Question

determine whether each sum is rational or irrational.
drag the appropriate word next to each sum.
\\(\sqrt{6}+\sqrt{6}\\)
\\(\sqrt{9}+1\\)
\\(\pi + 8\\)
\\(\sqrt{20}+\sqrt{5}\\)
rational
irrational

Explanation:

For \(\boldsymbol{\sqrt{6}+\sqrt{6}}\):

Step1: Simplify the sum

\(\sqrt{6}+\sqrt{6} = 2\sqrt{6}\)

Step2: Determine rationality

\(\sqrt{6}\) is irrational, and multiplying an irrational number by a non - zero rational number (2) still gives an irrational number. So \(2\sqrt{6}\) is irrational.

For \(\boldsymbol{\sqrt{9}+1}\):

Step1: Simplify \(\sqrt{9}\)

\(\sqrt{9}=3\) (since \(3\times3 = 9\))

Step2: Calculate the sum

\(3 + 1=4\), and 4 is a rational number (it can be written as \(\frac{4}{1}\)).

For \(\boldsymbol{\pi + 8}\):

Step1: Recall the nature of \(\pi\)

\(\pi\) is an irrational number (it has a non - repeating, non - terminating decimal expansion).

Step2: Determine the sum's nature

The sum of an irrational number (\(\pi\)) and a rational number (8) is irrational.

For \(\boldsymbol{\sqrt{20}+\sqrt{5}}\):

Answer:

• \(\sqrt{6}+\sqrt{6}\): Irrational
• \(\sqrt{9}+1\): Rational
• \(\pi + 8\): Irrational
• \(\sqrt{20}+\sqrt{5}\): Irrational