1) Tea A and tea B are mixed in the ratio x:y by weight. A costs $80/kg and B costs $100/kg. If cost of A is increased by 10% and that of B is decreased by 12%, the cost of the mixture per kg remains unchanged. Find x:y.
2) Find the range of values of x, if the lengths of the sides of the triangle are 2x, (x-1) and 5 respectively.
3) {{[開方(x^2-y^2)]+x}/{[開方(x^2+y^2)]+y}}/{{[開方(x^2+y^2)]-y}/{x-[開方(x^2-y^2)]}
4) [(開方26)+5]^98[(開方26)-5]^99
5) If [(開方5)/5-0.25]x=1, then x=?
Maths questions (20 points)
2007-01-08 10:51 pm
回答 (2)
2007-01-09 1:58 am
✔ 最佳答案
1) 80x+100y=80*(1.1)x+100*(1-0.12)y80x+100y=88x+88y
8x=12y
x:y=3:2
2)For triangles, the sum of the lengths of two sides must be larger than the length of the remaining side
Therefore,
(a) 2x+x-1 >5, 3x>6, x>2
(b) 2x+5>x-1, x>-6
(c) 5+x-1>2x, x<4
Therefore, 2<4
3) {{[開方(x^2-y^2)]+x}/{[開方(x^2+y^2)]+y}}/{{[開方(x^2+y^2)]-y}/{x-[開方(x^2-y^2)]}
={{[開方(x^2-y^2)]+x}*{x-[開方(x^2-y^2)]}/{[開方(x^2+y^2)]+y}*{[開方(x^2+y^2)]-y}}
=[x^2-(x^2-y^2)]/[(x^2+y^2)-y^2] [Using the identity (a+b)(a-b)=a^2-b^2]
=y^2/x^2
=(y/x)^2
4) [(開方26)+5]^98*[(開方26)-5]^99
=[(開方26)+5]^98*[(開方26)-5]^98*[(開方26)-5]
=(26-5^2)^98*[(開方26)-5] [Using the identity (a+b)(a-b)=a^2-b^2]
=(開方26)-5
5) [(開方5)/5-0.25]x=1
x=1/[(開方5)/5-0.25]
x=1/{[1/(開方5)]-[1/4]}
x=1/{[4-(開方5)]/[4(開方5)]}
x=[4(開方5)]/[4-(開方5)]
x=[21-8(開方5)]/9 [Rationalized]
參考: Me
2007-01-09 1:43 am
自己既功課應該自己做
收錄日期: 2021-04-13 13:47:11
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