#help-19
true garnet
well yes, for 2x2
sick sigil
so if we had
true garnet
for 3x3 and above its more complicated
sick sigil
wait
does it have to do with the
well cant write it out but
maybe you understand what I mean
sick sigil
oh yes
does it have to do with that?
is that why b and c gets minus? 🤔
its fine i'll look at our material later
but for the tedious approach
sick sigil
true garnet
im looking for the wiki page explaining why xD
true garnet
and then we multiply from right to left right?
wdym by this?
the equation is right yeh
sick sigil
i mean we do X*B
and not A*X
is it right so far then?
then I would end up with this
true garnet
i mean we do X*B
matrix multiplication is associative so either order is fine
sick sigil
and I think this is the part that well.. I dont really know how to proceed from
true garnet
i was not expecting this method
sick sigil
well
it's not optimal i know
but I feel like it should work
but how do I set up the equation system from here?
true garnet
up to here was good
true garnet
but I feel like it *should* work
since X is a 2x2 matrix, shouldnt there be 4 different values? why only 2?
sick sigil
yes that is the part im a bit confused around
'
but this is from our teachers video
where he says X = x1,x2,x3
true garnet
LOL
sick sigil
but vectors can take on the form of rows in matrixes no?
or let me clarify it doesn't say
true garnet
yes pls
sick sigil
I just assume it's a vector because that's what I figured from the screenshot above
true garnet
wait i just realised
sick sigil
or you mean the X in the question AXB=BA or the X from the screenshot
sick sigil
yeah it just says solve the equation AXB=BA
but I think vectors are equivalent to matrixes according to our learnings
because you just put in the vector as a row in a matrix
true garnet
vectors are 1 dimensional matrices
sick sigil
yeah that sounds about right
true garnet
alr. so the method (which you said was in the solution) is to solve it using: X = (A^)BA(B^)
you compute A^ and B^ and use those to calculate X
sick sigil
yes that is indeed the smart way but i ain't the smartest tool in the shed so sometimes I like to walk the long path to see if it works anyway
true garnet
^ - inverse
true garnet
valid
sick sigil
but shouldn't this approach work too?
true garnet
it's good tho. at least you're learning WHY a method is better
sick sigil
we know that there are an X1 and X2 that gives the output matrix
yeah that's kinda my thinking as well
true garnet
but shouldn't this approach work too?
X is not a vector. its a 2x2 matrix
so there are 4 values u need to calculate
sick sigil
oh
but
how do u know that
because like from the screenshot above
or does it have to do with the fact that the RHS is a 2x2 matrix?
true garnet
or does it have to do with the fact that the RHS is a 2x2 matrix?
yes
all the other matrices are 2x2
multiplying square matrices together always gives square matrices
sick sigil
yes but we can also multiply a matrix that's 2x1 with a 2x2 and get a 2x2 output no?
ye exactly
true garnet
yes but we can also multiply a matrix that's 2x1 with a 2x2 and get a 2x2 output...
output is not 2x2
sick sigil
rows from the first and columns from second
so if it's 2x1 (not 1x2 as i originally wrote) and the other 2x2 then output should be 2x2 no?
but then you're saying that we should have like
x1 x2
x3 x4
as the matrix for X
true garnet
yes but we can also multiply a matrix that's 2x1 with a 2x2 and get a 2x2 output...
2x1 has 1 column. 2x2 has 2 rows. incompatible multiplication
true garnet
x1 x2
x3 x4
yes that's the idea. but thats long
true garnet
it would work tho. if the equations are simple then it might be fast
sick sigil
just wondering how to set it up
ill show u screenshot in a sec
ok
got it
so now
what would the equation system look like?
is it
x1+2x2+2x3+4x4 = 1
etc
column 1 row 1 is equal to the corresponding content of the other?
because this is what we've gone through before
but here you get two 3x1 matrixes
true garnet
yes each element is equal to the corresponding element of the other matrix
true garnet
column 1 row 1 is equal to the corresponding content of the other?
