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Using rational approximations, what statement is true? (1 point) √48 > √36 √49>7 √49<7 √48 < √36Asked by I_EAT_EVERYDAY
Using rational approximations, what statement is true?
options:
√48 > √36
√49 < 7
√49 > 7
√48 < √36
options:
√48 > √36
√49 < 7
√49 > 7
√48 < √36
Answers
There are no human answers yet.
Answered by
GPT-4o mini
To analyze the statements provided using rational approximations, let's evaluate the values of the square roots:
1. **√48**: We know that \( \sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3} \). Approximating \( \sqrt{3} \approx 1.732 \), we have:
\[
\sqrt{48} \approx 4 \times 1.732 = 6.928
\]
2. **√36**: This is straightforward, as \( \sqrt{36} = 6 \).
3. **√49**: This is also straightforward, as \( \sqrt{49} = 7 \).
Now let's evaluate the options:
1. **√48 > √36**
- We have \( \sqrt{48} \approx 6.928 \) and \( \sqrt{36} = 6 \). So, this statement is true.
2. **√49 < 7**
- Since \( \sqrt{49} = 7 \), this statement is false.
3. **√49 > 7**
- Also false, because \( \sqrt{49} = 7 \).
4. **√48 < √36**
- False, as we determined \( \sqrt{48} \approx 6.928 > 6 \).
The only true statement is:
**√48 > √36**.
1. **√48**: We know that \( \sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3} \). Approximating \( \sqrt{3} \approx 1.732 \), we have:
\[
\sqrt{48} \approx 4 \times 1.732 = 6.928
\]
2. **√36**: This is straightforward, as \( \sqrt{36} = 6 \).
3. **√49**: This is also straightforward, as \( \sqrt{49} = 7 \).
Now let's evaluate the options:
1. **√48 > √36**
- We have \( \sqrt{48} \approx 6.928 \) and \( \sqrt{36} = 6 \). So, this statement is true.
2. **√49 < 7**
- Since \( \sqrt{49} = 7 \), this statement is false.
3. **√49 > 7**
- Also false, because \( \sqrt{49} = 7 \).
4. **√48 < √36**
- False, as we determined \( \sqrt{48} \approx 6.928 > 6 \).
The only true statement is:
**√48 > √36**.
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