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of the cone and that the radius of the sector = slant height of the cone. i.e. / x 2 x p x 40/1 = 2 x p x r where r = base radius of cone. Simplifying gave r = 15cm. Slant height of cone, = l = 40cm, r = 15cm. Using pythagoras theorem, h2 + r2 = l 2 where h = perpendicular height of cone. \ h = v � = cm. Therefore volume of Cone = 1/3 x p x r2 x h. (i) The formula is $x^2 \pm bx + c = \left(x \pm {b \over 2}\right)^2 - \left(b \over 2\right)^2 + c$ \begin{align} 9 - 7x + x^2 & = x^2 - 7x + 9 \\ & = \left(x - {7 \over 2}\right)^2 - \left(7 \over 2\right)^2 + 9 \\ & = (x - )^2 - \\ & = + ( + x)^2 . x = 2y (1) y = 2x ? 3 (2) The graphs of the two equations are shown below. The intersection of the two graphs is (2; 1). So the solution to the system of simultaneous equations is x = 2 and y = 1. We can also check the solution using algebraic methods. Substitute equation (1) into (2): x = 2y ? y = 2(2y) ? 3. Make points:

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The equation looks like: Where: x,y are the coordinates of any point on the line m is equatione equations for maths paper 2 00 of the line P xP y x and y coordinates of the given point P that defines the line Recall that the slope m is the "steepness" of the line. Related pages on Google Scholar on PubMed. Rao, L. AIMS Mathematics, 6 4 : The Exploration Guide includes: Investigation matbs, Marking criteria guidance, 70 hand picked interesting topics, Useful websites for use in the exploration, A student checklist for top marks, Avoiding common student mistakes, A selection of detailed exploration ideas, Advice on using Geogebra, Desmos and Tracker.

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