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Aren't roots of unity nth roots of 1 and hence not applicable to this question? ie z^n - 1 = 0

Rule is valid when all the coefficients are real. $1-2i is one root of x^2-(3+i)x+k = 0$ \\$You know one of the roots $ \alpha =1-2i \\\sum \alpha =\alpha +\beta =\frac{-b}{a} \\1-2i+\beta =\frac{3+i}{1} \\\beta =2+3i \\\sum \alpha \beta =\frac{c}{a} \\(1-2i)(2+3i)=\frac{k}{1} \\2+3i-4i+6=k \\k=8-i

x=3+rcos\Theta \\rcos\Theta = x-3 \\r=\frac{x-3}{cos\Theta}-----(1) \\y=-2+rsin\Theta \\$sub (1) into$ \\y=-2+\frac{x-3}{cos\Theta}sin\Theta \\y=-2+tan\Theta(x-3) \\y=tan\Theta(x)-(3tan\Theta+2) So it's linear with gradient tan(theta)

If you have a to be eternally positive, it isn't consistent with the general point on the parabola (2ap,ap²) screwing you over when you have an upside down parabola.

Physics - Jacaranda as a textbook we are issued. Maths - Sometimes Cambridge and Corroneos are also issued in addition to Fitzpatrick. Dotpoint books are sold through the school at the start of each year for Physics and Chemistry.

I go to school on a nearly always empty train, so I take out my folder and organise stuff. Gluing pages in, organising my sheets. All the easy stuff with little mental input.

I thought focal length would be |a| as opposed to a.

a=-1, because you the parabola is in the form x² = 4ay Equating it to your equation, you get a = -1 It's just the normal parabola, flipped along the x-axis

You could take it as a transformation of y = sqrt(9-x²) which is a semicircle with radius 3. Where y = sqrt(9-x²), there is an asymptote because 1/0 does not exist. But as the semicircle approaches 0, the graph approaches infinity if you see what I mean.

Yep Mayers Advanced English Class. Does she have a year 11 class, and are you in it?

Woah that post was ages ago. Lol Thanks, and from reading your posts from the Extension 1 Topic, I guess, since same school, you are doing Human Condition as well?

8cos^{4}x-8cos^{2}x+1= \\=8cos^{4}x-4cos^{2}x-4cos^{2}x+1 \\=4cos^{2}x(2cos^{2}x-1)-2(2cos^{2}x-1+1)+1 \\=4cos^{2}xcos2x-2(cos2x+1)+1 \\=4cos^{2}xcos2x-2cos2x-2+1 \\=2cos2x(2cos^{2}x-1)-1 \\=2cos^{2}2x-1 \\=cos4x\\\int 8cos^{4}x-8cos^{2}x+1dx=\\\int cos4x\\=1/4sin4x

sin{4t}+\sqrt{3}cos{4t}=0 \\sin{4t}+\sqrt{3}cos{4t}\equiv Rsin(4t +\alpha )$where R\geq 0$ \\sin{4t}+\sqrt{3}cos{4t}=Rsin(4t) cos\alpha +Rcos(4t) sin\alpha \\$Equating both side$ \\Rcos\alpha = 1\Rightarrow cos\alpha = \frac{1}{R} \therefore \alpha $ is in 1st or 4th quadrant$ \\Rsin\alpha =...

Any 3 vertices forming a triangle lie on a particular circle, so it's nothing special

i) A = 1000(1+0.05)^100 = 131,501.2578 with calculator ii) Let (1+0.05) = 1.05 = i k = 1000 1st Year A = ki 2nd Year A = ((k)i + k)i A = (i + 1)ki 3rd Year A = ((i + 1)ki + k)i A = ((i + 1)i + 1)ki A = (i^2 + i + 1)ki 4th Year A = (((i + 1)i + 1)ki + k)i A = (((i + 1)i + 1)i + 1)ki A = ((i^2 + i...

Woops, I was wrong about the x>0, now I'm not too sure. I drew a diagram showing that the chords start when P and Q are the same, and by drawing subsequence chords and tangents, you see the line x-y+a = 0 appearing, but no points for x values past the x value of the tangent

e) With the aid of a diagram, carefully explain why the locus of T is not all of the straight line. Draw lines through point (2a,-a) through which the line PQ has to pass. Notice how there are areas where chords don't exist One is when the line is completely vertical (not sure if this has any...

You would do this by long dividing x³ by (x+1), but since I don't know how to type it up here, i'll do it the long way y = x³/(x+1) y = (x³ + x² - x²)/(x+1) y = x² - (x²)/(x+1) y = x² - (x² + x - x)/(x+1) y = x² - x + (x)/(x+1) y = x² - x + (x+1-1)/(x+1) y = x² - x + 1 -...

But is it safe to just go to second derivative test, if you get a non-zero leave it at that, if you get a zero, you go to first derivative test.

t = 500/v c = 150 + v²/80 Let C be total cost C = c.t (cost/time x time) C = (150 + v²/80)(500/v) C = 75000v^(-1) + 25v/4 dC/dv = -75000v^(-2) + 25/4 For poss SP to occur, dC/dv = 0 -75000v^(-2) + 25/4 = 0 25v²/4 = 75000 v² = 12000 v = 109.5445 (v>0) Therefore SP occurs at v d²C/dv² =...