Completely simplify and rationalize the denominator: $$\frac{\sqrt{160}}{\sqrt{252}}\times\frac{\sqrt{245}}{\sqrt{180}}$$
\(\sqrt{160} = \sqrt{16} \times \sqrt {10} = 4 \sqrt{10}\)
\(\sqrt{252} = \sqrt{36} \times \sqrt 7 = 6 \sqrt 7\)
\(\sqrt{245} = \sqrt{49} \times \sqrt{5} = 7 \sqrt 5 \)
\(\sqrt{180} = \sqrt{36} \times \sqrt{5} = 6 \sqrt 5\)
\({4 \sqrt {10} \over 6 \sqrt 7} \times {7 \sqrt 5 \over 6 \sqrt5} = {4 \sqrt {10} \over 6 \sqrt 7} \times {7 \over 6} = {28 \sqrt {10} \over 36 \sqrt 7} = {28 \sqrt {10} \over 36 \sqrt 7} \times {\sqrt 7 \over\sqrt 7} = {28 \sqrt {70} \over 252} = \color{brown}\boxed{\sqrt{70} \over 9} \)
.\(\sqrt{160} = \sqrt{16} \times \sqrt {10} = 4 \sqrt{10}\)
\(\sqrt{252} = \sqrt{36} \times \sqrt 7 = 6 \sqrt 7\)
\(\sqrt{245} = \sqrt{49} \times \sqrt{5} = 7 \sqrt 5 \)
\(\sqrt{180} = \sqrt{36} \times \sqrt{5} = 6 \sqrt 5\)
\({4 \sqrt {10} \over 6 \sqrt 7} \times {7 \sqrt 5 \over 6 \sqrt5} = {4 \sqrt {10} \over 6 \sqrt 7} \times {7 \over 6} = {28 \sqrt {10} \over 36 \sqrt 7} = {28 \sqrt {10} \over 36 \sqrt 7} \times {\sqrt 7 \over\sqrt 7} = {28 \sqrt {70} \over 252} = \color{brown}\boxed{\sqrt{70} \over 9} \)