Certainly! Let's analyze the given problem step-by-step.
We need to compare the ratios [tex]\(\frac{a_1}{a_2}\)[/tex], [tex]\(\frac{b_1}{b_2}\)[/tex], and [tex]\(\frac{c_1}{c_2}\)[/tex].
Given values are:
- [tex]\(a_1 = 1\)[/tex], [tex]\(a_2 = 2\)[/tex]
- [tex]\(b_1 = 2\)[/tex], [tex]\(b_2 = 3\)[/tex]
- [tex]\(c_1 = 3\)[/tex], [tex]\(c_2 = 4\)[/tex]
First, we calculate the ratio [tex]\(\frac{a_1}{a_2}\)[/tex]:
[tex]\[ \frac{a_1}{a_2} = \frac{1}{2} = 0.5 \][/tex]
Next, we calculate the ratio [tex]\(\frac{b_1}{b_2}\)[/tex]:
[tex]\[ \frac{b_1}{b_2} = \frac{2}{3} \approx 0.6666666666666666 \][/tex]
Then, we calculate the ratio [tex]\(\frac{c_1}{c_2}\)[/tex]:
[tex]\[ \frac{c_1}{c_2} = \frac{3}{4} = 0.75 \][/tex]
So, the ratios are:
[tex]\[
\frac{a_1}{a_2} = 0.5
\][/tex]
[tex]\[
\frac{b_1}{b_2} \approx 0.6666666666666666
\][/tex]
[tex]\[
\frac{c_1}{c_2} = 0.75
\][/tex]
By comparing these ratios, we can conclude the numerical relationships among them:
- [tex]\(\frac{a_1}{a_2}\)[/tex] (0.5) is less than [tex]\(\frac{b_1}{b_2}\)[/tex] (approximately 0.6666666666666666).
- [tex]\(\frac{b_1}{b_2}\)[/tex] (approximately 0.6666666666666666) is less than [tex]\(\frac{c_1}{c_2}\)[/tex] (0.75).
Therefore, we have the order:
[tex]\[ \frac{a_1}{a_2} < \frac{b_1}{b_2} < \frac{c_1}{c_2} \][/tex]
This is the result of comparing the given ratios.
