a+b+c=0
[(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc]
[a^2+b^2+c^2+2ab+2ac+2bc=0]
[a^2+b^2+c^2=-(2ab+2ac+2bc)]
[a^2+b^2+c^2=-2(ab+ac+bc)] (i)
also
[a=-b-c]
[a^2=-ab-ac] (ii)
[-c=a+b]
[-bc=ab+b^2] (iii)
adding (ii) and (iii) ,we have
[a^2-bc=b^2-ac] (iv)
devide (i) by (iv)
[(a^2+b^2+c^2)/(a^2-bc)=(-2(ab+bc+ca))/(b^2-ac)]
Answer:
-1 2/3
Step-by-step explanation:
I think the answer is Neither x nor y will eliminate because, by looking at the two equations we can add them and y will be elminated and we will have x-value but we can substitute the x value into one of the equation and obtain y value again.Hope this will help.Sorry If I was wrong .
C is the most reasonable answer
Answer:
5^ -8 or 1/ 5^8
Step-by-step explanation:
(5^-2)^4
We know that a^b^c = a^(b*c)
(5^-2)^4 = 5^(-2*4) = 5^(-8)
We know that a^ - b = 1/ a^b
5^ -8 = 1/ 5^8
