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Chapter 1


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Section 1-1: The Language of Algebra


Section 1-1
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In this lesson variables, algebraic expressions, and algebraic sentences were discussed. You had to evaluate the expression by substituting for the variables and finding the result. An algebraic expression is a conbination of numbers and variables using arithmatic operations to combine them. An algebraic sentence is and expression related with verbs.(=,<,>). Using the order of operations is important in this lesson(GEMDAS). Grouping comes first, then powers, multiplication and division, and addition and subtraction.It is also necessary to understand the definition of a formula aswell.
NJ

Section 1-1 is meant to help students better understand the terms used in advanced algebra. A variable is a symbol (usually a letter) that "stands in" for a number or expression. Instead of putting the whole number or expression, you can just replace it with a variable. An algebraic expression is a combo of numbers and variables using arithmetic operations (+, -, x, and division) to combine them. An algebraic sentence is an expression that has a "verb" (like =) in it. When you are asked to evaluate an expression, they want you to substitute for variables and find the result. An easy way to remember the order of operations is to use GEMDAS (Golly Excuse My Dear Aunt Sally). 1. Grouping Symbols 2. Powers (exponents) 3. Multiply and Divide from left to right 4. Add and Subtract from left to right. An equation is a sentence when two expressions are equal to each other. A formula is a rule that says a variable will always equal a certain expression.
Abigail B.


Section 1-2: What is a Function?

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SUMMARY 1-2:

In this lesson we learned about dependent and independent
variables. A dependent variable is the value that depends on another variable.
An independent variable is the variable that determines the dependent
variable, and it relies on nothing. In addition, we learned about functions and the
statement "is a function of." A function is a pairing between two variables so that
the independent variable matches with only one dependent variable. When we have
a function, we say that the dependent variable is a function of the independent
variable.
To add to that we also learned about domain, range, input, and output.
The domain is anything that can possibly be substituted for the independent
variable while the range is any value that can be listed as the dependent variable.
Input is another word for the independent variable and output is another word
for the dependent variable. Lastly, we learned about common sets such as
natural numbers, integers, whole numbers, rational numbers, and real numbers.
-Paige

Lesson 2 chapter 1 we learned about different variables. We have the dependent and the independent variables. The dependent variable depends on another variable, so the independent variable relies on nothing but itself. Also in this lesson we learned about functions. A pairing between two variables so that the independent variable matches with only dependent variable. The statement of a function " Is a function of " which states: statement used when we have a function [dependent] is a function of [independent]. We learned about common sets. We have natural numbers, integers, whole numbers, rational numbers, and real numbers. The last thing we learned about was the domain, range, input, and output. The domain is anything that can possibly be substituted in for the independent variable. D={ } Range is any value that can be liste as the dependent variable. R= { } The input is another word for the independent variable which means that output would be the dependent variable.

Josh H.

Section 1-3: Function Notation



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1-3 Summary

In lesson three of chapter one, two different function notations were used.
One is called the Euler notation. The Swiss mathematician Leonhard Euler discovered this notation. The main symbol f(x) is read “f of x.” The “x” that is inside the parentheses is the independent variable. Some examples of the f(x) notation can include:

T(x) = x, H(x) = x/20, G(x) = x/20 + 5

The letters above, T, H, and G, are functions. The x is the argument, and the numbers T(x), H(x), G(x) are values. Here are some more examples:

S(x) = 34x where x equals 3.
Solution: S(3) = 34 x 3 (which then equals) S(3) = 102
V(x) = x^4 where x equals 5.
Solution: V(5) = 5^4 (which then equals) V(5) = 625
M(x) = 50 + 3(x)^3 where x equals 8.
Solution: M(8) = 50 + 3(8)^3 (which then equals) M(8) = 1586

The other type of function notation is the arrow (or mapping) notation.

