Let's compare each pair of values step-by-step by considering the true hierarchy of the comparisons provided.
1. Compare [tex]\(\sqrt{3} + 2\)[/tex] and [tex]\(\sqrt{3} + 6\)[/tex]:
[tex]\[
\sqrt{3} + 2 \quad \text{vs} \quad \sqrt{3} + 6
\][/tex]
Since [tex]\(\sqrt{3}\)[/tex] is common in both expressions, it is clear that adding 2 to [tex]\(\sqrt{3}\)[/tex] will result in a smaller value than adding 6 to [tex]\(\sqrt{3}\)[/tex]. Therefore,
[tex]\[
\sqrt{3} + 2 < \sqrt{3} + 6
\][/tex]
2. Compare [tex]\(\sqrt{7} - 1\)[/tex] and [tex]\(\sqrt{7} - 3\)[/tex]:
[tex]\[
\sqrt{7} - 1 \quad \text{vs} \quad \sqrt{7} - 3
\][/tex]
Since [tex]\(\sqrt{7}\)[/tex] is common in both expressions, subtracting 1 from [tex]\(\sqrt{7}\)[/tex] will result in a larger value than subtracting 3. Therefore,
[tex]\[
\sqrt{7} - 1 > \sqrt{7} - 3
\][/tex]
3. Compare [tex]\(2 \sqrt{6}\)[/tex] and [tex]\(2 \sqrt{7}\)[/tex]:
[tex]\[
2 \sqrt{6} \quad \text{vs} \quad 2 \sqrt{7}
\][/tex]
Again, since the coefficient 2 is common and [tex]\(\sqrt{7}\)[/tex] is greater than [tex]\(\sqrt{6}\)[/tex], multiplying 2 by [tex]\(\sqrt{7}\)[/tex] will yield a larger result than multiplying 2 by [tex]\(\sqrt{6}\)[/tex]. Therefore,
[tex]\[
2 \sqrt{6} < 2 \sqrt{7}
\][/tex]
4. Compare [tex]\(-5 \sqrt{3}\)[/tex] and [tex]\(-7 \sqrt{3}\)[/tex]:
[tex]\[
-5 \sqrt{3} \quad \text{vs} \quad -7 \sqrt{3}
\][/tex]
Since [tex]\(\sqrt{3}\)[/tex] is negative in both expressions, and [tex]\(-5\)[/tex] is greater than [tex]\(-7\)[/tex], the product of [tex]\(-5\)[/tex] with [tex]\(\sqrt{3}\)[/tex] will be more than the product of [tex]\(-7\)[/tex] with [tex]\(\sqrt{3}\)[/tex]. Therefore,
[tex]\[
-5 \sqrt{3} > -7 \sqrt{3}
\][/tex]
In summary, to complete each statement with [tex]\(<\)[/tex] or [tex]\(>\)[/tex]:
1. [tex]\(\sqrt{3} + 2 < \sqrt{3} + 6\)[/tex]
2. [tex]\(\sqrt{7} - 1 > \(\sqrt{7} - 3\)[/tex]
3. [tex]\(2 \sqrt{6} < 2 \sqrt{7}\)[/tex]
4. [tex]\(-5 \sqrt{3} > -7 \sqrt{3}\)[/tex]
