like that
copper star
you should never write it like that
how do i
glad parrot
Thats the form of u'/u
velvet osprey
2x/(x^2 + 1)
rough blade
,, \frac{2x}{x^2+1}
potent lotusBOT
<rajel />
copper star
yes is it possible
glad parrot
Thats the form of u'/u
.
copper star
to primitive this algrabric
sweet stag
primitive (2x) divided by x squared + 1
use pictures if u cant use words
copper star
use pictures if u cant use words
alr
velvet osprey
yes is it possible
yes it is.
copper star
i meant the antiderivative
glad parrot
[ln(u)]' = u'/u
copper star
or the integral
glad parrot
i meant the antiderivative
(French do use primitive thats why i started speaking fr)
copper star
(French do use primitive thats why i started speaking fr)
ahh nah im dutch
glad parrot
Not sure a calculator can do this
scenic briar
this guy is wrong right
calm coralBOT
this guy is wrong right
calm coralBOT
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static violet
https://codeforces.com/problemset/problem/1442/B
any hints on how to rephrase the problem in different simple manner by ignoring unecessary information ? 
calm coralBOT
https://codeforces.com/problemset/problem/1442/B
any hints on how to rephrase ...
@static violet Has your question been resolved?
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static violet
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static violet
.close
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modern briar
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modern briar
I'm stuck can someone help me?
oh i forgot to write x/2 on the second to last equation but thats noy important
velvet osprey
i think this wasn't a good step
why not divide by log_3(12) from the beginning to get
log_2(x) = 1 + log_2(12)/log_3(12)
and then turn 1/log_3(12) into log_12(3)
it will work out very nicely
kind axle
and then turn 1/log_3(12) into log_12(3)
this is my approach 
modern briar
and then turn 1/log_3(12) into log_12(3)
kind axle
well that was simpler
modern briar
this is my approach <:hmmcat:831508801646755850>
i wouldn't even think to flip yhe logarith thats so clever
no no i dont need simple Ineed to learn as many ways as possible
thank you for reminding me of the flipping yhing
velvet osprey
that b-looking six 💀
kind axle
ann u need to stop criticizing others for handwriting
hard valley
.open
cerulean violet
i need help
calm coralBOT
i need help
cerulean violet
wary forge
I believe the bearing is 197 degress Mister.
cerulean violet
how to do this?
wary forge
Where'd u get stuck?
cerulean violet
What are we given?
we are given angle oac is 23and cb and oa are parallel
cerulean violet
how though
wary forge
Im not too sure acutally
wary forge
What about OBA?
slender cove
Also wrong I think
cerulean violet
Also wrong I think
angle bca is 23?
cerulean violet
what angle theorem was it again
cerulean violet
From the parallel lines
what was the same segment theorem again?
slender cove
Co-alternating angles I believe
wary forge
what was the same segment theorem again?
Im losing you here, dabusy41 will continue to help
cerulean violet
Im losing you here, dabusy41 will continue to help
oh thanks for trying though
slender cove
Not quite
slender cove
what can you say about arc AB (this is same segment)
slender cove
as isocieles triangle
Good, now you need to put everything together
Find BOA with same segment and finish with isosclees triangle
cerulean violet
Good, now you need to put everything together
is boa 23? or no
cerulean violet
is 46 ?
slender cove
cerulean violet
i forgot the theorems again
slender cove
Now finis with isosceles
cerulean violet
so 180-46 divide 2
slender cove
yes
cerulean violet
lol
slender cove
He did get ABC wrong
cerulean violet
whats 67-23
slender cove
uh 44
slender cove
why?
wary forge
uh oh the guy was right
slender cove
<:jesse:872561931528065144>
You still got ABC wrong
wary forge
You still got ABC wrong
Ok but i only tried on OBA
cerulean violet
i got 146 as the answer also
wary forge
i got 146 as the answer also
Good
wary forge
I think ABC is 180-67
Can confirm
cerulean violet
yeah i so 113?
slender cove
yes
calm coralBOT
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ember root
Nice, I need help real bad
calm coralBOT
Nice, I need help real bad
ember root
I have a number 87.6
and basically I wanted to give me the value without the decimal point
glad parrot
You want to round
ember root
So I thought if I just round(87.6) it would give me 87
main marlin
rounding 87.6 gives 88
there's three functions that math has, flooring, rounding, and ceiling
floor always rounds down
round splits halfway
(typically 5 and above is up)
and ceiling rounds up
ember root
rounding 87.6 gives 88
Oh my God, bro thanks
Do you like some sort of math, master person
I I just gotta remember this
main marlin
think of a house, floor is below and ceiling is above
In total:
if we have 3.4: floor(3.4)=3, round(3.4)=3, and ceil(3.4)=4
if we have 3.5: floor(3.5)=3, round(3.5)=4, and ceil(3.5)=4
if we have 3.6: floor(3.6)=3, round(3.6)=4, and ceil(3.6)=4
ember root
I get it now I think a little bit
It’s kind of like a seesaw
The ceil seesaw we’ll go down for heavier
The round seesaw will go for lighter or heavier weights
And the floor seasew will go for lighter weight
And the tipping point for round is .50
but if that’s the case, what if something’s x.5?
