急!!!!!!!!!!!!F.5 MATH

24a)Value of car A after n years: $120000 ( 1 – 10% )n-1
= $120000 ( 0.9 )n-1
Value of car B after n years: $230000 ( 1 – 20% )n-1
= $ 230000 ( 0.8 )n-1
[ As the values given are after 1 year, so the values after n years will be ( n – 1 ) years after the 1st year.]
b) $120000 ( 0.9 )n-1 > $ 230000 ( 0.8 )n-1
( 0.9 / 0.8 )n-1 > 230000 / 120000
1.125n-1 > 23 / 12
( 1.125n )( 1.125-1 ) > 23 / 12
1.125n > 2.15625
log 1.125n > log 2.15625
n log 1.125 > log 2.15625
n > log 2.15625 / log 1.125
n > 6.52 ( cor. to 3 s.f. )
Hence the smallest value of n is 7.
26a)Suppose the length of AB is a, then BC = a + d and CD = a + 2d, so
a + a + d + a + 2d = 24
a + d = 8
a = 8 – d --- ( 1 )
Sum of areas = 105cm2, so
a2 / 2 + ( a + d )2 / 2 + ( a + 2d )2 / 2 = 105 --- ( 2 )
Put ( 1 ) into ( 2 ),
( 8 – d )2 / 2 + ( 8 – d + d )2 / 2 + ( 8 – d + 2d )2 / 2 = 105
( 8 – d )2 + 64 + ( 8 + d )2 = 210
64 – 16d + d2 + 64 + 16d + d2 = 146
2d2 = 18
d2 = 9
d = 3 or –3 ( rejected )
a = 8 – 3 = 5
So BC = a + d = 5 + 3 = 8cm
b)The required difference is actually the common difference, i.e. 3cm.
c)The lengths of the triangle are in the AP:
T ( n ) = 5 + ( n – 1 )( 3 )
= 2 + 3n
T ( 10 ) = 2 + 3 ( 10 ) = 32
Hence the area of P10: ( 32 )( 32 ) / 2 = 512cm2
27ai)The length of A1B: 3a ( 2 / 3 ) = 2a cm
The length of BB1: 3a ( 1 / 3 ) = a cm
So the length of A1B1: √[ a2 + ( 2a )2] ( Pyth. Theorem )
= √5 a cm
ii)The length of A2B1: √5 a ( 2 / 3 ) = 2√5 a / 3 cm
The length of B1B2: √5 a ( 1 / 3 ) = √5 a / 3 cm
Then by Pyth. Theorem again, A2B2 = 5a / 3 cm
iii)As in I) and ii), the length of A3B2 = 10a / 9 cm, the length of B2B3 = 5a / 9 cm
and A3B3 = 5√5 a / 9 cm
b)The lengths of the square are in G.P.:√5 a, 5a / 3, 5√5 a / 9 …
Then their areas are in G.P. as well: 5a2, 25a2 / 9, 125a2 / 9…
The common ratio = 5a2 / 9a2 = 5 / 9
So AnBnCnDn = arn-1
= ( 5a2 )( 5 / 9 )n-1
= ( 5a2 )( 5 / 9 )n ( 9 / 5 )
= 9a2 ( 5n ) / 9n
= ( 5n )a2 / 9n-1
c)If CD = 6cm,
3a = 6
a = 2
So A10B10C10D10 = ( 510 )( 22 ) / 910-1 = 0.10cm2 ( cor. to 2 s.f. )