http://ow.ly/8d87I – Solving 1 Step Equations – Newest post on my newest blog. Check it out.
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http://ow.ly/8d87I – Solving 1 Step Equations – Newest post on my newest blog. Check it out.
Filed under: Algebra 1, Pre-Algebra, Solving Equations | Tagged: Algebra 1, Pre-Algebra, Solving Equations | Leave a comment »
Filed under: Algebra 1, Algebra 2, Equations and Graphs, Solving Equations | Tagged: Algebra 1, Solving Equations | 2 Comments »
Please post all questions in the comment area.
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This math help video covers algebra 1 level probabilities of independent compound events. The definition of independent compound events is covered, as well as how to find said probabilities of independent compound events.
Filed under: Algebra 1, Compound Events, Independent Events, Probability, Theoretical Probability | Tagged: Algebra 1, Compound Events, Probability | 1 Comment »
In this math help video, I define and discuss experimental probability. The first two examples model how to use data that has been collected to find the experimental probability. The last two examples examine how to use experimental probability to predict the outcome of a larger sample space. Both of these examples deal with manufacturing of goods.
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This Algebra 1 video covers how to find the theoretical probability using the complement of an event. Also, how to find the odds for and against an event are defined and used to solve problems.
Filed under: Algebra 1, Odds, Probability, Theoretical Probability | Tagged: Algebra 1, Odds, Theoretical Probability | Leave a comment »
In this video, I define theoretical probability and model how to find the theoretical probabilities for different situations. If you have any questions, use the comments sections. Also, feel free to leave a comment if this video algebra 1 lesson helped you.
Filed under: Algebra 1, Probability, Theoretical Probability | Tagged: Algebra 1, Probability, Theoretical Probability | 1 Comment »
This post is for use in my Algebra 1 class. It is the answer key to the assignment that was collected and graded.
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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 1st 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 »
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:
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 »