main marlin
round of (x.5) is conventionally x+1
round(x.4) is conventionally x
there's a rhyme I learned in my elementary school "Four or less, let is rest; five or more, let it soar."
ember root
You learn this stuff in elementary 🤯
main marlin
I mean just rounding, not the floor or ceiling 😅
short prawn
Note that in many application, they may use other ways of rounding. The most interesting one is rounding towards the closest even number. This is used in some banking tasks.
rich steeple
Is this right? :3
calm coralBOT
Is this right? :3
rich steeple
Pls ping!
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ancient grotto
Is this right? :3
Everything from parts a to d looks good
nimble harbor
,w pi int from 0 to 8 (4 - 1/2x)^2 dx
potent lotusBOT
ancient grotto
Checking e right now
rich steeple
Everything from parts a to d looks good
Thank you
calm coralBOT
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ripe spade
i want help regarding finding the jordan decomposition of a matrix. its a 6x6 matrix, and ive found that it has just one eigenvalue of 2. there are 3 linearly independent eigenvectors associated with it.
ripe spade
so its jordan matrix should have 3 blocks.
but how do i know what their sizes must be?
also if the original L.I. eigenvectors are v1, v2, v3, when it comes to finding generalized eigenvectors w, do i look for w's such that (A - 2I)w = v1, (A - 2I)w = v2, etc.. or do i solve (A - 2I)^2 * w = 0 ? there must be a difference between these methods
please ping me
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lament steppe
not really a math question but when teacher marks a test question as either right or wrong based on final answer, whats the best strategy to prevent making miscalculations especially when a question requires multiple steps to solve?
bold hatch
not really a math question but when teacher marks a test question as either rig...
If it's a word problem, consider asking yourself whether the number even makes sense at all as an answer.
coarse anvil
also if there are any variables whose value you are calculating, after calculating please box them
so that you know they are valuable info and you can plugin values later easily
rest do the calculation neatly and properly
if possible ake a rough work column on the side of the page
lament steppe
yea i had a few word problems to solve trig and if your drawing was wrong then your final answer is basically wrong
and i knew how to solve it easily but i messed up the initial sketch
and that f'd me up
coarse anvil
yea i had a few word problems to solve trig and if your drawing was wrong then y...
look it depends on the teacher as well
but try to make correct figure
especially in trig
other it might become a little misleading
coarse anvil
and i knew how to solve it easily but i messed up the initial sketch
remember not to repeat it
lament steppe
i know but its so easy to make small mistakes that will give you a wrong final answer
you mark the wrong angle or side and the rest of your solution means nothing
void frost
not really a math question but when teacher marks a test question as either rig...
That sounds like a pretty dumb system, a miscalculation should cost you little in comparison to getting the idea right
lament steppe
what do you mean?
void frost
If a problem gave you 3 points when correctly solved and you got some error in your calculations along your way, changing all the values, the correct thing for the teacher to do would be revoke maybe 0.5 points and if everything else is correct, give you the remaining 2.5
I'm saying just looking at the answer and judging how many points to give based on that is stupid
lament steppe
but that's the thing....a question that may require multiple steps will be valued the same as a question that requires only a straight up solution
all questions are valued as one point
regardless of question type
that's why im screwed
and there were only like 9 queseitons on the test
and i f'd up probably llike half of them cause i made some minor miiscalculations
so i either barely passed or failed
but i know in my head i knew how to soluve the solution
and it's bugging me
void frost
With practice, you will make less mistakes; don't rush because that could cause errors, if you've practiced enough you will be quick regardless
lament steppe
i did practice, felt confident but still managed to screw it up
it makes no sense to me how a question that requires multiple steps to solve is valued the same as a straight up calculation question
it leaves no room for part marks
void frost
it makes no sense to me how a question that requires multiple steps to solve is ...
