Question

determine whether each number is rational or irrational.
rational irrational
0.\overline{82}
\frac{\sqrt{99}}{3}
\sqrt{3.7}
-\frac{3}{7}

Explanation:

Step1: Analyze \(0.\overline{62}\)

A repeating decimal is a rational number because it can be expressed as a fraction. Let \(x = 0.\overline{62}\), then \(100x=62.\overline{62}\), subtracting \(x\) from \(100x\) gives \(99x = 62\), so \(x=\frac{62}{99}\), which is a fraction of two integers. So \(0.\overline{62}\) is rational.

Step2: Analyze \(\frac{\sqrt{99}}{3}\)

Simplify \(\sqrt{99}=\sqrt{9\times11} = 3\sqrt{11}\), so \(\frac{\sqrt{99}}{3}=\frac{3\sqrt{11}}{3}=\sqrt{11}\). \(\sqrt{11}\) is irrational (since 11 is not a perfect square), so \(\frac{\sqrt{99}}{3}\) is irrational.

Step3: Analyze \(\sqrt{3.7}\)

\(3.7=\frac{37}{10}\), and \(\sqrt{\frac{37}{10}}\) is irrational because 37 and 10 are not perfect squares and the square root of a non - perfect - square fraction is irrational. Wait, no, wait: Wait, 3.7 is not a perfect square, so \(\sqrt{3.7}\) is irrational. Wait, but wait, the original problem's third number: Wait, no, let's re - check. Wait, the third number is \(\sqrt{3.7}\). Since 3.7 is not a perfect square, its square root is irrational.

Step4: Analyze \(-\frac{3}{7}\)

A fraction of two integers (where the denominator is non - zero) is a rational number. So \(-\frac{3}{7}\) is rational.

Now, for the first number \(0.\overline{62}\): mark the "Rational" circle.

For \(\frac{\sqrt{99}}{3}\): mark the "Irrational" circle.

For \(\sqrt{3.7}\): mark the "Irrational" circle.

For \(-\frac{3}{7}\): mark the "Rational" circle.

Answer:

For \(0.\overline{62}\): Rational (mark the circle under Rational).

For \(\frac{\sqrt{99}}{3}\): Irrational (mark the circle under Irrational).

For \(\sqrt{3.7}\): Irrational (mark the circle under Irrational).

For \(-\frac{3}{7}\): Rational (mark the circle under Rational).