2. Announcing the release of VERSION 6 This Demo shows just 20 of the 10,000 available slides and takes 7 minutes to run through. Please note that in the proper presentations the teacher controls every movement/animation by use of the mouse/pen. Click when ready  750 MB of Presentations750 MB of Presentations 500+ files/10 000+ Slides500+ files/10 000+ Slides New Functional Maths ElementNew Functional Maths Element 100’s of Worksheets100’s of Worksheets 100’s of teacher Q + A Sheets100’s of teacher Q + A Sheets 1000’s of Teaching Hours1000’s of Teaching Hours 1200 Interactive SAT/GCSE Boosters1200 Interactive SAT/GCSE Boosters Huge Enrichment AreaHuge Enrichment Area Database of PresentationsDatabase of Presentations

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4. Drawing the graph of a quadratic function? This is a short demo that auto-runs.

5. 012 3 4 -2 1 2 3 4 5 6 7 8 -2 -3 -4 -5 -6 -7 -8 5 y -3 -9 x y = x 2 - 2x - 8 x -3-2 012345 x 2 -2x -8 y 94 101 4 9 1625 6 4 20-2 -4 -6-8 -10 -8 7 0 -5 -8 -9 -8 -5 07 LoS Equation of Line of symmetry is x = 1 Drawing quadratic graphs of the form y = ax 2 + bx + c Example 1. Minimum point at (1, -9)

6. Look at graphs of some trig functions?

7. sinx + circle 90 o 180 o  0o0o 270 o 1 The Trigonometric Ratios for any angle 0 90 180 360270-90 -180 -270 -360 0 90180 270 -90 -180 -270 -360360 450 o 0o0o 90 o 180 o 270 o 360 o -90 o -180 o -270 o -360 o -450 o 360 o 0o0o 90 o 180 o 270 o

8. x 270 360 90 -360 180 y = f(x) 0 -90 -180 -270 1 2 -2 f(x) = cosx f(x) = cos2x f(x) = cos3x f(x) = cos ½ x

9. Introducing addition of fractions with different denominators?

10. + + = 2015 1612 9 86 43 Multiples of 3 and 4 12 is the LCM Equivalent

11. Probability for Dependent Events

12. Conditional Probability: Dependent Events When events are not independent, the outcome of earlier events affects the outcome of later events. This happens in situations when the objects selected are not replaced.

13. A box of chocolates contains twelve chocolates of three different types. There are 3 strawberry, 4 caramel and 5 milk chocolates in the box. Sam chooses a chocolate at random and eats it. Jenny then does the same. Calculate the probability that they both choose a strawberry chocolate. Conditional Probability: Dependent Events P(strawberry and strawberry) =3/12 x

14. A box of chocolates contains twelve chocolates of three different types. There are 3 strawberry, 4 caramel and 5 milk chocolates in the box. Sam chooses a chocolate at random and eats it. Jenny then does the same. Calculate the probability that they both choose a strawberry chocolate. Conditional Probability: Dependent Events P(strawberry and strawberry) =3/12 x 2/11 = 6/132 (1/22)

15. Enlarge on object?

16. D To enlarge the kite by scale factor x3 from the point shown. Centre of Enlargement Object A B C Image A/A/ B/B/ C/C/ D/D/ Enlargements from a Given Point 1. Draw the ray lines through vertices. 2. Mark off x3 distances along lines from C of E. 3. Draw and label image. No Grid 2

17. Do some Loci?

18. Q2 Loci (Dogs and Goats) Scale:1cm = 3m Shed Wall A B 1. Draw arc of circle of radius 5 cm 2. Draw ¾ circle of radius 4 cm 3. Draw a ¼ circle of radius 1 cm 4. Shade in the required region. Billy the goat is tethered by a 15m long chain to a tree at A. Nanny the goat is tethered to the corner of a shed at B by a 12 m rope. Draw the boundary locus for both goats and shade the region that they can both occupy.

