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Question

1.There are y apple in a box.The weight of the box of apples is 6060 g.It is given that the box weights 300g when it is is empty.If each apple weights 160g,find the value of y.
2.N is a 3-digit number.It is given that N=970+a,where a is an integer between 0 and 9.
(a) If the order of the first and the third digits of the number is reversed to form a new 3-digit number,find its new value in terms of a.
(b) Hence,find the difference between the two 3-digit numbers in terms of a.
(c) Hence,find the value of a if the difference between the two 3-dight numbers 
is 594.
3.Given A=1/2(a+b)h.If A=48,b=12 and h=6,find the value of a.
4.Peter has $k and he spends half of his money to buy a toy car.David has $10 fewer than four times the money that Peter has.If David spends all his money to buy the toy car,how many toy cars can David buy?
A.  (2k-20)÷k
B.  (4k-10)÷k
C.  (4k-20)÷k
D.  (8k-20)÷k
5.Expand and simplify the following expressions.
(a) 4(a+2b-5)
(b) (a-b)(a-2b)(a-ab+b)

一個青檸 · Accepted Answer

您好，我是 lop，高興能解答您的問題。

(1)

160y + 300 = 6060
160y = 6060 - 300
160y = 5760
y = 5760 ÷ 160
y = 36

Answer : The value of y is 36 .
 
(2a)

Let N = 100x + 10y + z which x , y , z are integers between 0 and 9 ,

then if N = 970 + a which 0 ≦ a ≦ 9 then 970 ≦ N ≦ 979 ,

x = 9 , y = 7 , z = a

Then its new value ( N[2] ) = its old value ( N[1] ) - 100x + 100z - z + x = 970 + a - 900 + 100a - a + 9 = 100a + 79

Answer : Its new value is 100a + 79 .
 
(2b)

The difference is :

970 + a - ( 100a + 79 )
= 970 + a - 100a - 79
= 891 - 99a

Answer : The difference is 891 - 99a .
 
(2c)

891 - 99a = 594
-99a = 594 - 891
-99a = -297
a = -297 ÷ -99
a = 3

Answer : The value of a is 3 .
 
(3)

Put A = 48 , b = 12 , h = 6 into

A =(a+b)h/2
48 = (a+12)6/2
48 = 3(a+12)
48 = 3a + 36
48 - 36 = 3a
3a = 12
a = 12 ÷ 3
a = 4

Answer : The value of a is 4 .

(4)

David has :

4k/2 - 10
= $(2k-10)

David can buy :

(2k-10) ÷ (k/2)
= 2(2k-10) ÷ k
= (4k-20) ÷ k
 
Answer : C
 
(5a)

4(a+2b-5)
= 4(a) + 4(2b) - 4(5)
= 4a + 8b - 20

(5b)

(a-b)(a-2b)(a-ab+b)
= [a(a-2b)-b(a-2b)](a-ab+b)
= (a²-2ab-ab-2b²)(a-ab+b)
= (a²-3ab-2b²)(a-ab+b)
= a²(a-ab+b) - 3ab(a-ab+b) - 3b²(a-ab+b)
= a³- a³b + a²b - 3a²b - 3a²b²+ 3ab²- 3ab²- 3ab³+ 3b³
= a³- a³b - 2a²b - 3a²b²- 3ab³+ 3b³

2011-12-04 18:51:03 補充：
corr. 001

(4)

David has :

$4k - 10

David can buy :

(4k-10) ÷ (k/2)
= 2(4k-10) ÷ k
= (8k-20) ÷ k

(5)

a^3 - 2ba^2 - ba^3 - ab^2 + 3(a^2)(b^2) + 2b^3 - 2ab^3

? · Answer

1 to 3) see others.

4.
Car's value = k/2
David has = 4k - 10
David can buy = (4k-10) / (k/2)
= (8k - 20) / k => D

5a.
4a + 8b - 20
5b.
(a-b)(a-2b)(a-ab-b)
= (a^2 -3ab + 2b^2)(a - ab - b)
= a(a^2 - 3ab + 2b^2) - ab(a^2 - 3ab + 2b^2) - b(a^2 - 3ab + 2b^2)
= a^3 - 3(a^2)b + 2a(b^2) - (a^3)b + 3(a^2)(b^2) - 2a(b^3) - (a^2)b + 3a(b^2) - 2b^3
= a^3 - (a^3)b + 3(a^2)(b^2) - 4(a^2)b + 5a(b^2)- 2a(b^3) - 2b^3

數學知識交流(5)

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