Fibonacci Beauty Ratio

Measurements of Friends:
Mia
Foot to Navel: 103 cm
Navel to top of Head: 63 cm
Ratio: 103/63= 1.635 cm
Navel to chin: 44 cm
Chin to top of head: 23 cm
Ratio: 44/23= 1.913 cm
Knee to navel: 57 cm
Foot to knee: 50 cm
Ratio: 57/50= 1.14 cm
AVERAGE: 1.563 cm

Eriq
Foot to navel: 109 cm
Navel to top to head: 65 cm
Ratio: 109/65= 1.677 cm
Navel to chin: 47 cm
Chin to top of head: 25 cm
Ratio: 47/25= 1.88 cm
Knee to navel: 58 cm
Foot to knee: 54 cm
Ratio: 58/54= 1.074 cm
AVERAGE: 1.544 cm

Sarahi
Foot to navel: 99 cm
Navel to top of head: 66 cm
Ratio: 99/66= 1.5
Navel to chin: 41 cm
Chin to top of head: 22 cm
Ratio: 41/22= 1.864 cm
Knee to navel: 50 cm
Foot to knee: 45 cm
Ratio: 50/54= .9259
AVERAGE: 1.430 cm

Kelsea
Foot to navel: 100 cm
Navel to top of head: 58 cm
Ratio: 100/58= 1.724 cm
Navel to chin: 46 cm
Chin to top of head: 21 cm
Ratio: 46/21= 2.190 cm
Knee to navel: 49 cm
Foot to knee: 49 cm
Ratio: 49/49= 1
AVERAGE: 1.638 cm

Jenny
Foot to navel: 95 cm
Navel to top of head: 56 cm
Ratio: 95/56= 1.696 cm
Navel to chin: 36 cm
Chin to top of head: 23 cm
Ratio: 36/23= 1.565 cm
Knee to navel: 47 cm
Foot to knee: 48 cm
Ratio: 47/48= .979 cm
AVERAGE: 1.413 cm

   Based on the calculations, Kelsea is the most beautiful since her average is close to the Fibonacci number or the Golden Ratio of 1.618. Her average calculations were 1.638 cm. Because she is considered the most beautiful based on the Golden Ratio, this means that her body is symmetrical and proportional. Not only her body is proportional, but also her face. When faces and bodies are in proportion, we, as human beings, recognize it as the body to be healthy and likewise, the body and face to be very appealing and attractive to the eye. However, in my perspective, I believe that to be beautiful depends on who the person really is. Yes, there are times where someone can catch your eye in a split second but by the end of the day, if that person has no sweet or great personality that attraction means nothing. To be beautiful is to have a beautiful heart. To be beautiful is to also have a kind and courageous thought; no hate.

Fibonacci Haiku: Damon Salvatore

Vampire

Diaries

Damon Salvatore

Most attractive vampire

He has beautiful eye color

He's the peanut butter to my grape jelly

http://images.wikia.com/vampirediaries/images/c/c0/Damon-Salvatore-damon-salvatore-16725155-1280-720-1-.jpg

SP #5: Unit J Concept 6: Partial Fraction Decomposition with repeated factors

Steps and Explanations!!

1. Like in Concept 5, decompose the problem and find your like terms and system.

2. The hardest part of this problems is canceling and knowing what to cancel. There's many different ways to cancel the variables but for now, try to follow my work with the whole canceling process. 

3. Once you found "A", plug it in to one of the systems and you'll also find "B". Repeat the same process with different systems plugged in to find variables "C" and "D". 

4. Once you found what your variables equal to, double check your answer with matrix and rref in your graphing calculator.

   The viewer needs to pay attention to the canceling process and understanding how and what is multiplied in order to cancel out a certain variable. As the viewer figures out that "A" is, try to keep the same denominator throughout the process (which the denominator should be 27.) Also, when using rref to check the answer, the graphing calculator will not give a fraction in the matrix but instead a decimal. The answer will still be correct if it corresponds to the fractions of A, B, C, and D.

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

Steps and Explanations!!

1. Compose the problem by multiplying the common denominator to each fraction (numerator).

2. Multiply the denominator to its common denominator also.

3. After multiplying the common denominator, write out the complete factors of the numerator and denominator.

4. Decompose the problem by rewriting the numerator and write the common denominator at the bottom.

5. To decompose, use letters A, B, and C. Find its common denominator and multiply.

6. Then set it equal to the previous answer that you found (the compose part).

7. Combine like terms to the appropriate number that you set equal to (ex. 9x^2 = something that has x^2).

8. Find the "system" by crossing out the x's.

9. Rewrite the system with the matrix format and plug it in to your graphing calculator. It should have the same answer that you started with in the problem.

       The viewer needs to pay attention to the common denominator since mistakes can easily be found when multiplied incorrectly. Make sure to know the difference between "compose" and "decompose". Matrix can help the viewer know if the answer is correct. Thus, it needs to match to what the problem started with.

SV #5: Unit J Concept 3-4: Solving Three Variable Systems (Matrix)

**Sorry everyone about the video! Something went
wrong with the downloading part of the video so that's why you can't 
watch it right side up. I apologize :(

    The viewer needs to pay attention to the rows and where each equation
best fits in which row. Remember, we are trying to get zero on a specific
row and term! Also, before you actually start the problem, it'll be easier
if you factor the equation before hand. Near the end of the matrix, the
viewer should have the stairstep of 1's (in this problem at least because
it is an consistent independent system). If you want to double check your
answer, you may use your graphing calculator to make sure if your answer
is correct. In this case, the viewer should use the Gauss-Jordan elimination
system on his or her graphing calculator. Thanks for watching!