9r2
12\(\sqrt{3r}\)
13r
\(\sqrt{13r}\)
Correct answer is C
\(\sqrt{160r^2 + \sqrt{71r^4 + \sqrt{100r^8}}}\)
Simplifying from the innermost radical and progressing outwards we have the given expression
\(\sqrt{160r^2 + \sqrt{71r^4 + \sqrt{100r^8}}}\) = \(\sqrt{160r^2 + \sqrt{71r^4 + 10r^4}}\)
= \(\sqrt{160r^2 + \sqrt{81r^4}}\)
\(\sqrt{160r^2 + 9r^2}\) = \(\sqrt{169r^2}\)
= 13r