-f((x-5)/6) - 3
#help-28
leaden ermine
so when it says go right 5
then you do (x-(5))
(x-5)
you translate it 5 units to the right
cold wadi
to come to -3
cold wadi
one sec
lemme do it on the mssage
3(2/3(x+2)^2-7 -----> -3(2/3(x+2)^2+7 ----> -3(1/9(x+2))^2+7 ---->-3(1/9(x-3))^2+4
-3(1/9(x-3))^2+4
is my answer
key says instead of (x-3) its (x+7)
im not exactly sure
since right C units is -C in this form
x+2-c c=(5)
X-3
🤷♂️
leaden ermine
after you do -f(x) you do -f(x/6) so you would get
-3(2/3(x/6+2)^2+7 factor now intead 1/6
-3(1/9(x+12)^2+7
Now do (x-5)
-3(1/9((x-5)+12)^2+7
becomes
-3(1/9(x+7)^2+7
you didn't factorize correctly
2/3 is not only being multiplied with x/6 but also with 2 if you choose to expand it
leaden ermine
no problem
leaden ermine
just leave it
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cold wadi
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sly carbon
Find the number of pairs $\left( p, q \right)$ where $p, q \in {1, 2, 3, 4}$, and there exists a real number $x$ such that
$$x^{2} - p{x} + q = 0$$
\textit{Note. The notation ${x}$ represents the fractional part of $x$}.
glossy valveBOT
raymondclie
opaque slate
**raymondclie**
I guess you can do D≥0 for roots to be a real number
D is discriminant
sly carbon
I guess you can do D≥0 for roots to be a real number
What do you mean
opaque slate
What do you mean
Well a quadratic equation is given , x means root of equation.
For x to be real, discriminant D should be greater than or equal to 0
You know roots of quadratic equation formula?
Ohhh what's {x} ??? Is p{x} different than px?
sly carbon
Ohhh what's {x} ??? Is p{x} different than px?
Yes, of course, {x} denotes the fractional part of x
$0 \leqslant {x} < 1$
glossy valveBOT
raymondclie
opaque slate
Ohhhh mb then, ignore my suggestion. Wait for some other helpers
sly carbon
Alright
fervent owl
**raymondclie**
i think here cus we have such a limited number of possibilites we can effectively just exhaust
sly carbon
i think here cus we have such a limited number of possibilites we can effectivel...
How so
fervent owl
wait hang on
fervent owl
**raymondclie**
right so i think you can basically use the fact that {x} = x - \floor(x)
so if x satisfies the equation u've written
then x^2 - px + q = integer
so x is a root of x^2 - px + q+stuff
and then you can probably get bounds on x
so it doesn't work for a lot of values
sly carbon
hmmm
$$x^{2} = p{x} - q$$
Since $0 \leqslant {x} < 1$, we know $ 0 \leqslant p{x} < p$. So,
$$-q \leqslant p{x} - q < p - q$$
glossy valveBOT
raymondclie
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@sly carbon Has your question been resolved?
sly carbon
<@&286206848099549185>
fervent owl
roots of x^2 - x - k are 'approximately' sqrt(k) but we need to shift by k
so we'll never get any more
u can probably do something similar for p=2 and p=3
soft kiln
**raymondclie**
Observe the equation, what you have is $x^2$, we know that $x^2 >=0$ for all real values i.e
$ p{x}-q >=0$
or since p,q > 0 and the maximum value that p{x} can take is {x} = 1;
sly carbon
for now i know that
p = q = 1 doesnt work
(p, q) = (2, 2), (3, 3), (4, 4) works for x = 0
fervent owl
for now i know that
p = q = 1 doesnt work
(p, q) = (2, 2), (3, 3), (4, 4) wor...
in fact, if our original quadratic x^2 - px + q has a root between 0 and 1 it works too
you can plug in x=0 and x=1 to discover that
sly carbon
i cant even read this
glossy valveBOT
ZaNXar2315
soft kiln
man
fervent owl
(2,1), (3,2), (3,1), (4,4), (4,3), (4,2), (4,1) all work
fervent owl
for now i know that
p = q = 1 doesnt work
(p, q) = (2, 2), (3, 3), (4, 4) wor...
how did u get (2,2) and (3,3)
sly carbon
p = 2, q = 2
x^2 - p{x} + q = 0
x^2 = 2{x} + 2
x = 0 works
$x \in \mathbb{R}$ so it works
glossy valveBOT
raymondclie
soft kiln
basically the solution is find all p,q where p >=q
fervent owl
p = 2, q = 2
x^2 - p{x} + q = 0
x^2 = 2{x} + 2
x = 0 works
why?
sly carbon
Analogously for p = q = 3, and p = q = 4
fervent owl
0 = 0 + 2?
soft kiln
@sly carbon
sly carbon
Observe the equation, what you have is $x^2$, we know that $x^2 >=0$ for all rea...
