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Things TFG Might Say During DC Job Interview
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<blockquote data-quote="4shotB" data-source="post: 976997" data-attributes="member: 844"><p>Let's set this up as a word problem (bad pun intended). In this instance, let's let "n" stand for the actual number of times you did actually listen before tuning out. We all know that the domain of n is an integer or whole number represented by either the value 0,1, or 2. {research has proven that most husbands believe that the actual value of n lies in the domain of 8,9 or 10 or even higher. But this is patently false no matter what you might "think" otherwise}. So, to your question, researchers are still quantifying the actual answer. From a purely mathematical standpoint, the minimum would (falsely) appear to be n + 1, but scientists are leaning toward an actual answer closer to (3n + 20) squared. This is coming from the math department at Cambridge (MIT) . The good folks over in Cambridge (UK) feel like the derivate of 2x^3y^2 - 2xyz^5 + 6acdx^6 with respect to x as x approaches 5 from the left is a better approximation. I am following the research closely as a Nobel Prize hinges on the outcome. i fully expect to share this research with my better half, in which case she will ultimately decide what the correct answer is and let me know.</p></blockquote><p></p>
[QUOTE="4shotB, post: 976997, member: 844"] Let's set this up as a word problem (bad pun intended). In this instance, let's let "n" stand for the actual number of times you did actually listen before tuning out. We all know that the domain of n is an integer or whole number represented by either the value 0,1, or 2. {research has proven that most husbands believe that the actual value of n lies in the domain of 8,9 or 10 or even higher. But this is patently false no matter what you might "think" otherwise}. So, to your question, researchers are still quantifying the actual answer. From a purely mathematical standpoint, the minimum would (falsely) appear to be n + 1, but scientists are leaning toward an actual answer closer to (3n + 20) squared. This is coming from the math department at Cambridge (MIT) . The good folks over in Cambridge (UK) feel like the derivate of 2x^3y^2 - 2xyz^5 + 6acdx^6 with respect to x as x approaches 5 from the left is a better approximation. I am following the research closely as a Nobel Prize hinges on the outcome. i fully expect to share this research with my better half, in which case she will ultimately decide what the correct answer is and let me know. [/QUOTE]
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