7 Feb 2008 at 17:06 #21 stevenazari1 stevenazari1 Permabanned Joined 21 Oct 2007 Posts 1,518 I bow to you sir.
7 Feb 2008 at 17:09 #22 Mansize_tissue Mansize_tissue Soldato Joined 20 Mar 2007 Posts 3,095 Location Norwich I'm guessing point T: (0.5, 0) and the area is 0.75 units squared. How wrong am I?
7 Feb 2008 at 17:11 #23 Fraggr Fraggr Soldato Joined 29 Jun 2006 Posts 3,372 Location Sheffield markyp23 said: SQA Advanced Higher So neither Fraggr Click to expand... Ah. lol I didn't look at your location.
markyp23 said: SQA Advanced Higher So neither Fraggr Click to expand... Ah. lol I didn't look at your location.
7 Feb 2008 at 17:16 #24 [TR]SweetSpotOz [TR]SweetSpotOz Soldato OP Joined 5 Mar 2006 Posts 6,183 Location everywhere . Last edited: 3 Apr 2018
7 Feb 2008 at 17:17 #25 Mansize_tissue Mansize_tissue Soldato Joined 20 Mar 2007 Posts 3,095 Location Norwich markyp23 said: Quite! Click to expand... Pfft, what's the answer then?
7 Feb 2008 at 17:19 #26 [TR]SweetSpotOz [TR]SweetSpotOz Soldato OP Joined 5 Mar 2006 Posts 6,183 Location everywhere . Last edited: 3 Apr 2018
7 Feb 2008 at 17:20 #27 Squark Squark Soldato Joined 11 Apr 2004 Posts 11,550 Location In Christ I can't think how to approach that question
7 Feb 2008 at 17:29 #28 [TR]SweetSpotOz [TR]SweetSpotOz Soldato OP Joined 5 Mar 2006 Posts 6,183 Location everywhere . Last edited: 3 Apr 2018
7 Feb 2008 at 17:43 #29 |Ric| |Ric| Associate Joined 28 Jun 2005 Posts 895 area of rect = (1 - t^2) * 2t d(area of rect)/dt = 2t - 6t^2 Maximal when d(area of rect)/dt = 0 t = +/- 1/sqrt(3) (ignore negative solution) So area is (1 - 1/3) * 2/sqrt(3) = 0.770 (3sf)
area of rect = (1 - t^2) * 2t d(area of rect)/dt = 2t - 6t^2 Maximal when d(area of rect)/dt = 0 t = +/- 1/sqrt(3) (ignore negative solution) So area is (1 - 1/3) * 2/sqrt(3) = 0.770 (3sf)
7 Feb 2008 at 17:51 #30 [TR]SweetSpotOz [TR]SweetSpotOz Soldato OP Joined 5 Mar 2006 Posts 6,183 Location everywhere . Last edited: 3 Apr 2018