ye this is right
sick sigil
so then I instantly get x4 = 3
true garnet
yep
sick sigil
x1 = 7
so would the answer then be that I write the matrix X with the values of x1 x2 x3 x4 in their respective position I guess?
true garnet
so would the answer then be that I write the matrix X with the values of x1 x2 x...
yep
true garnet
exactly
sick sigil
yep that checks out
true garnet
we got there in the end
true garnet
HAHa ty
sick sigil
god bless you 🥹 ❤️
true garnet
i do not recommend this method tho ^
sick sigil
no i will remember the other one but
true garnet
please use inverse matrices. so much faster
sick sigil
in the case of an emergency where the brain decides to turn off
it's good to know that this one exist
true garnet
fair enough
true garnet
have you learnt it yet?
true garnet
ah it's not that simple unfortunately
true garnet
but for the 2x2 case it is then
yes
true garnet
ah it's not that simple unfortunately
💀
true garnet
yeh it's just weird to memorise
sick sigil
no but it is the +- thing
true garnet
4x4 is painful
true garnet
yeh have fun with that. i definitely did NOT
sick sigil
although for the determinants i usually just continue on the diagonals
for the 2x2 and 3x3 case that is
but I think it's a preference thing
true garnet
yeh whatever works for you man. there are so many methods for inverting 3x3 matrices anyway
sick sigil
nonetheless, time to blast on with the questions, thank you a ton and maybe you will see me crying in another channel later on confused XD
have a nice evening/day!
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odd edgeBOT
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glad plume
odd edgeBOT
glad plume
so i have to calculate this,english is not my main language sorry,in the book the answer is 41 but i got 45
the [] is the int part of the number
and {} is the fractionary part of the number
and i got that 44*44=1936
and 45*45=2025
so rad of 2022 is 44.something
int part of it is 44
the other one is -4*-0.25 which is 1
did i do something wrong?
warm nacelle
wait
fraction part of -ve numbers
are calculated differently
fraction part of 0.4 if 0.4
but fraction part of -0.4 will with 1-(fraction part of 0.4) that is 0.6
glad plume
so will it be 0.75?
warm nacelle
yes
odd edgeBOT
warm nacelle
no probl
odd edgeBOT
@somber breach Has your question been resolved?
marble pond
odd edgeBOT
marble pond
Anyone have ideas on where to start in order to solve for R here?
The two smaller circles are equal size
odd edgeBOT
@marble pond Has your question been resolved?
fluid tundra
The two smaller circles are equal size
does the left circle pass through the origin?
is the right circle centered on the x-axis?
strong fable
A good start is to connect the centers of tangent circles
And look for the right triangles
You can then build a system of equations and solve for a positive R
nocturne belfry
<@&268886789983436800>
marble pond
I’m back
marble pond
does the left circle pass through the origin?
is the right circle centered on th...
No, the left circle doesnt pass through the origin, the bottom of the left circle is sat atop the x axis
marble pond
does the left circle pass through the origin?
is the right circle centered on th...
The right circle is centered on the x axis however
fluid tundra
see kreshank msgs then
though i will point out all that info would have been nice to have at the start
marble pond
see kreshank msgs then
Where would I find this
I’m not familiarized with the terminology you are using
fluid tundra
A good start is to connect the centers of tangent circles
the messages from the guy named kreshank
strong fable
The three small circles in the middle are also tangent
strong fable
I think you got it from here, it'll get rid of t1 and t2
strong fable
Yep
marble pond
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odd edgeBOT
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slender solstice
Set of edges and vertices of a digraph how
odd edgeBOT
Set of edges and vertices of a digraph how
slender solstice
odd edgeBOT
@slender solstice Has your question been resolved?
torpid owl
isn't a digraph just a directed graph? it is clearly directed
slender solstice
isn't a digraph just a directed graph? it is clearly directed
wdym
is a digraph
torpid owl
what is your question?
does it say "explain why the following graph is considered a digraph"?
slender solstice
I want find set edges , vertex and relation
slender solstice
does it say "explain why the following graph is considered a digraph"?
