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Question

1.Consider the sequence
6,13,21,30,a,b,c
Find the value of a,b,c.
2.There are x male rabbits and x female rabbits in a farm.Each female rabbit forms a family with one and only one male rabbit.Each family gives birth to a rabbit in a year.If there are 84 rabbits in farm the nest year,what is the value of 
x?
3.Simplify the following expressions.
(a) 6-5×a-3a
(b) -8+4a ÷2a-3a
(c) (4a-7b-6)-(5a-4b+9)

數學人 · Accepted Answer

1) T(n) = 1/2(n^2+11n)
    a = T(5) = 1/2(5^2+11(5)) = 40
    b = T(6) = 1/2(6^2+11(6)) = 51
    c = T(7) = 1/2(7^2+11(7)) = 63

2) x+x+x = 84
    x=26

3)
(a) 6-8a
(b) -8+2-3a = -3a-6
(c) 4a-7b-6-5a+4b-9 = -a-3b-15

2011-11-23 21:07:12 補充：
sor
it should be x=28

Crystal · Answer

1.
6 + 7 = 13
13 +8 = 21
21 + 9 = 30
30 + 10 = a = 40
40 + 11 = b = 51
51 + 12 = c = 63 2.
1 male + 1 female + 1 new rabbit = 3
x male + x  female + x new rabbit = 3x = 84
x = 283.
(a) 6-5xa-3a = 6 -8a = 2(3 -4a)
(b) -8+4a ÷2a-3a = -8+ 2 -3a = - 6 -3a = -3(2+a)
(c) (4a-7b-6)-(5a-4b+9) = 4a-7b-6-5a+4b-9 = -a -3b -15

數學知識交流(4)

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