Download presentation
Presentation is loading. Please wait.
Published byAnnice Shelton Modified over 9 years ago
1
Whiteboardmaths.com © 2004 All rights reserved 5 7 2 1
2
BBBB EEEE DDDD BBBB DDDD CCCC EEEE BBBB DDDD CCCC DDDD CCCC EEEE BBBB CCCC EEEE ALGEBRA NUMBER SHAPE SPACE & M SPACE & M HANDLING DATA
3
Back to board Answer What fraction of the rectangle is shaded green? (In simplest form)
4
Back to board What fraction of the rectangle is shaded green? (In simplest form) Explain? 1 3
5
Back to board What fraction of the rectangle is shaded green? (In simplest form) 1 3 = 4 12 4 out of 12 are green 1 3 Divide top and bottom by 4 (Highest common factor) 1 3
6
Back to board Answer What is 2 3 x 3 4
7
Back to board What is 2 3 x 3 4 1 2 Explain?
8
Back to board What is 2 3 x 3 4 1 2 6 12 1 2 = Multiplying numerators and denominators, then simplifying gives:
9
Back to board Answer Write the number below as a product of its prime factors. 60
10
Back to board Write the number below as a product of its prime factors. 60 2 2 x 3 x 5
11
Back to board Answer The mass of the Earth 6 x 10 24 kg. Jupiter is about 400 times heavier than the Earth. Write down an estimate for mass of Jupiter in standard form.
12
Back to board Explain? The mass of the Earth 6 x 10 24 kg. Jupiter is about 400 times heavier than the Earth. Write down an estimate for mass of Jupiter in standard form. 2.4 x 10 27 kg
13
Back to board The mass of the Earth 6 x 10 24 kg. Jupiter is about 400 times heavier than the Earth. Write down an estimate for mass of Jupiter in standard form. 2.4 x 10 27 kg 4 x 10 2 x 6 x 10 24 = 4 x 6 x 10 26 = 24 x 10 26 = 2.4 x 10 1 x 10 26 = 2.4 x 10 27
14
Back to board Answer 2 5 1 20 7 8 6 3 4 3n - 7 Input Output Calculate the output values for the given rule.
15
Back to board 2 5 1 20 7 8 6 3 4 3n - 7 Input Output Calculate the output values for the given rule. 8 - 4 53 14 17 11 2 5
16
Back to board Answer Find a whole number solution to : n 2 + 2n + 3 = 66
17
Back to board 7 Find a whole number solution to : n 2 + 2n + 3 = 66 Explain?
18
Back to board 7 Find a whole number solution to : n 2 + 2n + 3 = 66 7 2 + 2 x 7 + 3 = 66
19
Back to board Answer Expand: ( x + 3 )( x + 4 )
20
Back to board Explain? Expand: ( x + 3 )( x + 4 ) x 2 + 7x + 12
21
Back to board Expand: ( x + 3 )( x + 4 ) x 2 + 7x + 12 x x x = x 2 3 x 4 = 12 4x + 3x = 7x By inspection Or by single brackets x(x + 4) + 3(x + 4) = x 2 + 4x + 3x + 12
22
Back to board Answer V = 7p 2 +4p - 8 Evaluate V when p = - 3.
23
Back to board Explain? V = 7p 2 +4p - 8 Evaluate V when p = - 3. 43
24
Back to board V = 7p 2 +4p - 8 Evaluate V when p = - 3. 43 V = 7(- 3) 2 +4(-3) - 8 V = 7 x 9 - 12 - 8 V = 63 - 20
25
Back to board Answer 5 ½ m 2 m Calculate: (a) the area of the piece of wood (b) the perimeter of the piece of wood
26
Back to board 5 ½ m 2 m Calculate: (a) the area of the piece of wood (b) the perimeter of the piece of wood Area = 2 x 5 ½ = 11 m 2 Perimeter = 2 + 5 ½ + 2 + 5 ½ = 15 m
27
Back to board Answer Find the area of the triangle. 8 cm 7.5 cm
28
Back to board 30 cm 2 Find the area of the triangle. 8 cm 7.5 cm Explain?
