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21^2=x^2y^3, where x and y are integers
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25 Dec 2019, 00:46
25 Dec 2019, 00:46
Expert Reply
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Question Stats:
58% (01:19) correct
41% (01:32) wrong based on 77 sessions
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\(21^2=x^2y^3\), where x and y are integers
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.
Kudos for the right answer and explanation
Quantity A |
Quantity B |
|x-y| |
7 |
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.
Kudos for the right answer and explanation
Re: 21^2=x^2y^3, where x and y are integers
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25 Dec 2019, 23:48
25 Dec 2019, 23:48
1
Ans A.
21^2 *1^3=x^2*y^3
21^2 *1^3=x^2*y^3
Re: 21^2=x^2y^3, where x and y are integers
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26 Dec 2019, 07:53
26 Dec 2019, 07:53
2
Expert Reply
\(21^2 = 3^2 \times 7^2\)
\(3^2 \times 7^2 = x^2 y^3
\)
\(9 \times 49 = x^2 y^3\)
No number on the LHS gives you 9 and 49 raised to the third power. So on the THS y must be one
So
\(21^2=x^2\)
Making square root we do have that \(21=x\)
therefore, \(x=21\) and \(y=1\)
\(x-y=21-1=20
\)
\(20 > 7\)
A is the answer
\(3^2 \times 7^2 = x^2 y^3
\)
\(9 \times 49 = x^2 y^3\)
No number on the LHS gives you 9 and 49 raised to the third power. So on the THS y must be one
So
\(21^2=x^2\)
Making square root we do have that \(21=x\)
therefore, \(x=21\) and \(y=1\)
\(x-y=21-1=20
\)
\(20 > 7\)
A is the answer
