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Sunday, November 24, 2013

Fibonacci Haiku: The Lovely Judith

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Judith?
Who?
My friend?
The charming sister?
The one mirroring our friendship?
You wonder who's she and still don't know...
No matter who she is, I love darn Judith that’s the flipping problem...

Saturday, November 16, 2013

SP #5: Unit J Concept 6 - Partial fraction decomposition with repeated factors



     The viewer needs to pay special attention that when there are common factors, each of the repeated factor will have a power one more than the one before. One must count up the factors and include the factor as many times as the exponent. The viewers need to pay attention to their work and make sure to not make an error throughout the process. There are many places where errors could be made, especially when distributing/multiplying factors. And finally, the viewer needs to be aware that there are (4) systems, therefore there are four variables. Process of elimination is used to solve for a variable, followed by substitution to solve for the others. 

Friday, November 15, 2013

SP #4: Unit J Concept 5 - Partial fraction decomposition with distinct factors



     The viewer needs to pay special attention when multiplying factors and combining like-terms. There are many places where one can make a mistake and ruin the entire problem. (Part 1 and Part 2 are certain places where making mistakes can occur). The viewers also needs to pa attention to the least common denominator used to add these fractions (again, in Part 1 and 2). Grouping like-terms need to be organized properly to make 3 system and, if possible, simplified. Finally, the viewers need to make sure the final answer in Part 3 is identical to the initial problem in Part 1.

Monday, November 11, 2013

SV #5: Unit J Concept 3-4 - Solving 3-variable systems

     The viewer needs to pay special attention on the order and patter we are solving to get the 3 0's. We start with Row 3 Term 1, then Row 2 Term 1, and finally Row 3 Term 2. After the 3 0's are solved for, we continue solving by making a 1 stair-step pattern. We do this by multiplying the reciprocal of the leading coefficient, or by dividing by the leading coefficient. Probably the most important thing overall is that this is solving to get consistent independent solutions (one answer). Consistent independents solutions are a bit different than solving for consistent dependent solutions (infinite solutions).