This post is for use in my Algebra 1 class. It is the answer key to the assignment that was collected and graded.

Filed under: Algebra 1, Scatter Plots | Tagged: Algebra 1, Scatter Plots | Leave a comment »

Posted on September 19, 2010 by Mr. Pi

This post is for use in my Algebra 1 class. It is the answer key to the assignment that was collected and graded.

Filed under: Algebra 1, Scatter Plots | Tagged: Algebra 1, Scatter Plots | Leave a comment »

Posted on July 2, 2010 by Mr. Pi

Next Lesson

There are many ways to define the word variable. For now, a **variable** is a symbol, usually a letter that represents an unknown number. This number is known because it can change. Let’s say you work a part time job at the local corner store and you are paid $7.50 per hour. How do you find the amount of money you make? Oh course, you multiply the number of hours you work by the amount of money you are paid, in this case $7.50. There are very few jobs that you will work the same amount of hours week after week. Because the number of hours worked changes, h represents the number of hours work and *$7.50h *is the algebraic expression used to find your pay. An **algebraic expression **is a mathematical statement relating numbers and variables with mathematical operations. Algebraic expressions are sometimes called **variable expressions**.

The following table and examples model how to write algebraic expressions from given English or written phrase.

Write an algebraic expression for each phrase.

a) x increased by 12

*x + 12***“increased by” means to add 12 to the variable x**

b) 9 less than n

*n – 9***“less than” means to subtract 9 from the variable n. Notice the order of the number and variable, it switches**

c) triple y

*3y *“triple” means to multiply by 3. Can you think of other special words that mean to multiply?

d) the quotient of q and

**“quotient” means to divide. Notice the order q is 1 ^{st} and goes in the numerator.**

In each part of the first example, the variable was given in the verbal phrase, in the next example, there are no variables given, so we will need to * define a variable* before translating the English phrase into an algebraic expression.

Define a variable and write an algebraic expression for each phrase.

a) four times a number subtracted from twenty-one

**Highlight/Underline the Different Parts**: four times a number subtracted from twenty-one

**Define a Variable**: Let x = the number

**Translate Into Math**: 21 – 4 x

b) the sum of 3 and a number divided by another number

Relate

**Highlight/Underline the Different Parts**: the sum of 3 and a number divided by another number

**Define variables**: Let x = a number

Let y = another number

**Translate Into Math**: (3 + x) ÷ y or .

An **equation** is a mathematical sentence that includes an equal sign. An equation consists of two numerical expressions, one on each side of the equation. For an equation to be ‘true’ both sides must evaluate to the same value. A special equation, called an **open sentence**, contains at least one variable. To determine if an open sentence is true or false, values must be substituted in for the variable. Once each side is simplified, the truth value of the open sentence can be determined.** **

Write an equation to model the amount of income from selling cups of lemonade for $1.25 per cup.

When I first read this problem, I think about making a table. To create the table I ask myself which of the two values affects the other? Which value ‘depends’ on the other value? In this case, the money made depends on the number of cups sold. I do not know how much money is made, until I know how many cups I have sold. Therefore, I put the number of cups sold in the first column and the money made into the second column. Once I filled out the cups sold column, I filled in 1.25 across from cups sold 1, then I added 1.25 to 1.25 to get 2.50 in the money made column adjacent to cups sold 2. I continued to add 1.25 to the new value to get the next. I stopped a 5 cups sold.

As I was making the table, I realized that repeated addition is the same as multiplying. If I multiply 1.25 by the number of cups sold, then I get the amount of money made. To say this simply, the amount of money made is 1.25 times the number of cups sold.

**Highlight/Underline the Different Parts**

the amount of money made is 1.25 times the number of cups sold

**Define a Variable(s)**

c = number of cups sold

m = the amount of money made

**Translate Into Math**

m = 1.25c** **

Define variables and write an equation to model the data in the table.

In the previous example, I talked about the dependent relationship between the quantities. When using a table the dependent value is stored in the second column. Knowing that the dependent variable is in the second column, I can try dividing the values of the second column by the corresponding values in the first column. If I get the same value I have found common factor or the number I can multiply the first column by to get the second column.