fractional parts cant be 1
$0 \leqslant {x} < 1$
glossy valveBOT
raymondclie
glossy valveBOT
raymondclie
sly carbon
oh god i choke latex so bad
sly carbon
**ZaNXar2315**
You mean
Observe the equation, we know that $x^{2} \geqslant 0:\forall x \in \mathbb{R}$ i.e. $p{x} - q \geqslant 0$. Since $p, q > 0$
glossy valveBOT
raymondclie
soft kiln
for all p > q it the equation should hold true
fervent owl
(2,1), (3,2), (3,1), (4,4), (4,3), (4,2), (4,1) all work
and we already have that it works for all p > q
fervent owl
ignore this i made rearranging mistakes
sly carbon
I got $p \geqslant q$ from:
$$x^2 - p{x} + q = 0$$
$$\implies x^2 = p{x} - q$$
Bound
$$-q \leqslant p{x} - q < p - q$$
Since $x^{2} \geqslant 0 \implies p{x} - q \geqslant 0$
$$p{x} \geqslant q$$
$$p \geqslant q$$
glossy valveBOT
raymondclie
sly carbon
$p \neq q$, right?
glossy valveBOT
raymondclie
sly carbon
for all p > q it the equation should hold true
leqslant or leq for =<
geqslant or geq for >=
sly carbon
(2,1), (3,2), (3,1), (4,4), (4,3), (4,2), (4,1) all work
how does (4, 4) work?
fervent owl
yeah
fervent owl
**ZaNXar2315**
no bcus of this
sly carbon
what does maximum value that p{x} can take is {x} = 1 mean
fervent owl
they were just saying that p{x} < p
sly carbon
yea
x^2 - 3{x} + 3 = 0
x^2 = 3{x} - 3
oh yea
its obv
n{x} < n
my bad for not realizing
alright no solutions for all p = q
how so for p > q
fervent owl
plug in x=0 and x=1
sly carbon
**raymondclie**
and is this correct?
fervent owl
=> root between x=0 and x=1
sly carbon
u can change it to > tho
alright, noted.
ill just write it on my proofs then make it p > q, after showing n{x} < n
thank you
i checked with wolfram
(2, 1), (3, 1), (4, 1), (4, 2) all have real solutions
not (3, 2), (4, 3)
no output for (3, 2), complex solutions for (4, 3)
sly carbon
u can change it to > tho
does (3, 2) and (4, 3) not have real solutions, i checked with wolfram
fervent owl
oh (2,1), (3,2) and (4,3) don't work because 1 is a root
fervent owl
no output for (3, 2), complex solutions for (4, 3)
how tf did u get complex
sly carbon
how tf did u get complex
idk lol i used wolfram
for (2, 1)
x = 1?
x^2 - 2{x} - 1 = 0
x := 1
1^2 - 2(0) - 1 = 0
1 - 1 = 0
sly carbon
oh (2,1), (3,2) and (4,3) don't work because 1 is a root
(2, 1) works
x^2 - 2{x} + 1 = 0
x := 1
1^2 - 2(0) - 1 = 0
1 - 1 = 0
sly carbon
oh (2,1), (3,2) and (4,3) don't work because 1 is a root
wait is it the
sly carbon
this right?
fervent owl
this right?
yeah so u might have to use that to see if there is some weird construction that does work for (2,1), (3,2) and (4,3)
sly carbon
yeah so u might have to use that to see if there is some weird construction that...
yea, i checked with wolfram
no solutions for all 3
(3, 1), (4, 1), (4, 2) does have solutions
thank you so much
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static verge
can someone help? idk how they got the y=1
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can someone help? idk how they got the y=1
gritty rose
can someone help? idk how they got the y=1
Denominator is zero when?
static verge
Denominator is zero when?
x is 1 or x is -3
gritty rose
can someone help? idk how they got the y=1
Oh it's the limit as x goes to Infinity
static verge
yea
the asymptotes r
those
but i dont understand
the y=1
@gritty rose
wait nvm
it wa sjs a typo
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rugged laurel
having issues with this problem, I'm not sure what it wants me to put. I thought it was the difference quotient, but it doesn't accept that as a correct answer. What am I missing here?
plush egret
,w Expand[ (7-(x+h)^2) - (7-x^2 ) ]
glossy valveBOT
plush egret
@rugged laurel does that help
plush egret
this is the numerator of the difference quotient
glad whale
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umbral dome
he chose to add the terms on the right side to the left instead, it doesn't really matter which way you do it
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bleak gyro
is this correct
full forumBOT
is this correct
drowsy niche
is this correct
Well, first, notice you're multiplying a degree 3 a degree 2 and a degree 1 polynomial
bleak gyro
yea
drowsy niche
The degree of the result is the sum of those
bleak gyro
so adding not multiplying
drowsy niche
Yup
bleak gyro
so 6
drowsy niche
Yup
And for the y intercept
All you've gotta do is set x=0, so the y coordinate will be (2)^2 x (-4)^3 x (-3)
That'll give ya the y, but as you probably can see, it'll be positive and greater than 192
bleak gyro
(0,768)
drowsy niche
im kinda confused
With what?
bleak gyro
i think i understand was just confused for a sec
drowsy niche
what abt this
Alright, first of all the x intercepts are points for which the function equals 0
So what I'd do is actually equal it to 0, and solve for the possible values of x(it's easier than it seems)
That'll give you both the number of them and their coordinates
bleak gyro
there would be 3
drowsy niche
Yup
bleak gyro
(-2,0) is the other?
drowsy niche
(-2,0) is the other?