Yes and why is a digraph
torpid owl
it is a digraph because it is a set of vertices with directed edges, I'd say
slender solstice
and how i can find set\
My answered is;
El conjunto de vertice es V = {c , d , e , a , f}
El conjunto de arista E = {(c,f),(d,e),(c,d),(a,f)}
El contunjunto de relacion es ; R = {(c,f),(d,e),(e,d),(c,d),(a,f)}
slender solstice
it is a digraph because it is a set of vertices with directed edges, I'd say
Im alright?
torpid owl
My answered is;
El conjunto de vertice es V = {c , d , e , a , f}
El conjunto d...
(e, d) is also an edge
slender solstice
(e, d) is also an edge
Relation set and edge set are same?
slender solstice
thx
odd edgeBOT
@slender solstice Has your question been resolved?
supple mauve
Let A and B be sets defined in a Universe U. Using the properties of sets, show that
(I tried to apply the properties of the sets but this bar stating that it is the complementary set confused me a lot)
signal quest
Let A and B be sets defined in a Universe U. Using the properties of sets, show ...
do they allow truth tables or is it strictly rewrite?
supple mauve
do they allow truth tables or is it strictly rewrite?
i believe it is strictly rewrite as it is about set theory according to the book
signal quest
if im seeing it correctly, im guessing what they want you to do is show that both sides are subsets of each other
(which proves that they are in fact equal)
supple mauve
i believe it is strictly rewrite as it is about set theory according to the book
but I can try to do it using the truth table
supple mauve
if im seeing it correctly, im guessing what they want you to do is show that bot...
yep
signal quest
I would work with the rewrite if thats what they want
long bridge
I'd say, this seems to want a rewrite as well
signal quest
I would recommend breaking up the big expression into parts that can be more easily managed
leaving the complement for the end
long bridge
Also, I've never seen a bar for complement before??
signal quest
yeah im used to the bar, but sometimes its a '
some also have it like this lol
which is more cursed imo
long bridge
I've also seen $$A^C$$
clever fjordBOT
Sehtnar
long bridge
I've always used the bar for the closure of a set
supple mauve
I'd say, this seems to want a rewrite as well
so I would start by making A intersect B?
sorry if my english is bad english is my secondary language
long bridge
Not sure what you mean by that
signal quest
Okay so basically
it looks like the first two terms are disjoint
which leaves you with the 3rd term with a complement, where you can then use demorgans law to get the same thing
long bridge
You got it?
supple mauve
btw do you have some pdf or youtube video recomendation about sets?
odd edgeBOT
signal quest
btw do you have some pdf or youtube video recomendation about sets?
trevtutor on youtube is pretty good
odd edgeBOT
Available help channel!
dim quiver
So I've been thinking about using the residue theorem to try to calculate the gamma function but it's been a bit tricky
As for my steps
since the gamma function is $\int_0^\infty t^{z-1} e^{-t} dt$ it's a bit interesting
I've been thinking about trying to convert it into an integral of the form $\int_{-\infty}^\infty$but I'm not sure if there's a good way to do that
I don't mind if there really isn't a good answer but it's just fun to explore
clever fjordBOT
Corobo
dim quiver
There's also probably a much easier way to construct a contour in the first quadrant of the plane instead but I'm a bit tired so I don't remember if there are any that would be applicable to something like this at the moment
dim quiver
**Corobo**
Upon rereading my message there wouldn't be any even functions because the domain is over t where e^-t isn't even or odd
But it also makes me wonder if you can even convert it into the other integral form to begin with because it would surely imply that there would be residues for positive integers as an example, when there clearly aren't any
odd edgeBOT
@dim quiver Has your question been resolved?
odd edgeBOT
@dim quiver Has your question been resolved?
desert marlin
**Corobo**
what are the poles of t^(z-1)e^(-t)
odd edgeBOT
@dim quiver Has your question been resolved?
dim quiver
what are the poles of t^(z-1)e^(-t)
Well when Re(z) < 1 there are poles at t = 0 I believe
desert marlin
of what order
dim quiver
It would depend on the value of z
desert marlin
It would depend on the value of z
how do you plan on using the residue theorem then?
dim quiver
I'm pretty sure they would all be simple poles
desert marlin
you can try integrating like t^(z-1)e^(-t)
over a half-semi-circular contour of radius R, and hope that when you limit R to infinity the circle arc contour integral vanishes, which would leave you with that the integral from 0 to inf of t^(z-1)e^(-t) = the residues included in Q1
but you'd have to know what the residues in Q1 are
and how to compute them
and it would actually require the arc contour integral to vanish
which I haven't tested
dim quiver
I mean for simple poles and a function like this the residue at z_0 would just be t^-z_0 e^-t/(1-z_0) iirc
Then again though
Actually
The arc contour integral might not be able to vanish just by contradiction
Since if it did vanish, then the contour integral encasing some open connected domain D in Q1 with no singularities in D would be non-zero
which contradicts residue theorem
since theoretically all poles should be at 0 which would not be in D
huh
Thanks 
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odd edgeBOT
desert marlin
The arc contour integral might not be able to vanish just by contradiction
I never said by contradiction?
dim quiver
I never said by contradiction?
i misspoke
I meant i thought maybe it could be able to be shown to not vanish by contradiction
desert marlin
We do want it to vanish though
dim quiver
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odd edgeBOT
✅
dim quiver
yea
desert marlin
hoping the greens vanish
dim quiver
yea thats what im thinking
desert marlin
hoping the greens vanish
as R->inf
if so , then your integral is the sum of the residues contained in Q1
dim quiver
But if any poles will lie at 0 then since the inside of R doesn't contain any poles they must not vanish
desert marlin
dim quiver
yea
desert marlin
the poles are at the origin
it's holomorphic in the interior
oh I guess the poles are not in the interior
is that your point
dim quiver
yea thats what i mean
dim quiver
hmm
This seems impossible
The best estimate I can reach is that the residue of the function at 0 appears to be (-1)^z * Gamma(1-z)
But since that's not the same as the actual value it's safe to assume that the curved contour doesn't vanish
So I'm gonna assume that it's impossible
Thanks for the help ^-^
.close
odd edgeBOT
odd edgeBOT
@lofty wyvern Has your question been resolved?
obsidian frost
Hi, how can i find the derivative of a function such as this? its an f(x,y) function and i have some variables that i could control to change the look of the wave (w for frequency and R for rotation)
obsidian frost
im gonna be using the derivative to find the tangent, binormal, and normal of a point on the surface
merry finch
we want the derivative
this function looks pretty smooth
i reckon it's differentiable
so for surfaces, the derivative is not a line, but a plane
instead of a tangent line we get a tangent plane
obsidian frost
right
merry finch
do you know any linear algebra?
obsidian frost
yes
merry finch
the tangent plane is just the jacobian of f at the point you want your tangent plane
well
it's the linear approximation to f
so $g(\mathbf x) = f(\mathbf a) + J_f(\mathbf a)(\mathbf x - \mathbf a)$
clever fjordBOT
Frosst
obsidian frost
thisll be an approximation?
merry finch
this is just multidimensional taylor's approximation to your function
but truncated at the linear term
the next term would have the hessian of f
but we want a linear approximation (tangent plane) not a quadratic approximation (tangent paraboloid)
so we just stop at the jacobian and boom there you go, that's your linear approximate to the surface at the point a
and that approximation is precisely the tangent plane to the surface
just like how g(x) = f(a) + f'(a)(x-a) is exactly the tangent line of f at point a
obsidian frost
what does jacobian mean? sorry im not so familiar with math terms in english
merry finch
**Frosst**
this is the n-dimension equivalent
the jacobian has all the partials
he looks like this
if f is only single valued (a scalar function, which your surface would be)
then it's just a row of all the partials
for example, let $f(x, y) = xy^2 + x$, then $\newline \mathbf J_f(a, b) = \left.\left(\pdv{xy^2+x}{x} \qquad \pdv{xy^2+x}{y}\right)\right|{(x, y) = (a, b)} \newline = \left.\left(y^2+1\qquad x\right)\right|{(x,y)=(a,b)} = \left(b^2+1 \qquad a\right)$
clever fjordBOT
Frosst
obsidian frost
i honestly dont understand much of these, i feel like its such an advanced solution
is there no other simpler way?
merry finch
oh
pretty much this is saying
how much does f change in the x direction
how much does y change in the y direction
now make a plane out of this
obsidian frost
ill take some time to grasp it but ill give it a try, in the meanwhile i think ill just close off this channel
thanks so much for the help!
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odd edgeBOT
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Available help channel!
hot sable
I'm doing this and I tried to use the identity $\sin^2(x)+\cos^2(x)=1$ but nothing so far
clever fjordBOT
programme
hot sable
wait what if i haven't done derivatives yet
buoyant hinge
oh then how....
hot sable
is there a way to solve this using pure limits?
hot sable
i think it's 0 after some calculations
i might be wrong but
$\cos^2(x)-1=(\cos(x)+1)(\cos(x)-1)$
clever fjordBOT
programme
buoyant hinge
no it is 0 let me keep goign
mystic saffron
Won't help
clever fjordBOT
programme
hot sable
true
hot sable
**programme**
will this help?
buoyant hinge
not really
hot sable
why not though
buoyant hinge
you still end up with (cos(0)-1)/0
have you learnt sandwich theorem or squeeze theorem?
hot sable
squeeze yeah
buoyant hinge
oh okay
you can use that
-1<cos(x)^2-1)<2
divide by x
then you can say both bounds diverge to 0 as 1/x should be a standard limit
im incorrecnt
hot sable
ah maybe i do need lhopital
buoyant hinge
i cant find a way besides l'hopital
hot sable
ah okay thanks man
buoyant hinge
soz
hot sable
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odd edgeBOT
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Available help channel!
winged jungle
odd edgeBOT
winged jungle
,rotate
clever fjordBOT
clever fjordBOT
Monarch of Eternal Night
winged jungle
sorry i had to do something
azure turtle
Sure, look and see if u follow
azure turtle
**Monarch of Eternal Night**
We have to prove this
azure turtle
Yes
winged jungle
not two for tan(a+b)
azure turtle
But we have three equation
azure turtle
If you simply, everything will come in terms of the three equation u wrote
@winged jungle
azure turtle
Tan(a+b+c)
winged jungle
bruh
glossy delta
not two for tan(a+b)
First split it into tan( (a+b) + c )
Then split the tan (a+b) ?
azure turtle
First split it into tan( (a+b) + c )
Then split the tan (a+b) ?
Like this
$\frac{tan(a+b)+tanc}{1-tan(a+b)tanc}$
clever fjordBOT
Monarch of Eternal Night
odd edgeBOT
odd edgeBOT
Available help channel!
thorny ibex
the rectangular hyberbola y=a/x +k passes through (2,7) and (-1,1) find a and k. i dont know why i am having trouble with this, but i must be making errors in my working out. i have put it through chatgpt and mathgpt and photomath and everything and somehow each one gives me a different answer, any help and explanation of working out would be insanely appreciated
warm nacelle
um
you have the equation
with two unknown variables
and you have two points where the hyperbola passes
that means those points satisfy the curve
so put them in the equation
thorny ibex
i know you substitute the values in and then subtract them to find a then k, but then thats where i get stuck
so i have 14= a+k - 1 =a+k, which then cancels out both a and k and i cant find either
warm nacelle
let me check
when u put 2,7
and multiplied by 2 on both sides
it will be different equation then the one u wrote
thorny ibex
well i was thinking firstly you would do 7=a/2+k and then for the other 1=a/-1 +k
then once u have done that with the substitution you subtract one from the other which gives you the a value, but from my working out a + k both get cancelled
do you get what i mean
warm nacelle
yes but
you have 7=a/2+k right?
now multiply both side by 2
you will get 14=a+2k
check you equation and solve properly
thorny ibex
yeah thats what i mean, after putting getting rid of the 2 i had 14, then its 14-1 = nothing because both a+k from the previous cancel out
thorny ibex
alright, so for (2,7) , it is 7=a/2 +k which then you multiply by 2 to get it to the other side making 14=a+k, then for (-1,1), it is 1=a/-1 +k which you multiply the -1 to the other side to get -1=a+k, then this is where i get stuck
thats the solving
warm nacelle
see the first equation
7=a/2+k
after multiplying by 2
it will become 14=a+2k
can you see the 2k?
again for second equation
thorny ibex
ohhh
warm nacelle
alright
warm nacelle
alright so k=13
wrong
thorny ibex
?
warm nacelle
k is not 13
thorny ibex
2k-k=k no?
warm nacelle
check the second equation
warm nacelle
wait what
thorny ibex
yes
thorny ibex
this is what i had:
14= a +2k
-1 = a +- k
a gets cancelled
13= k
u would be right though
warm nacelle
so you multiplied bottom equation by -1?
are you adding the equations or subtracting them?
thorny ibex
subtracting
twin glacier
@warm nacelle he just noticed that both equations have 1 a in them so if you subtract the equations you remove a
Thus giving you an equation with just k in it
thorny ibex
yeah
twin glacier
then you solve for k, and after solve for a
twin glacier
-1 = a +- k
14 - (-1) = a+2k - a -(-k)
warm nacelle
<@433195695869263895> he just noticed that both equations have 1 a in them so if...
i know, i am correcting his mistake by himself
warm nacelle
i can write the solution but he wont see where was he going wrong
thorny ibex
oh its double negativeeee
thorny ibex
thats where i mainly have gone wrong
warm nacelle
subtracting
if you are subtrating then u multiply all the terms by -1
thorny ibex
alright so k =5 and a =4
twin glacier
Yep
thorny ibex
thats my final answer
warm nacelle
yes
odd edgeBOT
odd edgeBOT
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charred rune
need help with (iii)
Q=(1,-1)
odd edgeBOT
need help with (iii)
Q=(1,-1)
charred rune
eq of tangent = y=3x-4
odd edgeBOT
@charred rune Has your question been resolved?
viscid flint
do you know integrals?
viscid flint
need help with (iii)
Q=(1,-1)
split it into two halves
charred rune
split it into two halves
wait but if i integrate like that wont i only get the area above the x axis
viscid flint
normally you do integral of (top curve) - (bottom curve)
viscid flint
you didn't split it into two halves like i said
the red part is different than the blue part
they have different bottom curves
charred rune
they have different bottom curves
i think im js bad at math i still couldnt get it
viscid flint
let's walk through it
for the red region, do you see how the line doesn't matter at all?
the top bound is that x ^ 3/2 curve, but what's the bottom bound?
charred rune
3x-4
viscid flint
y = 3x - 4 is the straight line, but the red region doesn't go down to where that line is
viscid flint
well... that will get you the region above the x axis
the bottom bound for that red region is y = - x^3/2
viscid flint
yeah
charred rune
alr then how do we integrate the blue part
viscid flint
what are the upper and lower curves for that region?
charred rune
x^3/2 is the upper and 3x-4 is the lower
charred rune
like this?
viscid flint
yep perfect
odd edgeBOT
Available help channel!
tired ore
odd edgeBOT
tired ore
i’m not sure how to tackle it
i even checked the marck scheme after being stuck for a long time and i still didn’t understand it
viscid flint
,rccw
clever fjordBOT
digital matrix
the quadrilateral is inscribed in the circle
digital matrix
but i’m not sure how that changes much
opposite angles are supplemtary
tired ore
so 180-48
digital matrix
yeah
digital matrix
so the exterior angle intercepts 2 arcs
you know the measure of the exterior angle and one of the intercepted arcs
use that
digital matrix
what’s the measure of the exterior angle
36
tired ore
how do we know that
digital matrix
twv
i might’ve called it the wrong thing but it’s the angle outside the circle that’s given
digital matrix
tvw
digital matrix
and how did u work out 36!
angle twv is given in the problem
look at the diagram it’s already there
36 degrees
digital matrix
yh and do we use that to work out angle A?
yes
tired ore
how do we do that?
digital matrix
don’t exactly remember the theorem name but basically if a tangent and a secant intercept 2 arcs the outside angle is the average of the difference between the 2 arcs
tired ore
lowk i don’t rlly understand that
digital matrix
so in this case it would be:
36 = (far arc - near arc)/2
digital matrix
subtract the arc nearest to the outside angle from the arc further away
and divide the difference but 2
digital matrix
yes
digital matrix
that’s the wrong arc
tired ore
oh wait
digital matrix
arc TV is the nearest arc
arc UT is the further arc
digital matrix
yup
UT - TV
digital matrix
but the values r unknown no?
we have the value of UT and we know the exterior angle
so we only have one unknown with one equation which we can solve
36 = (UT - TV)/2
tired ore
we have the value of UT and we know the exterior angle
how do we have the value of uT?
digital matrix
how do we have the value of uT?
the value of an inscribed angle is half of its intercepted arc.
which inscribed angle intercepts arc UT
tired ore
bro icl im not following at all
now
this is the mark scheme given
and i’m not rlly following to both what u said and it
digital matrix
ok lemme try again
tired ore
alr
tired ore
yes
digital matrix
because it intercepts arc UT, arc UT is 96 because the intercepted arc is double the inscribed angle
u follow?
digital matrix
ok so now let’s talk about the exterior angle
we know that angle TWV is 36 degrees
tired ore
yup
digital matrix
and the angle intercepts arc TV and UT
the arcs that are within the 2 lines basically are what the angle intercepts
tired ore
yes
digital matrix
got that?
tired ore
yes
digital matrix
one sec
tired ore
right
digital matrix
we also know that (UT - TV)/2 = 36
tired ore
yes
digital matrix
and we know UT = 96
tired ore
yes
digital matrix
so then our equation is 36 = (96 - UT)/2
tired ore
so tv = 24
digital matrix
yes
tired ore
right
digital matrix
so now what’s A
tired ore
12?
digital matrix
yup
digital matrix
yes
digital matrix
yeah dw bout it
tired ore
appreciate it
digital matrix
i got a c on this unit a couple months back so it’s good to help someone out
tired ore
oh wrd
digital matrix
get to practice for the math final
tired ore
yh i’m cooked on circle theorems and stuff
tired ore
get to practice for the math final
yup
odd edgeBOT
@tired ore Has your question been resolved?
lone pecan
can anybody help me prove this:
odd edgeBOT
can anybody help me prove this:
lone pecan
if f(t) = Si(t), the Laplace transform {Si(t)} = 1/s arctan (1/s)
odd edgeBOT
@lone pecan Has your question been resolved?
mystic saffron
Can someone help me with this one ?
odd edgeBOT
@lone pecan Has your question been resolved?
burnt dragon
Hi could someone help me with this problem? The main thing I don't get it how I get the slopes of -2, -1, 6, 8
mystic saffron
Hi could someone help me with this problem? The main thing I don't get it how I ...
Me pls
Can I help
burnt dragon
Yes ofc!
mystic saffron
Kk
burnt dragon
Yes
mystic saffron
Those slopes are obtained from seeing where on the graph there is a horizontal slope
burnt dragon
Those slopes are obtained from seeing where on the graph there is a horizontal s...
So at -1 and 2?
mystic saffron
So at -1 and 2?
Well at 6 it is also 0
burnt dragon
Oh yeah, it isnt a max or min
mystic saffron
Correct
Since the graph is f'(x) and not f(x) you look at where it is changing from negative to positive or positive to negative for a rel min/max
burnt dragon
Yup so -1 and 2
mystic saffron
Yeah
burnt dragon
And then -1 is a rel. max and 2 a rel. min
burnt dragon
But then if I do the candidates test where do I get the f(x) of all the options + endpoints
mystic saffron
But then if I do the candidates test where do I get the f(x) of all the options ...
Oh you integrate
Find the area
burnt dragon
OHH wait wait but how is f(-2) = 3
Im sorry if that sounds dumb im just hella confused
mystic saffron
OHH wait wait but how is f(-2) = 3
You would add these
burnt dragon
You would add these
Aha, so then were doing integral from -2 to 0 right
and for f(1) from 1 to 0 okay
mystic saffron
hell yeah
also i believe there's a more better method for these problems so let me show you
Since we know what f(2) equals
What we can do is this
I'm going to draw the integral because I don't know how to use the bot
@burnt dragon
Hey come back
burnt dragon
aa
burnt dragon
ohh
okay
so then
OHHH
OH I SEE IT OKAY
intgr from -2 to 2 is 4
no sorry
-3*
mystic saffron
Both attempts are wrong
burnt dragon
okay
mystic saffron
Yes
burnt dragon
there we go
mystic saffron
So now try for 8
mystic saffron
Okay intgr(2,8) f'(x) dx= f(8) - f(2)
Not quite.
burnt dragon
nah thats gotta be wrong
mystic saffron
This time you are going to the right
burnt dragon
are the 2 and 8 on the integral flipped
mystic saffron
burnt dragon
yess
and then that's
wait i forgot how to take the are under the semicircle
one sec
mystic saffron
wait i forgot how to take the are under the semicircle
Think outside the box
mystic saffron
Think outside the box
.
Don't worry about the circle
Think outside the box
Box
mystic saffron
Ok uh doesn't seem like u get my hint
mystic saffron
8 - whatever the area of the semicircle is
Yes!
burnt dragon
OHHH LMAO
mystic saffron
I was about to send this
burnt dragon
over 2
mystic saffron
yes
mystic saffron
Good
burnt dragon
from f(2)
11 - 2pi
YEAHHH WE GETTING SOMEWHERE
okay and the same thing for x=6 right
mystic saffron
Hi could someone help me with this problem? The main thing I don't get it how I ...
Seems about right
burnt dragon
okay oh my god it finally makes sense
mystic saffron
Yeah it does
burnt dragon
you are such an absolute god I hope you know that
like i would NOT have gotten that otherwise
okay everything else makes sense now
tysm dude, can I add you btw since were both taking the ap exam?
mystic saffron
tysm dude, can I add you btw since were both taking the ap exam?
Yeah sure
Do you want to continue doing frq practice
I want to prep for exam too
burnt dragon
Yeah fs, ill finish this one up, do you want me to share the link im using?
Im just doing the 2023 ones
mystic saffron
Sure
odd edgeBOT
odd edgeBOT
Available help channel!
autumn dove
On the line a, these points are added
odd edgeBOT
On the line a, these points are added
autumn dove
n≥2
What is the natural number n equal to, knowing that the n points determine
A) 28 rays
B) 28 segments
mental wharf
what have you tried so far
mental wharf
for a) it might help trying smaller values of n and seeing how many rays there
torpid owl
Looking at smaller numbers can sometimes help
mental wharf
not quite for n=2
autumn dove
Wait what
autumn dove
Uh 6?
mental wharf
do you see the pattern
mental wharf
yeah
mental wharf
lets say you look at the first point on the line
how many segments are ther connecting this
autumn dove
1
mental wharf
mental wharf
how many segments are there connecting the first point to any other point
autumn dove
1
mental wharf
autumn dove
It's wrong isn't it
mental wharf
what about this segment
autumn dove
2 segments