In this example, dividing the total commission by the number of policies sold:

33 ÷ 1 = 33

66 ÷ 2 = 33

99 ÷ 3 = 33

132 ÷ 4 = 33

**Highlight/Underline the Different Parts**

the total commission earned is 33 times the number of policies sold

**Define a Variable(s)**

c = total commission earned

p = number of policies sold

**Translate Into Math**

c = 33p

When working with variables, it is important to remember what the variable means. That it why I define variables near the beginning of the process of writing an algebraic expression or an equation. The most important take away from this lesson is writing an equation to represent a table of data. You will want to look for a pattern. I will usually ask this question when looking at a table of values, “how can I change the first column into the second column?” I start with single operations such as addition or subtraction, looking for a common difference between the two. The exercises in this lesson are focused on one step equations, which means only one operation is involved.

Next Lesson

Filed under: Algebra 1, Using Variables | Tagged: Algebra 1, Modeling Data in a Table, Modeling Relationships with Equations, Modeling Relationships with Variables, Mr. Pilarski, Using Variables, Writing Algebraic Expressions | 3 Comments »

Posted on May 30, 2010 by Mr. Pi

**Inverse Variation** can be modeled after any equation in the form where . The constant of variation is *k.*

The table to the left models the time it would take to roller blade 36 miles at various speeds. As the rate increases the amount of time to travel 36 miles decreases. If the rate data is examined more closely and focus is placed on the times of 3, 6 and 12 hours, it should be come clear as the rate is doubled, the amount of time is cut in half.

Since the product is 36 for each pair of numbers, this is an inverse variation problem and the product is the constant of variation for this data. Letting x be the rate and y be the time, one equation to model this data is . A more useful equation is . Figure 2 is the graph of the inverse variation modeled in this problem.

Why is a more useful equation than ?

*Write your responses in the comments section.*

This problem involves finding the equation or function that models a given inverse variation. This is a two step process:

- Multiply the given x- and y-values
- Substitute the product into

Before just memorizing this two step process, it would be better to understand what these two steps are ‘doing’ mathematically. In step 1, multiplying the given x- and y-values calculates the **constant of variation**. The second step is to substitute the constant of variation into one of the acceptable forms of the inverse variation equation.

Suppose that x and y vary inversely and x = 7 and y = 5. Write an equation that models the inverse variation.

*Definition of Inverse Variation*

*Substitution Property of Equality – Step 1 from above
*

*Simplify*

*Symmetric Property of Equality*

*Step 2 from above*

Final Answer

When given a table of values, it is often important to be able to identify the equation modeling the data. To determine if a table of values is indeed an inverse variation, all of the data pairs must have the same product. This is because of the definition of an inverse variation and how the constant of variation is calculated. So, to identify a table of values as an inverse variation, one must test the products of all the available data. If the products are all equal, then the data models an inverse variation and by find the products, the constant of variation is already known.

Is the relationship between the variables in figure an inverse variation? If so, then write the function that models the inverse variation.

As state above, first find the product of all pairs of numbers:

.

Since all pairs of corresponding numbers form the same product, this table of values models an inverse variation with the constant of variation being 1.4. Therefore the function to models this data:

.

If I don’t write this at the end of every blog article, I should start. Use the comments section for any questions you may have.

Filed under: Algebra 2, Equations and Graphs, Inverse Variation | Tagged: Algebra 2, Identify Inverse Variation, Inverse Variation, Modeling Inverse Variation | 1 Comment »

Posted on May 28, 2010 by Mr. Pi

As I think about circles in a plane and intersecting lines, I can think of more triangle proofs and work with the angle and segment addition postulates. This will lead to both algebraical and geometric types of problems. By algebraic, I mean you will have to use a theorem or property to set up the equation. On the other hand, a geometric type of problem will consist of a proof of a stated given or already known theorem. Before getting into solving any problems, I should discuss the theorem of topic for the geometry article.

The measure of an angle formed by two lines that intersect inside a circle is half the sum of the measures of the intercepted arcs. Figure 1 illustrates the following equation to represent this theorem:

.

The first example ends up being a an easy three step equation to solve.

To set up the equation I HAVE to use geometry, in this case, **Two Lines Intersecting Inside of a Circle Theorem (TLIICT)**, then I solve the equation using an algebra 1 skill.

. *TLIICT*

(1). *Substitution Property of Equality*

(2) *Multiplication Property of Equality*

(3) * Subtraction Property of Equality*

(4) *Symmetric Property of Equality*

The solution to this problem may not present itself right away if one too focused on the current theorem. There is no variable for what you have to solve for first. To find the value of the variable, I must keep in mind that the measure of the angle I am solving for is supplementary to the measure of the angle I can find. Since it is not labeled in the original picture, in figure 2, I label what I can find with what is given, x.

Let x = to the value of the vertical angles formed not labeled y

In this post, I have discussed the Lines Intersecting Inside of a Circle Theorem and worked out two problems. I worked out the problems differently in each example. Which method do you prefer to read and why? I look forward to your answers.

