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Re: HSC 2013 2U Marathon $State the range and domain for$ $i)$ \ \ \ \sqrt{4-x^2}

Re: HSC 2013 2U Marathon 0 lol (x-x)

Re: HSC 2013 2U Marathon Ln(xroot3)^2=ln(6-7x) 3x^2=6-7x 3x^2+7x-6=0 X=2/3 or x=/=-3 Therefore x=2/3

Re: HSC 2013 2U Marathon Shut it

Re: HSC 2013 2U Marathon Correction: 8x^9sinxcosx

Re: HSC 2013 2U Marathon It's correct now

Re: HSC 2013 2U Marathon Sorry I read it as twelve not twenty T10= 39 t20=79 S= 11/2(39+79)=649

Re: HSC 2013 2U Marathon I) t17= 3+16*4= 67 ii) 129

Re: HSC 2013 2U Marathon Yeh that's correct

Re: HSC 2013 2U Marathon Tan3x + 3xsec^2(3x)

Re: HSC 2013 2U Marathon Yes true

Re: HSC 2013 2U Marathon Because the gradient isn't 6 it's 2.. Plus you can use that way.. Just with the gradient of 2 not 6

Re: HSC 2013 2U Marathon Use that only if you have the points and gradient but not the equation. For example finding tangent to the line or the normal equation. Here they give you the derivative of the curve. So just integrate find C by subbing the points and that's your equation; y=2x^3-6x-5

Re: HSC 2013 2U Marathon Differentiate: 2yln(y) __y

Re: HSC 2013 2U Marathon Yeh that's what I got... Mines just not simplified

Re: HSC 2013 2U Marathon Arc length= 5pi/3 = circumference of cone Hence radius of cone= 5/6 Radius of sector= 5 = slant height of cone Therefore height of cone, H^2= 5^2 - 25/36 h= root(875/36) Volume cone= 1/3pi R^2 h = 1/3 pi 25/36 x root(875/36) ?

Re: HSC 2013 2U Marathon Nope 2u

Re: HSC 2013 2U Marathon The quadratic equation x^2 + Lx + M = 0 has one root twice the other Prove that M= 2L^2 .........9

Re: HSC 2013 2U Marathon Ahha i see... Very deceptive

Re: HSC 2013 2U Marathon No soln ?