T: x ® x, H: x ® x/20, G: x ® x/20 + 5

Here are more examples:

S: x ® 34x where x equals 3.
Solution: S: x ® 34x x 3 (which then equals) S: x ® 102
V: x ® x^4 where x equals 5.
Solution: V: 5 ® 5^4 (which then equals) V: 5 ® 625
M: x ® 50 + 3(x)^3 where x equals 8.
Solution: M: 8 ® 50 + 3(8)^3 (which then equals) M: 8 ® 1586
These notations can be interchanged:
Z(x) = 55x (can change to) Z: x ® 55x

~ Jacob L ~



Chapter 1 lesson 3

In this lesson we learned about

Euler Notation , arguments, and values.

An

argument

is another word to represent the independent variable, this
can be filled with anything that is in the possible domain.


Mapping Notation or Arrow

Read, "A maps X onto" it still identifies the independent variable
(after colon) and dependent variable (A:X)


Value is simply the dependent variable

Section 1-4: Graphs of Functions


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SUMMARY 1-4:
In this lesson we learned what a horizontal and vertical axis are. The horizontal axis contains the independent variables. This means the vertical axis contains the dependent variables. We also learned about relations, which are any set of ordered pairs. Along with this comes the fact that every function is a relation but a relation is not always a function. When a function is a relation each first coordinate appears only one time out of all of the ordered pairs. We also learned how to tell if a graph is a function. To tell if a graph is a function you use the vertical-line test. To do this you draw a vertical line on a graph anywhere and it can only touch the graph once to be a function. If it touches more than one line the graph is not a function.
Tricia L.

1-4 summary
This lesson we looked over the vertical and horizontal Axis. we then learned that the vertical axis holds the dependent variable and the horizontal holds the independent variable. Any set of ordered pairs is a relation. Any relationship between two variables can be written as a ordered pair. functions are always relations but relations dont have to be functions and for a function to be a relation each of the first coordinates in the pairs cant be the same. last of all we looked at the vertical line test which says if a vertical line can be drawn anywhere on a graph and and it touches once its a function. if the line touches more than once it's not a function
Chris S.

Section 1-5: Solving Equations


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In Section 1-5, we learned about the Distributive Property and the Opposite of a Sum Theorem.

The Distributive Property states that for all real numbers a, b, and c, c(a+b) = ac + bc.
Example of the Distributive Property:
a=2, b=6, c=5.
5(2+6) = 5 multiplied by 2 + 5 multiplied by 6.
5(2+6) = 10 + 30.
5(2+6) = 40.

The Opposite of a Sum Theorem states that for all real numbers a and b, -(a+b)= -a + -b.
Example of the Opposite of a Sum Theorem:
a=4, b= -9.
-(4 + -9) = -4 + 9.

We also learned how to clear fractions and decimals. To clear a fraction, multiply both sides of the equation by a common denominator.
Example of clearing fractions:
1/6x + 2/5x = 273.
(common denominator = 30)
5/30x + 12/30x = 273.
5x + 12x = 273.
17x = 273.
----- -----
17 17
x = 273/17.

To clear a decimal, you essentially do the same thing (multiply both sides of the equation by a common denominator), but it can be viewed as just multiplying the decimals by 10, 100, 1000, 10,000, etc.
If you wanted to clear out the decimal .6, you would only multiply it by 10 because 10 has one zero and one is how many decimal places you want to move to the right.
6 multiplied by 10 = 6.
If you had the decimal .06, however, you would multiply it by 100 (Two zeros in one hundred -- two decimal places to the right).
.06 multiplied by 100 = 6.

-Emma R.