Yes
drowsy niche
it gave me this one
It's the same question, isn't it?
And I'm totally sure your answer's righr
bleak gyro
drowsy niche
Exactly 3 x intercepts, with the coordinates we said(that gray point near (3,0) isn't an x intercept, just is close to kt
bleak gyro
i have no idea what im doing wrong
drowsy niche
i have no idea what im doing wrong
If I'm being honest, I've no idea either
As you can see above, your answer is correct
Exactly three intercepts, and with exactly those coordinates
drowsy niche
This is your graph
Worst case send em the graph
bleak gyro
yeah im so confused
leaden ermine
,w plot (x+2)^2(x-3)^3(x-4)
bleak gyro
huh
glossy valveBOT
leaden ermine
am i tripping
bleak gyro
what the hell did I type in wrong
leaden ermine
guess the y scale is too high
leaden ermine
they all cross the axis so im not sure
if we go by specifically crossing the x-axis that's not true for x = -2
drowsy niche
Probably too zoomed in, cuz I also graphed mine on desmos
drowsy niche
if we go by specifically *crossing* the x-axis that's not true for x = -2
The question asks for x intercepts
They clearly want points for which the function equals 0
bleak gyro
leaden ermine
Number of x-intercepts at which f crosses the x-axis
leaden ermine
Obviously at x = -2 it's not crossing
bleak gyro
i just did (3,0)(4,0) and its still wrong
drowsy niche
Number of x-intercepts *at which f crosses the x-axis*
Alright, didn't notice that part of the question
Quite sneaky to add that to an otherwise typical exercise if ya ask me but whatever
bleak gyro
what did i do wrong in that question
drowsy niche
what did i do wrong in that question
The question's asking you for points in which the function touches the x axis
leaden ermine
I am noticing here it says flattens out
drowsy niche
And then changes sign
bleak gyro
so it would've just been (-2,0)
leaden ermine
So the multiplicity of the root is even.
bleak gyro
omf this is so annoying
leaden ermine
At least that's my interpretation.
drowsy niche
I swear, I hate poorly written questions
bleak gyro
drowsy niche
Ain't no way we've been here for ~10 minutes just because the guy didn't make clear what he was asking
leaden ermine
bleak gyro
thanks yall i wouldn't have gotten it
leaden ermine
But what's wrong with this part?
drowsy niche
Np(though for this last part I may've made it more confusing, really)
bleak gyro
there were 2 answers
leaden ermine
What is the second?
bleak gyro
i was too zoomned out
drowsy niche
But what's wrong with this part?
She wrote -3 there
drowsy niche
She wrote -3 there
Nvm, the function changed again, istg
leaden ermine
I guess it keeps changing, when you fail thrice
drowsy niche
does this look right tho
(x^2-3)=(x+sqrt(3))(x-sqrt(3))
drowsy niche
but i dont think you need to do that
Says “factor completely“
Which is quite ambiguous
leaden ermine
see this one of those things where i would be confused
but i agree with your argument technically
bleak gyro
i tried one similar but got it wrong so idk
leaden ermine
but i have a feeling it will be wrong
bleak gyro
yeah idk what else it would be
drowsy niche
We're basically playing mind games with her teacher, I swear
bleak gyro
i think im wrong tho
drowsy niche
Yeah, let's hope her original answer was right
bleak gyro
but idk what it would be
leaden ermine
else what prim suggested
bleak gyro
that?
leaden ermine
you are missing the other two factors
drowsy niche
this just simplifies to x²-3 <:kekw:752021756650455061>
I think imma go to sleep btw, you should be able to help her out just fine
drowsy niche
Good luck with it to both of you
leaden ermine
(x^2-3)=(x+sqrt(3))(x-sqrt(3))
true
but you are missing (x-5)(x-3)
[ \textcolor{cyan}{(x^2-3)} (x-5)(x-3) = \textcolor{cyan}{(x+\sqrt{3})(x-\sqrt{3})}(x-5)(x-3) ]
This was what Prim was pointing out to do, as the task suggests to factorize completely.
bleak gyro
oh
glossy valveBOT
bacc
leaden ermine
As obvious as it gets.
leaden ermine
I actually think you used a calculator anyway 
leaden ermine
degree should be obvious here
bleak gyro
stop sayinf that i keep getitng confsued wether i multiply or add
leaden ermine
huh
bleak gyro
obvious
leaden ermine
you just look at the biggest power
bleak gyro
what
leaden ermine
It was in the form of linear factors
(x-1)(x+2)² for example
here you would need to multiply everything out
bleak gyro
for the degree
bleak gyro
i swear it was the same typa problem
glossy valveBOT
bacc
leaden ermine
This is the rule he meant, but only because you need multiply the stuff out to figure the highest power
leaden ermine
Here in this example you can already tell the highest power, and that is x^4
Nothing can be expanded as opposed to the previous example
bleak gyro
so the degree is just 4
leaden ermine
yes
A polynomial of 4th degree.
