5識做數學暑假作業F1-F2.

1. 解 6y-3(5-y)=21
6y - 15 +3y = 21
9y = 36
y = 4


2.如果正方形的兩條鄰邊分別長 x cm和 (2x-3) cm,求正方形的周界.
x = 2x - 3
x = 3
正方形的周界 = 4 x 3 = 12 cm



3.小明有100元,她買了7隻陶瓷小貓,其中3隻比其餘的每隻貴3元.
如果她還餘下7元,問該兩種小貓每隻各售多少元?

設兩種小貓每隻各售x及y元
3x + 4y = 100 - 7
x = y + 3
hence,
x = 15
y = 12
兩種小貓每隻各售 $12 及 $15

4.茶葉A每公斤成本是$30,茶葉B每公斤成本是$20.如果把 w 公斤的
茶葉A和 (w+1) 公斤的茶葉B混合起來,使每公斤的成本變成$24,求w.
30w + 20(w+1) = 24(2w+1)
2w = 4
w = 2



5.已知△ABC~△DEF. 如果＜E=＜F=72,DE=8, EF=5, BC=12.5
a)求＜A
＜A = 180 - 72 x 2 = 36o

b)求AB
AB / BC = DE / EF
AB = 8 x 12.5 / 5
= 20

1. 6y-3(5-y)=21
6y-15+3y=21
9y=36
y=4

2. 2x-3=x
x=3
正方形的周界
4x=4x3=12cm

3. 4x+(x+3)+(x+3)+(x+3)=100-7
7x+9=93
7x=84
x=12
該兩種小貓每隻各售 x , (x+3)
12 , 15 元

4. 30w+20(w+1)=24
30w+20w+20=24
50w=4
w=0.08

5a. ＜E=＜F=72° (given)
in△DEF
<DEF+<EFD+<EDF=180° (< SUM OF △)
<DEF+<DEF+<EDF=180°
72°+72°+<EDF=180°
144°+<EDF=180°
<EDF=36°
since <A ~ <D
＜A=36° (AAA)

5b. DE=8 (given)
SINCE AB=DE
SO AB=8 (unit)

1. 6y-3(5-y)=21
6y-15+3y=21
6y+3y=21+15
9y=36
y=36/9
y=4
2.Find x:
x=(2x-3)
x=2x-3
x-2x=-3
-x=-3
x=3
find the perimemter:3x4=12cm.
3.Let the cheaper cat 's money be x.
so the more expensive cat 's money is x+3
(7-3)x+3(x+3)+7=100
4x+3x+9=100-7
7x=93-9
7x=84
x=84/7
x=12.
So the cheaper cat 's money is $12 and the more expensive cat 's money is $15.

4.30w+20(w+1)/2w+1=24
30w+20w+20/2w+1=24
50w+20/2w+1=24(2w+1)
50w+20=48w+24
50w-48w=24-20
2w=4
w=2.
5.Find <A
<B=<E=72°(corr.<s,~△s)
<C=<F=72°(corr.<s,~△s)
<A+<B+<C=180°(< sum of △)
<A+72°+72°=180°
<A=180°-72°-72°
<A=36°.
Find AB:
AB=DE=8(corr.sides,~△s)

Thank you!