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√2÷√5
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rationalize?
simplifying radicals
okay :) \[\sqrt{2}\over\sqrt{5}\]you can't have a radical in the denominator so what we do is multiply the denominator and numerator by the radical, basically multiplying by 1, but removing the radical (\(\sqrt{5}\over\sqrt{5}\) is the same thing as \(1\)):\[{\sqrt{2}\over\sqrt{5}}\times{\sqrt{5}\over\sqrt{5}}\]\[{\sqrt{2}\times\sqrt{5}}\over{\sqrt{5}^2}\]\[{\sqrt{10}\over5}\]
is √10/5 the answer
yup :)
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thanks
yw
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