Now for the y-intercept you literally do f(0)
and then your ordered pair looks (0, f(0))
bleak gyro
-45
bleak gyro
ok yay
would this only be 2
when i graph it two of the points are like 1,546464
theres 3,0 and 5,0 so would it only be those two
leaden ermine
**bacc**
All of them linear factors have an odd degree n = 1, so they all cross the x-axis.
bleak gyro
so what would i round it to
leaden ermine
when i graph it two of the points are like 1,546464
You can see it here too, that all 4 x-intercepts cross.
round?
bleak gyro
i wouldnt?
leaden ermine
It says to choose the amount of x-intercepts where f crosses the x-axis
Where does it say to round?
leaden ermine
It's 5 am dont play games with my mind lmao
leaden ermine
so id type that whole number in
You select the amount of x-intercepts that do that
bleak gyro
it dosent let me put that nuber in
bleak gyro
I dont understand you
bleak gyro
so theres only 2
leaden ermine
No
leaden ermine
when i graph it two of the points are like 1,546464
You can literally see what is going on
?
can't you select this?
leaden ermine
On the left, is there no square root symbol?
leaden ermine
then use it
bleak gyro
uh how
leaden ermine
i dont have your hands unfortunately
glossy valveBOT
bacc
leaden ermine
is that form not possible?
leaden ermine
😂 almost
glossy valveBOT
bacc
bleak gyro
yay
leaden ermine
1.5784 what lmao
bleak gyro
lmao im in the wrong math class idk how to do any of this
leaden ermine
damn girl
bleak gyro
what
leaden ermine
i mean damn girl
bleak gyro
the hell does that mean
leaden ermine
you are taking a test or some and you say you dont know anything
bleak gyro
leaden ermine
that's tough
bleak gyro
is this none
leaden ermine
ye
bleak gyro
they put me in the wrong math im supposed to be taking college algebra
leaden ermine
is that harder?
leaden ermine
oh ok
bleak gyro
youre supposed to take algebra before precalc
leaden ermine
well dont worry we will get you through
leaden ermine
3?
bleak gyro
what
glossy valveBOT
Result:
432
yea def. not
bleak gyro
idk what im doing
leaden ermine
-3 and 1 correct
bleak gyro
is that the only ones
leaden ermine
ye
bleak gyro
wrong
leaden ermine
it says counting multiplicity
bleak gyro
oh
leaden ermine
-3, 1, 1 should it be
bleak gyro
3i,-3i
bleak gyro
guess so
leaden ermine
I think it's only about real zeros...
bleak gyro
what.....
leaden ermine
try -3,1,1
bleak gyro
yes
leaden ermine
-1,-1,1,3i,-3i
leaden ermine
shut up lmao
leaden ermine
like you
bleak gyro
boy what
bleak gyro
wym
bleak gyro
oh wow so sweet thanks
crimson gust
Damn it's vegito the goat
bleak gyro
hi
leaden ermine
Damn it's vegito the goat
are you good at math
crimson gust
Maybe
leaden ermine
do me a favor
bleak gyro
r u leaving
crimson gust
I'm in calc III
bleak gyro
no
bleak gyro
excuse you
leaden ermine
or call you bitch when you say something cute
leaden ermine
skd i still cant undertsand anything
that's ok dear, good night
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wild sorrel
i put this in desmos but it was hard to find the values
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i put this in desmos but it was hard to find the values
wild sorrel
idk if this is right js wanna make sure
devout valley
If you're using Desmos, they can easily give you each of those points if you e.g. click after plotting it, and type them in
wild sorrel
If you're using Desmos, they can easily give you each of those points if you e.g...
i put in a function, can i js click around the relative extrema n it will tell me the coordinates of them?
devout valley
i put in a function, can i js click around the relative extrema n it will tell m...
Yep, for example
wild sorrel
ohhhh okay tysm
devout valley
I shall leave it to you to try the last one (because it's not what you have) 
wild sorrel
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whole fern
Suppose ( p ) is a limit point of ( A \cup B ). If ( I ) is an open interval containing ( p ), then by the definition of a limit point, ( I \cap (A \cup B) \setminus { p } ) is nonempty. Therefore, it contains a point ( x ). Then ( x ) is in ( I ) and ( x \in A \cup B ), and ( x \neq p ). Since ( x \in A \cup B ), either ( x ) is in ( A ) or ( x ) is in ( B ). In the first case, ( x \in A ). Then ( x ) is in ( I \cap A \setminus { p } ), so ( I \cap A \setminus { p } \neq \emptyset ). Therefore, ( p ) is a limit point of ( A ). Similarly, if ( x ) is in ( B ), ( p ) is a limit point of ( B ).
