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Module 5 Paper 2 Higher Tier June 2008 PowerPoint PPT Presentation


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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.

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Module 5 Paper 2 Higher Tier June 2008

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Module 5 paper 2 higher tier june 2008

Module 5

Paper 2

Higher Tier

June 2008


Module 5 paper 2 higher tier june 2008

Area: ½ x 10.4 x 5.5

= 28.6 cm2

2 Marks


Module 5 paper 2 higher tier june 2008

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


Module 5 paper 2 higher tier june 2008

Y= 2

2 Marks


Module 5 paper 2 higher tier june 2008

1 Mark


Module 5 paper 2 higher tier june 2008

1 Mark

Sum: 25 + 33 + 34 + 35 = 127


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 number

Even + Odd = Odd

6 marks


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


Module 5 paper 2 higher tier june 2008

5d = c - 2

2 Marks


Module 5 paper 2 higher tier june 2008

Need to extend line so it goes from x =-1 to x=5

X

X

X

5 Marks

x = 3.3

y = 1.7


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


Module 5 paper 2 higher tier june 2008

Equidistant from P and Q

3 marks


Module 5 paper 2 higher tier june 2008

Answer : x4

Answer: 6y7z5

Answer: 8p9r6

5 Marks


Module 5 paper 2 higher tier june 2008

OPP

Use TAN

Adj

A = Tan-1 0.6875

= 34.5⁰

3 Marks


Module 5 paper 2 higher tier june 2008

Adj

Hyp

Use Cosine

DF = 24 x cos 64

DF = 10.5cm

DF = 10.5(209.....)

4 Marks


Module 5 paper 2 higher tier june 2008

30 x 1.20 = 36

15 X 1.10 = 16.5

3 Marks

New area = 36 x 16.5 = 594cm2


Module 5 paper 2 higher tier june 2008

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

Answer: 3 ( y2 - 4z2 )

= 3 ( y - 2z ) ( y + 2z)

5 Marks


Module 5 paper 2 higher tier june 2008

Use Sine Rule

YZ = 14.65cm (14.7cm)

3 Marks


Module 5 paper 2 higher tier june 2008

MC = OA (=a) –equal and parallel

Hence AC = OM (and parallel)

If opposite sides are equal and parallel then OACM is a parallelogram

2 Marks


Module 5 paper 2 higher tier june 2008

Volume of cone: ⅓ πr2h = 2400

r2 = 81.85

r = 9.04cm (9.05cm)

3 Marks


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

x = 1 or x = 8

Factorise

5 Marks

( x - 1 ) ( x – 8 ) = 0


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