$\displaystyle \frac{1}{3}-\frac{1}{8}$

In order to solve this problem, we first need to make common denominators. 

$\displaystyle \frac{1}{8}\times\frac{3}{3}=\frac{3}{24}$

$\displaystyle \frac{1}{3}\times\frac{8}{8}=\frac{8}{24}$$\displaystyle \frac{1}{3}\times\frac{8}{8}=\frac{8}{24}$$\displaystyle \frac{1}{3}\times\frac{8}{8}=\frac{8}{24}$$\displaystyle \frac{1}{3}\times\frac{8}{8}=\frac{8}{24}$

$\displaystyle \frac{1}{3}\times\frac{8}{8}=\frac{8}{24}$$\displaystyle \frac{1}{3}\times\frac{8}{8}=\frac{8}{24}$

Now that we have common denominators, we can subtract the fractions. Remember, when we subtract fractions, the denominator stays the same, we only subtract the numerator. 

$\displaystyle \frac{8}{24}-\frac{3}{24}=\frac{5}{24}$

$\displaystyle \frac{1}{3}-\frac{1}{9}$

In order to solve this problem, we first need to make common denominators. 

$\displaystyle \frac{1}{3}\times\frac{3}{3}=\frac{3}{9}$

Now that we have common denominators, we can subtract the fractions. Remember, when we subtract fractions, the denominator stays the same, we only subtract the numerator. 

$\displaystyle \frac{3}{9}-\frac{1}{9}=\frac{2}{9}$

$\displaystyle \frac{2}{3}-\frac{1}{12}$

In order to solve this problem, we first need to make common denominators. 

$\displaystyle \frac{2}{3}\times\frac{4}{4}=\frac{8}{12}$

Now that we have common denominators, we can subtract the fractions. Remember, when we subtract fractions, the denominator stays the same, we only subtract the numerator. 

$\displaystyle \frac{8}{12}-\frac{1}{12}=\frac{7}{12}$

$\displaystyle \frac{4}{14}-\frac{1}{7}$

In order to solve this problem, we first need to make common denominators. 

$\displaystyle \frac{1}{7}\times\frac{2}{2}=\frac{2}{14}$

Now that we have common denominators, we can add the fractions. Remember, when we add fractions, the denominator stays the same, we only add the numerator. 

$\displaystyle \frac{4}{14}-\frac{2}{14}=\frac{2}{14}$

$\displaystyle \small \frac{2}{14}$ can be reduced by dividing $\displaystyle \small 2$ by both sides. 

$\displaystyle \small \frac {2}{14}\div\frac{2}{2}=\frac{1}{7}$

When multiplying fractions you simply multiply the numerators to find the numerator of the product, and multiply the denominators to find the denominator of the product.

$\displaystyle \small 3\cdot 4=12$

$\displaystyle \small 5\cdot 15=75$

So, our product is $\displaystyle \small \frac{12}{75}$.

Since both of these numbers are divisible by three, we simplify. 

$\displaystyle \small \frac{12}{3}=4$

$\displaystyle \small \frac{75}{3}= 25$

Our final answer is $\displaystyle \small \frac{4}{25}$.

Rewrite the mixed numbers as improper fractions, then multiply across:

$\displaystyle 3 \frac{1}{3} = \frac{3\times 3 + 1}{3} = \frac{10}{3}$

$\displaystyle 4 \frac{1}{5} = \frac{4 \times 5 + 1}{5} = \frac{21}{5}$

$\displaystyle 3 \frac{1}{3} \times 4 \frac{1}{5} = \frac{10}{3} \times \frac{21}{5} = \frac{2}{1} \times \frac{7}{1} = 14$

$\displaystyle 14 > 12 \frac{1}{5}$

Rewrite the factors as improper fractions, then multiply across:

$\displaystyle 6 \frac{5}{6} = \frac{6\times 6 + 5}{6} = \frac{36 + 5}{6} = \frac{41}{6}$

$\displaystyle 6 = \frac{6}{1}$

$\displaystyle 6 \frac{5}{6} \times 6 =\frac{41}{6}\times \frac{ 6}{1}=\frac{41}{1}\times \frac{ 1}{1} = 41$

(a) $\displaystyle 0.75 t = 0.3$

$\displaystyle 0.75 t \div 0.75 = 0.3 \div 0.75$

$\displaystyle t= 0.3 \div 0.75$

Divide by moving the decimal point right two places in both numbers:

$\displaystyle t= 30 \div 75 = 0.4$

(b) $\displaystyle \frac{7}{4} k = \frac{7}{10}$

$\displaystyle \frac{4} {7} \cdot \frac{7}{4} k =\frac{4} {7} \cdot \frac{7}{10}$

$\displaystyle k =\frac{4} {7} \cdot \frac{7}{10}$

Cross-cancel:

$\displaystyle k =\frac{2} {1} \cdot \frac{1}{5} = \frac{2} {5}$

$\displaystyle 0.4 = \frac{4}{10 } = \frac{4\div 2}{10\div 2 } = \frac{2}{5}$, so $\displaystyle k = t$

$\displaystyle \frac{3}{5} = 3 \div 5 = 0.6$, so 

$\displaystyle \left ( \frac{3}{5} \right )^{3} = 0.6^{3} = 0.6 \times 0.6 \times 0.6 = 0.36 \times 0.6 = 0.216$

$\displaystyle 0.6 ^{2} = 0.6 \times 0.6 = 0.36$

$\displaystyle 0.36 > 0.216$, so  $\displaystyle 0.6^{2} > \left ( \frac{3}{5} \right )^{3}$, making (B) greater.

$\displaystyle \left ( \frac{4}{5} \right )^{3} = 0.8 ^{3} = 0.8 \times 0.8 \times 0,8 = 0.64\times 0,8 = 0.512$

Since $\displaystyle 0.512 > 0.5$, $\displaystyle \left ( \frac{4}{5} \right )^{3} > 0.5$, so (B) is greater.