In Chapter 1 Lesson 5, we learned about solving equations. We began by restating the basic steps of solving a problem which are as follows:
1) Read the problem.
2) Make sure your solving exactly what is being asked.
3) Identify the variables that are being used or will be used in the problem.
4) Set up the problem, like in an equation, so that the answer can be found.
5) Solve the problem using the set up you created in step 4.
6) Be sure that you answer the question. Restating the answer in a sentence would help.
[7) would be to check your work for mistakes.]
After reviewing the basic steps of solving a problem, we move on to discussing the Distributive Property and the Opposite of a Sum Theorem. The Distributive Property states that for all real numbers a, b, and c: c(a+b)=ac+bc.
Ex: c=3, b=2, a=5: 3(5+2)=5*3+2*3
3*7 = 15+6
21 = 21

The Opposite of a Sum Theorem states that for all real numbers a and b: -(a+b)=-a+-b ( the +-b would be simplified for -).

Ex: a=-3, b= 1: -(-3+1)=3+-1 -(-2)=2 2=2

We also took a moment to look at clearing out fractions. In order to clear out fractions, you multiply both sides of the equation by the common denominator of the fractions. Try to use the lowest valued common denominator in order to keep the numbers you are using small and therefore easier to work with. This would help prevent mistakes.

Ex: 2/3b+15+5/6b=105
4/6b+5/6b+15-15=105-15
9/6b=90
9/6*6/9*b=90*6/9
b=60
-Trisha T.-

Section 1-6: Rewriting Formulas


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In this lesson we learned about solving for a variable “in terms of” other variables. We also learned how to simplify that process when a fraction is involved. You can just cross multiply over the equal sign, making sure you exchange the correct variables/numbers. From there you can try to simplify as much as possible. During class we also talked about how to solve for a variable in terms of other variables in an equation without a fraction. In this process you simply solve for the variable like any other equation, save for the fact that you will have more than one variable in the answer. Jacob R.



In this lesson we learned about solving for a variable "in terms of" other variables. Solving for a variable is just applying the rules of mathematics to a problem to isolate a certain variable like A. "In terms of" is simply when an equation is solved for a variable, the equation is written "in terms of" the remaining variable. We also learned in this lesson how to how to solve for a variable in fraction form.
In this equation, b=2A/h your solving for h in terms of b and a. To solve this you simply cross multiply so you can get h alone. So you can divide b on both sides so you get b/b=2A/bh. Since b/b, the left of the equal sign, can be canceled out you can know multiply h on both sides so you can get h alone by canceling out the h's on the right side of the equal sign, h=(2A/bh)h. Make sure that you remember parenthases. So know the equation is solved for h, h=2A/b.
Amanda T.

Section 1-7: Explicit Formulas for Sequences


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In this lesson we learned what an explicit formula for a sequence was. This section also informed us that numbers in a sequence are often called triangular numbers and that each number in the sequence is called a term. If you didn't know already, a sequence is a function whose domain is the set of natural numbers or the natural numbers from 1 to n. You could also use these explicit formulas to generate terms of a sequence. Used with sequences are special notations which read out to be "t sub 10 equals 55."

Arron H.

Section 1-8: Recursive Formulas for Sequences



In this lesson we learned about recursive formulas for sequences. A recursive or recursive definition for a sequence is a set of statements that a) indicates the first term or first few terms and b) tells how the nth term is related to one or more of the previous terms. If you can describe a sequence in words then you can use that description to write a formula for the sequence. You can also solve recursive formulas with a calculator.
-Brittani

Section 1-9: Notation for Recursive Formulas



In this lesson we learned about Recursive Formulas. We already learned how to do this in Lesson 1-8 but now we area substiting "previous term" with "T(n)-1" (which is read "T" sub "n" minus "1").
For Example:
{t(1)= 50
{t(n)= 1.05(previous), for int.≥2

Which is the same as saying:
{t(1)=50
{t(n)=1.05(t(n)-1), for int. ≥ 2

t(1)=50
t(2)=1.05(t(2)-1)=52.5
t(3)=1.05(52.5)=55.125
t(4)=57.88125

So that you understand:
t(1)=50
t(2)=52.5
t(3)=55.125
t(4)=57.88125

Jessica R. & Jeremiah B.

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