# Author: madissonr2016

## Week 17 – Sine Law

This week in pre-calc I learned how to use the sine law for triangles. This can be used when your triangle isn’t a right triangle, there needs to be an angle and on the side across from that angle, a side length and another side or angle given. If there isn’t these 3 clues, then you can’t use the sine law to solve your triangle for the missing side or angles.

When you have your triangle, for example <ABC, angle A, the side across from that is side a, across from angle B is side b and angle C is side c. The sine law is very easy. When solving for a missing side, you put side a on top and it’s being divided by angle A this is equal to side b divided by angle B and thats equal to side c divided by angle C. For a missing angle, the angle is on top and the side length is on the bottom. All the fractions are equal to each other so out of the 3 possible fractions, you only need to use 2, a left side equal to a right side.

With the sine law, you use the hints they give you, the angle with the side length and another clue. When you fill in the missing sides and angles, you can see which 2 fractions you need to use. From there, you need to isolate the variable by itself so the whole equation is trying to solve for that missing value. If the 3rd clue given is another angle, you can simply add the 2 angles together and subtract them form 180 because all triangles have to add up to 180 degrees.

## Journal 3

## Week 15 – Multiplying and dividing rational expressions

This week in pre-calc we learned how to multiply and divide rational expressions together.

You always want to state your non-permissible values first.

For multiplying, the first step to make simplifying your answer easier is to start by reducing your expression with a common factor. Doing this makes multiplying the numbers together a smaller answer and seems better. To reduce with a common factor, you need to use the numerator and find a common factor with a denominator from another expression.

After you’re done reducing your fractions, you can multiply straight across, numerator x numerator and denominator x denominator. Once you’re done that, you can see if you can simplify any more and get the lowest possible fraction.

If you can factor the expression, start by doing that and if you can cancel out terms that are exactly alike, you can make your multiplying easier.

When it comes to dividing, you want to state the non-permissible values of only the terms after the 1st one. After stating those, you want to flip the second fraction and then factor it out. Once you do that you can cancel exact terms and multiply throughout.

It’s important to remember you cannot factor and cancel out terms then flip the second fraction because you will get a completely different answer from the right one.

## Week 14 – Adding and subtracting rational expressions and equations

This week in pre-calc, we learned how to add and subtract ration equations and expressions in 4 easy steps.

**Determine the lowest common denominator (LCD)**. The lowest common denominator is the number 2 denominators have in common when they’re multiplied. So if 3 and 4 were in the denominator, the lowest common denominator is 12. Its like lowest common multiple. The first number they have in common when multiplied.**Rewrite each equation with the LCD in the denominator**. When you find the lowest common denominator, you multiple each fraction by the LCD until you get the same denominator for all the fractions. You also need to multiply the numerator as well so you can add or subtract all your numbers over one common denominator.**Solve.**After rewriting the question, you can simply the whole equation by collecting all your like terms and adding or subtracting them together all under your LCD.**Reduce if possible.**You can reduce your equation only if it can be reduced on the top and bottom by the same number.

## Journal 2

## Week 13 – Graphing Reciprocals of linear functions

This week in pre-calc, we learned how to the reciprocals of linear functions. A reciprocal is when you have 2 hyperbolas that are reflections of each other. So if you divide the graph in a slanted direction, you get a reflection of the 2 hyperbolas, one in the positive quadrants and one in the negative quadrants.

The equation for graphing the reciprocal of a linear equation is 1/x+1 (you just take your linear equation and put it over 1.

The first step when graphing the reciprocal of a linear function is to draw out the original linear function. Once you have your line, the second step is to go to 1 and -1 on your y-axis. When you get to 1 and -1, you go across on the x-axis and find the points where the linear equation lines up with 1 and -1.

After you find where they intersect, you can draw a line that is equal distance between the point at -1 and 1 to get your vertical asymptote. This is acting as a magnetic field or a boundary that you cannot cross. It’s acting as your ‘0’ value. Your horizontal asymptote will be ‘0’ overtime for what we’re learning. This means that when you’re graphing your reciprocals, you will not be able to cross over the horizontal and vertical asymptotes. You always get closer and closer to reaching ‘0’ but you will never reach ‘0’ when graphing reciprocals.

## Journal #1

## Week 12 – Solving systems of equations algebraically

This week in pre-calc, we learned about substitution and elimination. Using substitution or elimination can help us find the points of intersection(s) for lines or parabola’s if they intersect. Doing this gives us the co-ordinates of where they intersect and how many there are.

For 2 lines, there can only be 0, 1 or infinite. Parabola’s can have 0, 1, 2 or infinite. Parabola’s and lines can have 0, 1 or 2 solutions.

Using substitution is taking one of your two equations and getting one variable all by itself so you can input the solution for that variable and replace it with any other variable that has the same value.

Ex. *x – 2y = -10*

3x – y = 0

this becomes 3(2y-10) – y = 0

Then you want to gather like terms

6 y – 30 – y = 0

5y = 30

then you divide the ‘y’ value by it’s co-efficient on both sides to get the value of ‘y’

y=6

Now that you know your x-value, you can input this value into the first equation.

*x – 2y = -10*

x – 2(6) = 10

x – 12 = 10

x = 2

Now you know your co-ordinates for this system is (2,6)

Ex. *x – 2y = -10*

3x – y = 0

With elimination, you want to take your 2 equations, and make one variable equal to the other.

*x – 2y = -10*

6x -2y = 0

Now you can subtract the equations to make 1 equation.

-5x = -10

Once again, you just divide the ‘x’ term with its co-efficient on both sides and you get your ‘x’ value.

x= 2

Now you can input your ‘x’ value into the on of the equations and then you can simplify for ‘y’

6x -2y = 0

6(2) – 2y = 0

12 – 2y = 0

-2y = -12

y = 6

Co-ordinates = (2,6)

Both ways work for finding the point(s) of intersections it just depends on which method you prefer to use.

## Week 11 – Graphing linear inequalities in two variables

This week in pre-calc we learned how to graph linear equalities.

The first step when graphing, is the find your y-intercept. If your inequality doesn’t have a ‘c’ term in general form or a ‘b’ in the equation y=mx +b, then your y-intercept is going to be at 0.

After you point that down, you can look at you ‘mx’ and figure out your slope. If the number is positive, your slope will rise up and to the right on your graph. For the opposite slope, it will run down towards the left. If your slope is negative, your slope will rise up and to the left on your graph and the opposite slope will run down towards the right.

Once you graph this all down, you can take a look at your inequality sign. ≥ or in the opposite direction means that you have a solid line when you are connecting your points. This also means that all the points on the line are a solution to the inequality. < or > means that your line is broken and that the points on the line are not a solution. A solution for an inequality is to make the statement true.

A way to test to see which side of the line is the solution, you take a test point (0,0) is the easiest to test. If these 2 points make the statement true then you shade in the side of the solution. If it’s not true, you know to shade in the opposite side.