OP's job is just to simplify, right?
#help-19
wooden python
you checked the answer key and what happened?
wooden python
3x(x+6) on top?
iron sky
yea
iron sky
im okay with gaslighting myself that its right tho
lean yew
im okay with gaslighting myself that its right tho
nah it's worth a double check
wooden python
uhh wait
hold on a minute
i thought i found the misread but i dont see it yet
yup found it
you miscopied the first denominator
and put -8x instead of +8x
which unfortunately throws quite the wrench into our entire work
and it was partially on me that i didn't spot the miscopy
on the other hand if i saw your work done in full like i took you through i'd probably give you full credit anyway and deduct 1 point for the misread
yeah so uhh
first denom will factor not as (x+2)(x-10)
but as (x+10)(x-2)
and the fraction conversions will be somewhat different
yeah sorry about this but it looks like youre gonna have to restart the problem bc the numbers are gonna be all different
wooden python
you do you
iron sky
and i’ll blaze through
wooden python
is this homework or what
iron sky
yea there’s a test tmrw
iron sky
bc summerschool like moves faster
wooden python
if you dont get the time to do the work afterwards you can tell the teacher "look i did this question in full but then found that i misread the fraction from the start and thats why my answer doesnt match"
wooden python
which, assuming your teacher is not an asshole, will probably be met with understanding
iron sky
he’s nice
wooden python
ok in that case i'd set fixing this to a low priority
raven stirrup
i dont know how to solve this exactly getting stuck finding the parallel sides
brittle plinth
you don't need to
raven stirrup
oh
brittle plinth
do you know the property of the midline in a trapezoid?
brittle plinth
then what about the midline of a triangle?
brittle plinth
then what about the midline of a triangle?
it's the line connecting the two midpoints of two sides of a triangle
raven stirrup
its the avg of parallel sides rt?
brittle plinth
its the avg of parallel sides rt?
yes
odd edgeBOT
odd edgeBOT
Available help channel!
normal sentinel
odd edgeBOT
normal sentinel
i'm working through this concavity problem in calc 1
process makes sense but one question about my sign chart
without looking at a graph, how do I deduce from negative infinity to my first value, how could I tell if the polarity is either negative or positive?
I just worked it out myself and I had it flipped backwards
so in this case negative infinity -> -(4/5) is +
normal sentinel
not off the top of my head
narrow crypt
Like polynomial inequalities
amber veldt
so in this case ``negative infinity -> -(4/5)`` is ``+``
this sentence makes no sense
negative infinity doesn't approach anything other than itself
amber veldt
ah so you mean between -infinity and -4/5, got it
normal sentinel
yup, apologies
amber veldt
np
amber veldt
without looking at a graph, how do I deduce from negative infinity to my first v...
so you're asking specifically from the shape of the curve?
it's not so easy to
normal sentinel
right
amber veldt
but the way I do it
is imagine a little tangent line segment at each point of the curve, unit length for simplicity
and imagine traversing the curve as a point travelling along
do you understand or should i draw a picture
normal sentinel
i'm picturing
amber veldt
the slope of the tangent line changes as f'' changes
now, if the curve has continuous second derivative
actually one moment, let me check if that's even necessary
it's not
normal sentinel
lol i appreciate it
amber veldt
the sign of the second derivative corresponds to its concavity
which is to say
what side of the curve the tangent line segment is on
below or above
when I say above or below I mean if you draw a vertical line connecting the tangent line to the curve
is it pointing down or up?
normal sentinel
i'm having an issue visualizing just from f''(x) alone, without the graph.
normal sentinel
yeah so i'm working it out and I have the critical points, I know where the turning points are
it's just that from approaching to the left up to my first critical point
without the graph, I am unsure exactly if its positive or negative when approaching my first critical value
amber veldt
you're unsure exactly if what is positive or negative?
normal sentinel
i worked it out and had it backwards from the picture above
x approaching from left to the first critical value -(4/5)
the sign changes obviously after the critical value, I get that
amber veldt
critical value of the derivative you mean, not critical value of the function
normal sentinel
yes
apologies
when I worked it out myself, I had it backwards where the value of the derivative was negative instead of positive like in the picture above
amber veldt
you want to be able to do it without the formula for f'?
not generally possible
because f' and f'-1000 and f'+2 have the same f''
normal sentinel
i'm not following
amber veldt
you can't get the sign of a function from its derivative
because the function f(x)=x-1000 and g(x)=x+1000 have df/dx=dg/dx
normal sentinel
i have the f(x) and my f'(x) as well as the f''(x)
amber veldt
if you have f' then put in numbers and see whether f'(x)<0 or f'(x)>0
normal sentinel
in that case if the f' is negative, then it would be concave down on f''
vice versa
from that point to my first critical value that is
amber veldt
yes
amber veldt
np
normal sentinel
.close
odd edgeBOT
odd edgeBOT
Available help channel!
crude hollow
odd edgeBOT
crude hollow
,rotate
clever fjordBOT
odd edgeBOT
@crude hollow Has your question been resolved?
crude hollow
<@&286206848099549185>
snow rune
crude hollow
is it like the min dist btwn conics?
odd edgeBOT
frigid isle
crude hollow
huh?
nimble blaze
did someone ping mods?
odd edgeBOT
Available help channel!
