Question

The cube of a number is less than five times the square of the number. For what set of numbers is this true?

Answer 1

Answer:

$(-\infty,0)\cup (0,5)$.

Step-by-step explanation:

Note: In the given options it should be negative infinity instead of -0 because -0 does not make any sense.

It is given that cube of a number is less than five times the square of the number.

Let the unknown number be x.

$x^3$

$x^3-5x^2$

$x^2(x-5)$

This is possible when both factors $x^2\text{ and }(x-5)$ have different sign.

We know that $x^2$ is always greater than or equal to 0. So,

$x-5$

$x$

$x\in (-\infty,0)\cup (0,5)$

Because $x^2(x-5)$ is not true for x=0, so, 0 is not included in the solution set.

Therefore, the solution set is $(-\infty,0)\cup (0,5)$.

Answer 2

Answer:

C

Step-by-step explanation: