You're on a roll — Step 1: Simplify the expression $\left(\frac{4x^4y^{-2}z}{16x^2y^{-1}z^{-2}}\right)^{\frac{1}{2}}$. First, simplify the fraction inside the parentheses: $$\frac{4x^4y^{-2}z}{16x^2y^{-1}z^{-2}}$$ Simplify the coefficients: $$\frac{4}{16} = \frac{1}{4}$$ Simplify the $x$ terms using the rule $\frac{a^m}{a^n} = a^{m-n}$: $$\frac{x^4}{x^2} = x^{4-2} = x^2$$ Simplify the $y$ terms: $$\frac{y^{-2}}{y^{-1}} = y^{-2 - (-1)} = y^{-2+1} = y^{-1}$$ Simplify the $z$ terms: $$\frac{z^1}{z^{-2}} = z^{1 - (-2)} = z^{1+2} = z^3$$ Combine these simplified terms: $$\frac{1}{4} x^2 y^{-1} z^3 = \frac{x^2 z^3}{4y}$$ Now, apply the exponent $\frac{1}{2}$ to the entire simplified fraction. Remember that raising to the power of $\frac{1}{2}$ is the same as taking the square root. $$\left(\frac{x^2 z^3}{4y}\right)^{\frac{1}{2}} = \frac{(x^2)^{\frac{1}{2}} (z^3)^{\frac{1}{2}}}{(4)^{\frac{1}{2}} (y)^{\frac{1}{2}}}$$ Apply the power to each term: $$(x^2)^{\frac{1}{2}} = x^{2 \times \frac{1}{2}} = x^1 = x$$ $$(z^3)^{\frac{1}{2}} = z^{\frac{3}{2}}$$ $$(4)^{\frac{1}{2}} = \sqrt{4} = 2$$ $$(y)^{\frac{1}{2}} = \sqrt{y}$$ Combine these results: $$\frac{x z^{\frac{3}{2}}}{2 \sqrt{y}}$$ $$\boxed{\frac{x z^{\frac{3}{2}}}{2 \sqrt{y}}}$$ Step 2: Simplify $-6x^0 + 4x^2 + 3x^2$. Recall that any non-zero number raised to the power of 0 is 1, so $x^0 = 1$. Substitute $x^0 = 1$ into the expression: $$-6(1) + 4x^2 + 3x^2$$ $$-6 + 4x^2 + 3x^2$$ Combine the like terms ($4x^2$ and $3x^2$): $$-6 + (4+3)x^2$$ $$-6 + 7x^2$$ It is standard practice to write terms with higher powers first: $$7x^2 - 6$$ $$\boxed{7x^2 - 6}$$ Step 3: Simplify, and write as an Index notation: $3 \times 3 \times 3 \times 3 \times \frac{1}{3} \times \frac{1}{3}$. Group the terms: $$(3 \times 3 \times 3 \times 3) \times \left(\frac{1}{3} \times \frac{1}{3}\right)$$ Write the first group in index notation: $$3 \times 3 \times 3 \times 3 = 3^4$$ Write the second group in index notation: $$\frac{1}{3} \times \frac{1}{3} = \left(\frac{1}{3}\right)^2$$ Recall that $\frac{1}{a^n} = a^{-n}$, so $\frac{1}{3} = 3^{-1}$. Therefore, $\left(\frac{1}{3}\right)^2 = (3^{-1})^2 = 3^{-1 \times 2} = 3^{-2}$. Now multiply the two parts: $$3^4 \times 3^{-2}$$ Use the rule $a^m \times a^n = a^{m+n}$: $$3^{4 + (-2)} = 3^{4-2} = 3^2$$ $$\boxed{3^2}$$ Step 4: Evaluate $\left(\frac{13^0}{5}\right)^2$. Recall that any non-zero number raised to the power of 0 is 1, so $13^0 = 1$. Substitute $13^0 = 1$ into the expression: $$\left(\frac{1}{5}\right)^2$$ Apply the exponent to both the numerator and the denominator: $$\frac{1^2}{5^2}$$ Calculate the squares: $$\frac{1 \times 1}{5 \times 5} = \frac{1}{25}$$ $$\boxed{\frac{1}{25}}$$ Send me the next one 📸