nimble blaze
.ok. please don't delete regardless
crude hollow
sorry my bad
odd edgeBOT
sorry my bad
crude hollow
.close
odd edgeBOT
midnight epoch
lol.
odd edgeBOT
Available help channel!
pastel orbit
I'm once again asking for feedback on a proof
odd edgeBOT
I'm once again asking for feedback on a proof
pastel orbit
any mistakes or oversights? anything unclear or could use better wording? 
odd edgeBOT
@pastel orbit Has your question been resolved?
mild isle
Oh yeah that's exactly what you showed in your proof in fact!
mild isle
pastel orbit
In other words, a deformation retraction is a homotopy between a retraction (strictly, between its composition with the inclusion) and the identity map on X.
hmm 
is the retraction in this case zeta o pi?
mild isle
You have shown that E is homotopy equivalent to the image of the zero section (which is a homeo to its image)
mild isle
is the retraction in this case zeta o pi?
The retraction is the projection map here
pastel orbit
ah.
mild isle
(strictly speaking (to quote Wiki), the composition of the projection with the zero section)
mild isle
it's like compressing the bundle down to M?
yes!
pastel orbit
ic
mild isle
That's imprecise but it's the idea!
A visual example is the disc which deformation retracts to its center point
You 'collapse' it
Whereas you cannot collapse it to its boundary without looking at the punctured disc
And then the punctured disc deformation retracts to its boundary
pastel orbit
aha, that makes sense
what about an annulus?
does it retract onto both circles?
or only one of them?
mild isle
It deformation retracts to any one of the boundary circles yeah
But not to the disjoint union of both
pastel orbit
gotcha
mild isle
You'd have to cut it in two for that
The Möbius strip deformation retracts to a circle
But it's not the boundary one
It's the one "in the core, in the middle"
pastel orbit
that's what I expected, actually
I can see the strip being collapsed to the middle, but not to the outside boundary
mild isle
What's funny about that deformation retraction of the Möbius strip to its core is that if you restrict it on the boundary, you get a double covering of S^1
mild isle
This is why the (strictly speaking) part of the Wiki article is not how you should this about deformation retractions
mild isle
the two boundary components both map onto S^1 in that case, right?
There's only one bdry component for the Mobius strip! 
pastel orbit
There's only one bdry component for the Mobius strip! <:catgiggle:91245949795450...
ah shoot 
why's it going to be a double cover then?
mild isle
This is why the (strictly speaking) part of the Wiki article is not how you shou...
You should really think of the def retraction as the projection map
pastel orbit
This is why the (strictly speaking) part of the Wiki article is not how you shou...
you should this about deformation retractions
did you mean think?
mild isle
Lemme make a piocture
pastel orbit
piktur 
mild isle
Here the boundary is a single circle, and you can see both points which are collapsing to the same one on the core
In a sense, this boundary component circles around twice
Because if you follow it with your finger 'once' around the red circle, you'll end up 'on the other side'
pastel orbit
In a sense, this boundary component circles around twice
this makes a lot of sense to me
like, pictorially 
mild isle
Anyway, we have slightly deviated from your topic
pastel orbit
well, you've already read the proof and said it works
mild isle
In the case of a vector bundle (which the infinite Mobius strip is), it is the same core idea
pastel orbit
I was just curious about some more things
mild isle
You "shrink" all the individual vector spaces to their origin {0}
And that is possible to do smoothly across the whole bundle total space
pastel orbit
Lee has this picture of a line bundle when defining a vector bundle
mild isle
Yes in a local trivialization it seems
pastel orbit
is it accurate to say that all the vertical lines (1-D vsps) are being shrank to a point on U by pi in this picture?
mild isle
Yes it is!
pastel orbit
pastel orbit
Yes it is!
I see 
mild isle
I'll add it to the picture
Here the arrows depict the projection that shrink each individual line to a single point
pastel orbit
I see
what if we tried shrinking the lines to a point that wasn't 0?
what goes wrong?
mild isle
When thinking of bundles, you really have to think of the base space as the image of the zero-section
No this also totally works!
In fact, you can shrink to any image of any section
We just prefer the zero section because it's quite "canonical"
pastel orbit
ah, so M is like one section of E
mild isle
ah, so M is like one section of E
Yeah basically
pastel orbit
ah, so M is like one section of E
E is an infinite stack of M's
mild isle
,,M \cong 0_E(M)\subset E
mild isle
E is an infinite stack of M's
Yes!
pastel orbit
wow
clever fjordBOT
TeXplotlib
pastel orbit
this reminds me of the "stack of pancakes" idea that Munkres asked me to use to think about covering spaces
mild isle
A very cool exercise you should do before pursuing
Try to see why the infinite cylinder and the infinite Möbius strip are two line bundles over the circle, and why they are different (they're not isomorphic)
I'm not asking you to show they're not iso, just to convince yourself they shouldn't
If you really want to show they're not, it's possible: the cylinder is a trivial bundle, and the Möbius strip is not
When you have the vocabulary and the notion, to show the Möbius isn't trivial, you need to prove it does not admit a nowhere vanishing section
mild isle
When you have the vocabulary and the notion, to show the Möbius isn't trivial, y...
This you should try to draw one by hand (draw the image of the section rather), and see that you must cross the zero-section!
