This week we are learn about about Sine Law and Cosine Law.

The formulas are

Sine Law: to use the sine law you need to know two angles and one side of the triangle  or two sides and an angle opposite one of them. It’s used to find a side or an angle of a triangle. It has two versions of the formula, you use it depending on what it is that you’re looking for, a missing angle or a missing side.

or 

Cosine Law: you can use cosine law when the lengths of two sides and the measure of the included angle is known or the lengths of the three sides are known.  It’s used when you need to find a third side of a triangle, when the angle opposite to the side is given. If we also want to find the length of the third side, we can just change the formula to solve for the variable.

Thank you

This week in math we learned about angles in standard form. First, we need to know some terminology. The rotation angle is the angle formed between the initial arm and the terminal arm. We would draw a sketch of this on a graph, that has the x-axis and the y-axis. This initial arm is the arm that is on the right side of the graph, this is where we always start when drawing a sketch. The terminal arm is the line you draw that represents the angle of the sketch you’re drawing. The initial arm is 0°, then as we move around counter-clockwise, the next arm at the top is 90°, then on the left, 180°, then at the bottom, 270° and back to the initial arm.

Standard position is when the initial arm is on the positive x-axis, the rotation is about the origin. The reference angle is that of the terminal arm and where it creates a right-triangle with the x-axis. We can use this right-triangle to find more information about it, using sine, cosine, tangent, and Pythagorean theorem. The reference angle can be found by adding or subtracting your rotation angle from the arm’s degree that is closest to it (e.g. 180° or 360°). This will make more sense when we look into an example. Quadrant I is the top right corner, quadrant II is the top left corner, quadrant III is the bottom left corner, and quadrant IV is the bottom right corner. You will need to know the quadrants to tell which quadrant an angle is in. Lastly, angles with the same terminal arm are co-terminal angles. These angles add to 360° when they are both positive, but don’t forget that one of the angles is always negative and the other is positive, so make sure if you subtract your rotation angle from 360, then add a negative sign to your answer.

Adding and subtracting rational expressions with monomial numerators 

First, we should mention what a rational expression is that has a binomial denominator. A rational expression is that has a binomial denominator is an expression that has a binomial, in the place of the denominator. The binomial can be any numbers in addition or subtraction of any variable of any power.

This week in math, we learned how to multiply and divide monomials. The first step is always to factor if possible, we always need to take a look at what we’re doing; either multiplication or dividing. From previous math, we know that if it’s dividing then we need to reciprocate the fraction. The next step is to simplify all like terms, you always need to find what X can’t equal to, because the denominator cannot equal 0.

In the examples below, i forgot my non-permissible values, for the first one it’s b cannot equal 0, and same for the second one, except B and A cannot=0

This week in Pre Cal, we learned how to multiply and divide rational expressions. Rational expressions are equations or quotients that have polynomials on most likely both the bottom and top, each with their own variables. It would be very difficult if not impossible to divide a polynomial by another polynomial without using a calculator. This is why we have to take steps in finding and making binomials that are easy to cancel out on both the bottom and top.

Disclaimer: When canceling out polynomials or numbers, one must be on the top of the division and the other on the bottom. It does not matter if the polynomial or number is apart of the same quotient, as long as one is somewhere on the top, and the other somewhere on the bottom. Also, if you must divide by another fraction/quotient, flip the fraction around to make it a multiplication. Ex. 1/2 ÷ 2/3 –>  1/2 x 3/2

First, we must find the “non-permissible” values, which are essential ‘x’. Contrary to the last unit, instead of ‘x’ equalling only specific numbers, in this chapter, ‘x’ cannot equal specific number. Therefore, ‘x’ would be 0 making, for example, the equation 3/0 which is impossible. Next, you must find alike polynomials, mostly binomials, by factor them. We try and make it so that a polynomial on the top and the bottom both have the same polynomial so that they can cancel each other out, making the equation easier to solve. Watch out for negatives (-) or any small details because when you cancel out polynomials, they both have to be the exact same. Then after canceling out all that is possible, you should be left with regular, real numbers. If so, you are allowed to cancel those out too. Once finished those steps, you should be left with an easy multiplication that will give you your answer.

Q1.) 16/45 x 25/42 x 21/24 x 12/8

Ans.) 15/18

This week in math lesson 8.1 we learn how to solving absolute value equation algebraically. when solving with an absolute value of a square root you may come across extraneous solution which are solution that are not true solution  of the equation given algebraically we can find out whether a solution is extraneous by verifying and inserting the “x” value beck into the original equation to see if the equation is true or to see whether both sides equal each other.

Q1.) l 2x–3l =7

Ans.) 2x –3–7=0

2x–10=0

ii.) x=5 or x=–2

Ans.) 2(5)–3–7=0

10–3–7=0

10–10=0

This week we have been doing everything related to inequalities.

We first started with graphing linear inequalities. For this we returned to the idea of graphing regular linear equations.

If we take a look at the equation y=x-2 we know that the slope is 1/1 and the y-intercept is -2

Q1.) y ≥ 3x² – 4

a.) (–3,8)

Ans.) 3(–3)² –4

27–4

y≥23

b.) (0,5)

Ans.) y≥3(0)²–4

5≥0–4

c.) (4,44)

Ans.) 44 ≥ 48 –4

44≥44

In week 10 we were studying for the midterm. I studied unit 3 and a little of 4 because I did not do so well on those sections before.

Unit 3 was quadratic equations. The different ways to solve one is by factoring, quadratic formula and completing the square.

One thing that I did struggle with while studying was the word problems that made you put numbers into a quadratic equation.

Q1.) solve inequality quadratic.

i.) (x-2)(x-6)≤0

Ans.) y= -x² +10x-16

x>2,    x<8xER

Q2.) -5x²>17x-12

Ans.) -5x²-17x+12

(-5x-3)(x+4)

x<3/5 ,   x>-4