College Algebra

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Open Chapter Ch. 1: Graphs, Functions, and Models
Lesson #1 Graphing
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Lesson #2 Graphing Functions
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Lesson #3 Slope of Linear Functions
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Lesson #4 Linear Equations and Modeling
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Lesson #5 Linear Equations, Functions, Zeros and Applications
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Lesson #6 Linear Inequalities
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Open Chapter Ch. 2: More on Functions
Lesson #7 Analyzing Functions
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Exam Exam 1
Lesson #8 Algebra for Functions
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Lesson #9 The Composition of Functions
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Lesson #10 Symmetry
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Lesson #11 Transformations
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Lesson #12 Equations of Variation
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Open Chapter Ch. 3: Quadratic Functions and Equations; Inequalities
Lesson #13 Complex Numbers
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Lesson #14 Quadratic Equations, Functions, Zeros and Models
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Lesson #15 Analyzing Graphs of Quadratic Functions
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Lesson #16 Rational and Radical Equations
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Exam Midterm Exam
Lesson #17 Absolute Value Equations and Inequalities
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Open Chapter Ch. 4: Polynomial Functions and Rational Functions
Lesson #18 Polynomial Functions and Models
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Lesson #19 Graphing Polynomial Functions
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Lesson #20 Polynomial Division
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Lesson #21 The Zeros of Polynomial Functions
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Lesson #22 Rational Functions
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Lesson #23 Polynomial and Rational Inequalities
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Open Chapter Ch. 5: Exponential Functions and Logarithmic Functions
Lesson #24 Inverse Functions
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Lesson #25 Exponential Functions and Graphs
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Lesson #26 Logarithmic Functions and Graphs
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Exam Exam 3
Lesson #27 Properties of Logarithmic Functions
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Lesson #28 Solving Exponential Equations and Logarithmic Equations
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Lesson #29 Applications and Models: Growth and Decay; Compound Interest
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Open Chapter Ch. 6: Systems of Equations
Lesson #30 Systems of Equations in Two Variables
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Lesson #31 Systems of Inequalities and Linear Programming
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Lesson #32 Nonlinear Systems of Equations and Inequalities
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Lesson #33 Sequences and Series
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Exam Final Exam

Assignments:

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Lesson Objectives:

- Solve linear equations.
- Solve applied problems using linear models.
- Find zeros of linear functions.


A linear equation in one variable is an equation that can be written in the form `mx + b = 0`. Where m and b are real numbers and `m != 0`.

The addition principle. If a = b is a true statement, then `a+c = b+c`.

The multiplication principle. a = b is true, then `a*c = b*c`.


Let's solve the following linear equation for x.

`5x+2 = 0`. Go ahead and subtract 2 from both sides. We'll get `5x = -2`. And then divide both sides by 5 to get our answer of `x = -2/5`.



Solve for x, then check.

Start by distributing 8 into the parentheses and the negative sign into the parentheses. That gives us `24x+40 = 11-x-1`. `24x+40` and then combine the 11 and -1 to `10-x`. And then subtract 40 from both sides. `24x = -30-x`. And then add x to both sides. `25x = -30`. Divide both sides by 25. `x = -30/25`. Then divide the top and bottom by 5. So, x = -6/5.

So here we plug in -6/5 for x. So we get `8(-18/5+25/5) = 11-(-6/5+5/5)` or `56/5=11+1/5`. And since 11 is the same as 55/5, when we add 1/5, we get 56/5. So we have the true statement, `56/5 = 56/5`. And since its a true statement, we know that our solution of `x = -6/5` is correct.



Solve the following equation. `1/11-2x = -2x+3/11`.

Start by adding 2x to both sides. Notice how the x terms completely cancel and we're left with `1/11 = 3/11`, which is a false statement. Therefore, there's no solution.



Solve the following equation. `25-1/3x = -1/3x+25`.

So let's start by adding 1/3x to both sides. That'll cancel the x terms on both sides, and we're left with 25 = 25. Since this is a true statement, that means our solution is all real numbers, or in interval notation, `(-oo, oo)`.