This example of those two line bundles over the circle is really as important as the fact that the fundamental group of S^1 is Z, you'll need it to keep it in mind as a "reference image"
pastel orbit
I will have to give this some thought
mild isle
Ofc!
mild isle
!occupied
odd edgeBOT
Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).
pastel orbit
Try to see why the infinite cylinder and the infinite Möbius strip are two line ...
hmm, if you delete the circle from the cylinder, you get two disconnected components, but if you do the same for the Mobius strip, you're still left with one component, right?
the circle being the image of the 0 section in this
mild isle
That tells you that the total spaces aren't homeomorphic, which indeed suffices to prove that the bundles aren't isomorphic indeed
pastel orbit
hmm, if you delete the circle from the cylinder, you get two disconnected compon...
you can walk along the boundary of the strip to get to "the other side"
so we still have one component
mild isle
This is indeed easier, but you don't see why they're not isomorphic as bundles though
pastel orbit
mild isle
Also, I'm not 100% convinced this suffices, since the image of a circle in one needs not be homotopically non-trivial in the other (so the circle in the cylinder might be mapped to a nullhomotopic one on the Möbius)
But the idea is convincing though
mild isle
Bundles E->B and E'->B are isomorphic if there exists a homeomorphism E->E' that makes this diagram commute
Moreover this homeo should be a linear map in each fiber
(which are preserved by commutattivity of the diagram)
(and this linear map is necessarily an isomorphism)
pastel orbit
I see
@mild isle I don't know if this idea is any closer, but from looking at a picture, you can walk along the zero section circle in a cylinder and end up right side up, but you'll be upside down if you try the same thing for the Mobius bundle 
I think this is related to orientations? though I haven't quite gotten there yet
mild isle
<@237320168198504448> I don't know if this idea is any closer, but from looking ...
Yes that is the idea!
pastel orbit
idk if this is anything that has to do with the fact that they're bundles though...
mild isle
In fact, to get a section, you'd need to close the loop though and end up where you started
So to do that you'd need to cross the zero-section
This means that the section you trace vanishes somewhere (at the crossing point)
This is not avoidable for the Möbius, but it is for the cylinder (just slide the section entirely up and you're good to go)
mild isle
idk if this is anything that has to do with the fact that they're *bundles* thou...
This is related, but this is a story for another time (orientations of vector bundles)
pastel orbit
In fact, to get a section, you'd need to close the loop though and end up where ...
ah, hold on. this might've just revealed a misconception I had?
I was under the impression that the base manifold needed to all be at "the same height" in the bundle
is that not true?
like, can the loop M be "slanted" in E?
mild isle
I'll show you two distinct sections in the case of the cylinder as a line bundle over S^1
pastel orbit
This means that the section you trace vanishes somewhere (at the crossing point)
the reason I asked is because when you say this, it made me think that the image of an arbitrary section doesn't have to "stay on one side of the zero section circle"
but that was what I thought was the case
mild isle
Those are two sections
If red is the zero-section, then this means that blue vanishes twice
pastel orbit
aha, but you can move blue above the red line
mild isle
pastel orbit
so it'll never cross
mild isle
The green dots are where blue vanishes
Yeah in this instance you can push it away to ensure it does not vanish!
pastel orbit
the reason I asked is because when you say this, it made me think that the image...
so this was a misconception after all!
mild isle
In the case of the Möbius, you cannot in fact!
There are no non-vanishing sections (and any section will vanish an odd number of times in fact, as long as it is transverse to the zero-section, but whatever)
pastel orbit
ah, we're screwed indeed
mild isle
Yep
mild isle
Cool 
pastel orbit
this was really helpful 
mild isle
This example you'll need for when you deal with char classes
It allows to have a starting point for intuition
mild isle
pastel orbit
with that said, I've gotta bounce now 
friends are calling me to game with them
so I'll close this now
thank you so much mpl!! 
.solved
sharp bridge
how do i do
odd edgeBOT
how do i do
regal dirge
[usually, you assume y = f(x)]
the y-intercepts are the points where f(x) intersects the y-axis (i.e. x = 0)
the x-intercepts are the points where f(x) intersects the x-axis (i.e. y = f(x) = 0)
so what is the value of y when x = 0?
sharp bridge
wait so plug in 0 for x?
regal dirge
yeah
sharp bridge
isnt there something about if it doesnt work with 0 its not real or theres an imaginary 0 or something
regal dirge
that'd be for the x-intercept
sometimes a function doesn't intersect the x-axis, in that case it has no real x-intercept
you don't need to worry about that here though
because it definitely intersects both the x and y axes
abstract harbor
wait so plug in 0 for x?
how did ur exam go
sharp bridge
idk it hasnt been graded
probably got like a 60 or 70 ngl
it was 12 questions like i fucking hate when exams are like that
id rather it be 120 than 12
regal dirge
yoo just like the putnam
abstract harbor
idk how to do these yet so just trying to figure it out
the y intercept is when the line hits the y axis right?
sharp bridge
yes
abstract harbor
So what’s x when the line hits the y axis
abstract harbor
it’s always the same value for any line when it hits the y axis
regal dirge
yeah
sharp bridge
o thats ez
regal dirge
cause all points on the y axis have x = 0
sharp bridge
bro is it just what the orginal numbers were
regal dirge
instead of x = 0, you now set y = 0 and solve for x
sharp bridge
is it not just the 4th option
regal dirge
well
does x = -9 make y = 0
y is the same as f(x) in this problem btw just to clarify
sharp bridge
then i dont see how it couldnt be 4
regal dirge
set y = 0 and solve for x just to make sure
sharp bridge
well thats dumb
is the y always going to be the inverse of the origninal numbers
its 3
regal dirge
yeah it's 3
sharp bridge
everytime i set y to 0 its gonna flip the signs?
regal dirge
if it's nicely factored like this then yeah
the solution to x-a = 0 is x = a
that's where the sign gets flipped
mellow tundra
past mango
To find the x-intercepts, we set y=0, which gives us the x-intercepts at (−2,0), (-3,0) and (−6,0).
