Search results

Let F{n}=x^n

Maybe setting a question on it's derivation would be good as the induction is quite easy.

Yeah, just check it the solutions afterwards

(2x-1)/x >= x+1 x(2x-1)>= x^2(x+1) 2x^2-x>=x^3+x^2 x^3-x^2+x=<0 x(x^2-x+1)=<0 x((x-1/2)^2+3/4)=<0 x<0 since x E R OR -2x^2+x>=x^3+x^2 x^3+3x^2-x=<0 x(x^2+3x-1)=<0 0< x =< 0.5*(sqrt(13)-3) or x =< -0.5 (3+sqrt(13))

(1+cos@+isin@)/(1-cos@-isin@) =(1+2(cos*)^2-1+2sin*cos*)/(1-1+2(sin*)^2-2isin*cos*) where *=@/2 =icos*/sin*.(cis*/cis*) =icot* (z-1)^n+(z+1)^n=0 1+((z+1)/(z-1))^n=0 1+(-icot*)^n=0 cot*=i^(2/n-3) {sum} a0a1+a0a2+...+a{n-2}a{n-1}=(n n-2) Since roots are of form ia{k}...

AED's an angle. I didn't write Angle AED as it's too long. Yeah, thats what i did. I proved for D outside or inside the circle and lead to contradictions.

Not sure is this a very good proof by contradiction. Assume, ABC+ADC=180 deg but A,B,C,D aren't concyclic. This must mean either A,B,C,D lie outside the circle. Assume, A lies inside the circle. Since, D's inside the circle. AB or CD intersect the circle. Now, suppose AD intersects the...

dw about it. Close enough. Can you post another question? Preferably more difficult than previous as questions should get sequentially harder

equ of normal: px-y/p=c(p^2-(1/p^2)) At x=0: y=-cpk where k=p^2-p^-2 At y=0: x=ck/p Mid QR: (ck/2p,-cpk/2) x=ck/2p and y= -cpk/2 p^2=-y/x 2px=ck where k=y^2/x^2-x^2/y^2=y^4-x^4/(xy)^2 4p^2x^2=c^2k^2 4x^4.y^2(-y/x)=c^2(y^2-x^2)^2 4x^4.y^2(y/x)+c^2(y^2-x^2)^2=0...

Yeah, i solved a question in 4u a long time ago which basically used similar triangles and geometry opposed to algebra bashing.

Can we make a new rule? No conics questions

d/dx(dy/dx)=d/dx(uv'+u'v) =u'v'+uv"+u"v+v'u' y"=2u'v'+u"v+v"u d/dx(uvw)=u'vw+uv'w+uvw' Can't see how w variable somes into y'

Re: 2 Unit Revising Marathon HSC '10 I don't think i would've realised the negative solution for absolute value question

S (x-1)^n dx = S {sum} (nk) x^(n-k)(-1)^k dx (x-1)^(n+1)/(n+1) = {sum} (nk) x^(n-k+1-1)^k/(n-k+1) +C At x=0: C=(-1)^(n+1)/(n+1) (x-1)^(n+1)/(n+1) = {sum} (nk) x^(n-k+1-1)^k/(n-k+1)+(-1)^(n+1)/(n+1) n(n-1)(x-1)^(n-2)={sum}(nk) (n-k)(n-k-1)(-1)^k.(x-1)^(n-k-2) k=n-2,x=2: n(n-1)=(n...

I got something else for d) but i'm probably wrong. Can someone post up another question? Preferably not uni maths. EDIT: Yeah, i'm wrong about d) :(

Same here, I'm waiting for another question.

Re: 2 Unit Revising Marathon HSC '10 abs(y-3)=sqrt((x-3)^2+(y-4)^2) y^2-6y+9=x^2-6x+9+y^2-8y+16 2y=(x-3)^2+7

I've solved the question but idk if it's a correct method as i haven't learned 2nd ODE properly. Thus i won't be posting a "solution."

Let S(n) be the statement defined such that S(n): ln(n!)>n for {n E Z: n>=6} Skipping base case where n=6 Let n=k: Assume; ln(k!)>k Let n=k+1: LHS=ln((k+1)!) =ln(k+1)+ln(k!) >ln(k+1)+k >k+1 (As ln(k+1)>1) b) n!>e^n 1/n! <1/e^n c)...

4tan(a-b)=4((tan(a)-tan(b))/(1+tan(a)tan(b)) 4tan(a)-4tan(b)=3(tan(a))^2.tan(b)+3tan(a) tan(a)/(4+3(tan(a))^2)=tan(b) sin(2a)/(7+cos(2a))=2sin(a)cos(a)/(2(cos(a))^2+6) =sin(a)cos(a)/((cos(a))^2+3) =tan(a)/((sec(a))^2+3) =tan(a)/(4+3(tan(a))^2) =tan(b)