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x^3=27 and y^2=16
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25 Jan 2020, 06:54
25 Jan 2020, 06:54
Expert Reply
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Question Stats:
75% (00:19) correct
24% (00:13) wrong based on 41 sessions
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\(x^3=27\) and \(y^2=16\)
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 |
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: x^3=27 and y^2=16
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25 Jan 2020, 10:12
25 Jan 2020, 10:12
1
A cube root will result in a positive value but a square root results in both negative and positive values.
x = 3 and y = +4
when y = +4 it is greater than x
when y = -4 it is less than x
Hence, OA is D
x = 3 and y = +4
when y = +4 it is greater than x
when y = -4 it is less than x
Hence, OA is D
Carcass wrote:
\(x^3=27\) and \(y^2=16\)
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 |
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: x^3=27 and y^2=16
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03 Feb 2020, 11:11
03 Feb 2020, 11:11
2
x^3 = 27 --> x= 3 (x cannot be negative, otherwise it would -27)
y^2= 16 can be +4 or -4, since we apply the square root and therefore have to consider both options
--> y can be greater or smaller than x --> D
y^2= 16 can be +4 or -4, since we apply the square root and therefore have to consider both options
--> y can be greater or smaller than x --> D
