By J. Newbury PhD (auth.)

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**Example text**

What is its gradient? Answer 2. Now look at the points marked. Let us write down their co-ordinates: A : (0,1) B : (1,3) C : (2,5) D : (3,7) E : (4,9) You might like to add a few more points of your own. Can we now find a connection between the numbers in the 2nd column (the y co-ordinates) and those in the first (x co-ordinates)? The first thing to notice is that the x co-ordinate increases by 1 at each step as we go from A to E, and as we would expect the y co-ordinate increases by 2 each time, since 2 is the gradient of the line (remember gradient is the change in y for an increase of 1 in x).

To prepare for that, attempt the following questions. Exercise 71: What is the area of a rectangle whose length is 6 cm and whose width is 4 cm? Exercise 72: If you found the previous answer by saying that the area is 6 x 4 square cm, that is, 24 sq. cm, can you explain why you multiply 6 x 4 in order to get the answer? (Do not say that you did it 'because that is the rule'! ) Exercise 73: If a rectangle has length x cm and width y cm, what is its area? Exercise 74: If a square has a side of length x cm, what is its area?

C = 37. J(a2 + b2 ) ;a = -2W (R 2 - r2 ) " r a + 2b h + b) ; X = 3(a a 1 1 u v 26. - + - 2 = -f ; u 28. A - y'-2P2 _ Q2 29. A = 1T,y(h 2 + r 2 + 1Tr 2 ):h . d _ (p 2 _ 2Q2) . S3\ %y (g;) ; 8 31 e - y(Z); H 24. A=1Tr 2 y(h 2 +r 2 );h 25. T = H + - - ; (i) H 4 (ii) 1 27. M="4 (1-"2);1 19 H = y(3;); h 30. r = 32. E ; d ,P ~ + y(~2 + q2} = 2Hg - v2 • Wf ' ; q H 34. T = 21Ty (}12g ; k2) 36. E +~)= 3k(1-~);m 38. W 32x ' 2 5 Quadratic Expressions The type of equation I have so far dealt with is usually known as a 'linear' equation because of its connection with a straight line graph.