Yeah, it's stupid
lament steppe
im almost certain the teacher is only going to look at final answer and wont bother to look at the work
she explicitly said to highlight the final answer so it makes her job easier
void frost
lament steppe
is normal or what?
void frost
No, that's not normal
In Germany you get revoked 0.5 points for calculation errors
Problems that have multiple steps give 3-6 points
lament steppe
but that means it requires the teacher to actually look over the student's work to see how they solved the question
lament steppe
do you think i should bring it up to my teacher....if i happened to get a failing grade?
cause that is a real possibility
void frost
I mean, I'm not sure if there is much you can change if that's how your country's high school system is
lament steppe
i think it's more specific to this one teacher cause my other math related course, the instructor says he gives partial marks
she's just lazy af....everybody knows
void frost
Ah, yeah then sure, you could bring it up to her
lament steppe
ok but is there any actual strategies i can do next time so i dont fuck up?
besides just practicing?
void frost
With practice, you will make less mistakes; don't rush because that could cause ...
.
Set yourself a timer or something
So that you can practice under exam conditions
lament steppe
oh yea thats a good idea....
void frost
Aim for comfortably completing the mock exam in maybe 3/4 of the time, without rushing
void frost
Aim for comfortably completing the mock exam in maybe 3/4 of the time, without r...
Managing it in 1/2 of the time is optimal, I'd say, afterwards you can look over everything
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spiral geyser
but then i wondered whetehr (a,a ) is an interval or not
i asked ai, and it set that it's a edge case cuz' some do not consider it a real set because
it has 0 points
civic dirge
This might be book-dependent
civic dirge
I'll look at what Tao's analysis says lol
drifting seal
why wouldn’t it
if you define an interval in R to just be a set containing all points between the two endpoints and possibly the endpoints then what you described satisfies it vacuously because there are no points between
spiral geyser
hmmm... i believe u r right tbh
i think i am done
thanks very much
got some good thoughts tbf
thanks guys ! 😄
drifting seal
you’re welcome
ancient grotto
is Tao like a mathematician or so?
Arguably the best in the business currently
main marlin
that'd be him
swift laurel
I'll look at what Tao's analysis says lol
civic dirge
Dang you're faster than me
I did lazily get this:
https://math.stackexchange.com/questions/1228307/why-is-the-empty-set-considered-an-interval
spiral geyser
could u explain this, cuz' it went over my head, like examples and stuff
swift laurel
so taking this to be our definition of an interval (from the same book):
we have
so (a,a) would be {x in R : a < x < a}, or in words, all the real numbers such that a < x < a. but it can't be true that both a < x and x < a at the same time, so (a,a) must be the empty set
swift laurel
so we "allow" (a,a) to be an interval but call it "degenerate" because it's the empty set
spiral geyser
so we "allow" (a,a) to be an interval but call it "degenerate" because it's the ...
does it come from the fact that we can't generate them on the no. line?
swift laurel
but what about a>b one?
again we would have (a, b) be the set of all real numbers x, such that a < x < b. but in that case it's similarly impossible for a < x and x < b since a > b
spiral geyser
oh shit
Bro you are actually smart
THanks very much
is there like anything else which i might wanna know
or like whihc might be important to know?
drifting seal
not really
spiral geyser
understood mate, thanks very much to you too
I guess i am done then
THanks very much guys
You are really cool people
🤟
drifting seal
no problem sir
quiet island
Claim: Every open interval is connected.
Don't get this proof at all
quiet island
why is alpha in A?
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digital island
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celest thistle
What do you need to do with it?
rough blade
i got the common denominator as 3n^3
should be square
indigo grove
so how do i do it
3n
n^2
LCM is 3n^2
kind axle
common denominator is 3n^2
indigo grove
3n
nn
3nn
digital island
to get the common denominator
digital island
What is LCM (3n, n^2)
dont knwo
potent lotusBOT
<rajel />
digital island
**<rajel />**
i understand this
u have to multiply b by c
to get the common denominator
but why does that not apply
rough blade
its not 3n^3 lol
digital island
for my question
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zealous basin
need help with this question and finding whats the shortest method for doing ques like these
leaden thunder
find the highest growth in denominator and numerator separately
if the highest is numerator, then divide by the highest growth term from the denominator
velvet osprey
those fucking l's hovering above the rest of the letters...
does the dot in $3.4^{x-1}$ mean multiplication or decimal point?
potent lotusBOT
Ann
pulsar mirage
multiplication prolly look at the denom
velvet osprey
$\lim_{x \to 1} \frac{(\ln(1+x)-\ln(2))(3 \cdot 4^{x-1} - 3x)}{[(7+x)^{1/3} - (1+3x)^{1/2}] \sin(x-1)}$
potent lotusBOT
Ann
zealous basin
does the dot in $3.4^{x-1}$ mean multiplication or decimal point?
