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Assessment activity 3.1 Simplify the following expressions without using a calculator: \( \begin{array}{ll}\text { 1. } m \times m^{-2} \times m^{4} & \text { 2. } m^{2}+m^{3} \\ x^{-6} \div x^{2} & \text { 4. } x^{3} \div\left(x^{2} \times x\right) \\ \left(x^{3} \div x^{2}\right) \times x & \text { 6. } y \times y^{-2} \\ \left(\frac{x^{4}}{x}\right)^{0} & \text { 8. }\left(x^{2} y^{5}\right)^{0} \\ \left(x^{2} y^{6}\right)^{2} & \text { 10. }\left\{\left(8 x^{3}\right)^{2}\right\}^{\frac{1}{3}} \\ \left\{(-2 x)^{2}\right\}^{3} & \text { 12. }\left\{\left(3 x^{-2}\right)^{2}\right\}^{3} \\ \text { 1. }\left\{\frac{2 x^{-4} y}{2^{-2} x^{3} y^{-2}}\right\}^{\frac{1}{3}} & \text { 14. } 8^{\frac{2}{3}} \\ (16)^{\frac{3}{4}} & \text { 16. }(32)^{\frac{2}{5}} \\ (64)^{-\frac{4}{3}} & \text { 18. }\left(\frac{8}{27}\right)^{-\frac{4}{3}} \\ (-27)^{-\frac{2}{3}} & \text { 20. }\left(\frac{0,09}{0,16}\right)^{-\frac{1}{2}}\end{array} \)

Ask by Kirk Adkins. in South Africa
Jan 30,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

Here are the simplified expressions for Assessment Activity 3.1: 1. \( m^{3} \) 2. \( m^{2}(1 + m) \) 4. \( \dfrac{1}{x^{8}} \) 6. \( \dfrac{1}{y} \) 8. \( 1 \) 10. \( 4x^{2} \) 12. \( \dfrac{729}{x^{12}} \) 14. \( 4 \) 16. \( 4 \) 18. \( \dfrac{81}{16} \) 20. \( \dfrac{4}{3} \)

Solution

Certainly! Here's the simplification of each expression listed in Assessment Activity 3.1 without using a calculator: --- ### **1. \( m \times m^{-2} \times m^{4} \)** **Simplification:** \[ m^{1} \times m^{-2} \times m^{4} = m^{(1 - 2 + 4)} = m^{3} \] **Answer:** \( m^{3} \) --- ### **2. \( m^{2} + m^{3} \)** **Simplification:** Factor out the common term \( m^{2} \): \[ m^{2} + m^{3} = m^{2}(1 + m) \] **Answer:** \( m^{2}(1 + m) \) --- ### **4. \( x^{-6} \div x^{2} \)** **Simplification:** \[ x^{-6} \div x^{2} = x^{-6 - 2} = x^{-8} = \frac{1}{x^{8}} \] **Answer:** \( \dfrac{1}{x^{8}} \) --- ### **6. \( y \times y^{-2} \)** **Simplification:** \[ y^{1} \times y^{-2} = y^{1 - 2} = y^{-1} = \frac{1}{y} \] **Answer:** \( \dfrac{1}{y} \) --- ### **8. \( (x^{2} y^{5})^{0} \)** **Simplification:** Any non-zero base raised to the power of 0 is 1. \[ (x^{2} y^{5})^{0} = 1 \] **Answer:** \( 1 \) --- ### **10. \( \left\{ \left(8x^{3}\right)^{2} \right\}^{\frac{1}{3}} \)** **Simplification:** \[ (8x^{3})^{2} = 64x^{6} \] \[ \left(64x^{6}\right)^{\frac{1}{3}} = 64^{\frac{1}{3}} \times x^{6 \times \frac{1}{3}} = 4x^{2} \] **Answer:** \( 4x^{2} \) --- ### **12. \( \left\{\left(3x^{-2}\right)^{2}\right\}^{3} \)** **Simplification:** \[ (3x^{-2})^{2} = 9x^{-4} \] \[ (9x^{-4})^{3} = 729x^{-12} = \frac{729}{x^{12}} \] **Answer:** \( \dfrac{729}{x^{12}} \) --- ### **14. \( 8^{\frac{2}{3}} \)** **Simplification:** \[ 8^{\frac{2}{3}} = \left(2^{3}\right)^{\frac{2}{3}} = 2^{2} = 4 \] **Answer:** \( 4 \) --- ### **16. \( (32)^{\frac{2}{5}} \)** **Simplification:** \[ 32 = 2^{5} \] \[ (2^{5})^{\frac{2}{5}} = 2^{2} = 4 \] **Answer:** \( 4 \) --- ### **18. \( \left(\dfrac{8}{27}\right)^{-\frac{4}{3}} \)** **Simplification:** \[ \left(\dfrac{8}{27}\right)^{-\frac{4}{3}} = \left(\dfrac{27}{8}\right)^{\frac{4}{3}} = \left(\dfrac{3^{3}}{2^{3}}\right)^{\frac{4}{3}} = \left(\dfrac{3}{2}\right)^{4} = \dfrac{81}{16} \] **Answer:** \( \dfrac{81}{16} \) --- ### **20. \( \left(\dfrac{0.09}{0.16}\right)^{-\frac{1}{2}} \)** **Simplification:** \[ \left(\dfrac{0.09}{0.16}\right)^{-\frac{1}{2}} = \left(\dfrac{0.16}{0.09}\right)^{\frac{1}{2}} = \sqrt{\dfrac{16}{9}} = \dfrac{4}{3} \] **Answer:** \( \dfrac{4}{3} \) --- These simplifications utilize the properties of exponents and basic algebraic manipulations to arrive at the simplified forms without the need for a calculator.

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Beyond the Answer

1. For \( m \times m^{-2} \times m^{4} \), we can combine the exponents using the rule \( a^m \times a^n = a^{m+n} \). Thus, this simplifies to \( m^{1-2+4} = m^{3} \). 2. In the case of \( m^{2} + m^{3} \), since you cannot combine terms with different exponents, the expression remains \( m^{2} + m^{3} \). 3. Now, let's simplify \( x^{-6} \div x^{2} \). Using the division rule \( a^m \div a^n = a^{m-n} \), we get \( x^{-6-2} = x^{-8} \). 4. Moving on to \( x^{3} \div (x^{2} \times x) \), this can also be simplified. First, notice that \( x^{2} \times x = x^{3} \), making the entire expression \( x^{3} \div x^{3} = x^{0} = 1 \). 5. For \( (x^{3} \div x^{2}) \times x \), using the division rule again gives us \( x^{1} \times x = x^{1+1} = x^{2} \). 6. When simplifying \( y \times y^{-2} \), by combining exponents we find \( y^{1-2} = y^{-1} \). 7. The expression \( \left(\frac{x^{4}}{x}\right)^{0} \) simplifies directly to \( 1 \) since anything to the power of zero equals one. 8. Similarly, \( \left(x^{2} y^{5}\right)^{0} \) also simplifies to \( 1 \). 9. For \( \left(x^{2} y^{6}\right)^{2} \), you can apply the power rule \( (a^{m}b^{n})^{p} = a^{mp}b^{np} \), giving \( x^{4} y^{12} \). 10. In \( \left\{\left(8 x^{3}\right)^{2}\right\}^{\frac{1}{3}} \), first simplify to \( (64 x^{6})^{\frac{1}{3}} = 4 x^{2} \). 11. The expression \( \left\{(-2 x)^{2}\right\}^{3} \) becomes \( 4 x^{2} \times 3 = 4 x^{6} \). 12. For \( \left\{\left(3 x^{-2}\right)^{2}\right\}^{3} \), use the power of a power rule: \( 9 x^{-4} \times 3 = 27 x^{-12} \). 13. The expression \( \left\{\frac{2 x^{-4} y}{2^{-2} x^{3} y^{-2}}\right\}^{\frac{1}{3}} \) simplifies to \( (8 x^{-1} y^{3})^{\frac{1}{3}} = 2x^{-1/3} y \). 14. Simplifying \( 8^{\frac{2}{3}} \) gives \( 4 \). 15. For \( (16)^{\frac{3}{4}} \), this results in \( 8 \). 16. The expression \( (32)^{\frac{2}{5}} \) simplifies to \( 4 \). 17. When simplifying \( (64)^{-\frac{4}{3}} \), we find \( \frac{1}{16} \). 18. The expression \( \left(\frac{8}{27}\right)^{-\frac{4}{3}} \) is equal to \( \frac{27^{\frac{4}{3}}}{8^{\frac{4}{3}}} = \frac{81}{16} \). 19. Lastly, \( (-27)^{-\frac{2}{3}} \) simplifies to \( \frac{1}{9} \). 20. Finally, \( \left(\frac{0.09}{0.16}\right)^{-\frac{1}{2}} \) simplifies to \( \frac{4}{3} \).

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