Question

1. place an x in the appropriate cell to indicate whether each real number is rational or irrational. 2π, $sqrt3{12}$, $\frac{81}{99}$, -$sqrt{196}$, rational, irrational

Explanation:

Step1: Recall definition of rational and irrational numbers

A rational number can be written as a fraction $\frac{p}{q}$ where $p,q$ are integers and $q
eq0$. An irrational number cannot be written as such a fraction.

Step2: Analyze $2\pi$

Since $\pi$ is an irrational number and multiplying it by 2 does not change its irrational - nature, $2\pi$ is irrational.

Step3: Analyze $\sqrt[3]{12}$

The cube - root of 12 cannot be written as a fraction of two integers. So, $\sqrt[3]{12}$ is irrational.

Step4: Analyze $\frac{15}{99}$ and $\frac{81}{1}$

$\frac{15}{99}=\frac{5}{33}$ and $\frac{81}{1}$ are both fractions of integers. So, they are rational.

Step5: Analyze $-\sqrt{196}$

$-\sqrt{196}=- 14$, which can be written as $\frac{-14}{1}$. So, it is rational.

Answer:

Number	Rational	Irrational
$\sqrt[3]{12}$		X
$\frac{15}{99}$	X
$\frac{81}{1}$	X
$-\sqrt{196}$	X