WPP #9: Unit L Concept 4-8 - Possibilities with FCP, combination, permutation, and distinguished permutation

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SP #6: Unit K Concept 10 - Writing repeating number as rational number using geometric series

     The viewer needs to pay special attention that this number is INFINITE, meaning it is continuous and has a repeating integers. With infinite numbers, the infinite sign is needed on top of the repeating numbers. One must also need to pay attention that the process of infinite geometric series is being used, which includes  summation notation, plugging it to the formula (a sub one divided by one minus r), and the sum. One last thing to remember before finding the final sum is to add the value on the left side of the decimal (which we ignored from the start) from the original. 

Fibonacci Haiku: The Lovely Judith

https://encrypted-tbn1.gstatic.com/images?q=tbn:ANd9GcSksbuqyORTj7HaMSC957G4T36YckVdWUFZPkPx5qPeZbgV_-fa

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.

SV #3: Unit H Concept 7 - Finding logs with given approximations

          The viewer needs to pay special attention to the approximations that he/she would use and the ones on the video. There are several ways to factor the logs, and the approximations, itself, can be factored (for ex. in this video: 4 and 2). Remember that when expanding logs: multiplication corresponds to addition; division corresponds with subtraction; and exponents move to the front of the log and act like coefficients. And finally, one must know and remember that logbb=1 and logb1=0, which serves as two extra approximations.

SV #2: Unit G Concepts 1-7 - Finding all parts and graphing a rational function

     This problem is about graphing rational functions. The equations is a fraction with polynomials on both numerator and denominator. The problem contains several steps in order to graph the function and it's asymptotes. In order to graph it, we must find the domain, x-intercept(s), y-intercept(s), and a few points. This problem also contains:

• a numerator with a degree of 3
• a denominator with a degree of 2
• one hole
• one vertical asymptote

The viewer needs to pay special attention to the degrees to understand the rules of finding the asymptotes.(There is no horizaontal asymptote because the degree is bigger on the top. To find the slant asymptote,, long division is used. Factor and simplify the equation, and make the denominator equal to 0 to find the vertical asymptote. The common factors are equaled to 0 and are the holes.) The viewer also needs to pay attention the graph. In order for the graph to fit, the intervals were changed: x by 1s and y by 2s. Remember that DIVAH stands for Domain Is Vertical Asymptote(s) and Holes.To find the x-intercept(s), y equals to 0, and vice versa.

SV #1: Unit 7 Concept 10 - Finding all real and imaginary zeroes of a polynomial

      This problem is about solving for zeroes and factorizing the polynomial: -35x^4+89x^3-110x^2+63x-7. The zeroes for this type of polynomial will result to be real and imaginary/complex. The factorization will also result as complex for the most part. This problem will have multiple steps starting with finding the possible real/rational zeroes and finding the possible positive/negative real zeroes. From there, the possibilities make a more precise attempts to find the zeroes, and lead to factorization.

     The viewer needs to pay special attention in finding the possible zeroes: real/rational and positive/negative. Finding the real/ration zeroes involves p/q, which is the factored last term over the factored first term. Find the possible positive and negative zeroes involves Descartes' rule of signs. The positive and negative possibilities have a small different rule throughout the process. The last thing views should pay attention is that the quadratic formula is easier to use when the polynomial is reduced to the 2nd degree and there can be no negative in the radical.

SP #2: Unit E Concept 7 - Graphing a polynomial and identifying all key parts

     This problem is about graphing a polynomial with a degree of 4 and a positive leading coefficient. In order to do so, the polynomial must first be factored and the end behavior must be determined. From the factored equation, one must solve for x (by equaling the factored portion to 0 and solving) which will be the x-intercepts/zeroes with multiplicities for the graph. To solve for the y intercept, solve f(0) in the original equation. Have the points plot on the graph and sketch.
     Remember that the even/odd of a degree and the positive/negative of the leading coefficient will determine the end behavior of a graph (of a polynomial) Also note that the multiplicity of a number directs how the graph will "react" with the x intercepts/zeroes. An x value with the multiplicity of 1 will have the graph go THROUGH the x intercept; an x value with the multiplicity of 2 will BOUNCE off the x intercept; and finally, a multiplicity of 3 will CURVE through the x intercept. Remember that the graph must lead toward the end behavior, meaning don't make extra or minimize moves to reach the end behavior.

WPP #4: Unit E Concept 4 - Maximizing Area

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WPP #3: Unit E Concept 2 - Path of Ice-Cream Pint

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SP #1: Unit E Concept 1 - Graphing a quadratic and identifying all key parts

*Notice- the scale on the graph was modified for the function to fit. There is color coding to distinguish: blue- parent graph; yellow- vertex; orange- y intercept; purple- axis of symmetry: and pink- x intercepts.

Finding the Parent Graph: Solving 3x^2-18x-22=5

1. Add 22 to both sides and factor out a 3. 

2 a. Use the vertex form equation [(b/2)^2] to create a prefect square, which will end up with 9. 

b. 9 will be added to both sides in the boxes and the prefect square can be completed on the right hand side: 3(x-3)^2. 

c. Add/multiply the numbers on the left to get 54.

3. The equation 3(x-3)^2=54 is what is left. Subtract 54 to get the parent graph: f(x)= 3(x-3)^2-54

Solving 3x^2-18x-22=5 for X Intercept/s 

*Do the previous 1-2 steps.

.4. The equation left off was 3(x-3)^2=54. Solve for x algebraically to find the x intercepts. There will be three possibilities: two answers, one answer, or none (imaginary numbers). Ex: (answer value, 0)

Make the original standard from equation equal to 0 and solve to find the y intercept. The vertex will be (h,k) from the parent graph and the axis of symmetry will be x= "h".

This problem is about sketching quadratics in standard form (f(x)= ax^2+ bx+ c) by converting it into parent form (f(x)= a(x-h)^2+k) for simplicity. In order to do so, one must fully complete the square. With the parent graph, you'll need to find the vertex, y-intercept, axis of symmetry, and x-intercepts in order to graphs. Several algebraic steps will be needed to be processed to find few of the requirements.

Pay special attention that (h,k) from the parent function form are the (x,y) values of the vertex. For "k", simply shift up or down with the corresponding number value. For "h" make x-h=0 and solve (note: the value from h should be the opposite from the original value) Also remember that the quadratic function will have two, one, or none (imaginary numbers) x intercepts. And thirdly, if "a" is (+) positive, then the graph will have a minimum, since the vertex will be on the bottom of the curve. If "a" is (-), then the graph will have a maximum, since the vertex will be on the top of the curve. With the parent form, the sketches will be more accurate and detailed.

WPP #2: Unit A Concept 7 - Profit, Revenue, Cost

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WPP #1: Unit A Concept 6 - Linear Models

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