Week 3 – Precalculus 11

This week was short. Monday we had our test for Sequences and Series, Tuesday we had a snow day, Wednesday we started review on Roots and Radicals with a new element of Absolute Value and today we continued with those ideas and completed the Checkpoint questions. This week was pretty low-key. What I will be highlighting in my ShowMe, was what I struggled with most – When there is a fraction under a root sign. I enjoy attaching ShowMe’s as it allows my viewers to see the specific steps, along with hearing my voice; It has most learning styles included! Hopefully it works for you. As I said, you will be watching how to solve for a fraction being placed under a root sign, some mixed and entire radicals, along with a question or two of solving for an Absolute Value. 

Week 2 – Precalculus 11

This week wrapped up our first unit! It was generally based around Geometric Sequences and Series. For myself, it was a step higher from Arithmetic Sequences and Series and slightly more challenging. Due to formulas, solving the questions in the workbook came with ease… except word problems. Therefore, I thought I would demo one for my Blogpost this week!

Geometric Series

Week 1 – Precalculus 11

This week, being the first week of Precalculus, involved a lot of new feelings and knowledge. I found the first lesson fairly easy to grasp. Due to the formulas being provided, plugging in the corresponding information into that formula was the easy part… Because it was given to you. However, the most challenging piece I have been faced with over the last week would be when you had to solve for something that was not given to you directly at the beginning of the exercise. After numerous tries I finally got the hang of it. Provided bellows a question in which you have to solve for a part that is not there. Luckily with the help of algebra you can the answer!

Example:

t_1=3,d=4,and,t_n=59-find n

Step One : Plug in what you know!

t_n=t_1+(n-1)d

59 = 3 + (n-1)4

Step Two : Distribute the 4 into the brackets.

59 = 3 + (4n-4)

Step Three : Remove the 3. (add the opposites!)

59  + (-3) = 4n-4

56 = 4n-4

Step Four: move the -4 to the opposite side.

56 +4 = 4n

60 = 4n

Step Five : Divide!

\frac{60}4 = n

15 = n