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No, 0.9 (recurring)=1. Not approaches.

x-y axis are perpendicular because its easier to find distances between points. And, 0.9 (recurring)=1

That's how my uni tutor does it. I was going to do that method but it would've been hard to understand without LaTeX. It's analoguous to what Gurmies said.

Kinda looks like the ellipse equation.

Let u=j*4*pi*c-j*2*pi*f y= t: (-inf,inf) (1/4) (e^tu/u) = inf

Let A=atanh(x) and u=sinhA: sinhA=u sqrt(coshA^2-1)=u coshA=sqrt(u^2+1) u/sqrt(u^2+1)=tanh(A)=x u^2=x^2u^2+x^2 u^2(1-x^2)=x^2 u=x/sqrt(1-x^2) (sinh(atanh(x)))^2=x^2/(1-x^2) Same as above: tanh(A)=u coshA=x sqrt(1+sinh(A)^2)=x sinhA=sqrt(x^2-1) u=tanh(A)=sqrt(x^2-1)/x...

x^3+5x^2+8x+4 =x^3+x^2+4x^2+8x+4 =x^2(x+1)+4(x^2+2x+1) =x^2(x+1)+4(x+1)^2 =(x+1)(x^2+4x+4) =(x+1)(x+2)^2 Try to use the 8 and 4

y=x^2/8 y'=x/4 At P: y'=p y-2p^2=p(x-4p) y=px-2p^2 Similarly, equ of tangent at Q: y=qx-2q^2 px-2p^2=qx-2q^2 x(p-q)=2(p+q)(p-q) x=2(p+q) => y=2pq For focal chord: pq=-1 as tangents meet at 90 degrees y=-2

z=(-1+-sqrt(1^2-4))/2 =(-1+-isqrt(3))/2

Yeah realised that after writing that up. But i don't think it matters. ln(k+1)>ln(k) 1/k+ln(k+1)>ln(k) 1/k>ln(k/(k+1)) 1/k-ln(k/(k+1))>0 1/k+ln(1+(1/k))>0

1) I get confused with left and right very often. Sorry it should be RHS. 2) Do you agree that...1/k and ln(k/(k+1)) are positive for all k: k>=1? This can be verified by using the derivative and proving the functions are positive for k>=1 but decreasing. So, 1/k+ln(k/(k+1)) >= 0 [Adding...

S(n): 1/n =< (1+...+1/n)-ln(n) Base case is trivial Assume, 1/k =< (1+...+1/k)-ln(k) Let n=k+1: LHS=(1+...+1/k+(1/(k+1)))-ln(k+1) =1/k+ln(k)+1/(k+1)-ln(k+1) =(1/k+1/(k+1))+ln(k/(k+1)) >=1/(k+1) Additionally, 1/k, 1/(k+1) and ln(k/(k+1)) <1 for k<1 Thus, LHS=<1 1/n =<...

Always write your working down. In last year's 4u HSC exam, i only gave the answer using auxillary method for a 4marker but didn't write out the auxillary form ie. sqrt(...) is max value thus, something happens. opposed to full working.

i) 2p^2=iq ii) 2(p^2-1)=q I might put up working a little later.

Schoey93 for "Troll of the Year." HSC Physics is a lot worse than HSC Chem, HSC chem was alright excluding titration practical which was gay because of my pedantic teacher.

S (x-2)/sqrt(x+2) dx = S (u^2-4)/sqrt(u^2) 2udu = 2 S (u^2-4) du = 2[(1/3)u^3-4u]+C = 2sqrt(x+2).[(1/3)(x+2)-4]+C Yes, is the answer to your question where applicable

Didn't that cancel with the 2udu part?

Where?

No because -1 =< p =< 1

-sqrt(1-p^2)?