Identify each as rational or irrational by typing the word in the blank. These must all be correct to earn $2$ points: $\newline$ A) $\sqrt{144}$ is a/an $\qquad$ number. $\newline$ B) $\sqrt{\frac{4}{36}}$ is a/an $\qquad$ number. $\newline$ C) $\sqrt{8}$ is a/an $\qquad$ number. $\newline$ D) $\pi$ is a/an $\qquad$ number. $\newline$ E) $\sqrt{0.64}$ is $a / a n$ $\qquad$ number. $\newline$ F) $\qquad$ $1$ is a/an $\qquad$ number.

Full solution

Q. Identify each as rational or irrational by typing the word in the blank. These must all be correct to earn

2

points:

\newline

\sqrt{144}

is a/an

\qquad

number.

\newline

\sqrt{\frac{4}{36}}

is a/an

\qquad

number.

\newline

\sqrt{8}

is a/an

\qquad

number.

\newline

\pi

is a/an

\qquad

number.

\newline

\sqrt{0.64}

a / a n

\qquad

number.

\newline

\qquad

1

is a/an

\qquad

number.

A) Type of $\sqrt{144}$ : A) $\sqrt{144}$ is a/an ___ number. $\newline$ $\sqrt{144} = 12$ because $12 \times 12 = 144$ . $\newline$ Since $12$ is a whole number, $\sqrt{144}$ is rational.
B) Type of $\sqrt{\frac{4}{36}}$ : B) $\sqrt{\frac{4}{36}}$ is a/an ___ number. $\sqrt{\frac{4}{36}} = \frac{\sqrt{4}}{\sqrt{36}} = \frac{2}{6} = \frac{1}{3}$ .Since $\frac{1}{3}$ is a fraction, it's a rational number.
C) Type of $\sqrt{8}$ : C) $\sqrt{8}$ is a/an ___ number. $\sqrt{8} = \sqrt{4 \times 2} = \sqrt{4} \times \sqrt{2} = 2 \times \sqrt{2}$ . $\sqrt{2}$ is not a perfect square, so $\sqrt{8}$ is irrational.
D) Type of $\pi$ : D) $\pi$ is a/an ___ number. $\newline$ $\pi$ is a non-repeating, non-terminating decimal. $\newline$ Therefore, $\pi$ is irrational.
E) Type of $\sqrt{0.64}$ : E) $\sqrt{0.64}$ is a/an ___ number. $\sqrt{0.64} = 0.8$ because $0.8 \times 0.8 = 0.64$ . Since $0.8$ is a decimal that can be expressed as a fraction $\frac{8}{10}$ , it's rational.
F) Type of $\sqrt{200}$ : F) $\sqrt{200}$ is a/an ___ number. $\sqrt{200} = \sqrt{100 \times 2} = \sqrt{100} \times \sqrt{2} = 10 \times \sqrt{2}$ . $\sqrt{2}$ is not a perfect square, so $\sqrt{200}$ is irrational.