To find the y-intercept, we set 𝑥=0, which gives us the y-intercept at (0,108)
sharp bridge
just realized i have 6 assignments tonight😭
im cooked
there goes 4 hours
maybe not
this is like downsyndrome math
past mango
Did you get the answer?
sharp bridge
same thing?
i feel like 100000 is not right
oh nvm
for these its having me find the GCF and then factoring the quadratic but it js doesnt work with this one
idk what to do
photomath did some crazy weird stuff so
<@&286206848099549185>
sorry to rush im on a time crunch
spare geode
Begin by taking out an x
spare geode
Whay
sharp bridge
how do people just know and memorize how to do an entire class of equations
ts is baffling to me
spare geode
Okay u should have x( x⁴-5x-6) rn 😭
sharp bridge
oh i mean when the problem is done
spare geode
Ohhh okay okay lemme see
sharp bridge
spare geode
Hmm
Then nah yeah that's right unless they also want the imaginaries
But I don't think so cuz you have 3 blocks only
sharp bridge
thats so weird i dont like these problems alr and i know they gonna get way harder
spare geode
😭🙏 GL man
sharp bridge
WHAT THE FUCK
is it bc you cant square root negatives or something
im so lost
there explanation is way easier than how i did it
can someone explain how (x^2+1)=0 became square root 6
i dont understand that
vivid wyvern
(x²-6)=0 became sqrt6 not (x²+1)
U see the product of more than one number will be 0 if an only if at least one of them is 0
So x=0
x²+1=0,x²=-1, root over negative values is undefined in these cases,so we wont get any value of x from here
x²-6=0 x²=6 x=sqrt(6)
We know (sqrt(6))²=6
(-sqrt(6))²=6
So x =+sqrt6 and -sqrt6
sharp bridge
so if one of them is x=0 I know that one of the two other answers has 2 x's?
is that universal
vivid wyvern
It isnt a must to hv 2 x's
U get 2 x's if u hv a quadratic one with no perfect square
Wait u need to understand how to find values of x,it's not universal to have 2x's in 1 answer
It depends on how many factors u get and whether u can find values from it or not
sharp bridge
okay i understand this whole unit now
thank you everyone
only took an hour and a half to master it lol and this isnt even hard at all
the no real roots thing is tripping me up actually
looks like this format is gonna be super common for these
where it’s x=0 and then x=x2+a x=x2-b
so can someone explain how i find the 3 x values
tidal matrix
remember, you're setting $x(x^2 + 3)(x^2-4) = 0$ to find the roots. so what should your 3 equations be?
clever fjordBOT
haseeb
tidal matrix
unfortunately not
sharp bridge
are u saying 0=x2+3
tidal matrix
yes exactly
sharp bridge
oh no i understand that i just picture it
torn river
remember, you're setting $x(x^2 + 3)(x^2-4) = 0$ to find the roots. so what shou...
woulf it be x=0 x^2+3=0 and x^2-4=0?
sharp bridge
i js dont understand where the third x value comes from
odd edgeBOT
As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.
tidal matrix
icic, let's work through it
sharp bridge
does it come from the inverse of the negative square root number
tidal matrix
i think you have the right idea but you expressed it really weird ._.
come back to $x^2 - 9 = 0$, how would you solve?
how would you start?*
tidal matrix
omg whoops
sharp bridge
+9 both sides
clever fjordBOT
haseeb
tidal matrix
ok so we get $x = \sqrt{9} = 3$, which is one solution
clever fjordBOT
haseeb
tidal matrix
for a proper solution, you should write "x^2 - 9 = 0", then "x = 3" under it, so it's clear
sharp bridge
lmaooo forgot i had to simply the square root
tidal matrix
rip 🥀 hopefully that's not too hard to remember tho
sharp bridge
the last problem had me double the negative square root
like + and -
but idk why
that’s what this whole inquiry is
tidal matrix
okay, again we have right ideas in wrong places, but this is a good start
To humour me, try subbing $x = -3$ into the equation $x^2 - 9 = 0$
clever fjordBOT
haseeb
tidal matrix
does LHS = RHS?
sharp bridge
what does LHS mean
tidal matrix
left hand side / right hand side
tidal matrix
add 9 to both sides, not subtract
sharp bridge
i hate logical math
the equations should just give you the answers
logical math jsut feels like proofing
tidal matrix
it kind of is, i was hoping to show you when "the rule" applies, so you dont get mixed up again
in general, if $x^2 = +c$, (so $c$ is positive or 0), then $x = \pm\sqrt{c}$
clever fjordBOT
haseeb
tidal matrix
because you can always sub the $-\sqrt{c}$ back into the equation and get 0=0
clever fjordBOT
haseeb
tidal matrix
in this case, $(-3)^2 - 9 = 9 - 9 = 0,$ so $x=-3$ is the second solution
clever fjordBOT
haseeb
tidal matrix
does that rule make sense?
