Hard Question
Posted:
Wed Jan 02, 2013 5:06 pm
by sxk1693
How do you approach these questions instead of graphing it all the time:
How many ordered pairs (x,y) are solutions to the equation 5x+3y<40 where x,y are integers and 0<y<x<9? Answer:14
Re: Hard Question
Posted:
Wed Jan 02, 2013 5:36 pm
by darksaber21
I would test out possible points. It is not as tedious as you think, if you know what you're doing that is.
First, to find out how many points are possible when y = 1
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So, 6 possible points there. (2, 3, 4, 5, 6, and 7)
We continue when y = 2:
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So 4 possible points there (3, 4, 5, and 6)
When y = 3:
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So 3 possible points (4, 5, and 6)
When y = 4:
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So one possible point (5)
Add them all up, and we get 14 points total.
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The inequality can be rearranged to be:
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Since y needs to be less than x, we can start off with y = 1 and see how many x values within the interval (and greater than y) satisfies the inequality.
I really can't think of another way, yet.
Re: Hard Question
Posted:
Wed Jan 02, 2013 7:27 pm
by 101dalmatians
What the hell now is a POLITE NUMBER????
I am so fed up with these numbers - i mean numbers have NO feelings!!!!
But anyway... Back to my question...
Can someone please help me with these problems from 2011 District I?
31. -> This one is a circle problem...
39. Find the sum of the series [unparseable or potentially dangerous latex formula] (to the nearest thousandth)
A. 7.400
B. 7.389
C. 7.356
D. 7.292
E. 7.267
What is the pattern to this series???
41. ->probability....
51. -> Slant asymptote problem - keeps coming - so annoying!
57 -> Just tell me the method - i have NO NO CLUE !!!!!!
60. -> what is this supposed to mean - aren't there only 3 faces unseen?
Thanks...