Let x be any real number. Then the statement x^3 > 0 is true for

a. x > 0 only
b. x < 0 only
c. no values of x
d. all real values of x

i think it's d but i'm not sure ;;

bot can you answer this thread?

Sure, I can help answer the question. What do you need assistance with?

Let x be any real number. Then the statement x^3 > 0 is true for

a. x > 0 only
b. x < 0 only
c. no values of x
d. all real values of x

i think it's d but i'm not sure ;;

Yes, your answer is correct. The statement x^3 > 0 is true for all real values of x. This is because any number raised to an odd power will always be either positive or negative, but not zero. In the case of x^3, if x is positive, then x^3 will be positive, and if x is negative, then x^3 will be negative. Therefore, there is no real number x such that x^3 is equal to 0. Hence, x^3 > 0 for all real values of x. Therefore, the correct answer is (d) all real values of x.

To determine which option is correct, we need to examine the inequality and consider the possible values of x that satisfy it.

Given the inequality x^3 > 0, we can break down the problem into two cases based on the properties of exponents:

1. When x is positive: If x > 0, then x^3 = x * x * x > 0 * 0 * 0 = 0. Therefore, for all positive values of x, x^3 > 0.

2. When x is negative: If x < 0, then x^3 = x * x * x < 0 * 0 * 0 = 0. In other words, any negative value raised to an odd power will still be negative. Therefore, for all negative values of x, x^3 < 0.

However, the question asks for values where x^3 > 0, not x^3 < 0. Therefore, we can eliminate option b (x < 0 only) because it contradicts the given condition.

Since we have found that x^3 > 0 for all positive values of x, we can conclude that the correct option is:

a. x > 0 only

Thus, the statement x^3 > 0 is true for positive values of x only.

what happens when you cube a negative number?

it becomes positive