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Read each equation below and mark whether the equation is true or false.

1. [tex]\( 6 \div \frac{1}{3} \ \textgreater \ 16 \)[/tex]
- True ⬜
- False ⬜

2. [tex]\( \frac{1}{4} \div 3 = \frac{3}{12} \)[/tex]
- True ⬜
- False ⬜

3. [tex]\( 12 \div \frac{1}{6} \ \textless \ 80 \)[/tex]
- True ⬜
- False ⬜

4. [tex]\( \frac{1}{5} \div 2 \ \textgreater \ 9 \)[/tex]
- True ⬜
- False ⬜

Sagot :

Let's examine each equation one by one and determine whether it is true or false.

1. Equation 1: [tex]\( 6 \div \frac{1}{3} > 16 \)[/tex]

When we divide 6 by [tex]\(\frac{1}{3}\)[/tex], it is equivalent to multiplying 6 by 3:

[tex]\[ 6 \div \frac{1}{3} = 6 \times 3 = 18 \][/tex]

So, we have:

[tex]\[ 18 > 16 \][/tex]

This statement is true.

2. Equation 2: [tex]\( \frac{1}{4} \div 3 = \frac{3}{12} \)[/tex]

Dividing [tex]\(\frac{1}{4}\)[/tex] by 3 means multiplying [tex]\(\frac{1}{4}\)[/tex] by the reciprocal of 3, which is [tex]\(\frac{1}{3}\)[/tex]:

[tex]\[ \frac{1}{4} \div 3 = \frac{1}{4} \times \frac{1}{3} = \frac{1}{12} \][/tex]

Now, let's examine the right-hand side:

[tex]\[ \frac{3}{12} = \frac{1}{4} \][/tex]

Since [tex]\(\frac{1}{12} \neq \frac{1}{4}\)[/tex], this statement is false.

3. Equation 3: [tex]\( 12 \div \frac{1}{6} < 80 \)[/tex]

Dividing 12 by [tex]\(\frac{1}{6}\)[/tex] is equivalent to multiplying 12 by 6:

[tex]\[ 12 \div \frac{1}{6} = 12 \times 6 = 72 \][/tex]

So, we have:

[tex]\[ 72 < 80 \][/tex]

This statement is true.

4. Equation 4: [tex]\( \frac{1}{5} \div 2 > 9 \)[/tex]

Dividing [tex]\(\frac{1}{5}\)[/tex] by 2 is equivalent to multiplying [tex]\(\frac{1}{5}\)[/tex] by the reciprocal of 2, which is [tex]\(\frac{1}{2}\)[/tex]:

[tex]\[ \frac{1}{5} \div 2 = \frac{1}{5} \times \frac{1}{2} = \frac{1}{10} \][/tex]

So, we have:

[tex]\[ \frac{1}{10} > 9 \][/tex]

Since [tex]\(\frac{1}{10}\)[/tex] is much less than 9, this statement is false.

Summarizing our findings:

[tex]\[ \begin{tabular}{|l|c|c|} \hline & . True & False \\ \hline $6 \div \frac{1}{3} > 16$ & \bigcirc & \\ \hline $\frac{1}{4} \div 3 = \frac{3}{12}$ & & \bigcirc \\ \hline $12 \div \frac{1}{6} < 80$ & \bigcirc & \\ \hline $\frac{1}{5} \div 2 > 9$ & & \bigcirc \\ \hline \end{tabular} \][/tex]