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Yeah you're right. Books wrong

Isn't M-theory a combination of the main string theory perspectives not an extension?

geometrically it has symmetry of rotation around the origin, that would be better def but algebraic's more safer

Ignore the troll

Dude you're so fail...Your proofs based on an observation rather than a logical process.

I'm the opposite. I'd like my significant other to be less intelligent than me.

I've heard that they tell you you're doing it. But idk, if this is correct. I may contact UNSW tomorrow.

Yeah, same here. I found 4unit more fun than 3u as it was more challenging.

4unit's only a little harder than 3unit. Some 3unit questions can be really hard or tedious and for me personally, i would only make little algebraic or arithmetic errors in 3unit which would screw up my final answer. On the other hand, in 4unit, you really have to study opposed to 3unit and be...

2x+2yy'=0 -x/y=y' At (3,-4): y'=3/4 y+4=3/4.(x-3) y=(3/4)x-25/4 4y=3x-25 3x-4y=25 At x=5: y=-5/2 and At x=-5: y=-10 m1= (-5/2)/5 = -1/2 and m2= (-10)/(-5)=2 m1m2=-1

Nice, a lot better than my solution but i think i could've just cut-off a=+-sqrt(5) before hand as a solution

Let P(x)=z^4-2z^3+7z^2-4z+10 P(ai)=a^4+2a^3.i+7a^2-4ai+10 P(-ai)=a^4-2a^3+7a^2+4ai+10 P(ai)+P(-ai)=2(a^4-7a^2+10)=0 (a^2-5)(a^2-2)=0 a=+-sqrt(5) and +-sqrt(2) Let A,B,C,D be roots of P(z): A+B+C+D=2 C+D=2 and CD=5 or 2 C^2+D^2=(C+D)^2-2CD ie. C^2+D^2=-6 or 0 but we know C=/=+-iD as...

f'(x)=2cos(2x)-2cos2x+4x.sin2x =4xsin(2x) S{pi/4,0} xsin2x dx=(1/4)S{pi/4,0}4xsin2x dx =(1/4) S{pi/4,0} d/dx(sin2x -2x.cos2x) dx =(1/4) {pi/4,0}sin2x -2x.cos2x =(1/4)

4([tan(x)]^2+1)=3tan(x)+5 [from (sin(x))^2+(cos(x))^2=1] 4(tan(x))^2-3tan(x)-1=0 (4tan(x)+1)(tan(x)-1)=0 4tan(x)=-1 or tan(x)=1 x=-atan(1/4) or pi/4 now just input the cycles in 360 deg

f'(x)=-sin(x)+sin(x)+xcos(x) =xcos(x) Hence, S {x=0 to x=pi/2} xcos(x) dx = S {x=0 to x=pi/2} d/dx(cos x + x sin x) dx = {x=0 to x=pi/2} [(cos x + x sin x)] = pi/2-1

Has to be a straight line of you think about it. (y/x)-(x/y)=-2 Let's generalise: Let y/x=f where f is a function and -1=k where k is constant f-(1/f)=2k f^2-1=2fk f^2-2fk-1=0 (f-k)^2=1+k^2 f=+-sqrt(1+k^2)+k

Maybe solving as I did is the method to go but doesn't seem right. Additionally i tried: (x/y)-(y/x)=-2 which was my first thought. (x^2+y^2)(xy'-y)=0 ...maybe x=+-iy but same problem as before

Re(z^2)+Im(z^2)=0 Re(x^2+2ixy-y^2)+Im(x^2+2ixy-y^2)=0 x^2-y^2+2xy=0 y^2-2xy-x^2=0 y=(2x+-sqrt(4x^2+4x^2))/2 =x+-xsqrt(2) y=x(1+-sqrt(2)) ie y=x+sqrt(2).x OR y=x-sqrt(2).x Implicitly differenitiating i get: turning pt and a vertical asymptote at (0,0)

Try graphing calculators just enter arbitrary lines like y=2x or y=(1/2)x and let them be z and a

arg(z-a)=pi/2+arg(a) arg(z-a)-arg(a)=pi/2 Draw a parallelogram then construct a diagonal through z and a. arg(z-a)-arg(a) is the angle from (z-a) to a and this has to be equal to pi/2.