Assignment 32

"It is by the aid of statistics that law in the social sphere can be ascertained and codified." -- (Florence Nightingale, 1820-1910)

Math History Tidbit:

Claudius Ptolemy (Second century A.D.): He produced a cosmological model of the universe that had the earth surrounded by eight concentric spheres that carried the moon, the sun, the five planets known at the time, and the stars (on the outermost sphere). The model was embraced by Christian thinkers since (a) it made Earth the center of the universe (and the center of God's attention), and (b) provided room outside the sphere of stars for Heaven. The Christian Church accepted this earth-centered model for about 1500 years.

Interestingly, some early Greek mathematicians suggested the possibility that the earth revolved around a stationary sun, contrary to the model produced by Ptolemy. Beginning with Nicholas Copernicus (1473-1543), mathematicians and astronomers began to produce solid evidence that Earth was not the center of the universe. The Church was extremely reluctant to accept these findings.

 

Herkimer's Corner

What did one of Herkimer's pet fleas say to the other flea when it wanted to go outside?

Answer: "Shall we hop or take the dog."

Herky's friends:

CAL O'REE... he is always concerned about his weight.

DEA KLINE... she never accepts things offered to her.

ASSIGNMENT #32

Reading: Section 5.7, pages 299-303 (review).

Written: Page 304/31,32,33,34,35,36,37,51. SOLVE THE SYSTEMS GRAPHICALLY. WRITE NEATLY!

Items for reflection:

Mathematical word analysis:
PARALLEL: From the Greek word parallelos (beside one another). Parallel lines are indeed "beside one another."

Here is an example showing how one can logically solve an algebraic inequality:

Solve for x: x2 + 3x - 10 > 0.

Solution: x2 + 3x - 10 > 0

==> (x - 2)(x + 5) > 0

==> (x - 2) > 0 and (x + 5) > 0 OR (x - 2) < 0 and (x + 5) < 0

==> x > 2 and x > -5 OR x < 2 and x < -5

==> x > 2 OR x < -5.

Conclusion: The conditional statement is true if x is a number greater than 2 OR x is a number less than -5.

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Returning to quadratic equations, here is a TI-83 program that will produce real number solutions for the quadratic equation Ax2+Bx+C = 0 if you input values for A, B, and C. This program does not identify complex number roots.

PROGRAM: AQUAD
: Prompt A,B,C
: B^2-4*A*C -> D
: If D < 0
: Then
: Disp "NO SOLUTIONS"
: Else
: (-B+÷
(D))/(2*A) -> P
: (-B-÷
(D))/(2*A) -> Q
: Disp "ROOTS ARE ",P," AND ",Q
: End

Problem: Solve x2 - 7x < 0.

Solution (with communication):

x2 - 7x < 0

==> x(x - 7) < 0.

Since x - 7 is less than x, the only way the conditional statement can be true is if

x - 7 < 0 or x > 0

==> x < 7 or x > 0.

In other words, the conditional statement is true if x < 7 or if x > 0.

Problem: Describe the solution set defined by

x2 - 4 > 0 and 4 - x2 < 0

Solution (with communication):

Graph the parabolas y = x2 - 4 and y = 4 - x2. The requested solution set is all points above the parabola y = x2 - 4 and below the parabola y = 4 - x2.