whole fern
there is something wrong with this proof I can't find it
bleak gyro
whoops
coarse elbow
You can't conclude that p is a limit point of A until you've checked both cases.
whole fern
could you elaborate please I still dont see it
coarse elbow
The structure of what you need to prove is $(\forall I(I \cap A \setminus {p}) \neq \varnothing) \lor (\forall I(I \cap B \setminus {p}) \neq \varnothing)$ while what you've proven is $\forall I(I \cap A \setminus {p}) \neq \varnothing \lor (I \cap B \setminus {p}) \neq \varnothing$.
whole fern
what is BA
glossy valveBOT
Morrow
coarse elbow
A typo :)
You've shown that for each I there's some x that works for either A or B, while what you need to show is either (for each I there's some x that works for A) or (for each I there's some x that works for B).
whole fern
Why cant the I be the same for both?
coarse elbow
The issue is that some Is are (potentially) only going to get an x that works for A while other I s will only get an x that works for B.
whole fern
but from the definition we know p is within I so why does it matter if x only works for A or B as long as p is in the same A or B
coarse elbow
The problematic step in your proof is "Therefore, ( p ) is a limit point of ( A )."
In order to conclude that, you must prove that for all I containing p, the set I ∩ A \ {p} is nonempty.
That means, you must take an arbitrary interval I containing p and show that there is some x in I ∩ A \ {p}.
You used the given to show that there is some x such that either x is in I ∩ A \ {p} or x is in I ∩ B \ {p}. At that point you're stuck, because what you need to prove is that I ∩ A \ {p} is nonempty.
You can try instead to prove that for all I containing p the set I ∩ B \ {p} is nonempty. You take an arbitrary interval I containing p and wish to show that there is some x in I ∩ B \ {p}. You get your x from the given and run into a similar problem.
You cannot take an arbitrary interval I and show that either I ∩ A \ {p} is nonempty or I ∩ B \ {p} is nonempty. That's not an equivalent statement.
$\forall z(P(z) \lor Q(z))$ is not equivalent to $(\forall z, P(z)) \lor (\forall z, Q(z))$, in general.
glossy valveBOT
Morrow
whole fern
thank you so much I have been stuck in this for longer than I would care to admit
coarse elbow
Glad I was able to help
whole fern
analysis is a b*tch lol
coarse elbow
one that keeps you coming back for more, hehe
whole fern
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worn widget
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worn widget
I'm trying to prove
For any random point P on Circle O
angle PAO will ALWAYS be less than or equal to 90 degrees
When A is a point outside circle O
Does anyone have an idea on how to approach this? Anything would help!!!!
limpid moat
worn widget
Right beacuse when a triangle is an Isoscles triangle, The two same angles can't be over 90 degrees, as if it were, it can't be a triangle!!!
limpid moat
right!
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worn widget
right!
Hi sorry! one last question
How would I use the isoscles method if point P was touching the circle?
As line segment AP would be touching the circle
limpid moat
do you wonder about the case where AP is a tangent line ?
limpid moat
if an angle of a triangle is 90 degrees then the other two must be stricly less than 90
worn widget
o h.
OOOPS.
why didn't I think of that!!! silly me
Thanks again, it helped a lot!!
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stuck patrol
Need help understand how to approach:
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bold mica
stuck patrol
?
pure dock
stuck patrol
So it doesn't converge then, that makes sense to me as well. Wolfram Aloha thinks it doesn't though, which is odd.
So what would be enough reason to prove this.
That lim x to 0 (from positive side) goes to inf?
glacial pasture
So it doesn't converge then, that makes sense to me as well. Wolfram Aloha think...
?
wolfram said it diverges
which means it doesnt converge
why is that odd?
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stuck patrol
Okay i see to miss understand wolframs answer
Instill need to give a reason why it diverges
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rare pine
find the volume of a cone with radius r and height h, using integration
rare pine
how do i start?
urban patio
draw a graph first
hasty compass
you know element taking?
echo marsh
take disc elements
hasty compass
urban patio
what is element taking?
rare pine
you know element taking?
? sorry i might know this might i dont really speak english
hasty compass
what is element taking?
i mean taking elements
rare pine
maybe i use similar triangles?
echo marsh
? sorry i might know this might i dont really speak english
like where u take a disc of thickness dx at a distance x
rare pine
like where u take a disc of thickness dx at a distance x
oh idk this then
i know that i should let the tip be 0,0 and find S(x)
oh wait
i got it
nvm
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thick hedge
I'm lost
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trim gazelle
Why do you have to multiply by the reciprocal when dividing fractions?
quick thicket
imagine dividing a value by 2, this equal the value being multiplied with 1/2, wich is the reciprocal of 2.
lets take 50
50 / 2 = 25 and 50 * (1/2) =25
thats how fractions work
exotic condor
yeah you need to balence the equation
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trim gazelle
50 / 2 = 25 and 50 * (1/2) =25
Oh i see so like if you divided by 4
then 1/4 = /4
but u would multiply with 1//4
and divide with 4
so 50/4 = 50 * 1/4
quick thicket
yes
trim gazelle
that makes so much sense now
fossil wedge
Can someone explain please? I don't understand why the first sum converges and the second sum diverges... arent they both telescopic series?