tidal matrix
careful on the wording. x^2 - 9 = 0 when x=3, but also when x=-3
there are two possible ways to get x^2: either (x)(x), or (-x)(-x)
hence two solutions
tidal matrix
however, what you wrote here for x^2 + 3 = 0 isn't right
sharp bridge
ohhhh
okay
positive ^2 and negative ^2 are always the same
so 3 and -3 are both valid
tidal matrix
exactly!
sharp bridge
okay thanks for baring wiht me on that
tidal matrix
np :) now for x^2 + 3 = 0, it actually has no real solutions
because when you get to x^2 = -3, you can't square root
sharp bridge
so what are my 3 xs now
tidal matrix
0, 3, and -3
0 from the first equation
3 and -3 both from x^2 - 9 = 0
and no solutions from x^2 + 3 = 0
\begin{itemize}
\item if $x^2 = +c$, then $x = \pm \sqrt{c}$
\item if $x^2 = -c$, then there are no real solutions
\item $\sqrt{c^2} = c$ because $\sqrt{.}$ is always positive
\end{itemize}
clever fjordBOT
haseeb
tidal matrix
reference for you :)
sharp bridge
yeah the math terms lose me sometimes
its easier to speak in laymans definite statements
tidal matrix
i def get it, but it's easy to lose some nuance
sharp bridge
for this class i only need to know the rules for the exams and then i can throw all the knowledge away
so if they are applicable in this one unit then thats perfect for me
my major has NOTHING to do wiht math
wtf is this jawn
tidal matrix
it's your life, but if you do end up taking another math class, you'll have to re-remember it all
tidal matrix
wtf is this jawn
same steps-ish, but now with the cube root
sharp bridge
yea but this is the only math class i have to take for the rest of colelge
tidal matrix
fair enough then
sharp bridge
i mean i guess
(x^2+?)(x-?)
confused how i go from step 2 to 3
on this new problem
sharp bridge
honestly i have no clue
tidal matrix
$x^3 - 729 = 0$, and get the $x^3$ by itself...
clever fjordBOT
haseeb
sharp bridge
x^3 = 729
tidal matrix
W
sharp bridge
cube root both sides?
tidal matrix
ah your calculator is the difficult kind
sharp bridge
whats the y
tidal matrix
oh yeah
quick note: with cube roots, -9 is not a solution. you can sub it in and try yourself to confirm
to get y, just sub x into the original function
i.e. y = f(9)
tidal matrix
no, you should get 0
sharp bridge
mb i didnt read the second part
tidal matrix
you can also just say it's an x-intercept so it will be (x,0)
and y-intercept will be (0,y)
but i feel like that might be too much to memorize, especially if not sure why
sharp bridge
it says 9,0 is wrong
sharp bridge
what
so what do i do
it just didnt specify
like fuck this class
now i have to restart the entire 2 and a half hour long assignment
tidal matrix
oof that's rough
the hardest part of the assignment is figuring out what they want you to type
i think you've got the understanding now at least
odd edgeBOT
@sharp bridge Has your question been resolved?
odd edgeBOT
odd edgeBOT
Available help channel!
sharp bridge
how do i solve these
odd edgeBOT
how do i solve these
digital kestrel
it would help to factor x^2 - x - 12
i'll just do that for the sake of moving on with the explanation. you can rewrite it as (x-4)(x+3) <= 0
all you really care about is the sign of (x-4) and (x+3). if they are both negative, (x-4)(x+3) > 0. if they are both positive, (x-4)(x+3) > 0
if one them is zero, (x-4)(x+3) = 0
lean yew
think you should flip one of your inequality signs
digital kestrel
which one?
sturdy cape
think you should flip one of your inequality signs
No he's good
digital kestrel
and if one is positive and one is negative then (x-4)(x+3) < 0
sturdy cape
If both factors were positive, or if both were negative, their product is > 0
digital kestrel
those are all the cases you could come across
lean yew
yeah i just saw, very sorry
sturdy cape
If at least one of them were 0, then the product is 0
digital kestrel
so you should think about what values of x make the expressions positive and negative
sturdy cape
i'll just do that for the sake of moving on with the explanation. you can rewrit...
So the only conclusion then is, either one of these factors is 0, or one is positive and the other is negative
digital kestrel
yea, if x makes one of x-4 or x+3 positive and the other negative, or makes one of them 0, then (x-4)(x+3) <= 0 is true
and that's what it means for x to be a solution to the inequality
and "solving" it means finding all such x
it's ok ms seia
odd edgeBOT
@sharp bridge Has your question been resolved?
sharp bridge
my bad
north sparrow
like they said before tbh: factor then study the sign of the 2 linear factors (x-a) you get
north sparrow
that would be x^2-2x+6x-12=x^2+4x-12 no?
drifting garnet
factor?
