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rs = \sqrt 3
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15 Jul 2019, 02:53
15 Jul 2019, 02:53
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\(rs = \sqrt{3}\)
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
\(\frac{2r \sqrt{12}}{r^2s \sqrt{72}}\) |
\(\frac{14rs^2}{42s}\) |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
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rs = \sqrt 3
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15 Jul 2019, 09:26
15 Jul 2019, 09:26
3
Carcass wrote:
\(rs = \sqrt{3}\)
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
\(\frac{2r \sqrt{12}}{r^2s \sqrt{72}}\) |
\(\frac{14rs^2}{42s}\) |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
\(\sqrt{12}=(\sqrt{4})(\sqrt{3})=2\sqrt{3}\)
\(\sqrt{72}=(\sqrt{36})(\sqrt{2})=6\sqrt{2}\)
We get:
QUANTITY A: \(\frac{2r \sqrt{12}}{r^2s \sqrt{72}}=\frac{(2r)(2 \sqrt{3})}{(r)(rs)(6\sqrt{2})}=\frac{(4r)(\sqrt{3})}{(r)(\sqrt{3})(6\sqrt{2})}=\frac{4}{6\sqrt{2}}=\frac{2}{3\sqrt{2}}\)
QUANTITY B: \(\frac{14rs^2}{42s}=\frac{14(rs)(s)}{42s}=\frac{rs}{3}=\frac{\sqrt{3}}{3}\)
At this point, we can use matching operations
Multiply both quantities by 3 to get:
QUANTITY A: \(\frac{2}{\sqrt{2}}\)
QUANTITY B: \(\sqrt{3}\)
Multiply both quantities by \(\sqrt{2}\) to get:
QUANTITY A: \(2\)
QUANTITY B: \(\sqrt{6}\)
Since \(\sqrt{6}>2\), the correct answer is B
RELATED VIDEO
rs = \sqrt 3
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29 May 2021, 08:37
29 May 2021, 08:37
GreenlightTestPrep wrote:
Carcass wrote:
\(rs = \sqrt{3}\)
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
\(\frac{2r \sqrt{12}}{r^2s \sqrt{72}}\) |
\(\frac{14rs^2}{42s}\) |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
\(\sqrt{12}=(\sqrt{4})(\sqrt{3})=2\sqrt{3}\)
\(\sqrt{72}=(\sqrt{36})(\sqrt{2})=6\sqrt{2}\)
We get:
QUANTITY A: \(\frac{2r \sqrt{12}}{r^2s \sqrt{72}}=\frac{(2r)(2 \sqrt{3})}{(r)(rs)(6\sqrt{2})}=\frac{(4r)(\sqrt{3})}{(r)(\sqrt{3})(6\sqrt{2})}=\frac{4}{6\sqrt{2}}=\frac{2}{3\sqrt{2}}\)
QUANTITY B: \(\frac{14rs^2}{42s}=\frac{14(rs)(s)}{42s}=\frac{rs}{3}=\frac{\sqrt{3}}{3}\)
At this point, we can use matching operations
Multiply both quantities by 3 to get:
QUANTITY A: \(\frac{2}{\sqrt{2}}\)
QUANTITY B: \(\sqrt{3}\)
Multiply both quantities by \(\sqrt{2}\) to get:
QUANTITY A: \(2\)
QUANTITY B: \(\sqrt{6}\)
Since \(\sqrt{6}>2\), the correct answer is B
RELATED VIDEO FROM MY COURSE
for quantity A why not just divide by 12 to 72 so it will become root(1/6)
