1. a) the equation of a line is y=A^x. A is the point where the line intersects the y axis. Find A. then Q15 https://revisionmaths.com/sites/mathsrevision.net/files/imce/1MA1_3H_QP.pdf

a) on the y axis, x=0. sub x=0 into y=A^xtherefore y=A^0 = 1 so A is at (0,1)
15) WILL USE DIAGRAM
so firstly, we need an equation linking the area of the triangle to the information we are given(two sides and the opposite angle) so area of a triangle= 0.5 x b x c x sinA(check all units are the same- metres)so 6 root 2= 0.5 x (x+3)(2x-1) x sin45 12root2= sin45 x (x+3)(2x-1) divide both sides by 0.512root2/(sin45)= (x+3)(2x-1) divide both sides by sin4524 = (x+3)(2x-1) type the left side into calculator to simplify24 = 2x^2 - x + 6x -3 multiply out the brackets using moon/beak technique24 = 2x^2 + 5x -3 collect like terms on right hand side(x terms) 0 = 2x^2 +5x - 27 to solve a quadratic equation(Ax^2 + Bx + C) must make it =0 by -24 use quadratic formula to solve... a=2,b=5,c=-27gives two figures: x=2.6310..., negative value is no solution as gives a negative length.