During the 20th century, about 90,000 new words were added to the English words. This is about 80% more than the number of new words added during the 17th century. How many new words appeared in the English language during the 17th century?

Let's let x = the number of new words added to the English language during the 17th century. Since 90,000 words in the 20th century = 80% more than the number of words in the 17th century, we can construct a linear equation. `90000 = 80%*x + x`. So if we convert 80% to decimal form, that's .80. And then if we combine .80x + x, we'll get 1.8x. Now we just divide by 1.8 on both sides. Therefore, `50000 = x`.

About 50,000 new words appeared in the English language in the 17th century.



In 2012, Samsung received 1907 more patents than Cannon. Together they received 8255 patents. How many patents did each company receive?

Let x represent the number of patents that Cannon received. Then `1907+x = "Samsung's patents"`. Then given the total number of patents as 8255, we could setup a linear equation.

`8255 = (1907+x)+x`. Combine the two x's. Now subtract 1907 from both sides, then divide both sides by 2. Samsung received 5081 patents and Cannon received 3174.



The motion formula. The distance, d, traveled by an object at a rate, r, in a time, t, is given by `"distance" = "rate"*"time"`.



Two delivery trucks leave adjacent sorting facilities in Wasilla, Alaska and travel 312 miles due north along AK-3 N to reach Fairbanks, Alaska. If the red truck leaves first, traveling 45 miles an hour, and the blue truck leaves an hour later traveling 80 miles an hour, then how long will it take the blue truck to overtake the red truck?

So we start by setting up a table, with the truck, the distance, the rate, and the time. We have a red truck and a blue truck. And since they're both traveling the same distance, we'll let their distances equal d. Now `"distance" = "rate"*"time"`. The rate for the red truck is 45 miles an hour, and the rate for the blue truck is eighty miles an hour. Let's let the total time that it takes for the blue truck to overtake the red truck equal t. And that will mean that the red truck traveled for `t+1` hours before being overtaken. So now the distance for the red truck is going to equal `45(t+1)` and this will equal the distance of the blue truck, which is `80t`. Now distribute on the left, and we get `45t+45` and all of this is equal to `80t`. Now we need to solve for t. Subtract 45t on both sides, and this will give you 35t on the right is equal to 45. Now divide both sides by 35. This cancels the 35 on the right, and we get 1.29 hours is equal to t.

So this means that it takes 1.29 hours for the blue truck to overtake the red truck.



Victoria invested a total of $2,500, part at 3.5% simple interest, and part at 5% simple interest. At the end of one year, the investments had earned $175 in interest. How much was invested at each rate?

Let's start with a column for loans, principal, rate, time, and interest. The simple interest formula is `I = prt`. Let's let x = the amount that she borrowed at 3.5%. Then since she borrowed a total of 2500, the amount that she borrowed at 5% is 2500-x. Now let's convert each of her interest rates into decimal form. So, 3.5% is the same as .035, so 5% is 0.05. Since she invests both of her investments for one year, time is 1 in both rows. Then the amount of interest is going to be `p*r*t`, so the interest for her 3.5% loan is `0.035x`. And then the amount of interest for her 5% loan is `0.05(2500-x)`. Now if we add all of her interests up, we can let it equal to 175.

Now we have the following linear equation to solve for our applied problem. First distribute .05 into your parentheses and get `125-.05x`, then combine your x terms to get .085x and then subtract 125 from both sides to get 50.

The amount invested at 3.5% is $588.24. And the amount invested at 5% is $1,911.76.



Zeros of functions.

An input, c, for a function's output `f(c) = 0`, is called the zero of the function.

In this section, we'll only be considering zeros of linear functions.



Find the zero of the linear function, `f(x) = 15-x`.

Start by letting `f(x) = 0`. So we get `15-x = 0`. And then solve for x by subtracting 15 from both sides. So `-x = -15` or `x = 15`. 15 is the zero of the function.

If we graph the function of f(x), we get a straight line which intersects the x-axis at the point (15, 0). So notice that the zero of your function is also the x-intercept.