Is the first question wrong?
There are a integers between 5.5 and its reciprocal, and there are b integers between 5.5 and its reciprocal, and | A+ 1 |+B-5 |-| C-2 | = 0. Try to find the value of A+B+C b+c.
The reciprocal of 5.5 is 2/ 1 1, and the reciprocal of 5.5 is-5.5;
From 2/ 1 1 to 5.5, there are 1, 2,3,4,5; a = 5;
The range from -5.5 to 5.5 is -5, -4, -3, -2,-1, 0, 1, 2, 3, 4, 5 respectively; b = 1 1;
|a+ 1|+|b-5|-|c-2|=0
|a+ 1|+|b-5|=|c-2|
5+ 1+ 1 1-5 = | c-2
|c-2|= 12
c 1= 14,c2=- 10
When c= 14
a+b+c = 5+ 1 1+ 14 = 30
When c=- 10
a+b+c = 5+ 1 1- 10 = 6
(2)
Suppose that the distance between a and b is x.
(x-36)÷( 10-8)=(36+36)÷( 12- 10)
x-36 =(36+36)÷( 12- 10)*( 10-8)
x=72+36
x= 108
A: The distance between a:AB and Beijing is108km.
Reference/question/19398058.html? si=3
(3)
Setting: Wang planted eggplant × mu and tomato × mu.
1700 * x+ 1800 *(25-x)= 44000
1700 x+45000- 1800 x = 44000
100x= 1000
x= 10
25- 10= 15 mu
2400 *10+2600 *15 = 63000 (yuan)
A: The Wang family made a net profit of 63,000 yuan.