 55 lessons
 1 quizzes
 10 week duration

Numerical Computation
Here you'll be introduced to the bare basics of mathematics. Topics include commonly used words and phrases, symbols, and how to follow the order of operations.

Measurements
An introduction to numerical computation. Emphasis is placed on scientific and engineering notation, the rule of significant figures, and converting between SI and Imperial units.

Trigonometry with Right Triangles
Here we focus on right angle triangles within quadrant I of an xy plane. None of the angles we evaluate here are greater than 90°. A unit on trigonometry with oblique triangles is covered later.

Trigonometry with Oblique Triangles
This unit is a continuation of trigonometry with right triangles except we'll extend our understanding to deal with angles *greater* than 90°. Resolving and combining vectors will be covered at the end of this unit.

Vector Analysis

Introduction to Algebra

Factoring

Solving Equations

Functions and Graphs

Geometry
This unit focuses on analyzing and understand the characteristics of various shapes, both 2D and 3D.
 Identify, measure, and calculate different types of straight lines and angles
 Calculate the interior angles of polygons
 Solve problems involving a variety of different types of triangles
 Calculate the area of a variety of different types of quadrilaterals
 Solve problems involving circles
 Calculate the areas and volumes of different solids

Introduction to Statistics
Multiply and Divide Algebraic Fractions
A key component to multiplying and dividing algebraic fractions is knowing how to do it to ordinary fractions. That being said, you’re first expected to review how it’s done before continuing. If after a few examples you feel confident enough, you may skip it.
Multiplying Algebraic Fractions
Just as you would multiply ordinary fractions as discussed above, algebraic fractions are no different. Take, for example, the expression:
$\frac{2a}{3b}\xb7\frac{5c}{4a}\phantom{\rule{0ex}{0ex}}$To find the product of these monomials, you multiply the numerators separately from the denominators.
Numerator product: 2a × 5c = 10ac (confused? review this lesson)
Numerator product: 3b × 4a = 12ab
$\frac{10ac}{12ab}=\frac{5\overline{)a}c}{6\overline{)a}b}=\frac{5c}{6b}\phantom{\rule{0ex}{0ex}}$Let’s try another example:
$\frac{x}{x+2}\xb7\frac{{x}^{2}\u20134}{{x}^{3}}\phantom{\rule{0ex}{0ex}}$After multiplying the numerators and denominators, you should end up with the following expression ↓. You might be tempted to expand x(x² − 4), but instead, you could cancel out this x with 1 of 3 x‘s found in the denominator (x³).
$=\frac{x\left({x}^{2}\u20134\right)}{\left(x+2\right){x}^{3}}\phantom{\rule{0ex}{0ex}}=\frac{\overline{)x}\left({x}^{2}\u20134\right)}{\left(x+2\right){x}^{\overline{)3}2}}\phantom{\rule{0ex}{0ex}}=\frac{{x}^{2}\u20134}{x+2}$
You may think you’re done at this point, but recall that x² − 4 is a difference of squares. Therefore:
$\frac{\left(x\u20132\right)\left(x+2\right)}{x+2}\phantom{\rule{0ex}{0ex}}=\frac{{\displaystyle \left(x\u20132\right)\overline{)\left(x+2\right)}}}{{\displaystyle \overline{)\left(x+2\right)}}}\phantom{\rule{0ex}{0ex}}=x\u20132\phantom{\rule{0ex}{0ex}}$Try this final example on your own:
$\frac{{x}^{2}+x\u20132}{{x}^{2}\u20134x+3}\xb7\frac{2{x}^{2}\u20133x\u20139}{2{x}^{2}+7x+6}\phantom{\rule{0ex}{0ex}}$Answer:$\frac{\stackrel{\overline{)1}}{\stackrel{\u23de}{{x}^{2}+x\u20132}}}{\underset{\overline{)3}}{\underset{\u23df}{{x}^{2}\u20134x+3}}}\xb7\frac{\stackrel{\overline{)2}}{\stackrel{\u23de}{2{x}^{2}\u20133x\u20139}}}{\underset{\overline{)4}}{\underset{\u23df}{2{x}^{2}+7x+6}}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
 (1) and (3) factor by trialanderror.
 (2) and (4) factor by decomposition.
Your final answer should be 1.
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Dividing Algebraic Fractions
As with any fractions being divided, the second fraction in the expression is always reciprocated (flipped) and the sign changed to multiplication.
$\frac{a}{b}\xf7\frac{c}{d}\phantom{\rule{0ex}{0ex}}=\frac{a}{b}\times \frac{d}{c}\phantom{\rule{0ex}{0ex}}=\frac{ad}{bc}\phantom{\rule{0ex}{0ex}}$Let’s look at an easy example containing algebraic terms.
$\frac{x}{y}\xf7\frac{x+2}{y\u20131}\phantom{\rule{0ex}{0ex}}$Here we’re expected to divide and simplify. We first reciprocate the second fraction, then change the symbol to multiply. You’ll also notice that further simplification isn’t needed.
$\frac{x}{y}\times \frac{y\u20131}{x+2}\phantom{\rule{0ex}{0ex}}=\frac{{\displaystyle x\left(y\u20131\right)}}{{\displaystyle y\left(x+2\right)}}\phantom{\rule{0ex}{0ex}}$Here’s another example:
$\frac{{x}^{2}+x\u20132}{x}\xf7\frac{x+2}{{x}^{2}}\phantom{\rule{0ex}{0ex}}$You might notice that x² + x – 2 can be factored by trialanderror into (x + 2)(x – 1). Also, after flipping the second equation, x² cancels out with x and (x + 2) in the denominator with (x + 2) in the numerator:
$\frac{\left(x+2\right)\left(x\u20131\right)}{x}\times \frac{{x}^{2}}{x+2}\phantom{\rule{0ex}{0ex}}=\frac{{\displaystyle \left(x+2\right)\left(x\u20131\right){x}^{2}}}{{\displaystyle x\left(x+2\right)}}\phantom{\rule{0ex}{0ex}}=\frac{{\displaystyle \overline{)(x+2)}\left(x\u20131\right){x}^{\overline{)2}1}}}{{\displaystyle \overline{)x}\overline{)(x+2)}}}\phantom{\rule{0ex}{0ex}}=x\left(x\u20131\right)\phantom{\rule{0ex}{0ex}}$Simplify Complex Fractions
Fractions that have only one fraction line are called simple fractions. Fractions with more than one fraction line are called complex fractions. Examples are shown below:
$\overline{)1}\frac{{\displaystyle \frac{a}{b}}}{{\displaystyle \frac{c}{d}}}\mathrm{or}\overline{)2}\frac{{\displaystyle \frac{a}{b}+\frac{c}{d}}}{{\displaystyle \frac{e}{f}}}\phantom{\rule{0ex}{0ex}}$The general method to tackling these types of fractions is to first convert them into a format that’s understandable to you. Example 1 can be rewritten as:
$\frac{a}{b}\xf7\frac{c}{d}\phantom{\rule{0ex}{0ex}}$and example 2 as:
$\left(\frac{a}{b}+\frac{c}{d}\right)\xf7\frac{e}{f}\phantom{\rule{0ex}{0ex}}$Notice how the numerator was placed in parentheses. This suggests that you have to combine the terms in the numerator before you can start dividing. Let’s look at examples with a thorough stepbystep explanation: