Mimi_x3
If the chord of contact of the tangents to a parabola x^2=4ay, from an external point T(x1,y1) meets the directrix at R, prove that RT subtends a right angle at the focus.
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perl
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mmmmm, you werent kidding around
Mimi_x3
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Is this easy for you ? xD
perl
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i will have to think about it
perl
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the directrix is that line under the parabola
Mimi_x3
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lols, you were thinking that its easy ? xD
Mimi_x3
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Yup, y=-a
perl
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what class is this ?
Mimi_x3
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I live in Australia, so its kinda different, there's only a maths class xD
perl
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sorry wrong question, have you done calculus
Mimi_x3
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Yes I have, how did you know ? xD
Mimi_x3
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What grade are you in ?
perl
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it will definitely help
perl
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im a sophomore in college
Mimi_x3
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Cool~ college, and you havent learnt that ?
perl
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not specifically, but i bet i can do this
Mimi_x3
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lols, then try hehe
perl
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im trying to make a diagram, im not even sure i understand the question yet
perl
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ok you have an external pt, and you want the tangent through it
Mimi_x3
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Yeah, I don't really know where's the external point is, thats why im lost =/
perl
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ok first thing is first. the chord of contact means something special
Mimi_x3
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Something special ? What do you mean ?
perl
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well given an external point there are two tangents from that point to the parabola
perl
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now where the tangents intersect the parabola, now draw a chord through those two points
perl
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where the tangents meet the parabola, the two tangents
Mimi_x3
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wow~ nice pdf file, its detailed on parametrics ty
Mimi_x3
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But where is the external point ?
perl
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anywhere
Mimi_x3
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Can't it have to be specific
perl
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in the pdf, the point is below the parabola
Mimi_x3
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It can be but it doesn't make sense when it says RT subtends a right angle at the focus.
perl
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like this
Mimi_x3
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like this ?
perl
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|dw:1319452686670:dw|
Mimi_x3
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I know but where's the external point ?
perl
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|dw:1319452750463:dw|
Mimi_x3
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Imaginary ?
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|dw:1319452790390:dw|
Mimi_x3
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Is the parabola imaginery or somethinbg ?
Mimi_x3
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imaginary* something*
perl
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|dw:1319452894222:dw|
Mimi_x3
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I don't get your diagrams ? sorry
perl
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|dw:1319453015665:dw|
Mimi_x3
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I don't get it ? >_<
perl
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yeah the diagram is a beast
Mimi_x3
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lols, why is there like 3 digrams ? i dont get it >_<
perl
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|dw:1319453240562:dw|
Mimi_x3
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lols, this is not 3D trig xD
perl
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can you do better ?
Mimi_x3
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Nope, i dont know how to start, the problem is where is T
perl
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just a point below the parabola
perl
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like in that pdf
Mimi_x3
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But it won't make sense, and i dont know how to prove
perl
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have faith
Mimi_x3
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have faith ?
perl
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i think they mean the segment R T
perl
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yes have faith when you draw it that it will all come into place, the problem will unravel
perl
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i am atheist, what did you think i mean ?
Mimi_x3
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idk, never mind then i will work it out myself
perl
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|dw:1319453605908:dw|
Mimi_x3
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ty (:
perl
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im working on it
perl
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would be nice if they gave you a diagram
Mimi_x3
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Yeah i know, none of the hard questions have diagrams =/
perl
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i think RT is the line segment though
Mimi_x3
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hm idk
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|dw:1319454238070:dw|
agdgdgdgwngo
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\[\text{An equation of the parabola with focus }(0,p)\text{ and directrix }y = -p\text{ is}\]
\[x^2=4py\]
agdgdgdgwngo
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so the directrix would be y = -a in your question :-D
perl
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yes
Mimi_x3
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Yeah, but im having problem proving
agdgdgdgwngo
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If the chord of contact of the tangents to a parabola x² = 4ay, from an external point T(x1, y1) meets the directrix at R, prove that RT subtends a right angle at the focus.
chord of contact of tangents to parabola is xx1=2a(y+y1)
directrix is y=-a substitute into that, x=[2a(-a+y1)]/x1 y=-a R([2a(-a+y1)]/x1,-a)
let S be the focus
d/dx RS = 2a/ -[2a(-a+y1)]/x1= x1/(a-y1)
d/dx TS=(y1-a)/x1
d/dx RS * d/dx TS= -1
Mimi_x3
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lols, you got it from boredofstudies right ?
agdgdgdgwngo
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yes... :-P
perl
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how did you get chord of tangents equation
perl
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hey, can you draw a diagram then of whats going on, please post the link
Mimi_x3
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Yeah but the problem is that I don't understand the solution
agdgdgdgwngo
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it helps to draw a diagram.
perl
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can you post the link to the solution please , so i can read it
Mimi_x3
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But what's xx1 ? im trying to figure it out
Mimi_x3
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oh i get it, but where did y=1/2(p+q)x-apq come from ?
agdgdgdgwngo
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xx1 must mean something like \[x_1\times x= 2a(y+y_1)\]
Mimi_x3
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where did 2a come from ?
agdgdgdgwngo
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good question
agdgdgdgwngo
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\[parabola: x^2 = 4py\]\[a = \frac{1}{4p}\]
Mimi_x3
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no, isn't it x^2 = 4ay
y = x^2/4a
y' = 2x/4a = x/2a
perl
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|dw:1319455562515:dw|
perl
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i wish there were other colors, but from T draw two tangents to the parabola
agdgdgdgwngo
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perl's diagram works.... but how do we get there using calculus and derivatives if we don't even know the equation for chord of contact of the tangents to a parabola
Mimi_x3
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oh the equation of the tangent to the parabola is y=px-ap^2 right ?
perl
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|dw:1319455702848:dw|
perl
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the chords of tangents formula
agdgdgdgwngo
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now it all makes sense
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|dw:1319455885258:dw|
perl
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i dont get that parametrisation, is it because they want x^2 = 4ay ? , what is p
Mimi_x3
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Doesn't it involve parameters as well P(2ap,ap^2)
agdgdgdgwngo
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they want something like the quotient of the slopes between 2 lines is equal to -1
agdgdgdgwngo
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and one of those lines should be RT
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|dw:1319456032235:dw|
perl
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crap, it missed the R it is the right bottom corner
Mimi_x3
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Im confused, is parameters involved ?
agdgdgdgwngo
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no parameter seems to be involved :-(
perl
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actually a is the parameter, you have to be careful, this word parameter is used in different senses
perl
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, for instance in y = mx + b , m and b are parameters . now another sense of parameter is used when we 'parametrize' an equation. this means we introduce a third variable, often t
Mimi_x3
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isn't parameters P(2ap,ap^2)
So there must be a point for the parameters ?
agdgdgdgwngo
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meh, I assumed m and b to be constant
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right, but they are considered 'parameters'
Mimi_x3
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but its not a straight line so it cant be y=mx + b
perl
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in the sense that they change your line .
so for x^2 = 4ay, the parameter is a since that produces different parabolas by tweaking a
perl
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think of in this sense a parameter is your tweaking constant
agdgdgdgwngo
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:-(
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the parameter is 'a'*
agdgdgdgwngo
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I thought a was just a slope
Mimi_x3
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Im lost =/ Parameter P(2ap,ap^2)_
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changing 'a' changes your parabola. but for the line y = mx + b, changing m,b changes your line
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these are your tweaking values .
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I dont like their use of p, scratch that and use t instead
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ok they are using 'parametric equations' , x = f(t), y = g(t)
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so x = 2at, y = at^2
agdgdgdgwngo
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I like to use matrices and vectors.
Mimi_x3
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Yeah thats in cartesian form
perl
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ok, use a matrix or vector, id like to see that
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mimi, when you remove t, you get cartesian equation
Mimi_x3
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matrix or vector ?
Mimi_x3
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Oh yeah my bad i forgot xD
Mimi_x3
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I haven't learnt that complicated stuff yet =/
agdgdgdgwngo
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you will learn it soon.
Mimi_x3
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Hope not looks hard already =/
perl
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thats curve fitting, no big deal
agdgdgdgwngo
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you will find it easier than conics
Mimi_x3
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Whats conics ?
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its a quick way to fit a quadratic sequence, cubic, etc
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mimi, what we are doing is conics, oh dea
agdgdgdgwngo
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conic sections: parabolas, circles, ellipses, hyperbolas, stuff.
perl
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oh dear *
agdgdgdgwngo
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you do them along parametric equations usually.
perl
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agdg, put the proper definition on board ;)
agdgdgdgwngo
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well, they're what you get when you cut a cone.
Mimi_x3
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LOL, this is not conics
perl
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a conic has a specific algebraic equation in x and y , quadratic
Ax^2 + Bxy + cY^2 + dy + ex + f = 0
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yes it is, a parabola is a conic
perl
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a conic is a family of curves, as agdg pointed out
Mimi_x3
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But Ax^2 + Bxy + cY^2 + dy + ex + f = 0 ?
agdgdgdgwngo
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forget the algebra and xs and ys and f(x)s, you will understand them really well once you picture them geometrically
perl
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right
Mimi_x3
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Um, this doesnt look like a conic
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agdg, whats the nth term of this quadratic sequence
3,5,10,18...
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lets test your matrix skills
agdgdgdgwngo
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hmmm... is it infinity?
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no, the nth term
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give me a formula for the sequence
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obviously it diverges, ;)
agdgdgdgwngo
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its \[2^n + 2^\frac{n}{2}\]
perl
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no, lol
agdgdgdgwngo
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wait! it's \[\frac{2^n+2^{\frac{n}{2}}}{2}\]
perl
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its a quadratic sequence, like in your matrix
agdgdgdgwngo
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how do we find it :-D
agdgdgdgwngo
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so what's the 5th term
perl
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ok so first we find the differences
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the first row of differences is 2 , 5, 8, 11 , ...
the second row of differences is 3, 3, 3, ... so you know its a quadratic sequence
agdgdgdgwngo
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that looks like a long matrix
perl
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3, 5, 10, 18, 29 ... thats the sequence
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it should fit f(n) = an^2 + bn + c
agdgdgdgwngo
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I guess that works.
agdgdgdgwngo
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how do we multiply matrices?
perl
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are you asking in general ?