Question

find the root. assume that the variable represents a positive number.
\sqrt{\frac{49y^{10}}{16}}

\sqrt{\frac{49y^{10}}{16}} = \square

Explanation:

Step1: Apply square - root property

We know that for a fraction under a square root, \(\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}\) (where \(a\geq0\) and \(b > 0\)). So, \(\sqrt{\frac{49y^{10}}{16}}=\frac{\sqrt{49y^{10}}}{\sqrt{16}}\).

Step2: Simplify numerator and denominator

• For the numerator, we use the property \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\) (where \(a\geq0\) and \(b\geq0\)). We know that \(49 = 7^{2}\) and \(y^{10}=(y^{5})^{2}\) (since \((y^{m})^{n}=y^{mn}\), here \(m = 5\) and \(n = 2\), so \((y^{5})^{2}=y^{10}\)). So, \(\sqrt{49y^{10}}=\sqrt{49}\cdot\sqrt{y^{10}}=\sqrt{7^{2}}\cdot\sqrt{(y^{5})^{2}}\). Since the variable \(y\) represents a positive number, \(\sqrt{7^{2}} = 7\) and \(\sqrt{(y^{5})^{2}}=y^{5}\). So, \(\sqrt{49y^{10}} = 7y^{5}\).
• For the denominator, \(\sqrt{16}=\sqrt{4^{2}} = 4\) (because \(4^{2}=16\)).

Step3: Combine the results

Now, we substitute the simplified numerator and denominator back into the fraction \(\frac{\sqrt{49y^{10}}}{\sqrt{16}}\). So, \(\frac{7y^{5}}{4}\).

Answer:

\(\frac{7y^{5}}{4}\)