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 PresentationTRANSCRIPT
Module 5Paper 2Higher TierJune 2008
Area: ½ x 10.4 x 5.5
= 28.6 cm2
2 Marks
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
Y= 2
2 Marks
1 Mark
Sum: 25 + 33 + 34 + 35 = 127 1 Mark
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
Exterior angle of regular Octagon = 360 ÷ 8 = 45⁰
Exterior angle of regular Pentagon = 360 ÷ 5 = 72⁰
Angle a = 45 + 72 = 117⁰ 4 Marks
5d = c - 2
5
2c
d
2 Marks
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
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
Equidistant from P and Q3 marks
Answer : x4
Answer: 6y7z5
Answer: 8p9r6
5 Marks
OPP
Adj
Use TAN
6875.016
11TanA
A = Tan-1 0.6875 = 34.5⁰
3 Marks
Adj Hyp
Use Cosine
2464cos
DF
DF = 24 x cos 64
DF = 10.5(209.....)
DF = 10.5cm
4 Marks
30 x 1.20 = 36
15 X 1.10 = 16.5
New area = 36 x 16.5 = 594cm23 Marks
Answer: ( 5x + 1 ) ( x + 7 )
Answer: 3 ( y2 - 4z2 ) = 3 ( y - 2z ) ( y + 2z)
5 Marks
Use Sine Rule
102sin
25
35sin
YZ
35sin102sin
25YZ
YZ = 14.65cm (14.7cm)
3 Marks
MC = OA (=a) –equal and parallelHence AC = OM (and parallel)If opposite sides are equal and parallel then OACM is a parallelogram
2 Marks
Volume of cone: ⅓ πr2h = 2400
28
324002
r
r2 = 81.85
r = 9.04cm (9.05cm) 3 Marks
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