Answer:
hmmm...? im gonna guess 2
Step-by-step explanation:
i finger plus another finger gives me......ahhhhh ik the answer its 2
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:
ill take some brainliest
Step-by-step explanation:
Answer:
-m^2 |n|^3 sqrt(n)
Step-by-step explanation:
-sqrt( m^4n^7)
We know that sqrt(ab) = sqrt(a) sqrt(b)
-sqrt(m^4) sqrt(n^7)
The sqrt(m^4) = m^2
The sqrt(n^7) = sqrt(n^6) sqrt(n) = |n|^3 sqrt(n) We take the absolute value because if n is negative, n^6 is postive, but n^3 is negative
-sqrt(m^4) sqrt(n^7)
-m^2 |n|^3 sqrt(n)
Answer:
x=2, x=1
Step-by-step explanation:
Multiply both sides by x, so you get (x^2)+2=3x, which comes from 
Solve for (x^2)+2=3x
(x^2)-3x+2=0
(x-2)(x-1)=0, use box method or FOIL if needed
x=2,x=1