Assignment 42
"Numbers are intellectual witnesses that belong only to mankind."  (Honore de Balzac, 17991850)
World loses ten days: In 1582, scientific thinkers concluded that the Julian calendar (established by Roman ruler Julius Caesar in 46 B.C.) was ten days out of line with the seasons. So Pope Gregory XIII, using advice and assistance from German mathematician Christoph Clavius (15371612) ordered a new calendar (the Gregorian calendar) put into effect. When people went to bed on October 5, 1582, they woke up the next morning on October 15, 1582. The Christian world actually lost ten calendar days.
Not all people use the same calendar. Even today, some branches of the Eastern Orthodox Church still determine their holidays according to the Julian calendar, which is now 13 days different from the Gregorian. The Hebrew calendar counts years from the day Judaism considers the day of creation, which some ancient scholars say took place in the year 3761 B.C. The Muslim calendar, called the Hijra, numbers its years from A.D. 622, the year the Muslim religion was founded. The Hijra year contains only 354 days because it is based on the phases of the moon rather than of the sun.

Why did Herkimer faint when he looked at a picture of Mickey Mouse?
Answer: He had a Disney spell.
Herky's friends:
MR. REE...this guy loved suspense stories.
MYRTLE BEECH...she loved to visit seaside resorts in South Carolina. 
ASSIGNMENT #42
Reading: Review Section 6.9, as necessary.
Written: On page 383, use your calculator to find the cubic polynomial for the graphs in problems 14, 15, 16. On page 385, do problems 47,48,49. Optional: Solve the investment problem stated in ITEMS FOR REFLECTION. (NOTE: We will not use the concept of finite differences.)

Mathematical word analysis: SYMMETRY: From the Greek roots sum and metros. The prefix refers to things which are alike, while metros is the Greek word for measure. 
Here are some (x,y) values:
x 
1 
2 
3 
4 
5 
6 
y = f(x) 
1 
1 
3 
7 
5 
9 
If you will do a scatterplot on your calculator, you should be able to observe that neither a linear or a quadratic function would be a good fit for a data. The data does display two turns, which suggests that a cubit function might fit well. It turns out that if you do fit a cubic to the data, you get
y = ax^{3} +bx^{2}+c+d a = 1 b= 9 c= 22 d= 15 R^{2}= 1
In this case, the cubic function y = f(x) = x^{3}  9x^{2} + 22x  15 contains all of the indicated points.
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Problem: I want to be able to withdraw an amount of $10,000 eight years from now, an amount of $5,000 nine years from now, and an amount of $2,000 ten years from now. What amount, x, must I deposit right now in order to make the indicated withdrawals if my deposit earns an annual interest rate of 6%?
Withdrawals 








$10,000 
$5,000 
$2,000 
Deposits 
x 










Year 
0 
1 
2 
3 
4 
5 
6 
7 
8 
9 
10 
Solution (with communication):
If x is the required amount, then
x = $10,000(1.06)^{8} + $5,000(1.06)^{9} + $2,000(1.08)^{10} = $10,350.41. 
