1. The first expression is $(x^{-2} x^{-3})^4$. Using the rule $(a^m a^n)^p = a^{(m+n)p}$, we get: $$ (x^{-2} x^{-3})^4 = (x^{-5})^4 = x^{-20} $$ This matches your answer $x^{-20}$. 2. For $(x^4)^{-3} \cdot 2x^4$, apply the power rule $(a^m)^n = a^{mn}$: $$ (x^4)^{-3} = x^{4 \times (-3)} = x^{-12} $$ Then multiply: $$ x^{-12} \cdot 2x^4 = 2x^{-12+4} = 2x^{-8} $$ This matches your answer $2x^{-8}$. 3. For $(n^3)^3 \cdot 2n^{-1}$: $$ (n^3)^3 = n^{3 \times 3} = n^9 $$ Multiply: $$ n^9 \cdot 2n^{-1} = 2n^{9-1} = 2n^8 $$ This matches your answer $2n^8$. 4. For $(2v)^2 \cdot 2v^2$: $$ (2v)^2 = 2^2 v^2 = 4v^2 $$ Multiply: $$ 4v^2 \cdot 2v^2 = 8v^{2+2} = 8v^4 $$ This matches your answer $8v^4$. 5. For $\frac{2x^2 y^4 \cdot 4x^2 y^4 \cdot 3x}{3x^3 y^2}$: Multiply numerator: $$ 2 \cdot 4 \cdot 3 = 24 $$ Add exponents for $x$: $$ 2 + 2 + 1 = 5 $$ Add exponents for $y$: $$ 4 + 4 = 8 $$ So numerator is: $$ 24x^5 y^8 $$ Divide by denominator: $$ \frac{24x^5 y^8}{3x^3 y^2} = 8x^{5-3} y^{8-2} = 8x^2 y^6 $$ Your final answer $8x^8 y^6$ is incorrect; the correct is $8x^2 y^6$. 6. For $\frac{2y^3 \cdot 3xy^3}{3x^2 y^4}$: Multiply numerator: $$ 2 \cdot 3 = 6 $$ Add exponents for $y$: $$ 3 + 3 = 6 $$ So numerator is: $$ 6x^1 y^6 $$ Divide by denominator: $$ \frac{6x y^6}{3x^2 y^4} = 2x^{1-2} y^{6-4} = 2x^{-1} y^2 $$ Your answer $2 x^1 y^2$ is incorrect; the correct is $2x^{-1} y^2$. 7. For $\frac{x^3 y^3 \cdot x^3}{4x^2 y^2}$: Multiply numerator: $$ x^{3+3} y^3 = x^6 y^3 $$ Divide by denominator: $$ \frac{x^6 y^3}{4x^2 y^2} = \frac{1}{4} x^{6-2} y^{3-2} = \frac{1}{4} x^4 y^1 $$ You wrote only $x^4$, missing $y^1$ and the denominator factor $\frac{1}{4}$. 8. For $\frac{3x^2 y^2}{2x^{-1} \cdot 4yx^2}$: Multiply denominator: $$ 2 \cdot 4 = 8 $$ Add exponents for $x$ in denominator: $$ -1 + 2 = 1 $$ So denominator is: $$ 8x^1 y^1 $$ Divide: $$ \frac{3x^2 y^2}{8x y} = \frac{3}{8} x^{2-1} y^{2-1} = \frac{3}{8} x y $$ Your answer $3x^4 y / 8$ is incorrect; the correct is $\frac{3}{8} x y$. 9. For $\frac{x^4 y^3}{4x}$: Divide: $$ \frac{x^4 y^3}{4x} = \frac{1}{4} x^{4-1} y^3 = \frac{1}{4} x^3 y^3 $$ 10. For $(2x^0)^2$: Recall $x^0 = 1$: $$ 2x^0 = 2 \times 1 = 2 $$ Square: $$ (2)^2 = 4 $$ Your answer $x/4$ is incorrect; the correct is $4$. 11. For $\frac{2m^{-4}}{(2m^{-4})^3}$: Calculate denominator: $$ (2m^{-4})^3 = 2^3 m^{-12} = 8 m^{-12} $$ Divide: $$ \frac{2m^{-4}}{8 m^{-12}} = \frac{2}{8} m^{-4 - (-12)} = \frac{1}{4} m^{8} $$ Your final answer $m^8 / 4$ is correct. Summary: Most answers are correct except problems 5, 6, 7, 8, and 10 where errors were found.