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Re: r=7 and s=-7 [#permalink]
r=-s, s=-r
Q1. 2r+2r+r^2=4r+r^2
Q2. -2r-2r+r^2=-4r+r^2
Now add 4r and subtract r^2 to compare, Q1. 8r and Q2. 0
We are given r=7. Hence, Q1. 8*7 and Q2. 0
Q1>Q2. Answer is A.
Carcass wrote:
\(r=7\) and \(s=-7\)

Quantity A
Quantity B
\(2r-2s+r^2\)
\(2s-2r+s^2\)



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.
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Re: r=7 and s=-7 [#permalink]
1
Carcass wrote:
\(r=7\) and \(s=-7\)

Quantity A
Quantity B
\(2r-2s+r^2\)
\(2s-2r+s^2\)



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.


Given:
QUANTITY A: \(2r-2s+r^2\)
QUANTITY B: \(2s-2r+s^2\)

If \(r=7\) and \(s=-7\), we can substitute those values into the quantities...
QUANTITY A: \(2(7)-2(-7)+(7)^2\)
QUANTITY B: \(2(-7)-2(7)+(-7)^2\)

Evaluate:
QUANTITY A: \(14+14+49\)
QUANTITY B: \((-14)-14+49\)

This point, we can probably see that Quantity A must be greater

Answer: A
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Re: r=7 and s=-7 [#permalink]
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