29
Back to board 30 cm 2 Find the area of the triangle. 8 cm 7.5 cm 8 x 7.5 2 A = = 30 cm 2
30
Back to board Answer Find the length of the hypotenuse. 5 cm 12 cm x Not to scale
31
Back to board Explain? Find the length of the hypotenuse. 5 cm 12 cm x 13 cm Not to scale
32
Back to board 13 cm Find the length of the hypotenuse. 5 cm 12 cm x x 2 = 5 2 + 12 2 (Pythagoras) x = (5 2 + 12 2 ) = 169 = 13 Not to scale
33
Back to board Answer One of the formulae shown is for the surface area of a cylinder. Use dimensions to show which one it is. r h
34
Back to board Explain? One of the formulae shown is for the surface area of a cylinder. Use dimensions to show which one it is. r h
35
Back to board One of the formulae shown is for the surface area of a cylinder. Use dimensions to show which one it is. r h length x length = Area
36
Back to board One counter is drawn from the bag at random. What is the probability that it is blue? Answer
37
Back to board 3/12 = (1/4) One counter is drawn from the bag at random. What is the probability that it is blue? Explain?
38
Back to board 3/12 = (1/4) One counter is drawn from the bag at random. What is the probability that it is blue? 3 of the counters are blue out of a total of 12 counters.
39
Back to board Answer f 5 10 15 0 1020 30 40 50 60 70 Height (cm) Heights of Flowers How many flowers are between 20 and 50 cm tall?
40
Back to board f 5 10 15 0 1020 30 40 50 60 70 Height (cm) Heights of Flowers How many flowers are between 20 and 50 cm tall? 27 Explain?
41
Back to board f 5 10 15 0 1020 30 40 50 60 70 Height (cm) Heights of Flowers How many flowers are between 20 and 50 cm tall? 27 8 6 13
42
Back to board Answer RedBlueWhiteSilverBlack 2017153810 Prof did a survey about the colour of cars passing underneath a motor way bridge. He recorded the information in the table below. (a) Estimate the probability that the next car to pass underneath the bridge will be silver. (b) How many red cars would you expect if 300 cars passed underneath the bridge?
43
Back to board RedBlueWhiteSilverBlack 2017153810 Prof did a survey about the colour of cars passing underneath a motor way bridge. He recorded the information in the table below. (a) Estimate the probability that the next car to pass underneath the bridge will be silver. (b) How many red cars would you expect if 300 cars passed underneath the bridge? Explain? (a) 38/100 = 0.38 (b) 60
44
Back to board RedBlueWhiteSilverBlack 2017153810 Prof did a survey about the colour of cars passing underneath a motor way bridge. He recorded the information in the table below. (a) Estimate the probability that the next car to pass underneath the bridge will be silver. (b) How many red cars would you expect if 300 cars passed underneath the bridge? (a) 38/100 = 0.38 (b) 60 (a) Since the table shows a total of 100 cars and 38 are silver, we can use this relative frequency, 38/100 as an estimate for the probability of the next car being silver. (b) 20/100 = 1/5 and 1/5 of 300 = 60.
45
Back to board Answer Peter and Rebecca each have a bag containing red and blue beads as shown. They each remove a bead at random from their bags. Peter selects his bead first. What is the probability that both beads will be red?
46
Back to board Explain? Peter and Rebecca each have a bag containing red and blue beads as shown. They each remove a bead at random from their bags. Peter selects his bead first. What is the probability that both beads will be red? 18/72 = 1/4
47
Back to board Peter and Rebecca each have a bag containing red and blue beads as shown. They each remove a bead at random from their bags. Peter selects his bead first. What is the probability that both beads will be red? 18/72 = 1/4 P(red and red) = 3/8 x 6/9 = 18/72
Similar presentations
© 2025 SlidePlayer.com. Inc.
All rights reserved.