mathematics f.2 - f.3 quick .

1.
let X be the weight of Rachel,
Y be the weight of Maggie.
X+Y=88-------1
X=(6/5)xY-------2
from 2:
5X-6Y=0---------3
1x5:
5X+5Y=440--------4
4-3:
11Y=440
Y=40
sub Y=40 into 1
X=48

2.
100x(-q/5q^6)^3
=100x(-q/5)^3{1/(q ^6)}^3
=(100/125)x(-q^3)x{1/(q^18)}
= - (4/5){1/(q^15)}

1)Let Rachel is R kg . Maggie is M kg:

R+M=88...(1)
6/5 * R = M...(2)

from(2):
R = 5M/6 ,sub it to(1):

5M/6 + M = 88

M(1+5/6) = 88

M = 48

So the weight of Maggie is 48kg

2)100x(-q/5q^6)^3

=100x [(-1) / 5q^5 ]^3

= -100x / 125q^15

= - 4x / 5q^15

2009-03-04 21:29:43 補充：
In Q.2, if x is 「*」,the answer = - 4 / 5q^15

2009-03-04 21:38:40 補充：
vin_orzv_v :
是7 x 2^n+1 - 2 x 2^n+1 / 2^2(n+1) 還是

(7 x 2^n+1 - 2 x 2^n+1) / 2^2(n+1) ?

1) Let x be the weight of Maggie.
6/5x + x = 88
6x + 5x = 440
11x = 440
x = 40
Maggie's weight is 40kg

Or

Let x be the weight of Maggie.
Let y be the weight of Rachel.
6/5x=y …1
x + y =88 …2
Sub. 6/5x = y into 2
x + 6/5x = 88
x = 40

2) no solution.
because the denominator cannot be the unknown number.(q)