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candid sail
Can someone explain please? I don't understand why the first sum converges and t...
In the first series if the sum is taken till n terms, you get 1/2 - 1/(n+2)
To which if you apply limit n—>infinity, you get 1/2 as 1/n+2 becomes 0
But in second one, sum upto n terms is log(n+1) - log1, to which if you apply limit n—>infinity, you get infinity as log(n+1) goes to infinity
fossil wedge
In the first series if the sum is taken till n terms, you get 1/2 - 1/(n+2)
To w...
Okay makes very sense now
Thanks very much 🙂
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hoary phoenix
to answer this question you have to rewrite the equation 2x + y = 6 into y = mx + c
which gives you y = -2x + 6 for the line thats parallel to the line you actually want
then you put that with the values 5 = -2(2) + c to find the equation of the line no?
hoary phoenix
what 2x2 + 5 = c?
hasty compass
alternatively u can just write 2x+y=c
yea now put c here
hoary phoenix
the thing that im wondering is I must've done something wrong because
oh no nvm
im tripping
agile ledge
to answer this question you have to rewrite the equation 2x + y = 6 into y = mx ...
use the given line to obtain slope which will be the same for the line we require
and now use point slope form
hoary phoenix
yeah c = 9
hasty compass
yeah c = 9
yea put that into 2x+y=c
agile ledge
(y-y1)=m(x-x1)
hoary phoenix
yea put that into 2x+y=c
well its asking for the equation of the line
hasty compass
you should be fine both the ways ig
supple jay
u * dx = u , but u' * dx = du ? I guess because u' = du/dx so dx cancel but why u * dx = u . does someone understand?
supple jay
wouldnt u * dx = u * dx
its from writing formula for partial integration starting from product rule
gritty rose
The notation in the red box is trash
gritty rose
u * dx = u , but u' * dx = du ? I guess because u' = du/dx so dx cancel but why ...
Don't do any of this
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paper bane
Anyone able to help with Instantaneous Acceleration?
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Anyone able to help with Instantaneous Acceleration?
glacial pasture
around t=35 v is practically a straight line, getting the slope of said line should be sufficient for your acceleration
in general you can just draw a tangent line and find its slope though if you dont have the function
paper bane
How does that work/
languid junco
Find the slope of that line at t=35, rise over run
paper bane
paper bane
Find the slope of that line at t=35, rise over run
Its a single point on it though isnt it?
apparently -8.5 is the instant velocity
torn jolt
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leaden ermine
impossible that the last one is wrong
You said that the last time too
torn jolt
$\int 4e^x\sin(y)+4y dx = 4e^x\sin(y) + 4xy + h(y)$
leaden ermine
wrt which variable?
glossy valveBOT
Derivative
torn jolt
wrt which variable?
x
leaden ermine
yea i see now 😄
torn jolt
for M(x,y)
leaden ermine
It also says 1 of the answers, so it could be any of them.
torn jolt
$F(x,y) = 4e^x\sin(y) + 4xy + h(y)$
glossy valveBOT
Derivative
torn jolt
It also says 1 of the answers, so it could be any of them.
yeah but chances are its the last one
leaden ermine
impossible that the last one is wrong
You just said that lol
torn jolt
yes
but maybe it is wrong 🤣
and im just stupid
so now i need to find Fy(x,y) (partial derivative with of y)
torn jolt
$F_y(x,y) = 4e^x\cos(y) + 4x + h'(y) = 4x+4e^x\cos(y)$
glossy valveBOT
Derivative
leaden ermine
So h'(y) = 0?
leaden ermine
forgot the constant?
torn jolt
no I said h'(y) was 4 - 4y
i made a mistake on the differentiation
it wont happen again 🫡
thanks again you are a life saver as usual
🫡
leaden ermine
Of course (I literally did nothing)
torn jolt
moral support is adequate
leaden ermine
You too.
torn jolt
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oak tulip
I wanted to confirm that I am doing some math right.
I want to calculate the probability that a person shares a birthdate (year + day) at least someone else in a group of 38,313 people. Assuming 80 year period with equal distribution of years and birthdays:
Probability that the person shares with another individual at random: 1/(80*365) = 1/29,200.
