Answer:
2x^2(x^3+3x+8)
Step-by-step explanation:
Factor out the gcf which in this case is -2x^2 because that's the greatest factor all of the terms have in common this results in -2x^2(x^3+3x+8)
Btw the ^ represents the exponent
The function is definately defined at x=0 but not x=1.
But its just one part of the coordinate (x,y).
If the value of y or f(x) is considered, you'll see that it is never possible to attain f(x)=0. In other terms (x,y)= (0,0) is not a defined point in the graph of the function because the graph doesnt pass through that point.
Now I hope you understood what I meant!
Conclusion- The above function is not defined at all points in the space having the abscissa or x=1 in the coordinate and also at ordinate or y=0 in the coordinate.nation:
B is false, since by the inclusion/exclusion principle,
By independence, we have 
, which is zero if either of 
or 
is 0, which isn't guaranteed.
Answer:
A) 6
Step-by-step explanation:
n/3+(-4)=-2
n/3-4=-2
n/3=-2+4
n/3=2
n=2*3
n=6
36v - 12 = 12 .....add 12 to both sides
36v - 12 + 12 = 12 + 12...simplify
36v = 24...divide both sides by 36
(36/36)v = 24/36
v = 2/3 <===
