數學呀 教我!!!好急急急急急(10)plz

Please pay attention to your TENSE next time!
(My computer points out many mistakes from your question)

2010-07-12 23:23:19 補充：
1) Ann had $100. She used part of her money to buy 7 ceramic cats, in which 3 of them are more expensive than the rest by $3. If she had $7 left, how much do the two types of ceramic cat most?

A. Let the cost of one more expensive ceramic cat be $x.

100-3x-(7-3)(x-3) = 7
100-3x-7x+21+3x-9 = 7
112-7x = 7
105 = 7x
x = 15

So, the cost of one more expensive ceramic cat is $15, the cost of one less expensive ceramic cat = 15-3 = $12

2) The cost of tea A is $30 per kg, the cost of tea B is $20 per kg. If w kg of tea A and (w+1) kg of tea B are mixed together, and the cost of the mixed tea is $24 per kg. Find w.

A. [30w+20(w+1)]/(w+w+1) = 24
50w+20 = 48w+24
2w = 4
w = 2

Solution: Total cost of tea A = 2*30 = $60
Total cost of tea B = (2+1)*20 = $60
Total weight of mixed tea = 2+2+1 = 5kg

3) Ann had a bag of glass beads. She used half of the beads to make a necklace.
She and her sister then use 1/6 and 1/7 of the original number of beads respectively to make two bracelets. If there are 16 glass beads left, find the original number of glass beads.

A. Let the original number of glass beads be x.
x/2 - x/6 - x/7 = 16
42(x/2 - x/6 - x/7) = 16*42
21x-7x-6x = 672
8x = 672
x = 84
So, the original numbers of glass beads are 84.

4) Last month, there were 30 male students and 20 female students in the class.
The number of male students decreases by r% and the number of female students increases by r% in this month. If the numbers of male and female students are the same in this month, find r.

A. 30(1-r%) = 20(1+r%)
30(1-0.01r) = 20(1+0.01r)
30-0.3r = 20+0.2r
10 = 0.5r
r = 20

5) In the figure, the length of rectangle ABCD is 3 times its width.
A (6, m +1)
B (6, 10 - 2m)
C (1 - m, 10 - 2m)
D未知
a) Find the value of m

A. m+1-(10-2m) = 3(1-m-6)
3m-9 = -3m-15
6m = -6
m = -1

b) Find the coordinates of D
D(1-m, m+1)
= D(1-(-1), (-1)+1)
= D(2, 0)

6) The area of trapezium ABCD is 25 sq.units
A 未知
B (1, 8)
C (2, 3)
D (-6, 3)
Find the length of AB

A. (AB+2-(-6))*(8-3)/2 = 25
5AB+40 = 50
AB = 2

Find the coordinates of A

A(1+2, 8)
= A(3, 8)

2010-07-12 23:24:43 補充：
For Q.7-9, I'll do it tomorrow.

2010-07-13 16:41:22 補充：
I haven't done Q.7-9...

2010-07-13 18:41:20 補充：
7) The length of the base AB of triangle ABC is two times the height.
A(a,-1) B(8,-1) C(5,4) c在最頂 b在右底 a在左底
a) Find the value of a

A. 8-a = 2(4-(-1))
8-a = 2*5
-2 = a
a = -2

b) Find the area of triangle ABC

A. area
= (8-(-2))*(4-(-1))/2
= 10*5/2
= 25sq. units

2010-07-13 18:41:34 補充：
8) If the area of triangle PQR is 21 sq.units and the coordinate of R are (4, k)
P (-4,-3) Q (3,-3). Find the two possible values of k.

A. (3-(-4))*(k-(-3))/2 = 21
7*(k+3) = 42
k+3 = 6
k = 3

(3-(-4))*(-3-k)/2 = 21
7*(-3-k) = 42
3+k = 42/-7
3+k = -6
k = -9

2010-07-13 18:41:39 補充：
9) Given two points P (a, 3) and Q (-5, b), and PQ is a vertical line.
a) Find the value of a

A. a = -5

b) If the distance between PQ is5 units, find two possible values of b.

A. 3-b = 5
b = -2

b-3 = 5
b = 8