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sqrt of 2*sqrt of 6* sqrt of 7
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20 Mar 2025, 01:17
20 Mar 2025, 01:17
Expert Reply
00:00
Question Stats:
66% (00:32) correct
33% (00:44) wrong based on 3 sessions
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Quantity A |
Quantity B |
\( \sqrt{2} \times \sqrt{6} \times \sqrt{7} \) |
\(9\) |
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sqrt of 2*sqrt of 6* sqrt of 7
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20 Mar 2025, 13:14
20 Mar 2025, 13:14
1
The product of the square roots of multiple numbers is equal to the square root of the product of those numbers.
I.e., \(\sqrt{2} * \sqrt{6} * \sqrt{7} = \sqrt{2*6*7} = \sqrt{84}\)
You can use the calculator to find that \(\sqrt{84}\) is about 9.165, which is greater than 9.
Also 84 > 81 (the square of 9), its square root must be greater than 9.
I.e., \(\sqrt{2} * \sqrt{6} * \sqrt{7} = \sqrt{2*6*7} = \sqrt{84}\)
You can use the calculator to find that \(\sqrt{84}\) is about 9.165, which is greater than 9.
Also 84 > 81 (the square of 9), its square root must be greater than 9.
Re: sqrt of 2*sqrt of 6* sqrt of 7
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26 Mar 2025, 04:30
26 Mar 2025, 04:30
Expert Reply
We need to compare the value of $\((\sqrt{2})(\sqrt{6})(\sqrt{7})\)$ with 9
Column $A$ has $\((\sqrt{2})(\sqrt{6})(\sqrt{7})=\sqrt{2 \times 6 \times 7}=\sqrt{84}\)$ which is to be compared with $\(9=\sqrt{31}\)$
Clearly $\(\sqrt{84}>\sqrt{81}\)$ i.e. column A has higher quantity.
Hence the answer is (A).
Column $A$ has $\((\sqrt{2})(\sqrt{6})(\sqrt{7})=\sqrt{2 \times 6 \times 7}=\sqrt{84}\)$ which is to be compared with $\(9=\sqrt{31}\)$
Clearly $\(\sqrt{84}>\sqrt{81}\)$ i.e. column A has higher quantity.
Hence the answer is (A).