Filed under: Circles, Geometry, Lines Intersecting Inside of a Cirlce | Tagged: Circle Geometry, Geometry, Lines Intersecting Inside of a Circle | 3 Comments »

Posted on May 26, 2010 by Mr. Pi

In this video there are two problems modeled. The first example works with common logs and the second example models an equation with a natural log. The keys to solving any type of logarithmic equation: being able to write the given equation in exponential form or being able to take the log of each side.

To be able to write any logarithmic function in exponential form knowing the definition of logarithm and natural logarithm is key. So here they are…

Definition: **Logarithm – **The log to the base b of a positive number y i s defined:

If , then .

Definition: **Natural Logarithm –** If , then which is also written as . The natural logarithmic function is the inverse, written as .

In other words, if , then .

Now that you have reviewed the definition of log and natural log, read through these two examples, then watch the embedded algebra 2 video.

This a problem that I assigned to my Algebra 2 class for a review before a quiz on solving logarithmic equations:

*Given*

*Take the log of each side*

*Product Property*

*Division Property*

*Addition Property*

*Use a calculator*

As you can see in this example, you can easily solve an equation with multiple log in it. All you need to do is use the properties of logarithms and that taking the log of each side is a legal mathematical move. The properties for common logs are used with natural logs.

Solve the following equation.

*Given*

*Power Property*

*Simplify*

*Product Property*

*Write in Exponential Form*

*Division Property*

* Square Root Property*

*Use a calculator*

Solving logarithmic equations with natural log is easy if you can use the properties of natural logs and you know how to write a natural logarithmic equation in exponential form.

Filed under: Algebra 2, Equations and Graphs, Logarithmic Equations | Tagged: Algebra 2, Logarithmic Equations, Natural Logarithm | 3 Comments »

Posted on May 3, 2010 by Mr. Pi

Here is a drawing I did to show to my **High School Geometry** classes. At the time of this post, the class is starting a unit on **3-Dimensional shapes**. Prior to this chapter, most of our work has been in a single plane. Learning how to draw simple 3-D shapes in both one point and two point perspective makes the transition to working with intersecting planes smoother for some students. Other students would do just as well without actually making the drawings. The image to the right is my little sketch after I drew my box. Be sure to click the image to see it in full size. Below is a picture I drew back in 2008 sometime.

**How to Draw a Box in Two-Point Perspective:**

- On a sheet of paper, draw horizontal line the entire length of a sheet of paper, place one point to the left of center and one point the right of center. Preferably closer to the side of the paper.
- Draw a vertical line of about 1 inch long above or below the box. The bigger the vertical line, the bigger the box.
- Draw light line segments connecting each end of the vertical line to both points. You should draw 4 line segments in this step.
- Draw two vertical segments, one on each side of the first vertical segment.
- Connect the new vertical segments two the vanishing points. You should draw four line segments for this step.
- Clean up the unnecessary lines and used dashed segments for the hidden edges.

Filed under: 2 Point Persective Drawing, 3-Dimensional, Geometry | Tagged: 3-D, 3-Dimensional Shapes, Geometry, Point Point Perspective | 4 Comments »

Posted on April 7, 2010 by Mr. Pi

Finding the **area of regular polygons** is the topic of this geometry math video.

First, the video discusses the parts of a regular polygon: apothem and radius. The **center** of the polygon is the same as the center of the circumscribed circle. The **radius** of the polygon is the distance from the center of the polygon to its vertex. The number of radii is determined by the number of sides. The **apothem** it the perpendicular distance from the center of the polygon to a side.

The first example models how to find the different angles of a regular polygon formed by the radius and the apothem. First, you divide the number in interior angles into 360 degrees. Once you have that you can figure out the other two angles quite easily.

Before applying the the **area formula of a regular polygon**, the video reviews the formula:

It is good to note that a = length of the apothem and p = perimeter of the polygon. The perimeter may not be calculated. In that case you multiply the number of sides by the length of each side.

This example is similar to the video, but it is different:

What is the area of a polygon with sixteen 36 in. sides and an apothem of

Questions for the comment section:

Comparing example 2T on the video and the problem above, how are the problems alike? How are the problems the same?

Example 3T is finding the area of a regular polygon, but you need to find the length of the apothem using the 30-60-90 triangle relationship.

Filed under: Apothem, Area Formulas, Geometry, Regular Polygons | Tagged: Area Formulas, Geometry | 5 Comments »

%d bloggers like this: