Simultaneous Equations:
1. There are some birds and dogs in the pet shop. If there are 36 heads and 128 legs in total, how many birds and dogs does the pet shop have?
不等式:
1.若三個連續的偶數之和小於270，求這三個偶數的最大可能值。
2.小明有40張紙幣，其中只有$20紙幣和$50紙幣兩種。若其總值最少等於$1000，問小明最少有多少張$50紙幣?
please~
數學不等式+Simultaneous Equations 急
2012-12-26 9:19 pm
回答 (3)
2012-12-26 9:35 pm
✔ 最佳答案
1 Let there are x birds and y dogsx + y = 36...(1)
2x + 4y = 128...(2)
Sub. (1) into (2)
2x + 4(36 - x) = 128
-2x + 144 = 128
x = 8
y = 28
There are 8 birds and 28 dogs
1 令最小的偶數是x
x + (x + 2) + (x + 4) < 270
3x + 6 < 270
x < 88
因此三個偶數的最大可能值是86,88,90
2 設有x張$20紙幣和y張$50紙幣
x + y = 40...(1)
20x + 50y >= 1000...(2)
代(1) 入 (2)
20(40 - y) + 50y >= 1000
800 + 30y >= 200
y >= 200/30 = 6.67
小明最少有7張$50紙幣
2012-12-28 1:47 am
1.
Let x and y be the number of birds and dogs respectively.
x+y=36---------(1)
2x+4y=128----(2)
2*(1) 2x+2y=72---(3)
(2)-(1) (2x+4y)-(2x+2y)=128-72
2y=56
y=28
Sub.y=28 into (1)
x+(28)=36
x=8
∴The pet shop has 8 birds and 28 dogs.
1.86,88,90
2.7
Let x and y be the number of birds and dogs respectively.
x+y=36---------(1)
2x+4y=128----(2)
2*(1) 2x+2y=72---(3)
(2)-(1) (2x+4y)-(2x+2y)=128-72
2y=56
y=28
Sub.y=28 into (1)
x+(28)=36
x=8
∴The pet shop has 8 birds and 28 dogs.
1.86,88,90
2.7
2012-12-26 9:36 pm
1. b+d=36，2b+4d=128
2. 66, 68, 134
3. t+f=40，20t+50f=1000
2. 66, 68, 134
3. t+f=40，20t+50f=1000
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