Grade 11 Zero and Negative Exponents Worksheets 2024 - Free Printable

Problem: Simplify the given expressions, ensuring that your answers contain only positive exponents.

We will solve each problem step by step using the rules of exponents:
1. Zero Exponent Rule: \( a^0 = 1 \) (for any nonzero \( a \)).
2. Negative Exponent Rule: \( a^{-n} = \frac{1}{a^n} \).
3. Product and Quotient Rules: Combine like terms and simplify as needed.

---

#### Problem 1: \( 2^0 \)
- Using the Zero Exponent Rule: \( a^0 = 1 \).
- Therefore, \( 2^0 = 1 \).

Answer: \( \boxed{1} \)

---

#### Problem 2: \( 2^{-1} \)
- Using the Negative Exponent Rule: \( a^{-n} = \frac{1}{a^n} \).
- Therefore, \( 2^{-1} = \frac{1}{2^1} = \frac{1}{2} \).

Answer: \( \boxed{\frac{1}{2}} \)

---

#### Problem 3: \( (-2)^{-1} \)
- Using the Negative Exponent Rule: \( a^{-n} = \frac{1}{a^n} \).
- Therefore, \( (-2)^{-1} = \frac{1}{(-2)^1} = \frac{1}{-2} = -\frac{1}{2} \).

Answer: \( \boxed{-\frac{1}{2}} \)

---

#### Problem 4: \( (-5)^{-1} \)
- Using the Negative Exponent Rule: \( a^{-n} = \frac{1}{a^n} \).
- Therefore, \( (-5)^{-1} = \frac{1}{(-5)^1} = \frac{1}{-5} = -\frac{1}{5} \).

Answer: \( \boxed{-\frac{1}{5}} \)

---

#### Problem 5: \( (9s)^{-2} \)
- Using the Negative Exponent Rule: \( a^{-n} = \frac{1}{a^n} \).
- Therefore, \( (9s)^{-2} = \frac{1}{(9s)^2} \).
- Simplify the denominator: \( (9s)^2 = 9^2 \cdot s^2 = 81s^2 \).
- Thus, \( (9s)^{-2} = \frac{1}{81s^2} \).

Answer: \( \boxed{\frac{1}{81s^2}} \)

---

#### Problem 6: \( 2u^0v^{-3} \)
- Using the Zero Exponent Rule: \( u^0 = 1 \).
- Using the Negative Exponent Rule: \( v^{-3} = \frac{1}{v^3} \).
- Substitute these into the expression: \( 2u^0v^{-3} = 2 \cdot 1 \cdot \frac{1}{v^3} = \frac{2}{v^3} \).

Answer: \( \boxed{\frac{2}{v^3}} \)

---

#### Problem 7: \( 2p^{-1}q^0 \)
- Using the Zero Exponent Rule: \( q^0 = 1 \).
- Using the Negative Exponent Rule: \( p^{-1} = \frac{1}{p} \).
- Substitute these into the expression: \( 2p^{-1}q^0 = 2 \cdot \frac{1}{p} \cdot 1 = \frac{2}{p} \).

Answer: \( \boxed{\frac{2}{p}} \)

---

#### Problem 8: \( 3a^{-3}b^0 \)
- Using the Zero Exponent Rule: \( b^0 = 1 \).
- Using the Negative Exponent Rule: \( a^{-3} = \frac{1}{a^3} \).
- Substitute these into the expression: \( 3a^{-3}b^0 = 3 \cdot \frac{1}{a^3} \cdot 1 = \frac{3}{a^3} \).

Answer: \( \boxed{\frac{3}{a^3}} \)

---

#### Problem 9: \( a^{-4}b^{-4} \)
- Using the Negative Exponent Rule: \( a^{-4} = \frac{1}{a^4} \) and \( b^{-4} = \frac{1}{b^4} \).
- Substitute these into the expression: \( a^{-4}b^{-4} = \frac{1}{a^4} \cdot \frac{1}{b^4} = \frac{1}{a^4b^4} \).

Answer: \( \boxed{\frac{1}{a^4b^4}} \)

---

#### Problem 10: \( a^{-4}b^2 \)
- Using the Negative Exponent Rule: \( a^{-4} = \frac{1}{a^4} \).
- Substitute this into the expression: \( a^{-4}b^2 = \frac{1}{a^4} \cdot b^2 = \frac{b^2}{a^4} \).

Answer: \( \boxed{\frac{b^2}{a^4}} \)

---

#### Problem 11: \( 2u^{-1}v^{-1} \)
- Using the Negative Exponent Rule: \( u^{-1} = \frac{1}{u} \) and \( v^{-1} = \frac{1}{v} \).
- Substitute these into the expression: \( 2u^{-1}v^{-1} = 2 \cdot \frac{1}{u} \cdot \frac{1}{v} = \frac{2}{uv} \).

Answer: \( \boxed{\frac{2}{uv}} \)

---

#### Problem 12: \( 2x^{-1}b^0 \)
- Using the Zero Exponent Rule: \( b^0 = 1 \).
- Using the Negative Exponent Rule: \( x^{-1} = \frac{1}{x} \).
- Substitute these into the expression: \( 2x^{-1}b^0 = 2 \cdot \frac{1}{x} \cdot 1 = \frac{2}{x} \).

Answer: \( \boxed{\frac{2}{x}} \)

---

Final Answers:

1. \( \boxed{1} \)
2. \( \boxed{\frac{1}{2}} \)
3. \( \boxed{-\frac{1}{2}} \)
4. \( \boxed{-\frac{1}{5}} \)
5. \( \boxed{\frac{1}{81s^2}} \)
6. \( \boxed{\frac{2}{v^3}} \)
7. \( \boxed{\frac{2}{p}} \)
8. \( \boxed{\frac{3}{a^3}} \)
9. \( \boxed{\frac{1}{a^4b^4}} \)
10. \( \boxed{\frac{b^2}{a^4}} \)
11. \( \boxed{\frac{2}{uv}} \)
12. \( \boxed{\frac{2}{x}} \)