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# Author: daphnev2017

## Invertebrate project

## Projet final

## Pre Calculus CC reflection

## Week 15- precalc 11

## Week 14-precalc 11

## Week 13- precalc 11

## Week 12- precalc

## Week 11- precalc

## Week 10-pre calc 11

## Week 9-Precalc

One thing we learned about this week was the sine and cosine law. These are used to find side lengths and angles in triangles without 90 degree angles or without the necessary information to use the primary triganomic ratios.

**Sine law:**

For this to work, you need a complete fraction as well as one element (either a side length of fraction

The sine law to find a side length:

The sine law to find an angle:

To solve for a side length or angle, you need to isolate the variable we are trying to find.

Example for finding a side length:

Find the length of side a.

To find an angle example:

this week we started trigonometry. One key idea was **coterminal arms**

Coterminal arms are angles with the same terminal arm. So for example, an angle of 50, 410, and 770 degrees are all coterminal. The reason for this can be shown with a diagram

Basically, we measure this from the horizontal line upwards. If it is a positive angle, we move counter clockwise. To reach 50 degrees, you have a rotational degree of 50. Then, we can move all the way around to reach that same coterminal arm by turning an extra 360 degrees. When you add these together, you get 410. And if you add another 360, you reach 770 degrees.

One thing we learned this week was how to solve rational equations. This was not too difficult, as we had the necessary skills from the previous lessons of rational expressions. The difference is that there is an = sign, and so we need to actually solve to find the value of the variable.

There are multiple ways of solving a rational equation, based on the type of equation.

Ex:

for this equation, all of the terms are monomials and there is no addition/subtraction/multiplying/dividing. Therefore we can go ahead and cross multiply. **Note: this only works with equations**

this ends up being **4d=40. **Then, you can simplify, and **d=10. **Lastly, add restrictions (in this case, d cannot equal 0)

For more difficult ones, (with more operations) you cannot just cross multiply. I prefer to simplify the portion on one side of the = sign and then go from there, or find a common denominator between all of the terms and then proceed.

Ex:

first, find a common denominator (in this case it is **x(x-3)**) then multiply each numerator by the denominators missing factors

This becomes:

**remember to add the non permissible values: x cannot equal 0 or 3**

In a rational equation, when the denominators are the same you can cancel them out and are left with the numerator.

Then you can simplify both sides if needed:

After this, solve like any rational equation. To do this, move everything to one side and make it equal to zero.

then we factor it:

(x-1)(x-3)=0

after this, we can find the solutions:

x=1 and x=3.

then we check with our non permissible values for x. X cannot equal zero, so our answer is 1.

This week we learned about adding and subtracting rational expressions. This required some thinking and focus but was actually not that hard.

These are the steps you use for all addition. When you do substraction, the only part that changes is that you need to distribute the negative into the expression and then evaluate like normal. And when you add variables in the eliminator, treat them like normal factors but do not forget restrictions!!!!

This week we learned about graphing quadratic equations to find solutions. At first this was hard for me to understand.

We started with linear equations. For example

In this, i moved everything to one side. Then you graph it.

To make the statement true we need to find which part of the line makes the statement true. In this case, the upper section is greater than zero, so we look at the x value that corresponds with it which is -8. This is our answer.

one main thing we focused on this week was **modelling problems with quadratic functions**

One type of question was:

two numbers have a sum of 20 and their product is a maximum. What are the numbers?

Here are the steps i took:

1)

2)

3)

The line of symmetry between these points was 12. And that was my answer. 12×12= 144 which is the maximum area