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Credit Maths

Credit Maths. Revision Chapters 1 - 4. Standard Form/Scientific Notation. 50 200. 5.02  10 4 =. 4.7  10 7. 47 000 000 =. 0.00031. 3.1  10 -4 =. 7.04  10 -5. 0.0000704 =. Large Numbers. a. Change to standard form. 9.9  10 4. i. 99000 =. 6.104  10 8.

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Credit Maths

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  1. Credit Maths Revision Chapters 1 - 4

  2. Standard Form/Scientific Notation 50 200 5.02  104 = 4.7  107 47000000 = 0.00031 3.1  10-4 = 7.04  10-5 0.0000704 =

  3. Large Numbers a. Change to standard form 9.9  104 i. 99000 = 6.104  108 ii. 610400000 = 5.223  106 iii. 5223000 = b. Change to normal form i. 6.03  103 = 603. 6030. 60.3 6.03 ii. 8.15  105 = 815. 815000. 8150. 81500. 81.5 8.15 iii. 2.7  106 = 270. 2700. 2700000. 27000. 270000. 27. 2.7

  4. Small Numbers a. Change to standard form 3.9  10-3 i. 0.0039 = 5.02  10- 4 ii. 0.000502 = 4  10-7 iii. 0.0000004 = b. Change to normal form i. 3.07  10-3 = 3.07 0.307 0.0307 0.00307 ii. 5.91  10-5 = 5.91 0.591 0.0591 0.00591 0.000591 0.0000591 iii. 8.6  10-6 = 8.6 0.86 0.086 0.0086 0.00086 0.000086 0.0000086

  5. 50 700 5.07  104 = 4.57  108 457000000 = 31000000 3.1  107 = 9.07  106 9070000 =

  6. 1 3 3 1 1 2 1 1 E.g. Fractions Non calc

  7. 2 5 5 1 1 4 1 1 Ex Fractions Non calc

  8. 6cm 6cm x cm 15cm 5cm y cm 2cm 8cm 5 cm Similar Shapes Enl s.f. = Red s.f. = x = 4 × 5 = 20cm y = 0.333. × 6 = 2cm

  9. 40 m 20m x m 22 m Ex 1 Find x Red s.f. = x = 0.5 × 22 = 11m

  10. 8cm 6 cm 20cm x cm Ex 2 Find x Red s.f. = x = 0.75 × 20 = 15cm

  11. 20 m 12m 2 m x m Find x to 1 d.p.

  12. Find x to 1 d.p. 20 m 22 m 12m x m Enl s.f. = x = 1.1 × 12 = 13.2m

  13. e.g.1 e.g.2 ?m2 21m2 ?m3 5m 8m 12m 60m3 9m Similar Areas and Volumes

  14. 25 cm2 144cm2 10 cm ? cm Find the missing dimension Enlargement Area Scale factor = k2 = Linear Scale Factor = k = Missing Length = 2.4 x 10 = 24 cm

  15. 1. ? cm2 15cm 30cm2 5cm Enlargement Linear Scale Factor = k = Area Scale factor = k2 = 32 = 9 Missing area = 9 x 30 = 270 cm2

  16. 2. ?mm2 80 mm2 15 mm 30mm Enlargement Linear Scale Factor = k = Area Scale factor = k2 = 22 = 4 Missing area = 4 x 80 = 320 mm2

  17. 3. ? mm2 540 mm2 6 mm 18 mm Reduction Linear Scale Factor = k = Area Scale factor = k2 = Missing area =

  18. 4. 0·5 m2 5.12 m2 ? m 7·5 m Reduction Area Scale factor = k2 = Linear Scale Factor = k = Missing Length = 0.3125 x 7.5 = 2.34 cm

  19. 5. V= x cm3 V= 20cm3 5cm 8cm Enlargement Linear Scale Factor = k = Volume Scale factor = k3 = Missing Volume = 20 x 4.096 = 81.92cm3

  20. D T S Distance Speed Time

  21. STARTER QUESTIONS a. D = 2km and S = 5m/s T= ? b. D = 272km and T = 4 hrs 15 mins S = ? c. T = 3mins and S = 4m/s D = ?

  22. 1. A bus leaves the airport at 19.24 and arrives in the city centre at 20.11. If the journey is 17 miles calculate its average speed in mph to 1 decimal place

  23. 2. Stewart drove 40 km from Laurencekirk to the golf course in Montrose. If he averaged 72km/h calculate the time of his journey to the nearest minute 3. Mr Sim drove to Dover in the holidays. He left at 06.30 and arrived at 14.15. If he averaged 70mph, how far did he travel?

  24. Positive and Negative Numbers With Algebra

  25. 1. Basic Rules: Pos and Neg = Neg Neg and Neg = Pos Examples a. 2 + (5)  ( 4) b. 4 × (2) c. 24 ÷ (6) = 8 = 4 = 2  5 + 4 = 1 Exercise a. 3  (1) + ( 7) b. 42 ÷ (7) c. 6 +(15) ÷ 3

  26. 2. Simplify: Pos and Neg = Neg Neg and Neg = Pos Examples a. 2a + (5b)  ( 4a) b. 2a2 +(3a)(4a2)(5a) = 2a2 3a + 4a2 + 5a = 2a  5b + 4a = 6a  5b = 6a2+ 2a Exercise a. 3x + (y)  (5x) b. 5m2 (3n2)(4m2)

  27. 3. Simplify: Pos and Neg = Neg Neg and Neg = Pos Multiplying makes powers Examples a. 2a × 3a × a b. 3c × (c) × (2c) × 4c × (c) = 6 a3  = 24 c5 Exercise a. 2y × 3y × y × y × y b. 2m × (5m) × (7m) × 2

  28. 4. Multiply out: Pos and Neg = Neg Neg and Neg = Pos Examples a. –3(x + 4) b. – 4(2y – 3) c. –p(3 – p) = –3x – 12 = –8y + 12 = –3p + p2 Exercise a. –5(w + 2) b. 3(2m  5) c. 3p(p  4)

  29. 5. Simplify: Pos and Neg = Neg Neg and Neg = Pos Examples a. –3(x + 2) + 8 b. 2(y + 2) + 4(2y – 3) c. 2p(4p – 3) – 3p(p – 3) = –3x – 6 + 8 = 2y + 4 + 8y – 12 = 8p2 – 6p – 3p2 + 9p = –3x + 2 = 10y – 8 = 5p2 + 3p Exercise a. 2(3c – 5) + 4(2 + c) b. 3(p – 4) – 2(p  5)

  30. 6. Evaluate: a = 3, b = -2, c = -5 Examples A) a2 – b B) abc + 2c C) b(2a – c) = 32 – (–2) = 3 × (–2) × (–5) + (–10) = –2(6 – (–5)) = 9 + 2 = 30 – 10 = –2(6 + 5) = 11 = 20 = –2(11) = –22 Exercise A) 4a + c B) c2 – b C) ab – 3b

  31. 7. Solve Examples a. 5x + 2 = x + 26 b. 2(3y – 1) = 2y + 10 c. 3m – 4 = –2(m – 4) 5x  x = 26 – 2 6y – 2 = 2y + 10 3m – 4 = –2m + 8 4x = 24 6y – 2y = 10 + 2 3m + 2m = 8 + 4 x = 6 4y = 12 5m = 12 y = 3 m = 12/5 Exercise a. 7w + 2 = w + 14 b. 3(2m – 1) = m + 16 c. 2(3c – 5) = 4(2 + c) d. 3(p – 4) = –2(p  5)

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