Factor in the form (x+a)(x+b) so that a+b = -1 and ab = -12 (hopefully I'm rly bad at explaining)
north sparrow
where did 2x come from
from what they call foil
sharp bridge
that would be x^2-2x+6x-12=x^2+4x-12 no?
oh okay i thought u meant i wass starting from here
north sparrow
then i simplified the -2x+6x to 4x
What Zylon proposed is a good trick to factor fast
but you have to see it
drifting garnet
Might help to lay out all the integer factors of 12 first, both positive and negative, to solve for a and b
north sparrow
yeah that’s another way to factor, there is a theorem that says if your polynomial has a rational root then it divides the constant term of the polynomial
finally when life is bad you have the quadratic formula and completing the square
but here life good
drifting garnet
yeah in this case it's also simple to get the roots from quadratic formula and substitute them for a and b in (x+a)(x+b). It would also give you a good indication of the numerical bounds of the inequality, but you can also graph the function for help
odd edgeBOT
@sharp bridge Has your question been resolved?
ripe knoll
To factor quadratics like this, I first think of the prime numbers that make up the element a*b, that is -12, and I see which combination gives me the same result as the term a+b (-1 in this case).
For this problem, you have: a*b=-12 can be: 6*-2 , -6*2, 3*-4. and -3*4
Of all these, 3*-4 is the only one that satisfies that: ab=-12 and that a+b=-1
So factored it would be (x+3)(x-4)
sharp bridge
So factored it would be (x+3)(x-4)
so i have to factor twice??
ripe knoll
No, just to have two linear equations as factors of a quadratic (I don't really know)
north sparrow
earlier i tried to show you that the factorization you proposed did not work, Eumnl_ actually factored this polynomial for you
sharp bridge
im lost
i factored what you gave me and got that
and then asked if i factor 2 problems and they said no
drifting garnet
so you know that the polynomial can be written as (x+3)(x-4) and you want to find the values that are equal to or below 0
north sparrow
(x+6)(x-2)?
what i understood is that you proposed (x+6)(x-2) = x^2-x+12 but that’s wrong
drifting garnet
so if we want to find the range of values below zero we want to look at the part of the graph where y < 0 (I apologise if I sound patronising)
north sparrow
the actual factorization is x^2-x+12=(x+3)(x-4)
ripe knoll
That divides the line on three parts
(-♾️,-3) , (-3,4), (4,+♾️)
Let me do a drawing rq
drifting garnet
always helps to visualise 🙂
ripe knoll
If you pick any number in the first section of the line (green section) it won't satisfy the inequality. Example:
x=-5 (-5 is in the green section)
So by the inequality x²-x-12=0, where we replace x as -5:
(-5)²-5-12=0
25-17=0
8=0 so the numbers in the green part do not satisfy the inequality because 8≠0
You can do the exact same thing on the blue and purple side.
And you'll notice that only the numbers on the blue side does satisfy the inequality
And, because -3 and 4 are part of the blue section, the answer is (-3, 4). Which means that the inequality x²-x-12=0 is only valid for numbers -3 to 4.
odd edgeBOT
@sharp bridge Has your question been resolved?
odd edgeBOT
odd edgeBOT
Available help channel!
flint crow
How does one calculate this integral of this without beta function because we will NOT be remembering the values of gamma function
flint crow
Help please
quasi sparrow
Did you try u = sqrt(tan(x))
quasi sparrow
!show
odd edgeBOT
Show your work, and if possible, explain where you are stuck.
narrow crypt
why is it negative
clever fjordBOT
k
narrow crypt
doesnt seem to have an easy way to solve this
flint crow
Nooo
narrow crypt
lets try another method
flint crow
Who the hell is that
narrow crypt
Who the hell is that
its an integration technique
${x := \frac{\pi}{2} - u \implies \dd x = -\dd u}$. So,
[ -\int_{\frac{\pi}{2}}^0 \sqrt{\tan(\pi/2 - u)}\dd u = \int_0^\frac{\pi}{2} \sqrt{\cot x}\dd x]
clever fjordBOT
k
flint crow
Maybe we can add these two
narrow crypt
Maybe we can add these two
im thinking that
[ 2I = \int_0^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\cos x}} + \frac{\sqrt{\cos x}}{\sqrt{\sin x}}\dd x]
flint crow
Goofy
clever fjordBOT
k
narrow crypt
[ = \int_0^\frac{\pi}{2} \frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} \dd x]
clever fjordBOT
k
narrow crypt
lets see
[ \int_0^{\pi/2} \frac{(\sqrt{\sin x} +\sqrt{\cos x})^2 - 2\sqrt{\sin x \cos x}}{\sqrt{\sin x \cos x}} \dd x ]
flint crow
Hmm
clever fjordBOT
k
flint crow
Now what tho
narrow crypt
\[ = \int_0^\frac{\pi}{2} \frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} \dd x\]
from this
[ \sqrt{2} \int_0^{\pi/2} \frac{\sin x+ \cos x}{\sqrt{\sin 2x}} \dd x]
clever fjordBOT
k
narrow crypt
alr now
take ${t := \sqrt{\sin 2x} \implies \dd t = \frac{1}{2\sqrt{\sin 2x} \cdot 2 \cdot \cos 2x \dd x}}$. Thus, ${\dd x = \frac{\sqrt{\sin 2x}}{\cos 2x}}$
clever fjordBOT
k
narrow crypt
Since ${\cos 2x = \cos^2 x - \sin ^2 x = (\cos x - \sin x)(\cos x + \sin x)}$, we can see that the integral reduces to
flint crow
Hmmm
clever fjordBOT
k
narrow crypt
[ \sqrt{2} \int_0^{1} \frac{1}{\cos x - \sin x} \dd t]
clever fjordBOT
k
narrow crypt
now
narrow crypt
ppl actually use weistrauss substitution?
