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x ≥ 0 y ≥ 0
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15 Apr 2018, 04:39
15 Apr 2018, 04:39
Expert Reply
00:00
Question Stats:
78% (00:36) correct
21% (01:09) wrong based on 65 sessions
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x ≥ 0
y ≥ 0
A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.
y ≥ 0
Quantity A |
Quantity B |
\(\sqrt{x^{12}}-y\) |
\((x^3-\sqrt{y})(x^3+\sqrt{y})\) |
A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.
Drill 1
Question: 8
Page: 511
Question: 8
Page: 511
Re: x ≥ 0 y ≥ 0
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18 Apr 2018, 01:21
18 Apr 2018, 01:21
1
qty B is in the form \((x -y) (x + y)\) hence it can be transformed into \(x^2 -y^2\)
qty B will become \((x^3)^2 - y = x^6 - y\)
qty A becomes \(x^6-y\) when square root is removed hence both quantities are equal
qty B will become \((x^3)^2 - y = x^6 - y\)
qty A becomes \(x^6-y\) when square root is removed hence both quantities are equal
Re: x ≥ 0 y ≥ 0
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22 Apr 2018, 12:20
22 Apr 2018, 12:20
Expert Reply
Explanation
In Quantity A, \(\sqrt{x^{12}}-y=\sqrt{(x^{6})^{2}}-y\).
In Quantity B, you may recognize one of the common quadratics: \((a + b)(a - b) = a^2 - b^2\).
If not, FOIL; either way, Quantity B is \(x^6 - y\). Thus, the two quantities are equal.
In Quantity A, \(\sqrt{x^{12}}-y=\sqrt{(x^{6})^{2}}-y\).
In Quantity B, you may recognize one of the common quadratics: \((a + b)(a - b) = a^2 - b^2\).
If not, FOIL; either way, Quantity B is \(x^6 - y\). Thus, the two quantities are equal.
