Answer :
To find the quotient of [tex]\(-\frac{3}{8}\)[/tex] and [tex]\(-\frac{1}{3}\)[/tex], we can follow these steps:
1. Identify the fractions:
- The two fractions we need to divide are [tex]\(-\frac{3}{8}\)[/tex] and [tex]\(-\frac{1}{3}\)[/tex].
2. Recall the rule for dividing fractions:
- Dividing by a fraction is equivalent to multiplying by its reciprocal. Therefore, dividing by [tex]\(-\frac{1}{3}\)[/tex] is the same as multiplying by [tex]\(-3\)[/tex].
3. Set up the multiplication:
- [tex]\[ -\frac{3}{8} \div -\frac{1}{3} = -\frac{3}{8} \times -3 \][/tex]
4. Perform the multiplication:
- When multiplying fractions, multiply the numerators together and the denominators together:
[tex]\[ -\frac{3}{8} \times -3 = \frac{3 \times 3}{8 \times 1} = \frac{9}{8} \][/tex]
5. Simplify the result:
- The fraction [tex]\(\frac{9}{8}\)[/tex] is already in its simplest form. It can also be expressed as a mixed number:
[tex]\[ \frac{9}{8} = 1 \frac{1}{8} \][/tex]
Thus, the quotient of [tex]\(-\frac{3}{8}\)[/tex] and [tex]\(-\frac{1}{3}\)[/tex] is [tex]\(\boxed{1 \frac{1}{8}}\)[/tex].
1. Identify the fractions:
- The two fractions we need to divide are [tex]\(-\frac{3}{8}\)[/tex] and [tex]\(-\frac{1}{3}\)[/tex].
2. Recall the rule for dividing fractions:
- Dividing by a fraction is equivalent to multiplying by its reciprocal. Therefore, dividing by [tex]\(-\frac{1}{3}\)[/tex] is the same as multiplying by [tex]\(-3\)[/tex].
3. Set up the multiplication:
- [tex]\[ -\frac{3}{8} \div -\frac{1}{3} = -\frac{3}{8} \times -3 \][/tex]
4. Perform the multiplication:
- When multiplying fractions, multiply the numerators together and the denominators together:
[tex]\[ -\frac{3}{8} \times -3 = \frac{3 \times 3}{8 \times 1} = \frac{9}{8} \][/tex]
5. Simplify the result:
- The fraction [tex]\(\frac{9}{8}\)[/tex] is already in its simplest form. It can also be expressed as a mixed number:
[tex]\[ \frac{9}{8} = 1 \frac{1}{8} \][/tex]
Thus, the quotient of [tex]\(-\frac{3}{8}\)[/tex] and [tex]\(-\frac{1}{3}\)[/tex] is [tex]\(\boxed{1 \frac{1}{8}}\)[/tex].