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...

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. 

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.

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). 

WPP #4: Unit I Concept 3-5 - Investment application problem

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SV #4: Unit I Concept 2 - Graphing logarithmic equations

   The viewers need to pay special attention on how the asymptote is found. The asymptote equation is x=h and h comes from the "base of" portion of the log and is equal to 0. In this video for example, the "base of" is x+2, it is equal to 0 and solve algebraically to get the asymptote of x=-2. The viewer also need to pay attention to how the x and y intercepts are solve. For the x intercept, the equation is equaled to 0 and solved algebraically, however exponentiate is required. For the y intercept, x is 0 and solved algebraically; in this video log base b of b=1 property is used. And finally, the range is always (-infinity, infinity), but the domain is (asymptote, infinity).

SP #3: Unit I Concept 1 - Graphing exponential equations w/out x-intercept

     The viewer needs to pay special attention to on how the aymtptote was found and the restrictions on the range. Remember "EXPONENTIAL YaK Died", which means that in an exponential graph, y=k and there are no restriction on the domain. The range (y-intervals) depends on the asymptote. The viewer also needs to pay special attention to why this graph does not have an x-intercept. There can be no natural log of a negative number, which results to no solution/no x-intercept. For y-intercept, make y equal 0. And finally, remember that if +a will place the graph above the asymptote, and -a below; if |b|<1 the graph is close to the asymptote to the right side and |b|>1 is on the left; x+h shifts the graph left and right; and fourthly, "k" is the asymptote.