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Which equation can be represented using the number line?

A. [tex]\(\frac{3}{4} \div \frac{1}{8} = 6\)[/tex]

B. [tex]\(\frac{1}{8} \div \frac{3}{4} = 6\)[/tex]

C. [tex]\(6 \div \frac{3}{4} = \frac{1}{8}\)[/tex]

D. [tex]\(E \div \frac{1}{2} = \frac{3}{1}\)[/tex]


Sagot :

Let's understand how these equations work and then determine which one can be represented on a number line effectively.

### Option 1: [tex]\(\frac{3}{4} \div \frac{1}{8} = 6\)[/tex]

To verify this equation, we calculate:
[tex]\[ \frac{3}{4} \div \frac{1}{8} = \frac{3}{4} \times \frac{8}{1} = \frac{3 \times 8}{4 \times 1} = \frac{24}{4} = 6 \][/tex]
So this equation holds true.

### Option 2: [tex]\(\frac{1}{8} \div \frac{3}{4} = 6\)[/tex]

To verify this equation, we calculate:
[tex]\[ \frac{1}{8} \div \frac{3}{4} = \frac{1}{8} \times \frac{4}{3} = \frac{1 \times 4}{8 \times 3} = \frac{4}{24} = \frac{1}{6} \][/tex]
This equation does not hold true as [tex]\(\frac{1}{8} \div \frac{3}{4} = \frac{1}{6}\)[/tex], not [tex]\(6\)[/tex].

### Option 3: [tex]\(6 \div \frac{3}{4} = \frac{1}{8}\)[/tex]

To verify this equation, we calculate:
[tex]\[ 6 \div \frac{3}{4} = 6 \times \frac{4}{3} = \frac{6 \times 4}{3 \times 1} = \frac{24}{3} = 8 \][/tex]
This equation does not hold true as [tex]\(6 \div \frac{3}{4} = 8\)[/tex], not [tex]\(\frac{1}{8}\)[/tex].

### Option 4: [tex]\(E \div \frac{1}{2} = \frac{3}{1}\)[/tex]

This equation is general and involves an unknown value [tex]\(E\)[/tex]. To find [tex]\(E\)[/tex]:
[tex]\[ E \div \frac{1}{2} = 3 \implies E = 3 \times \frac{1}{2} = \frac{3}{2} \][/tex]
So [tex]\(E\)[/tex] must be [tex]\(\frac{3}{2}\)[/tex] for the equation to hold true.

### Conclusion

Among the given options, the first equation:
[tex]\[ \frac{3}{4} \div \frac{1}{8} = 6 \][/tex]
is true and can be effectively represented on a number line. This is because dividing [tex]\(\frac{3}{4}\)[/tex] by [tex]\(\frac{1}{8}\)[/tex] scales the value by a factor of 8, resulting in 6. This operation is visually intuitive using a number line, showing increments and how multiple parts contribute to the division result.