Module 5 Paper 2 Higher Tier June 2008

21
Module 5 Paper 2 Higher Tier June 2008

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Module 5 Paper 2 Higher Tier June 2008. Area: ½ x 10.4 x 5.5. = 28.6 cm 2. 2 Marks. 5x - 3x = 7 + 4. 2x = 11. x = 5.5. 13 - 5y = 12. 13 – 12 = 5y. 5y = 1. 13 - 5y = 4 x 3. 6 Marks. y = ⅕ or 0.2. Y= 2. 2 Marks. 1 Mark. 1 Mark. - PowerPoint PPT Presentation

Transcript of Module 5 Paper 2 Higher Tier June 2008

Page 1: Module 5 Paper 2 Higher Tier June 2008

Module 5Paper 2Higher TierJune 2008

Page 2: Module 5 Paper 2 Higher Tier June 2008

Area: ½ x 10.4 x 5.5

= 28.6 cm2

2 Marks

Page 3: Module 5 Paper 2 Higher Tier June 2008

5x - 3x = 7 + 4

2x = 11

x = 5.5

13 - 5y = 4 x 3

13 - 5y = 12

13 – 12 = 5y

5y = 1

y = ⅕ or 0.26 Marks

Page 4: Module 5 Paper 2 Higher Tier June 2008

Y= 2

2 Marks

Page 5: Module 5 Paper 2 Higher Tier June 2008

1 Mark

Page 6: Module 5 Paper 2 Higher Tier June 2008

Sum: 25 + 33 + 34 + 35 = 127 1 Mark

Page 7: Module 5 Paper 2 Higher Tier June 2008

n + 8 n + 9 n + 10

n + n + 8 + n + 9 + n + 10 = 4n + 27

4n is always even.27 is an odd numberEven + Odd = Odd 6 marks

Page 8: Module 5 Paper 2 Higher Tier June 2008

Exterior angle of regular Octagon = 360 ÷ 8 = 45⁰

Exterior angle of regular Pentagon = 360 ÷ 5 = 72⁰

Angle a = 45 + 72 = 117⁰ 4 Marks

Page 9: Module 5 Paper 2 Higher Tier June 2008

5d = c - 2

5

2c

d

2 Marks

Page 10: Module 5 Paper 2 Higher Tier June 2008

x 1 2 4

y -3 -1 3

X

X

X Need to extend line so it goes from x

=-1 to x=5

x = 3.3 y = 1.75 Marks

Page 11: Module 5 Paper 2 Higher Tier June 2008

A

B Area of A is 3 x 2 = 6cm2

Area of B is 9 x 2 = 18cm2

Area of cross section is 18 + 6 = 24cm2

Volume of block is area of cross section x length:24 x 65 = 1560 cm3 5 Marks

Page 12: Module 5 Paper 2 Higher Tier June 2008

Equidistant from P and Q3 marks

Page 13: Module 5 Paper 2 Higher Tier June 2008

Answer : x4

Answer: 6y7z5

Answer: 8p9r6

5 Marks

Page 14: Module 5 Paper 2 Higher Tier June 2008

OPP

Adj

Use TAN

6875.016

11TanA

A = Tan-1 0.6875 = 34.5⁰

3 Marks

Page 15: Module 5 Paper 2 Higher Tier June 2008

Adj Hyp

Use Cosine

2464cos

DF

DF = 24 x cos 64

DF = 10.5(209.....)

DF = 10.5cm

4 Marks

Page 16: Module 5 Paper 2 Higher Tier June 2008

30 x 1.20 = 36

15 X 1.10 = 16.5

New area = 36 x 16.5 = 594cm23 Marks

Page 17: Module 5 Paper 2 Higher Tier June 2008

Answer: ( 5x + 1 ) ( x + 7 )

Answer: 3 ( y2 - 4z2 ) = 3 ( y - 2z ) ( y + 2z)

5 Marks

Page 18: Module 5 Paper 2 Higher Tier June 2008

Use Sine Rule

102sin

25

35sin

YZ

35sin102sin

25YZ

YZ = 14.65cm (14.7cm)

3 Marks

Page 19: Module 5 Paper 2 Higher Tier June 2008

MC = OA (=a) –equal and parallelHence AC = OM (and parallel)If opposite sides are equal and parallel then OACM is a parallelogram

2 Marks

Page 20: Module 5 Paper 2 Higher Tier June 2008

Volume of cone: ⅓ πr2h = 2400

28

324002

r

r2 = 81.85

r = 9.04cm (9.05cm) 3 Marks

Page 21: Module 5 Paper 2 Higher Tier June 2008

Multiply every term by x ( x – 2 )

4 ( x – 2 ) + 3x = x ( x – 2 )

4x - 8 + 3x = x2 - 2x

Collect terms together on one side and simplify

x2 - 9x + 8 = 0

Factorise

( x - 1 ) ( x – 8 ) = 0

x = 1 or x = 8

5 Marks