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2007 AMC 10A Problems/Problem 2

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Problem

Define $a@b = ab - b^{2}$ and $a\#b = a + b - ab^{2}$ . What is $\frac {6@2}{6\#2}$ ?

$\text{(A)}\ - \frac {1}{2}\qquad \text{(B)}\ - \frac {1}{4}\qquad \text{(C)}\ \frac {1}{8}\qquad \text{(D)}\ \frac {1}{4}\qqu...$

Solution

$\frac{6 @ 2}{6 \# 2} = \frac{(6)\times (2) - (2)^2}{(6) + (2) - (6) \cdot (2)^2} = \frac{8}{-16} = \frac{-1}{2} \Rightarrow \...$

See also

2007 AMC 10A (Problems • Resources)
Preceded by Problem 1	Followed by Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2007_AMC_10A_Problems/Problem_2"

Category: Introductory Algebra Problems

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