Use the properties of indices to evaluate the following, giving the answers as fractions or whole numbers. a) 6(3⁻¹) + 30 b) 3⁻¹ imes 62 + 60. Calculate the following giving the answers as fractions

We use the properties $a^{-n} = \frac{1}{a^n}$ and $a^0 = 1$.

$$3^{-1} = \frac{1}{3}$$ $$3^0 = 1$$

$$3$$

The final answer is $\boxed{\text{3}}$.

b) $3^{-1} \times 6^2 + 6^0$

$$3^{-1} = \frac{1}{3}$$ $$6^2 = 36$$ $$6^0 = 1$$

$$13$$

The final answer is $\boxed{\text{13}}$.

QUESTION 2. Calculate the following giving the answers as fractions.

a) $\frac{3^{-2}}{2^{-2}} - \frac{2^{-3}}{3^{-3}}$

$$3^{-2} = \frac{1}{3^2} = \frac{1}{9}$$ $$2^{-2} = \frac{1}{2^2} = \frac{1}{4}$$ $$2^{-3} = \frac{1}{2^3} = \frac{1}{8}$$ $$3^{-3} = \frac{1}{3^3} = \frac{1}{27}$$

The least common multiple of 9 and 8 is 72.

$$\frac{4 \times 8}{9 \times 8} - \frac{27 \times 9}{8 \times 9} = \frac{32}{72} - \frac{243}{72}$$

$$\frac{32 - 243}{72} = \frac{-211}{72}$$

The final answer is $\boxed{\text{-}\frac{211}{72}}$.

b) $5(2^3)^{-1} + 4^{-2}$

$$2^3 = 8$$ $$(2^3)^{-1} = 8^{-1} = \frac{1}{8}$$ $$4^{-2} = \frac{1}{4^2} = \frac{1}{16}$$

The least common multiple of 8 and 16 is 16.

$$\frac{5 \times 2}{8 \times 2} + \frac{1}{16} = \frac{10}{16} + \frac{1}{16}$$

$$\frac{10 + 1}{16} = \frac{11}{16}$$

The final answer is $\boxed{\frac{11}{16}}$.