Probability that the person does not share with another individual: 29,199/29,200
Probability that the person does not share with any other individual: (29,199/29,200)^38,313
Probability that the person shares with at least someone else: 1 - (29199/29200)^38313 ~= 0.7307?
rare dock
you want the probability for one specific person? not just the probability that some 2 people share a birthday?
oak tulip
Yes. In other words that is the probability that bob has the same birthdate (not birthday) with at least one other person
rare dock
ignoring leap years that looks ok
but if bob is part of the 38313 people then it should be 1 - (29199/29200)^38312
oak tulip
oh yeh true you are right. Should subtract one to not include bob himself
Also as well it is correct to say that around 73% of people in the group share a birthdate with at least one other person?
rare dock
if you want to incorporate leap years (which seems reasonable (or maybe not idk)) you can just add 20 to the total days
wild sleet
Also as well it is correct to say that around 73% of people in the group share a...
probably not true
oak tulip
probably not true
Any specific reasoning why not? Seems like if you have a 73% chance then overall there should be 73% of people in which this holds true?
rare dock
it sounds reasonable to me but i need to think about it more
oak tulip
Well I am in no rush
rare dock
ok i think i will have an answer soon
yea
i agree with 73% (approx) of people will have a shared birthday
oak tulip
Okay. Ill probably think about it more myself and ask clarification from others as well but best i can tell it is correct
rare dock
i think it's just linearity of expectation. i also just simulated and am getting 73%
oak tulip
oh right i should write a python program to simulate it that would be rather easy
rare dock
from collections import Counter
ppl_amt = 38313
days_amt = 80*365
days = [randint(0,days_amt-1) for ppl in range(ppl_amt)]
A = Counter(days)
def check_shared(i):
day = days[i]
if day not in A or A[day] == 1:
return 0
return 1
N(sum(check_shared(i) for i in range(ppl_amt))/ppl_amt)```it's in sage but close enough to python
just remove the N
rare dock
oh right i should write a python program to simulate it that would be rather eas...
i always do that when i can to make sure my math is right
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void dagger
ΔABC is located at A (2, 0), B (4, 4), and C (6, 3). Zackery says that ΔABC is an isosceles triangle, while Verna says that it is a right triangle. Who is correct?
void dagger
geometry
devout valley
Do you know how to check whether a general triangle would be isosceles/right-angled?
void dagger
Do you know how to check whether a general triangle would be isosceles/right-ang...
nope i forgot
devout valley
Do you know what an isosceles triangle is?
void dagger
Do you know what an isosceles triangle is?
yes
i put it in a graph and it is a right triangle. Verna, because segment BC is perpendicular to segment AC
Verna, because segment AB is perpendicular to segment BC
Im just confused on which one it is to prove its a right triangle
devout valley
Well, which pair is the perpendicular one?
(and are those the options given to you? It is possible that both of them can be right - see 45-45-90 triangles, which are both isosceles and right angled)
void dagger
void dagger
(and are those the options given to you? It is possible that both of them can be...
This is what i got
glossy valveBOT
Result:
-12
(you might want to show your work though...)
void dagger
(you might want to show your work though...)
No what im saying is i dont know if thats correct or not. I just know its a right triangle and only one is correct
devout valley
Well, show the plot you have, and which one you think is a right angle, then...
void dagger
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compact sonnet
Im having trouble proving the following formula (the one at the bottom of the proof-tree). I think this is akin to proving de morgan's law for the negation of a conjunction (but with added negations). Im really not sure where to go from here, or if thats even a good start. Tagging @narrow cradle because you helped me earlier with a similar topic. Thanks!
compact sonnet
maybe something like this? but it seems like a dead-end...
natural deduction with gentzen tree-proofs, its entry-level stuff but Im having a hard time wrapping my head around it
yeah youll be fine dw
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narrow cradle
Im having trouble proving the following formula (the one at the bottom of the pr...
You'll need some form of classical logic here since that implication doesn't hold constructively.
What rules does your system have other than the usual introduction and elimination rules?
compact sonnet
AFAIK, these are all the rules
compact sonnet
You'll need some form of classical logic here since that implication doesn't hol...
so, those are the usual introduction and elimination rules I imagine
narrow cradle
AFAIK, these are all the rules
Ok, calling double negation elimination (the rule in the bottom left) just negation elimination is a little evil
But that's the key here
Leaving out that rule you end up with the intuitionistic logic where what you're trying to prove doesn't hold
compact sonnet
ok, so the double negation rule is the classical logic part I see
narrow cradle
So a hint would be to try and prove the double negated form of your goal statement
compact sonnet
hmm ok that will help. I will give it a shot and get back to you if Im still having issues
compact sonnet
So a hint would be to try and prove the double negated form of your goal stateme...
I must thank you, your help has been greatly appreciated
compact sonnet
So a hint would be to try and prove the double negated form of your goal stateme...
just to be sure, you mean starting the proof like so?
narrow cradle
Ah, no you still start out like you usually would with implication introduction.
I.e. assume ¬(a ∧ ¬b), prove ¬a ∨ b.
Double negation elimination will help you prove ¬a ∨ b, since you may prove its double negated form instead
compact sonnet
ahhh ok gotcha, thanks
humble dock
is someone here online?
compact sonnet
@narrow cradle
phew what a brain-teaser lol
but I think that's right
thanks for setting me on the right track!
narrow cradle
<@220941391054897153>
Yep, that looks right! 
compact sonnet
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marsh star
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marsh star
Is this correct? Can someone help explain please?
devout valley
Which part?
marsh star
The solution
devout valley
Where is the first part of the solution you don’t understand, though? Or do you not get anything at all they’ve said?
