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Mathematics Lesson 03/05/07

Mathematics Lesson 03/05/07. Objectives (WALT). Revise Autumn term work. Negative Numbers (4,5). If a = 2, b = -1, c = -3, d = 10, what is (a) a + b + c (b) 3c – 2a (c) 5d – 2bc (d). a + b + c = 2 -1 -3 = -1 3c – 2a = 3(-3) – 2(2) = -9 -4 = -13 5d – 2bc = 5(10) -2(-1)(-3) = 50 -6 = 44

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Mathematics Lesson 03/05/07

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  1. Mathematics Lesson 03/05/07 Objectives (WALT) • Revise Autumn term work.

  2. Negative Numbers (4,5) • If a = 2, b = -1, c = -3, d = 10, what is • (a) a + b + c (b) 3c – 2a (c) 5d – 2bc (d) • a + b + c = 2 -1 -3 = -1 • 3c – 2a = 3(-3) – 2(2) = -9 -4 = -13 • 5d – 2bc = 5(10) -2(-1)(-3) = 50 -6 = 44 • cd = -3 (10) = -30 = 15 • ab 2(-1) -2 • F = 9(-50) + 32 = 9(-10) + 32 = -90 + 32 = -58 • 5 • (b) F = 9(-19)/5 + 32 = -171/5 + 32 = -34.2 + 32 = -2.2

  3. Solving equations by trial and improvement (6) A solution to the equation x3 +5x – 10 = 0 lies between 1.4 and 1.5. Use the method of trial and improvement to find the solution correct to 2 decimal places. Try x = 1.45 1.453 +5(1.45) -10 = 0.2986 (too big) Try x = 1.44 1.443 +5(1.44) -10 = 0.1859 (too big) Try x = 1.43 1.433 +5(1.43) -10 = 0.074 (too big) Try x = 1.42 1.453 +5(1.45) -10 = -0.0367 (too small) Therefore answer lies between 1.42 and 1.43. Try x = 1.425 1.4253 +5(1.425) -10 = 0.0186(too big) Therefore answer is smaller than 1.425 and so X = 1.42 to 2 decimal places.

  4. Nth term of linear and quadratic sequences (9) Find the nth term of the following sequences (a) 7, 10, 13, 16, …… (b) 2, 8, 16, 26, 38, ….. • 7 10 13 16 • 3 3 3  Common difference of 3 • Therefore nth term is 3n + zero term. Zero term is +4 and so • Nth term is 3n + 4 • 2 8 16 26 38 • 6 8 10 12 • 2 2 2  Common 2nd difference • 2nd difference = 2a = 2 Therefore a=1 • 1st difference = 3a+b = 6 Therefore b = 3 • 1st term = a+b+c = 2 Therefore c = -2 • Nth term is n2 +3n -2

  5. Pythagoras’ Theorem and area of a circle (10, 16) • Diameter PQ • PQ2 = PR2 + QR2 • PQ2 = 122 + 52 • PQ2 = 144 + 25 • PQ2 = 169 • PQ = 169 = 13cm (b) Area = r2. r = 13/2 = 6.5 Therefore Area =  (6.5)2 = 132.7323cm = 132.7cm2 (1dp) (c) Circumference = d = 13 = 40.8407cm = 40.8cm (1dp)

  6. Pythagoras’ Theorem and area of a circle (10, 16) (continued) 2) The circumference of a circle is 50cm2. Calculate the radius of the circle. Circumference = 2r. Therefore r = =50/(2) = 7.0577cm = 7.06cm (2dp) (3) AC2 = AB2 + BC2 11.72 = 9.82 + BC2 BC2 = 11.72 – 9.822 BC2 = 136.89 – 96.4324 BC2 = 40.4576 BC = 6.36m (2 dp)

  7. Equation of straight lines (14) 1) What is the equation of the line that passes through the points (3, 0) and (-2, 10) ? x1=3, y1 = 0 x2=-2, y2=10 Gradient = Therefore y = -2x + c. When y=0, x = 3 Therefore 0 = -6 +c c = 6 Equation is y = -2x + 6

  8. Equation of straight lines (14) (continued) 2) A straight line PQ is parallel to y = -3x + 12. The point (6,2) lies on the line PQ. What is the equation of the line PQ? If PQ is parallel, then gradient is -3 (same as other equation). When x = 6, y = 2 Equation is y = -3x +c 2 = -18 + c c = 20 Therefore equation of PQ is y= -3x + 20

  9. Solving simultaneous equations (19) Solve the following equations by the algebraic method 1) 2x + 3y = 9 2) 4x + 2y = 17 x + 4y = 7 5x + 3y = 23 Answers 1) x = 3, y=1 2) x=2.5, y=3.5

  10. Your job today…… • Work through revision exercises (on A4 lined paper) • Complete standard form exercise (in book) • Complete past paper (on the paper). Homework – 1) Revision of Spring term work

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