multiplication
velvet osprey
ok so then my typesetting-unfuck was correct
note that this is a product of four factors all of which approach 0
well
ok some are in the denominator
but my point stands nonetheless
zealous basin
if the highest is numerator, then divide by the highest growth term from the den...
wdym
velvet osprey
it would be nice if we could pair them with something nice, like sin(x-1) would go well with (x-1)
cause then you could exploit the known sin(t)/t limit as t->0
do you get my idea
zealous basin
do you get my idea
wait im caught up in smth irl
velvet osprey
it's ok if you don't see how to implement it yet
zealous basin
wait im caught up in smth irl
mb for that but its an emergence
zealous basin
need help with this question and finding whats the shortest method for doing que...
here's a clearer image
velvet osprey
my TeXit image is clear enough
and it doesn't have the weird typesetting defect with the l's that yours does
anyway ping me when you've dealt with the emergency
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scenic briar
but this on a number line is different
i dont do graph method
i plot number line
and it is x < 1.8 and x > 7
pulsar mirage
but this on a number line is different
i dont do graph method
i plot number lin...
What's your question i don't understand
scenic briar
See this
scenic briar
What's your question i don't understand
To get quadratic inequalities final answer
I use number line method
-1.8 and 7 overlap
He used graph method
But my answer is wrong
pulsar mirage
Looks to me like you just took the opposite cases
He took the cases where the expressions are greater than 0
And you took the cases where the expressions are less than zero
calm coralBOT
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bitter prism
The first line is mentioned almost in passing "there is a point x in F that is moved freely by G." I don't see why this should be true, it is not intuitive for me.
And it's even false in the example I drew (where the action is the graph's automorphism group, or $\mathbb{Z}_2 \oplus \mathbb{Z}_2$). In the example, it is indeed the case that no element of G takes F to itself, and yet each point has orbit of size at most 2, so has a stabiliser size at least 2 (i.e. each point has a nontrivial stabiliser and is thus not moved freely)
potent lotusBOT
Tymon M.
bitter prism
for points on the left half, the nontrivial stabiliser is the flip on the right half, while for points on the right half, the nontrivial stabiliser is the flip on the left half
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bitter prism
.reopen
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tall moon
find all integers $x$ such that $x^3-1$ is a square
calm coralBOT
find all integers $x$ such that $x^3-1$ is a square
potent lotusBOT
skissue.in.a.teacup
tall moon
$x^3-1=y^2$
potent lotusBOT
skissue.in.a.teacup
leaden thunder
Do you know how to factor the left side
tall moon
Do you know how to factor the left side
(x-1)(x^2+x+1)?
if x is even, then -1≡y^2 mod 8, but -1≡7 isnt a quadratic residue mod 8, thus x must be odd
formal pulsar
find all integers $x$ such that $x^3-1$ is a square
I think there is a slick solution to this
try the substitution, x = 3a + 1
tall moon
,w expand (3a+1)^3-1
potent lotusBOT
formal pulsar
right and factoring out the 9a gives us
tall moon
9a(3a^2+3a+1)
formal pulsar
right, then for this to be a square what do we need the gcd of these tw factors to be
ie, if we want this to be a square number. then what does that mean about 9a?
tall moon
a must be a square?
formal pulsar
right
kinda
we also need that the gcd of the two factors is 1. That makes sense right?
tall moon
well gcd(3,3a^2+3a+1)=1 and gcd(a,3a^2+3a+1)=1 so gcd(9a,3a^2+3a+1)=1
formal pulsar
yep. so now we have that since 9 is already a square. we need a = b^2 for some int b
so then substitute that into the second factor
tall moon
so 3b^4+3b^2+1 is a square
formal pulsar
so 3b^4+3b^2+1 is a square
but you see that b^2 is then odd
sorry, not b
jsut for convience, let z^2 = 3b^4+3b^2+1
then z^2 is odd
so z is then odd
oh fuck, im running late for work.
sorry, I gotta skaddadle
but the proof should just be a bit of number theory to get to the point where there is only one solution
from memory, I saw the proof ages ago, but pretty sure theres nothing too crazy after this
tall moon
huh i found a proof online
but how do you know that y+i can be written as a cubed integer complex number?
calm coralBOT
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formal pulsar
but how do you know that y+i can be written as a cubed integer complex number?
Because they're co prime.