Are you aware of how you’re meant to construct the matrix of a linear transformation with respect to given bases on each side?
marsh star
I’m not rly getting anything over here
Couldn’t understand lecturer, the professor’s English is horrible
rugged raft
Is this correct? Can someone help explain please?
They're correct and they already explained it well
marsh star
I’m not sure how a R2 matrix transfers into R4 matrix
rugged raft
You can also just ask them lol
Shortly, You can just think of n as the numbers of the row
marsh star
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muted gulch
Can anyone pls help me it’s math hw
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Can anyone pls help me it’s math hw
muted gulch
Anyone can help me?
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Can anyone pls help me it’s math hw
No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/
unborn glen
Just send your question and stop opening up channels to not ask it
muted gulch
-5 7/8 + (-3 3/14) + 5 7/8
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strange marlin
does e^x^2 = e^2x
(I'm rlly asking if the only difference is when it's LITERALLY e^(x^2)
balmy rose
e^(x^2)=e^2x implies x^2=2x which is not always true
stray helm
i think you have an ambiguity in notation; e^(x^2) is not necessarily equal to e^(2x), but (e^x)^2 = e^(2x).
balmy rose
(e^x)^2=e^(2x) however is true
strange marlin
i think you have an ambiguity in notation; e^(x^2) is not necessarily equal to e...
yeah (ambiguity/I have..) when it's ONLY e^x^2
so ye
I can re-write it as (e^x)^2
stray helm
yeah you can write e^(2x) as (e^x)^2
strange marlin
ye
so it will alwyas be equal
in that context
if not "specified"
or not written as e^(x^2)
stray helm
yeah they're the same expression (do you know why?)
strange marlin
yeah they're the same expression (do you know why?)
e^x^2=(e^x)^2=e^2x
e^(x^2) is just not gonna change
cause parenthesis encapsulate the square on x I suppose
I just know it can't-
strange marlin
yeah they're the same expression (do you know why?)
y
hey
hey
y
y cant I move ^2 out of the parenthesis
strange marlin
you're asking why e^(x^2) is not always equal to (e^x)^2?
yes
stray helm
yeah
stray helm
e^(x^2)=e^2x implies x^2=2x which is not always true
so it's what martin was saying earlier here
we already established that (e^x)^2 = e^(2x)
so if e^(x^2) = (e^x)^2 = e^(2x), then we would have that x^2 = 2x
strange marlin
we already established that (e^x)^2 = e^(2x)
ofc
stray helm
so if e^(x^2) = (e^x)^2 = e^(2x), then we would have that x^2 = 2x
do you see a problem with that result?
strange marlin
hm
yes
yes I do
wait
so I was solving for integrating factor
and p(x)=x^2-x
so it was e^ int(x^2-x)
and I was doing it
and I was like
e^x^2e^-x
so then
(e^x)^2e^-x
e^2xe^-x
e^(2x-x)
=e^x
can I assume that there are no parenthesis around the integral
because if I do, I get the right answer
(if I don't I need to do integration by parts I suppose)
muted gulch
Can some make help channel for me
muted gulch
Can u help me?
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low rose
Hello, I've been trying to solve this for a little while now (it's an extra credit assignment problem) and honestly I just keep running into walls. Could someone help me figure this out or show me how they might go about solving this?
Nevermind- I was overcomplicating it for myself and friend figured it out.
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paper bane
Anybody have a good resource for learning integration?
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Anybody have a good resource for learning integration?
dark moat
At what level?
paper bane
Enough for calc based physics
Typically you do Calc I before physics but Im doing Calc I while doing it so theres some stuff im supposed to know already
dark moat
https://tutorial.math.lamar.edu these notes are pretty good if u work thru it
The calculus section Ofc
paper bane
Theyre my teachers notes. The issue is it doesnt really explain whats going on.
dark moat
Do u want to understand the theory behind integration, or more what the integral represents?
paper bane
the latter
It seems like it represents delta in some way
based on that
have no clue what happens with time though
dark moat
Have u tried 3blue1browns video series on calculus?
paper bane
dark moat
It’s good for intuition and understanding
Probably best existing online explanation to get intuition
paper bane
Should I learn derivatives first?
dark moat
Yes
Integrals revolve around the idea of finding a function whose derivative is the function ur integrating
Derivative laws are easy and algorithmic
Integration is abt using those laws to undo them in some sense
Which is harder
paper bane
Oh, wait I do know some derivative stuff. Only recently learned limits but I do know it now at least lmao.
dark moat
Anyway, watching thru these videos will help with simply understanding what is ur doing by taking an integral
Tho questions are most important to be good at actually integrating, although slightly less emphasised if ur a physicist
torn jolt
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hot knoll
Line 3
When you subtract 2 log x from both sides
You already had -4logx there
You did minus 2 more
hot knoll
right answer: root of 3
If you fix that you should get the correct answer