Download e-book for kindle: MEI AS Pure Mathematics (3rd Edition) by Roger Porkess, Val Hanrahan, Peter Secker

By Roger Porkess, Val Hanrahan, Peter Secker

ISBN-10: 0340813970

ISBN-13: 9780340813973

The highly-acclaimed MEI sequence of textual content books, helping OCR's MEI established arithmetic specification, absolutely fit the necessities of the requisites, and are reknowned for his or her pupil pleasant process.

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Extra resources for MEI AS Pure Mathematics (3rd Edition)

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Iii) Use the equation to find the distance from the tip at which the diameter is 15 mm. 1x + 9. 1x = 6 x = 60 EXERCISE 2C 1 Find the equations of the lines (i) – (x) in the diagrams below. Exercise 2C ⇒ the diameter is 15 mm at a point 60 cm from the tip. y 6 (iii) (ii) 4 2 –4 –2 O (i) 2 4 8 x 8 x (iv) –2 –4 6 (v) y 6 (vi) 4 (x) 2 –4 (vii) –2 O 4 –2 –4 2 2 6 (ix) (viii) Find the equations of the following lines. (i) (ii) (iii) (iv) (v) (vi) parallel to y = 2x and passing through (1, 5) parallel to y = 3x – 1 and passing through (0, 0) parallel to 2x + y – 3 = 0 and passing through (–4, 5) parallel to 3x – y – 1 = 0 and passing through (4, –2) parallel to 2x + 3y = 4 and passing through (2, 2) parallel to 2x – y – 8 = 0 and passing through (–1, –5) 51 C1 2 3 Find the equations of the following lines.

8). Solving quadratic equations -b ! 9). 10). 10 23 Basic algebra C1 1 More on completing the square The process of completing the square can be used not only for solving quadratic equations, but also as a source of valuable information about a quadratic expression and its graph. Take, for example, the expression x 2 – 5x + 9. 75. 5 = 0. 5. 11. 11 24 1 2 3 4 5 6 x EXERCISE 1D 1 Factorise the following expressions. al + am + bl + bm (ii) px + py – qx – qy (iii) ur – vr + us – vs (iv) m 2 + mn + pm + pn (v) x 2 – 3x + 2x – 6 (vi) y 2 + 3y + 7y + 21 (viii) q 2 – 3q – 3q + 9 (x) 6v 2 + 3v – 20v – 10 (vii) z 2 (ix) 2 2x 2 + 2x + 3x + 3 Multiply out the following expressions and collect like terms.

This is a very useful form of the equation of a straight line. Two positions of the point (x1, y1) lead to particularly important forms of the equation. 16. 17. 17 Find the equation of the line with gradient 3 which passes through the point (2, –4). 6 SOLUTION The equation of a straight line Using y – y 1 = m(x – x1) ⇒ y – (–4) = 3(x – 2) ⇒ y + 4 = 3x – 6 ⇒ y = 3x – 10. C1 2 (ii) Given two points, (x1, y1 ) and (x2, y2 ) The two points are used to find the gradient: y -y1 m= 2 . x 2 -x1 This value of m is then substituted in the equation y - y 1 = m `x - x 1j .

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MEI AS Pure Mathematics (3rd Edition) by Roger Porkess, Val Hanrahan, Peter Secker


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