Rewrite the expression in the form $y^{n}$.
Write the exponent as an integer, fraction, or an exact decimal (not a mixed number).
$$
\sqrt[4]{y^{\frac{7}{3}} y^{\frac{1}{3}}}=
$$
$\square$

Question

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To simplify the given expression, we will use the properties of exponents. ‖ ‖ step 2 ⋮ The expression inside the radical can be combined using the property $y^{a} \cdot y^{b} = y^{a+b}$. ‖ ‖ step 3 ⋮ Adding the exponents $\frac{7}{3}$ and $\frac{1}{3}$ gives us $y^{\frac{7}{3} + \frac{1}{3}} = y^{\frac{8}{3}}$. ‖ ‖ step 4 ⋮ The fourth root of $y^{\frac{8}{3}}$ can be written as $y^{\frac{8}{3} \cdot \frac{1}{4}}$. ‖ ‖ step 5 ⋮ Multiplying the exponents gives us $y^{\frac{8}{3} \cdot \frac{1}{4}} = y^{\frac{8}{12}} = y^{\frac{2}{3}}$. ‖ ∻[1] Answer∻ ⚹ $y^{\frac{2}{3}}$ ⚹ ∻Key Concept∻ ⚹Properties of Exponents⚹ ∻Explanation∻ ⚹To simplify an expression involving radicals and exponents, we combine like bases using the property of exponents that states $y^{a} \cdot y^{b} = y^{a+b}$, and then apply the property that the nth root of $y^{m}$ is $y^{\frac{m}{n}}$.⚹ ∻Solution by Steps∻ ‖ step 1 ⋮ To determine the sign of $f(x)$ on the interval $-\frac{1}{5}-0.2$ since $-0.2$ is its zero and the leading coefficient is positive. ‖ ‖ step 3 ⋮ The factor $(4x-8)$ is positive for $x>2$, but since our interval is $-\frac{1}{5}-6$, which includes our entire interval. ‖ ‖ step 5 ⋮ Since two factors are positive and one is negative within the interval, their product $f(x)$ is negative for $-\frac{1}{5}

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