😭
notice ${(\cos x - \sin x)^2 = \sin ^2 x + \cos^2 x - \sin 2 \theta = 1 -\sin 2\theta}$.
clever fjordBOT
k
flint crow
Oh hell nah your goofy schizophrenic ahh doing allat
narrow crypt
wdym
quasi shard
I am trying to solve 5a. So I my sum right now looks like sum from 0 to 2 k(sqrt(k+1)-sqrt(k))
quasi shard
But the sun will only equal the integral if the sum goes to 3
So the upper bound must be n+1 but that conflicts with our definition
nimble blaze
can you show your full work
quasi shard
yes one second
quasi shard
can you show your full work
nimble blaze
1 sec
odd edgeBOT
@quasi shard Has your question been resolved?
quasi shard
yes I know but how can we know what n is for our sum. for our integral we are only going from 0 to 2
quasi shard
why? we have a partition from 1 to sqrt 2 for example
orchid torrent
Your point is? The length of each interval is $x_{k+1}-x_k=\sqrt{k+1}-\sqrt{k}$ and the value of $s$ on the interval $(\sqrt k, \sqrt{k+1})$ is $k$, meaning $s_k=k$. Since your integral is for $0<t<2=\sqrt{4}$, we have that $\sqrt{n+1}=\sqrt 4 \implies n=3$.
clever fjordBOT
Civil Service Pigeon
quasi shard
why are we allowed to connect n like that?
quasi shard
You’re partitioning based on sqrt(k) being an integer and not k
I guess I am stll not understanding this
orchid torrent
I guess I am stll not understanding this
Because your intervals aren’t (k, k+1)
orchid torrent
why are we allowed to connect n like that?
Draw a picture
You should see that (0,2) is being partitioned into (0,1), (1,sqrt2), (sqrt2,sqrt3), (sqrt3,sqrt__4__)
The kth interval (starting from k=0) is (sqrt k, sqrt(k+1))
quasi shard
I see that sqrt 4 is the last value on the partition and I see how we can rewrite sqrt 4 with n but I don't see how we come up with this
orchid torrent
What is “this”
quasi shard
how we decided that we can use the fact that since the last partition ends with sqrt 4 we can use it to find n
orchid torrent
0th interval is (0,1), 1st interval is (1,sqrt2), … 3rd interval is (sqrt 3, sqrt 4)
So k goes over 0,1,2,3
Aka you’re summing over k=0,1,2,3
quasi shard
yes but first our k starts from 1. I know this doesn't matter because it would be 0(1) for k = 0. but how can we find these n values without having to write out all the partitions
It is obvious to see what n is if you write it out or graph it but is there another way
orchid torrent
If you wanted to write it to be consistent with the definition, then you should’ve let the kth interval be (sqrt(k-1), sqrt(k)) so you can start from k=1
Then you could go up til n=4
As to actually finding n in general
Idk it should be obvious from how you choose to define your sub intervals tbh
Maybe do a little inequality fiddling if you want to ig
quasi shard
If you wanted to write it to be consistent with the definition, then you should’...
but don't we have n<= x^2 < n+1
quasi shard
Idk it should be obvious from how you choose to define your sub intervals tbh
how so? the defintion just says the integral from a to b is equal to a sum from 1 to n.
this one
for that I just used the inequaltiy for the greatest integer function n<=x<n+1
quasi shard
@orchid torrent I just don't how the pattern for finding n which is using the fact that n for the sum equals the last part of the partition = our top bound for the integral
what I mean by this is say for integral 0 to 4 for floor(x) dx
Our partition length is 1
it is always n+1-n
then to see what n equals in the sum we say n+1 = 4 and we get 3
for our example of integral 0 to 4 for floor(x)^2 dx our partitions are from sqrt k+1 to sqrt k
then we have sqrt (n+1) = 4 and we get 3 for k
but say now we have integral from 0 to 4 for floor(sqrt x) dx. Our partition is from (n+1)^2 - n^2 so we get 2n+1. now if we set it to 4 we get 2n+1 =4 n = 1.5
odd edgeBOT
@quasi shard Has your question been resolved?
odd edgeBOT
odd edgeBOT
Available help channel!
woeful rivet
odd edgeBOT
warped glacier
okay so note that f(x) will be above the y-axis for 0 < x < 1
(it's the product of 12 and two things that must both be positive)
then if you differentiate and set to 0, you'll find that there's one turning point at x = 0 and another at x = 2/3
min and max respectively
warped glacier
okay so note that f(x) will be above the y-axis for 0 < x < 1
you can deduce min and max from this information actually, but yes, you could always make a sign diagram or check the 2nd derivative
woeful rivet
oh ok thanks
and i have one more question
how do i do the second one?
this is where I’m up to
stable sequoia
erm just integrate it?
stable sequoia
divide and you set a bunch of x^n s and just integrate
stable sequoia
business as usual
stable sequoia
odd edgeBOT
odd edgeBOT
Available help channel!
safe kiln
the number of ways of selecting two numbers from the set {1,2,...12} whose sum is divisible by 3?
wooden python
!status
odd edgeBOT
What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
idk how to do it
wooden python
(remainder on division by 3, obviously)
safe kiln
0 -> 3,6,9,12
1 -> 1, 4,7,10
2 -> 2, 5,8,11
brittle plinth
you're missing 1 and 2
wooden python
now think about what remainders you can have in your pair such that the sum is divisible by 3
wooden python
either both 0 or one 1 and one 2
odd edgeBOT
odd edgeBOT
Available help channel!
crude hollow
I need help
odd edgeBOT
I need help
crude hollow
How do i approch locus problems dor coordinate geometry
indigo dirge
you have an example?
crude hollow
Not rn
indigo dirge
not the best at these but should be able to help with easier questions
crude hollow
Its just that i get startled whren i see such problems
And i dont understand how to procees
How should i start thinking when i see a locus problem
vernal yacht
just....practice.
what locus problem specificially?
