Team x and team y have a tug of war. from their starting positions team x pulls team y forward 3 meters, and team x is then pulled forward 5 meters. team y then pulls team x forward 2 meters. if the first team to be pulled forward 10 meters loses, how many more meters must team y pull team x forward to win?

 In any tug-o-war that I have participated in, the losing team has to be pulled a certain distance forward FROM THE STARTING POINT. 

If that is the case in this instance, then 

a) team X pulls Y forward 3m, therefore X's position is X - 3. 

b) team Y pulls X forward 5m, therefore X's position is now X - 3 + 5 

c) team Y pulls X forward another 2m, therefor X's position is X - 3 + 5 + 2 = X + 4. 

That means that X is 4m in front of the starting position. So they must be pulled another 6m if Y is to win.

i = 2 / - 5

step-by-step explanation:

evaluation of ( 2 - 4 ) ( p + q ) when p = 2 and q = 5 i, by putting values of p and q,                                                                    

0 = ( 2 - 4 ) ( p + q )0 = ( 2 - 4 ) ( 2 + 5 i ),   by multiplying these0 = ( 4 + 10 i - 8 - 20 i )0 = 4 - 8 + 10 i - 20 i0 = - 4 -10 i   0 = 2 ( - 2 - 5 i )0 / 2 = - 2 - 5 i- 2 - 5 i = 0- 5 i = 2i = 2 / - 5

Dis the answer for this question