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In the early days, one orange, 3 oranges, ..etc. any of the natural numbers, the negatives of these numbers, or zero. natural numbers (a positive integer). Root, Radix, Radicals. Real Number VS. Imaginary (complex) Number.

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In the early days, one orange, 3 oranges, ..etc.

any of the natural numbers, the negatives of these numbers, or zero

natural numbers (a positive integer)

Root, Radix, Radicals

Real Number VS. Imaginary (complex) Number

Ninth-century Arab writers called one of the equal factors of a number a root, and their medieval translators used the Latin word radix (“root,” adjective “radical”).

Real number = integer-part + fractional-part

Surds: an irrational root such as √3

lacking sense : IRRATIONAL; absurd

fractional index

4 + 2i

all have to be either -2 or +2

Radicals become easier if you think of them in terms of indices. Think instead of

real part

imaginary part

http://www.itc.csmd.edu/tec/GGobi/index.htm

Rational Numbers VS. Irrational Number

100.3; 1/6 = .16666; 2/7 = .285714285714

Number that can’t be expressed as p/q. Not a quotient of two integers

2½ = 1.4142135623730950488016887242097….

 (3.1416…) http://mathworld.wolfram.com/Pi.html

Approximating irrational number by rational numbers: number theory


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How do you represent large multiples such as 2x2x2x2  takes too much space to print 2x2x2x2 = 24 the birth of exponential notation (base, exponent or index (indices))

Now we need a set of rules to figure out what things such as is 22 x 23 Or 23 x 32

Properties of exponents

Logarithms: Math based on the exponents themselves, invented in the early 17th century to speed up calculations. Also from the result of the study of arithmetic and geometric series.

(study tip: the exponent is the logarithm).



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8 x 9 = 4 x 2 x 9 = 4 x 18  18 – 4 = 14


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Page 29 example 9

1 x 25 = 1 x 5 x 5  5 + 5 = 10

16 x 1 = 4 x 4  4 + 4 = 8

Page 30 example 12, 13

1 x 12 = 3 x 4  3 + 4 = 7

2 x 15 = 2 x 3 x 5 = 6 x 5  6 - 5 = 1


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Page 40 example 7

Page 39 example 6


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Page 42 – Example 9

Page 42 – Example 10

Combine the numerator terms


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Climbing the mountain on a straight slope

On the 2nd day You climbed 4 miles vertically

On the 2nd day, You covered 3 miles horizontally

3rd day end point

2nd day start point

How far did we walk ?


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Slope of a Straight Line

tangent

 Equation of a straight line

  • We need to know two points (locations)

    • The second day’s starting location and ending location

  • Can you identify the right-triangle in the previous slide?

  • Can you identify the right-angle?

  • It is customary to denote the slope of a straight line by “m”

Now that we know there is a right-triangle, how far did we walk ?


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Given two point on a line, what is the distance between the two points?

Vertical distance

B

4

A

We were only given points A and B. Using A and B we could simply figure out point C. Point C is same height as point A but it is (10 – 4) or 6 units away from A

C

(10,3)

6

Horizontal distance

We can find AB using the Pythagoras’ theorem

Mid-point of line segment AB 


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Equation of a line two points?

slope

y-intercept

x and y are variables -- various points along the line

Slope of a line joining points (0,c) and (x,y)

x can’t be 0

Point (0,c) lies on y axis


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Properties of two points?

Set y = 0 to find the x-intercept

Set x = 0 to find the y-intercept

When m = 0

no incline, line is parallel to x-axis. No x intercept

y = c

M can’t be 0

y = c

No gradient (undefined), straight up, perpendicular to the x axis. No y intercept. Parallel to y axis. x = k.

c is y-intercept

Ex: (1,2), (-1,2), (5,2)…


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Typical problems involving straight lines? two points?

  • Find whether 4 points form a parallelogram.

    • Method1: Calculate the distances between them to see if AB = DC and CB = DA

    • Method2: Using mid-points

      • If the mid-points of the diagonals AC and BD bisect each other then ABCD is a parallelogram

    • Mehtod3: Using gradients

      • If the gradients of AB and DC are same

Matlab: plot([-1,1,5,3,-1],[-2,1,3,0,-2])


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Example: given gradient, and a point on the line, find the line’s equation

Slope of the line is given

The lines passes through (2,1)

P(x,y)

 y=2x-3

A(2,1)

Try this: (-2,3); m = -1 y = -x + 1

Example: given two points on a line, find the line’s equation

Step2: once m is known, use the same equation and one of the points to find the equation

Step1: given two points, it is easy to find m

Try this: (3,4), (-1,2)  2y = x + 5