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Looks like we both made the same mistake lol

oh damn, yeah sorry.

1) A= pi.(2)^2/4 [Area of a quarter of a circle as sqrt(4-x^2) is a semicircle but area is only defined as 1st quadrant] =pi 2) S {x=-a to x=a} x^3 dx = (1/4) [x^4] {x=-a to x=a} A = (1/4) (a^4+(-a)^4) = (1/4) (a^4+a^4) = a^4/2

I did but i forgot most formulae from the HSC. It should be in any good textbook.

Yeah, think they're right. For c): Use (m1-m2)/(1+m1m2)=tan@ where m1=p and m2=q and @=45 deg.

I always look out for the little things. Just needed to realise the 1 and -3 coeffs and its easy from there.

Yeah, sorry. Usually i say: Assume < statement> (for n=k) but i didn't think it was necessary as i'm just writing out the proof for a person and not formally doing it in a test.

Consider (x+y)^3: (x+y)^3=(x+y)^2.(x+y) =(x^2+2xy+y^2)(x+y) =x^3+2x^2.y+xy^2+2xy^2+y^3 =(x^3+y^3)+2xy(x+y)+xy(x+y) 1=19+3xy xy=-6 (x+y)^2=x^2+y^2+2xy 1=x^2+y^2-12 13=x^2+y^2 For the next one, just try to get each variable as a subject of the other two and sub in. It should workout...

NOTE: (1+...+k)=(k(k+1))/2 by sum of Arithmetic Progression. First case for n=1 is trivial Let n=k: 1^3+...+k^3=(1+...+k)^2 Let n=k+1: LHS=1^3+...+k^3+(k+1)^3 =(1+...+k)^2+(k+1)^3 =k^2(k+1)^2/4+(k+1)^3 =RHS RHS=((1+...+k)+k+1)^2 =(1+...+k)^2+2(k+1)(1+...+k)+(k+1)^2...

x^2+3x-ix-3i=0 x(x+3)-i(x+3)=0 (x+3)(x-i)=0 x=-3,i

Yeah, your deduction's correct.

In parametrics, the aim is to eliminate the third variable.

Try to do well in english. If you don't have a good teacher for english then try to change classes. My biggest regret is not changing classes at the end of yr11 as I probably would've gotten a band 6 for english opposed to a band 5.

Yeah, nw

Let n=k: Assume, x^k-1=M(x-1) where M,x E Z Let n=k+1: x^(k+1)-1=x^k.x-1 =xM(x-1)+x-1 from assumption =(x-1)[Mx+1] =N(x-1) where N E Z: N=Mx+1 Let n=k: Assume, k(k+1)(k+2)=3M where M E Z Let n=k+1: LHS=[(k+1)(k+2)](k+3) =k(k+1)(k+2)+3(k+1)(k+2) =3M+3(k+1)(k+2) =3N where...

yeah, repeat it. This time you know most of the stuff just clarify more of the theory and practice a lot of papers. Then you could get 95 or so for 2unit.

(x+iy)(x-iy)=(a+ib)^2.(a-ib)^2 [As conj((x+iy))=x-iy=(a-ib)^2] x^2+y^2=((a+ib)(a-ib))^2 x^2+y^2=(a^2+b^2)^2

Yeah my bad lol. Sorry

I think your 4th to 5th line is wrong.

x=(t-(4/t)) x^2=(t-(4/t))^2 x^2=(t^2+(16/t^2))-8 x^2+8=t^2+16/t^2 (x^2+8)/2=(t^2)/2+8/t^2 (x^2+8)/2=y x^2+8=2y x^2=2(y-4)