Btw that's not the proof I was thinking of.
tall moon
Btw that's not the proof I was thinking of.
yeah i was guessing so
formal pulsar
I think this is the one I was thinking of
If it isn't its kinda similar
Pretty slick nonetheless
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tall moon
https://math.stackexchange.com/questions/955967/does-an-elementary-solution-exis...
oo thank you
calm coralBOT
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vernal bramble
Does anyone know what a typical value for (Standard Deviation) SD is?
vernal bramble
I got data showing SD of 0.1
which is apparently extremely low and therfore very good
cobalt gate
from lil bit of knowlege i know
isnt less Sd mean values are very close to eachother
vernal bramble
isnt less Sd mean values are very close to eachother
ye
little variability
which is good
cobalt gate
yeh
vernal bramble
because it means that the data is accurate
pulsar mirage
Does anyone know what a typical value for (Standard Deviation) SD is?
i think it varies depending on what experiment you're performing
pulsar mirage
in some experiments 0.1 would be considered great in others that's a big margin of error
cobalt gate
it doesnt have a specific range
vernal bramble
i think it varies depending on what experiment you're performing
see the graph in #help-47
pulsar mirage
sorry i can't help lmao i have no data about this whatsoever, this is more geology
vernal bramble
sorry i can't help lmao i have no data about this whatsoever, this is more geolo...
It's alg mate
pulsar mirage
you cant tell what sd should be by looking at the graph
only can tell if youve performed the experiment before
vernal bramble
you cant tell what sd should be by looking at the graph
its presnted in the graph by the error bars
cobalt gate
you cant tell what sd should be by looking at the graph
yeh and the graph is weird too
vernal bramble
only can tell if youve performed the experiment before
I see
vernal bramble
yeh and the graph is weird too
In what sense?
cobalt gate
SD
pulsar mirage
its presnted in the graph by the error bars
oh i didnt see that
pulsar mirage
yeh and the graph is weird too
yes there are no markings for the error margin
cobalt gate
at first i thought that was a kinematic graph
pulsar mirage
it's just a bar between two numbers, no way to measure it
cobalt gate
in before question
vernal bramble
yes there are no markings for the error margin
I'll have to add the value on
I sort of forgot about that
vernal bramble
it's just a bar between two numbers, no way to measure it
There's no option to actually
vernal bramble
.close
calm coralBOT
calm coralBOT
Available help channel!
graceful thistle
Help me simplify 5√-45
calm coralBOT
Help me simplify 5√-45
rough blade
Help me simplify 5√-45
huh
cobalt gate
complex
rough blade
you mean $5\sqrt{-45}$ or $5\sqrt{45}$
potent lotusBOT
<rajel />
graceful thistle
you mean $5\sqrt{-45}$ or $5\sqrt{45}$
-45
unkempt drift
$\sqrt{-1} = i$ so it doesn't matter
potent lotusBOT
south
rough blade
so you know that $i^2 = -1$
potent lotusBOT
<rajel />
unkempt drift
My test is soon and I forgot radicals
$\sqrt{5 \cdot 9}$, can you find a way to break the square root into two square roots?
potent lotusBOT
south
graceful thistle
15i√5
unkempt drift
you can take sqrt(9) = 3 out and another sqrt(-1) = i
unkempt drift
Oops that's the answer, how do I simplify 3 ³√24
same reasoning
so $\sqrt[3] {8 \cdot 3} = \sqrt[3] {8} \cdot \sqrt[3] {3}$
potent lotusBOT
south
graceful thistle
Okay
graceful thistle
so $\sqrt[3] {8 \cdot 3} = \sqrt[3] {8} \cdot \sqrt[3] {3}$
Continue
What now
IMA tweak
My test is in 10 minutes
@unkempt drift
tall moon
i think its pretty clear how to continue it?
tall moon
try to think about one of the terms, can you simplify it?
unkempt drift
No pls tell me
what's the cube root of 8
graceful thistle
try to think about one of the terms, can you simplify it?
³√8?
tall moon
yes
tall moon
find the value to it?
graceful thistle
Oh my God and then I multiply them
unkempt drift
Oops that's the answer, how do I simplify 3 ³√24
so all in all, you have $3 \cdot 2 \cdot \sqrt[3] {3}$ as claimed
yeah!
potent lotusBOT
south
unkempt drift
that's the number such that x^3 = 8
cool so you just need to remember this definition
see the parallel with square roots?
x = sqrt(25) or 25^(1/2) is the number such that x^2 = 25
x = 64^(1/3) is the number such that x^3 = 64
graceful thistle
x = 64^(1/3) is the number such that x^3 = 64
Im still fucked for this test but thanks to you I might pull a 70, it's a quarterly test
unkempt drift
npnp!