plz help me to answer the question,If you do not know the answer,plz answer no.I need the step~Thank for your help~
1) -(10/2+x)-(5x/x+2)
2)Given the formula q= r(x-1)-(x/p),find the value od x when r= -4,p= -3,q= -2.
3)In each of the following,make the letter in the square brackets the subject of the formula.
(a) y=(1+2x/2-x)+3 [x]
(b) 3b-1/a-2=2b+1/a-3 [a]
F2 math plz help~
2007-10-21 11:33 pm
回答 (1)
2007-10-22 1:41 am
✔ 最佳答案
1) -(10/2+x)-(5x/x+2)= -5-x-5-2
= -x-12
2) q= r(x-1)-(x/p),
-2=-4(x-1)-(x/-3)
-2= -4x+4+(x/3)
-6= (-12x+x)/3
-18= -11x
x= 18/11
3(a) y=(1+2x/2-x)+3 [x]
y-3= (1+2x/2-x)
(y-3)( 2-x) = 1+2x
2y-xy-6+3x = 1+2x
1-2y+6 = 3x-xy-2x
7-2y= x(3-y-2)
(7-2y)/ (1-y) = x
x = (7-2y)/ (1-y)
(b)3b-1/a-2=2b+1/a-3 [a]
(3b-1)(a-3)= (2b+1)(a-2)
3ab-9-a+3 = 2ab-4b+a-2
3ab-a-2ab-a = 6-4b-2
ab -2a = 4-4b
a(b-2) = 4-4b
a= (4-4b) / (b-2)
收錄日期: 2021-04-13 14:04:20
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