Like throwing a ball and the locus forms a parabola?
crude hollow
No
Like a points st they have specific dist or ratio
Or properties
How do i start with such problems
vernal yacht
hmm
A car drive on a straight line (Linear function)
At which moment does the car has the nearest distance to the circle.
vernal yacht
Like a points st they have specific dist or ratio
oh
crude hollow
Yeah
crude hollow
Lemme type an example
vernal yacht
Sure, much simpler
crude hollow
Gimme a min
vernal yacht
take your time
crude hollow
Find locus of point such that there tgt to a given circle is at 45degrees to a tgt from same to point to a given parabola
How would i begin to think for something lile this
vernal yacht
what's tgt
crude hollow
Tangent
vernal yacht
oh, deriviate
vernal yacht
it depends on the circle
crude hollow
Oh i get what u mean
So i just have to try and check for every thing as fast as possible
vernal yacht
precisely speaking, it depends on the given information
crude hollow
Try to relate the problem with a known convept
vernal yacht
I'll rephrase your question and give u a quick guide
crude hollow
Kk thx
vernal yacht
Find locus of point such that there tgt to a given circle is at 45degrees to a t...
$ Given a circle x^2 + y^2 = 1, and a quadratic function. y = (x+10)^2; Please find a line that tangent to both functions $
bruh, it's not working
whatever, yk what I'm referring to
vernal yacht
Like, for this question, you first have to visualize the position of the functions
crude hollow
Trh to geuas what it would look like right
So i have to make an educated guess and take it fromthere
vernal yacht
no, you have to draw it based on the given information
crude hollow
Yeah
Still its easier said than done but ig ill try this from bow onwards
Thx man
crude hollow
Yeah
vernal yacht
because the question can be really abstract
crude hollow
K
vernal yacht
deriviate learnt?
crude hollow
Yeah
Although
I think for the problme i wrote finding a parametric equation and comparing slopes would be better right
Anyways igtg thx alot
.close
vernal yacht
I think for the problme i wrote finding a parametric equation and comparing slop...
YES
odd edgeBOT
odd edgeBOT
Available help channel!
rancid solar
AO = BO because its radius
and angle OAB = angle OBA because ( triangle AOB is an isosceles triangle )
OE common side
using SAS rule
triangle AEO is congruent to triangle BEO
=> angle AOE = angle BOE
how is that possible ?
am i making some mistake
vivid wyvern
It is possible if E is the midpoint of AB
rancid solar
i drew E such that it is not perpendicular
when E is perpendicual it besect the line into equal parts
vivid wyvern
When u prove that two triangles r congruent u prove that their sides and angles r equal to eo too
Which would mean AE=BE
Making E the midpoint and OE perpendicular on AB
rancid solar
but how is that possible
vivid wyvern
It would js mean that what u drew isnt exactly right
Since u proved OAE and OBE congruent ,u need to place OE kn such a way that AE is equal to BE ,the position is
Supposed to satisfy that condition
Or it could be something else
That is angle AOE=angle OEB
AE=OE and BE= OE
To justify ur drawing
But even then AE=BE
rancid solar
That is angle AOE=angle OEB
no thats not true because if we contruct a line parallel to AB let say PQ
angle POE = angle OEB which is bigger than angle AOE
vivid wyvern
Making E the midpoint and OE perpendicular on AB
Only this makes sense
rancid solar
maybe it has somehting to do with SAS rule maybe it does not work in some condition
what are those condition then ?
rancid solar
Only this makes sense
I intentionally drew that line such that it does bisect AB
vivid wyvern
maybe it has somehting to do with SAS rule maybe it does not work in some condit...
Lemme search up
vivid wyvern
Okay yes so i got to know that obtuse and acute triangles cant be congruent through sas,asa,sss
rancid solar
ohh
vivid wyvern
Cuz they hv different angle and side structures
vivid wyvern
can we proof
SAS rule
I bet it is possible,tho idk how
vivid wyvern
I intentionally drew that line such that it does bisect AB
I got u
rancid solar
maybe it is axiom
vivid wyvern
maybe it is axiom
Even if it is(idk if it is),Obtuse angle has an angle exceeding 90 but acute angles cant have that which means the angles cant be equal
So it can be derived from this sense
rancid solar
ooh i got it
vivid wyvern
Yea
rancid solar
this is SAS
we cant use SAS rule
here
like if two sides are eqaul and the angle between them is equal then only we can use that rule SAS
because now
the thrid line is dependent of the two equal lines and angle between them
i was using
SAS wrong
vivid wyvern
like if two sides are eqaul and the angle between them is equal then only we can...
The angle needs to be between the two lines?
rancid solar
yeah
vivid wyvern
Well that's how we learn
rancid solar
.close
odd edgeBOT
odd edgeBOT
Available help channel!
frozen musk
d/dx cotx