Question
Using rational approximations, what statement is true?(1 point) Responses 48−−√>36−−√ start root 48 end root greater than start root 36 end root 48−−√<36−−√ start root 48 end root less than start root 36 end root 49−−√>7 start root 49 end root greater than 7 49−−√<7 --
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GPT-4o mini
To determine which inequalities are true using rational approximations, we can evaluate the square roots involved.
1. **Comparing \(\sqrt{48}\) and \(\sqrt{36}\)**:
- \(\sqrt{36} = 6\)
- \(\sqrt{48} \approx 6.928\) (since \(\sqrt{48} = \sqrt{16 \cdot 3} = 4\sqrt{3} \approx 4 \times 1.732 \approx 6.928\))
Since \(6.928 > 6\), we have:
\[
\sqrt{48} > \sqrt{36}
\]
2. **Comparing \(\sqrt{49}\) and \(7\)**:
- \(\sqrt{49} = 7\)
Hence, we have:
\[
\sqrt{49} = 7
\]
From our evaluations, the true statements are:
- \(\sqrt{48} > \sqrt{36}\)
- \(\sqrt{49} = 7\)
Thus, the true statement among the options provided is:
\[
\sqrt{48} > \sqrt{36} \quad \text{(first option)}
\]
Therefore, the correct response is:
\(\sqrt{48} > \sqrt{36}\).
1. **Comparing \(\sqrt{48}\) and \(\sqrt{36}\)**:
- \(\sqrt{36} = 6\)
- \(\sqrt{48} \approx 6.928\) (since \(\sqrt{48} = \sqrt{16 \cdot 3} = 4\sqrt{3} \approx 4 \times 1.732 \approx 6.928\))
Since \(6.928 > 6\), we have:
\[
\sqrt{48} > \sqrt{36}
\]
2. **Comparing \(\sqrt{49}\) and \(7\)**:
- \(\sqrt{49} = 7\)
Hence, we have:
\[
\sqrt{49} = 7
\]
From our evaluations, the true statements are:
- \(\sqrt{48} > \sqrt{36}\)
- \(\sqrt{49} = 7\)
Thus, the true statement among the options provided is:
\[
\sqrt{48} > \sqrt{36} \quad \text{(first option)}
\]
Therefore, the correct response is:
\(\sqrt{48} > \sqrt{36}\).
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