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Old 2009 July 9th, 10:30 PM   #17 (permalink)
MBA2010HereWeGo
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Agree that answer to 10 is E. I approached this question by subsituting numbers into the function. And choice E was the correct one. This might be a longer process but using smaller number makes the calculation takes less time. I used a=1 and b=2.

Question 2, IMO C. Because we need to figure out what is the value of either a or b.

y=(x+a)(X+b) or y=(x)^2+x(a+b) + ab

stmt 1 a+b=1 wouldn't help alone (we need 2 unique equations)
stmt 2 y intercept is (0,6) wouldn't help on its own either

but combined we know

0=(x)^2+x(1)+6 or x=2 or -3 thus intersect at (2, 0) or (-3, 0).

Question 3

since order doesn't matter thus 5C3*7C1
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