Week 12 – Intercepts

This week in math we learned about intercepts. Intercepts are the points on the graph where the line touches either the x or y axis. You can also find the x or y intercepts given a equation. This is done by replacing either the x or y values with zero and solving for x or y.

To find the x intercept, you replace the value of y in your expression with zero. If you are finding the y intercept, you replace the value of x with zero. Then solve the expression.

Example equation: 3x + 2y = 24

X- intercept:

3x + 2(0) = 24

3x = 24

x= 8

X-intercept = (8, 0)

Y- intercept:

3(0) + 2y = 24

2y = 24

y = 12

Y-intercept = (0, 12)

Week 11 – Factoring Polynomial Expressions

This week in math we learned about the steps for factoring polynomial expressions. These steps can be remembered using the acronym CDPEU. It’s not necessary to remember it but if it works for you, use it.

C – Common Factors

D – Difference of squares

P – Patterns within the polynomial

E – Easy – Is the leading coefficient 1 and is the expression easy to factor

U – Ugly – The expression is not easy to factor and will require further work

Also always remember to fully factor/simplify your expressions.

Example: 8$x^4$ + 10$x^2$-3

(4$x^2$ -1) (2$x^2$ + 3)

This can then be simplified to:

(2x + 1) (2x – 1) (2$x^2$ + 3)

Week 10 – Difference of squares

This week in math we looked over the difference of squares, a a concept we discovered in week 9 where you use the method as a way of skipping the distributive method while evaluating your expression.

The method works as follows:

(a+b) (a-b)

= $a^2$ + ab – ba – $b^2$

= $a^2$$b^2$

This can be used to work backwards as well, when factoring expressions.

Example:

75$x^2$$y^2$ -3

First we remove the common factor,

3 (25$x^2$$y^2$ -1)

Then we factor.

3 (5xy – 1) (5xy + 1)

Week 9- Recognizing Polynomial Patterns

This week in math 10 we learned how to recognize patters within polynomial expressions and evaluate them. These patterns can be remembered and then applied to skip over distributive property steps and evaluate the expressions quicker.

The patterns:

(a+b) (a-b)

= $a^2$ + ab – ba – $b^2$

= $a^2$$b^2$

$(a+b)^2$ = (a+b) (a+b)

= $a^2$ + ab + ba + $b^2$

= $a^2$ + 2ab + $b^2$

$(a-b)^2$ = (a-b) (a-b)

= $a^2$ -ab -ba + $b^2$

= $a^2$ -2ab + $b^2$

Week 8- F.O.I.L

This week in math 10 we learned about multiplying polynomials, distributive property, and the acronym FOIL.

The acronym FOIL stands for First terms, Outside terms, Inside terms, and Last terms. This is in regards to which terms you multiply and in what order, before you combine like terms.

Distributive property (FOIL) with binomials would look like: (a+b) (c+d) = ac + ad + bc + bd.

Example: In this expression, you begin by multiplying 3x and 2x, the FIRST terms, then 3x and 5, the OUTISDE terms, then 2 and 2x, the INSIDE terms, and finally 2 and 5, the LAST terms.

Week 7 – Calculating unknown angles (theta)

This week in math we learned how to calculate unknown angles of triangles given the sine, cosine and tangent ratios. The easiest way to remember these is the acronym SOH CAH TOA. The first letter stands for the ratio, and the second and third letters represents what lengths you put into the expression. When the expression is written, the second letter is in the numerator, and the third goes in the denominator.

To use the ratios you have to have your triangle labelled correctly. Then you can use the opposite, adjacent, and hypotenuse lengths to find your angle. These are the lengths represented by O, A, and H in SOH CAH TOA.

Example: In the triangle (see below) angle x is our unknown angle. Because we have at least 2 side lengths, we can find it. In this case the opposite and adjacent lengths will be used, so the tangent ratio will be the one we use. Then you simply set up the equation: $tan^-1$ (5/6), and plug it into your calculator. This gives us roughly 39.8° which will be rounded to 40°.

Week 6 – Slant Height, Pyramids, and Cones

This week in math 10 we learned about right pyramids, cones, and how to calculate slant height.

To calculate the slant height of right pyramids, you need to have the height and half of the length. Then you can use the Pythagorean theorem to find the hypotenuse (your slant height). Then you can use your slant height when you calculate the total surface area or volume.

Example: When you have the height (3 in this case) and the side length (8), you can calculate the slant height. Take half the side length (4) and use it and the height in the Pythagorean theorem. In this case $3^2$ + $4^2$ = 25 = $c^2$. Therefore 5 = c, and 5 is your slant height. You can then plug the slant height into the surface area and volume equations.

The Pythagorean Theorem principle is also used when calculating the slant height of a cone. The only difference is instead of using half the side length and the height to find the hypotenuse (slant height), you use the radius of the base and the height.

Week 5 – Converting Units of Measurement

This week in math we learned how to convert units of measurement in a simple way between both longer/shorter units within the same system, and between the metric and imperial systems.

Easy to do, you simply need to know how many of one unit is equal to the other unit.

Example: The units that you wish to convert go on the left (6780cm), the units that you wish to convert to go in the numerator position (1m) and however many units of one equal the other (100cm) in the denominator.

The same can be done for converting units between the metric and imperial systems.

Example: If you know 1ft is equal to 0.3048m, then you repeat the same steps again.

And evaluate.

Week 4 – Integral Exponents

This week in math 10 we learned about integral exponents (negative exponents) and how to get rid of them.

To get rid of negative exponents you need to take the base and flip it.

Example: Whatever your variable or base is, it on it’s own is equal to itself over 1. You just don’t write it as such. So when you flip it, you take whatever is in the numerator, and move it to the denominator, or the other way around. This switch makes the exponent positive.

This same principle of moving a base with an integral exponent to the other position applies to solving other expressions as well. This includes expressions where there are bases/exponents in both the numerator and denominator.

Example: Using the same principle of moving a base with a negative exponent to the other position, we move our negative exponent from the numerator to the denominator, changing the negative to a positive, and then evaluating as you would a normal expression.