What is the solution set for the inequalities x^3<0 or 4x^4>x^2?

  (x < 0) ∪ (x > 1/2)

Step-by-step explanation:

First inequality:

  x^3 < 0

  x < 0 . . . . . take the cube root

Second inequality:

4x^4 > x^2

4x^4 -x^2 > 0

x^2(4x^2 -1) > 0

Any non-zero x will make the first factor positive. The product will be positive only when the second factor is positive, for x^2 > 1/4

  x^2 > 1/4

  |x| > 1/2 . . . . . take the square root

  x < -1/2   or   x > 1/2

The first part of this solution is already covered by the solution to the first inequality. Hence the solution set for the system of inequalities is ...

  (x < 0) ∪ (x > 1/2)

no not all of them.

step-by-step explanation:

some of them have a square base.