Which expression is a perfect cube? A)x^8 B)y^24 C)m^28 D)s^64

2 answers:

6 0

The answer is B
Y^24

Aliun [14]3 years ago

5 0

Y in the power of 24 is a perfect cube

Not helping huh

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vfiekz [6]

<h3>Answer: Choice B</h3>

The set notation includes all values from -5 to 0, but the domain only includes the integer values

eg: something like -1.2 is in the second set, but it is not in the set {-5,-4,-3,-2,-1}

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Further explanation:

Let's go through the answer choices one by one

A. This is false because 0 does not come before -5, but instead -5 is listed first. The order -5,-4,-3,-2,-1,0 is correct meaning that $-5 \le x \le 0$ is the correct order as well.
B. This is true. A value like x = -1.2 is in the set $\left\{x| -5 \le x \le 0\right\}$ since -1.2 is between -5 and 0; but -1.2 is not in the set {-5, -4, -3, -2, -1, 0}. So the distinction is that we're either considering integers only or all real numbers in this interval. To ensure that we only look at integers, the student would have to write $\left\{x| x\in\mathbb{Z}, \ -5 \le x \le 0\right\}$ . The portion $x \in \mathbb{Z}$ means "x is in the set of integers". The Z refers to the German word Zahlen, which translates to "numbers".
C. This is false. The student used the correct inequality signs to indicate x is -5 or larger and also 0 or smaller; basically x is between -5 and 0 inclusive of both endpoints. The "or equal to" portions indicate we are keeping the endpoints and not excluding them.
D. This is false. Writing $-5 \ge x \ge 0$ would not make any sense. This is because that compound inequality breaks down into $-5 \ge x \ \text{ and } \ x \ge 0$ . Try to think of a number that is both smaller than -5 AND also larger than 0. It can't be done. No such number exists.