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Intermediate 2 PowerPoint Presentation
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Intermediate 2

Intermediate 2

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Intermediate 2

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  1. Intermediate 2 You must learn the formulae and rules thoroughly. Some will be given to you on your examination paper, but you should only use this list to check that you have remembered the formula correctly.

  2. 2 significant figures Significant Figures 56 000 000 45 0.0062 to nearest million 3 significant figures to nearest whole 378 000 to 4 dec pl 27.3 0.005 32 4 significant figures to nearest 1000 to nearest tenth to nearest hundred thousandth 7002 to 1 dec pl 5.308 to 5 dec pl 56.0 3050 0.004 07 to 3 dec pl to nearest tenth to nearest ten to 5 dec pl

  3. Percentage used to indicate the RATE at which something is paid. Percentages Simple Interest eg 4.5% pa on £400 for 5 months. Int for 1 yr = 4.5% of 400 = 0.045 x 400 = 18 So for 5 months Interest = 18 ÷ 12 x 5 = 7.50 Compound Interest / Appreciation / Depreciation Each time the interest is added there is a new balance for the interest calculation eg compound interest on 5000 at 3%pa After 1 year Int = 3% of 500 = 15, so new Balance = 515 After 1 year Balance = 1.03 x 500 = 515 After 2nd year Balance = 1.03 x 515 = 530.45 and so on ...

  4. Appreciation / Depreciation Appreciation ~ value has increased Depreciation ~ value has decreased Calculate the new value ~ remember to add for appreciation / subtract for depreciation The value of a painting appreciated each year by 10% In 1990 it was valued at £500 000. What was its value in 1995? After each year Value = 110% of its value in the previous year. After 1 year (91) Value = 1.10 x 500 000 = 550 000 After 2nd year (92) Balance = 1.10 x 1.10 x 500 000 After 3rd year (93) Balance = 1.10 x 1.10 x 1.10 x 500 000 After 4 th year (94) Balance = 1.104x 500 000 After 5 th year (95) Balance = 1.105x 500 000 = 1.61051 x 500 000 = 805 255

  5. Appreciation / Depreciation Appreciation ~ value has increased Depreciation ~ value has decreased Car has depreciated by 8% p a Was 100% Now 100 - 8 = 92% Car was valued at £12 000 After 1 year Value = 0.92 x 12 000 After 2 year Value = 0.92 x 0.92 x 12 000 After 3 year Value = 0.923x 12 000 Original value = £400 Appreciated! By £28 Value now = £428 28 = 0.07 = 7% original 400 So the £400 item has appreciated by 7%

  6. Basic Shapes A = 1/2x b x h A = lx b Areas A = 1/2 a.b.sin C A= 1/2x d1x d2 A = π r 2

  7. Volume h V = A x h b l Prisms Where A is the area of the cross section of the prism V = l x b x h Volume of cylinder V = π r 2 h

  8. Volume Other special objects Volume of cone V = 1/3 π r 2 h Volume of sphere V = 4/3 π r 3

  9. Algebra Multiplying out brackets Factorising 3(y - 4) 7x - 21 = 3y - 12 = 7(x - 3) = x 2 + 8 x + 15 ( x + 5 ) ( x + 6 ) = x ( x + 6 ) + 5 ( x + 6 ) = ( ) ( ) x x + 5 + 3 = x 2 + 6 x + 5 x + 30 = x 2 + 11 x + 30

  10. Multiplying out brackets Factorising v2 - 49 ( x - y ) ( x + y ) = v2 - 72 = x2 - y2 = (v + 7)(v - 7) [difference of 2 squares] x2 - 3x - 108 -108 ( x - 7 ) ( x + 6 ) subtract ( ) ( ) x + 9 x - 12 = x2 - 7x + 6x - 42 1 x 108 2 x 54 = x2 - x - 42 3 x 36 NB - 7 + 6 = -1 4 x 27 6 x 18 9 x 12 +9 - 12 = -3

  11. Factorising Question 1: Is there a common factor? 6x - 9 Question 2: Is it a difference of 2 squares? x2 - 81 Question 3: Is there still 2 ?: brackets and find factors! 4 ( x2 - 18x + 81)

  12. radius Circles sector arc What fraction of …….. ? x˚ x arc sector = = 360 C A

  13. Circles Isosceles triangles symmetry x˚ Diameter / Line of symm Bisects the chord at right angles x˚

  14. Circles Tangent meets radius at right angles 90˚ angle in semicircle 90˚ is 90˚

  15. = Linear Relationships Straight Lines vertical Gradient horizontal vertical (x2,y2) horizontal (y2 - y1) gradient m = (x2 - x1) (x1,y1) Gradient is positive Gradient is negative

  16. Equation of Straight Line Need gradient m Need point on line (a, b) y - b = m ( x - a ) Points (2, 6) and (3, 10) (10 - 6) 4 m = = 4 = 4 (3 - 2) 1 • 1 Line through (2, 6) with gradient 4 y - 6 = 4 ( x - 2 ) y - 6 = 4 x - 8 y = 4 x - 2 (0, -2)

  17. Trigonometry Triangle Measure : sides and angles! S o h opposite Pythagoras Theorem c sin A = c2 = a2 + b2 a hypotenuse b Soh Cah Toa adjacent C a h cos A = sides hypotenuse sides and angles T o a opposite tan A = If the triangle has a right angle adjacent

  18. Trigonometry More triangle Measure : sides and angles! C If not right-angled then Using the Sine Rule a b a b c = = sin A sin B sin C A c B To find an angle sin A sin B sin C = = a b c

  19. Trigonometry More triangle Measure : sides and angles! C If do not know an angle and side opposite then Using the Cosine Rule a b a2 = b2 + c2 - 2.b.c cos A • A c B b 2 + c 2 - a 2 cos A = To find an angle 2.b.c

  20. A = 1/2 a b sinC Area of Triangle A = 1/2 base x height A b h • a C B

  21. Y X Simultaneous Equations Two equations in two variables are true at the same time: Graphical Solution y = 2x + 5 (2, 9) y = 5x - 1 5 Checking! y = 2 x 2 + 5 -1 x = 2 y = 4 + 5 y = 9 y = 5 x 2 - 1 y = 9 y = 10 - 1 y = 9

  22. By substitution 2x + 4y = 24 …. x -1 …. x 2 y = 2x + 5 3x + 2y = 8 y = 5x - 1 6x + 4y = 16 5x - 1 = 2x + 5 - 2x - 4y = - 24 3x - 1 = 5 Add 4x = - 8 3x = 6 x = - 2 x = 2 3x(-2) + 2y = 8 y = 2x2 + 5 2y = 14 y = 9 y = 7 x = - 2 y = 7