1. **State the problem:** Simplify the algebraic expressions and write the answers with positive exponents only. 2. **Expression 1:** $4q^2 \times 4p^4 q^{-3}$ - Multiply coefficients: $4 \times 4 = 16$ - Multiply like bases by adding exponents: $q^{2} \times q^{-3} = q^{2 + (-3)} = q^{-1}$ - So, expression becomes: $$16 p^{4} q^{-1}$$ - Rewrite with positive exponents: $$16 p^{4} \frac{1}{q} = \frac{16 p^{4}}{q}$$ 3. **Expression 2:** $\left(-\frac{2x}{y}\right)^{-2} \div \left(-\frac{3x}{y^{-2}}\right)$ - First, simplify each part: - For $\left(-\frac{2x}{y}\right)^{-2}$, use the rule $\left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n$: $$\left(-\frac{2x}{y}\right)^{-2} = \left(-\frac{y}{2x}\right)^2 = \frac{y^2}{4x^2}$$ - For $\left(-\frac{3x}{y^{-2}}\right)$, rewrite $y^{-2}$ as $\frac{1}{y^2}$: $$-\frac{3x}{y^{-2}} = -3x \times y^{2} = -3x y^{2}$$ - Now divide: $$\frac{\frac{y^2}{4x^2}}{-3x y^{2}} = \frac{y^2}{4x^2} \times \frac{1}{-3x y^{2}} = \frac{y^2}{4x^2} \times \frac{1}{-3x y^{2}}$$ - Cancel $y^2$: $$\frac{\cancel{y^2}}{4x^2} \times \frac{1}{-3x \cancel{y^{2}}} = \frac{1}{4x^2} \times \frac{1}{-3x} = \frac{1}{-12 x^{3}} = -\frac{1}{12 x^{3}}$$ 4. **Expression 3:** $\left(-\frac{y}{2x}\right)^2 + (-3xy^2)$ - Square the first term: $$\left(-\frac{y}{2x}\right)^2 = \frac{y^2}{4x^2}$$ - So sum is: $$\frac{y^2}{4x^2} + (-3xy^2) = \frac{y^2}{4x^2} - 3xy^2$$ 5. **Expression 4:** $-\frac{7}{4}x^2 + -\frac{1}{3}xy^2$ - Combine as is, no like terms to combine further. 6. **Expression 5:** $-\frac{1}{12} x^3$ - This is already simplified. **Final answers:** - Expression 1: $$\frac{16 p^{4}}{q}$$ - Expression 2: $$-\frac{1}{12 x^{3}}$$ - Expression 3: $$\frac{y^2}{4x^2} - 3xy^2$$ - Expression 4: $$-\frac{7}{4}x^2 - \frac{1}{3}xy^2$$ - Expression 5: $$-\frac{1}{12} x^3$$