To determine which inequality is correct, we will evaluate each inequality step-by-step.
Inequality 1: [tex]\(\frac{2}{3} < \frac{3}{5} < \frac{5}{7}\)[/tex]
1. Compare [tex]\(\frac{2}{3}\)[/tex] and [tex]\(\frac{3}{5}\)[/tex]:
[tex]\[
\text{Cross-multiply:} \quad 2 \cdot 5 = 10 \quad \text{and} \quad 3 \cdot 3 = 9
\][/tex]
[tex]\[
\text{Since } 10 > 9, \quad \frac{2}{3} > \frac{3}{5}
\][/tex]
This makes the inequality false.
Inequality 2: [tex]\(\frac{2}{3} < \frac{5}{7} < \frac{3}{5}\)[/tex]
1. Compare [tex]\(\frac{2}{3}\)[/tex] and [tex]\(\frac{5}{7}\)[/tex]:
[tex]\[
\text{Cross-multiply:} \quad 2 \cdot 7 = 14 \quad \text{and} \quad 3 \cdot 5 = 15
\][/tex]
[tex]\[
\text{Since } 14 < 15, \quad \frac{2}{3} < \frac{5}{7}
\][/tex]
2. Compare [tex]\(\frac{5}{7}\)[/tex] and [tex]\(\frac{3}{5}\)[/tex]:
[tex]\[
\text{Cross-multiply:} \quad 5 \cdot 5 = 25 \quad \text{and} \quad 7 \cdot 3 = 21
\][/tex]
[tex]\[
\text{Since } 25 > 21, \quad \frac{5}{7} > \frac{3}{5}
\][/tex]
This makes the inequality false.
Inequality 3: [tex]\(\frac{3}{5} < \frac{2}{3} < \frac{5}{7}\)[/tex]
1. Compare [tex]\(\frac{3}{5}\)[/tex] and [tex]\(\frac{2}{3}\)[/tex]:
[tex]\[
\text{Cross-multiply:} \quad 3 \cdot 3 = 9 \quad \text{and} \quad 5 \cdot 2 = 10
\][/tex]
[tex]\[
\text{Since } 9 < 10, \quad \frac{3}{5} < \frac{2}{3}
\][/tex]
2. Compare [tex]\(\frac{2}{3}\)[/tex] and [tex]\(\frac{5}{7}\)[/tex]:
[tex]\[
\text{Cross-multiply:} \quad 2 \cdot 7 = 14 \quad \text{and} \quad 3 \cdot 5 = 15
\][/tex]
[tex]\[
\text{Since } 14 < 15, \quad \frac{2}{3} < \frac{5}{7}
\][/tex]
This makes the inequality false.
Inequality 4: [tex]\(\frac{3}{5} < \frac{5}{7} < \frac{2}{3}\)[/tex]
1. Compare [tex]\(\frac{3}{5}\)[/tex] and [tex]\(\frac{5}{7}\)[/tex]:
[tex]\[
\text{Cross-multiply:} \quad 3 \cdot 7 = 21 \quad \text{and} \quad 5 \cdot 5 = 25
\][/tex]
[tex]\[
\text{Since } 21 < 25, \quad \frac{3}{5} < \frac{5}{7}
\][/tex]
2. Compare [tex]\(\frac{5}{7}\)[/tex] and [tex]\(\frac{2}{3}\)[/tex]:
[tex]\[
\text{Cross-multiply:} \quad 5 \cdot 3 = 15 \quad \text{and} \quad 7 \cdot 2 = 14
\][/tex]
[tex]\[
\text{Since } 15 > 14, \quad \frac{5}{7} < \frac{2}{3}
\][/tex]
This makes the inequality true.
So, the inequality that is correct is:
[tex]\[
\boxed{\frac{3}{5} < \frac{5}{7} < \frac{2}{3}}
\][/tex]