19. Investigate some Properties of Pascal’s Triangle

20. Pascal’s Triangle 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 33 464 5510 6156 20 7 721 35 88 2856285670 9 3684 126 84 36 9 10 45 210 252 210 120 45 120 10 11 55 330 462 330 165 55 1112 66 495 792924 792 495 220 66 12 1716 13 78 286 715 1287 1716 1287 715 286 78 13 Pascal’s Triangle 1. Complete the rest of the triangle. Blaisé Pascal (1623-1662) Counting/Natural Numbers Triangular Numbers Tetrahedral Numbers Pyramid Numbers (square base)

21. Fibonacci Sequence 1716 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 33 464 5510 6156 20 7 721 35 88 2856285670 9 3684 126 84 36 9 10 45 210 252 210 120 45 120 10 11 55 330 462 330 165 55 11 12 66 495 792924 792 495 220 66 12 13 78 286 715 1287 1716 1287 715 286 78 13 Add the numbers shown along each of the shallow diagonals to find another well known sequence of numbers. 1 23813 21 5589 15 34 144 233 377 The sequence first appears as a recreational maths problem about the growth in population of rabbits in book 3 of his famous work, Liber – abaci (the book of the calculator ). Fibonacci travelled extensively throughout the Middle East and elsewhere. He strongly recommended that Europeans adopt the Indo-Arabic system of numerals including the use of a symbol for zero “zephirum” The Fibonacci Sequence Leonardo of Pisa 1180 - 1250

22. 8 7 6 5 4 9 3 2 1 27 26 25 24 23 28 22 21 20 29 17 16 15 14 13 18 12 11 10 19 37 36 35 34 33 38 32 31 30 39 47 46 45 44 43 48 42 41 40 49 National Lottery Jackpot? 49 balls choose 6 Choose 6Row 49 13 983 816 There are 13 983 816 ways of choosing 6 balls from a set of 49. So buying a single ticket means that the probability of a win is 1/13 983 816 49 C 6 Row 0

23. The Theorem of Pythagoras?

24. 625 576 49 7 2 + 24 2 = 25 2 49 + 576 = 625 7 24 25 A 3 rd Pythagorean Triple 7, 24, 25 In a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides.

25. Perigal’s Dissection The Theorem of Pythagoras: A Visual Demonstration In a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides. Draw 2 lines through the centre of the middle square, parallel to the sides of the large square This divides the middle square into 4 congruent quadrilaterals These quadrilaterals + small square fit exactly into the large square Henry Perigal (1801 – 1898) Gravestone Inscription

26. Look at one of the 6 proofs of the Theorem from the Pythagorean Treasury.

27. We first need to show that the angle between angle x and angle y is a right angle. This angle is 90 o since x + y = 90 o (angle sum of a triangle) and angles on a straight line add to 180 o Take 1 identical copy of this right-angled triangle and arrange like so. Area of trapezium = ½ (a + b)(a + b) = ½ (a 2 +2ab + b 2 ) Area of trapezium is also equal to the areas of the 3 right-angled triangles. = ½ ab + ½ ab + ½ c 2 So  ½ (a 2 +2ab + b 2 ) = ½ ab + ½ ab + ½ c 2  a 2 +2ab + b 2 = 2ab + c 2 2 + b 2 = c 2 QED a b c xoxo yoyo a b c xoxo yoyo Draw line:The boundary shape is a trapezium To prove that a 2 + b 2 = c 2 President James Garfield’s Proof (1876)

28. Sample some material from the Golden section presentation.

29. Constructing a Golden Rectangle. 1. Construct a square and the perpendicular bisector of a side to find its midpoint p. 3. Set compass to length PM and draw an arc as shown. 2. Extend the sides as shown. L M 4. Construct a perpendicular QR. O N Q THE GOLDEN SECTION LQRO is a Golden Rectangle. 1 P R

30. THE GOLDEN SECTION Johannes Kepler 1571- 1630 "Geometry has two great treasures: one is the Theorem of Pythagoras, and the other the division of a line into extreme and mean ratio; the first we may compare to a measure of gold, the second we may name a precious jewel."

31. Or just simply ride your bike!

32. Wheels in Motion Wheel The Cycloid It’s true! The point at the bottom of a moving wheel is not moving! Choose Order Forms/New for V5 to view latest material and other catalogues.

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