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Powers of 10

September 26th, 2011
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Logarithms to Exponentials and Back Again

April 13th, 2009

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FOIL

March 9th, 2009

 

 

This video covers the process of FOILing two binomials. Multiply: First, Outer, Inner, Last. A trick is given to help you remember the order.

 


Transcript:

 

Welcome to Your Tutor Online video lessons. Today we are going to look at how to FOIL. FOILing is what you do when you multiply two binominals together. FOIL is an acronym which stands for First, Outer, Inner, and Last.

 

When we talk about the first terms, we are going to look at each respective parenthesis. The first term in each parenthesis is the first term. The outer terms are for the outside of the entire thing, our 2x and our negative 5 are both our outside terms. The inside is just the opposite, so the two closest to each other. And then last, the two terms that are last in each of the parenthesis. So, 3 and negative 5.

 

And now we just follow our acronym in order: First, Outer, Inner, Last. First, we will multipy together our two first terms, our 2x and our x which gives us 2x squared. Next is outer, 2x and negative 5. 2x times negative 5 is negative 10x. Our inside terms are 3 and x which gives us positive 3x. And our last terms 3 and negative 5 gives us negative 15.

 

Now, almost always you will end up with your two middle terms being like terms so you just want to combine those two as your final answer. 2x squared, combine our two like terms (-10x and positve 3x) gives us negative 7x and minus 15.

 

Here’s a tip for you to remember to do the FOILing in the correct way. If you look back up here, how I drew the lines, I have almost a little smile face. Here is the smile, the nose, and two eye-ball looking things. If you get the smile face at the end that means you drew your FOIL correctly and you multiplied the terms together correctly.

 

Alright, I am going to show you how to draw this face one more time so you have a good check for when you have to do FOIL on your own. We are going to follow in order: First, Outer, Inside, Last. For the first term, we will start above the problem. Connect the first two terms with an arc. So in this example, 3x and 2x and that will give us our first eye brow. Now the outside, we will connect our two outside terms 3x and 1, which gives us our smile. The inside terms, the 2 and the 2x, gives us our nose. And our last terms, negative 2 and positive 1, give us our last eyebrow. And now if you did it correctly you will see a face in here. And now you know you did your problem correctly.

 

I hope you found this lesson useful. And, as always I’m glad to accept your donations. I am asking $1 per lesson. That is the cheapest tutoring that is out there. You can head over to my website at YourTutorOnline.com and click the donate button at the top of the page. I want to thank everybody for all of their comments they leave over at YouTube, keep them coming. I appreciate the comments and questions. And if you need more help, I am available for private tutoring. Just head on over to my webpage at YourTutorOnline.com and click on the tutoring link at the top of the page for more information. You could all help me out by writing a review of my podcast at PodcastAlley.com or write a review in iTunes. I will see you all next time, class dismissed.

 

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Exponents with Parenthesis

March 4th, 2009

This video looks at the exponent rules involving parentheses.


Transcript: coming soon.

Solution Problems

February 12th, 2009

This videos looks at problems involving solutions. Example: You have a 15mL solution which is 10% salt, how much water do you have to add to make it 8%?


Transcript coming soon.

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Simplify Complex Rational Expressions

February 2nd, 2009

This video shows how to simplify fractions within fractions (known as “complex rational expressions). Just find the Lowest Common Denominator (LCD) and multiply all the fractions by it.


Transcript: coming soon!

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Add & Subtract Positive & Negative Numbers

January 12th, 2009

This lesson teaches a method for adding and subtracting positive and negative numbers. (Other than the number-line method)


Transcript: coming soon!

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Factor a Quadratic Equation with a Leading Coefficient

December 8th, 2008

This video explains how to factor a quadratic equation with a leading coefficient. (an equation that looks like ax^2 + bx + c).


Transcript coming soon!

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Matrix Addition

December 1st, 2008

This video covers the basics of matrices and how to add them together. I am trying my hand at Adobe Flash. Let me know how you like this new style by clicking “Comments >>” below.


Transcript:

 

Welcome to Your Tutor Online video lessons. Today we will look at matrices and matrix addition.

 

A matrix is a way to organize numbers. It has rows and columns. This gives us our dimensions. In this example, this matrix is a 2 by 3 matrix because it has two rows and three columns.

 

In order to add matrices they must have the same dimensions. Here, we have two matrices that have 2 rows and 2 columns each. All we have to do is add up the corresponding numbers in each matrix and put that answer in a new matrix in the same spot. So, we’ll start with the numbers in the first row, first column; add those together and put that answer in the first row, first column. We’ll repeat with the number in the first row, second column and so on for each number until you’ve got your answer.

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Solve Systems of Linear Equations: Substitution

November 3rd, 2008

Also in this series:


Transcript:

Welcome to Your Tutor Online video lessons. This is the last of a 4 part series on systems of equations. In this video we will look at how to solve systems of equations by substitution.

 

Substitution just means you’ll find out what one variable is equal to and you plug it into the other equation. In this first example we already see that x is the same thing as 2y minus 3. So that means where-ever we see an x in the other equation we can put in this for it instead. And that way we will only have ys to deal with and be able to solve the equation.

 

So I’ll begin to re-write the second equation: 3y plus 2 x, and instead of x we are going to put in what we know x is equal to. X is the same thing as 2y minus 3; is equal to 8.

 

Now that we only have one variable, we can solve for y. 3y and distribute the 2; plus 4y minus 6 is equal to 8. Combine like terms and get 7y minus 6 is equal to 8. Add 6 to both sides. 7y is equal to 14. And divide both sides by 7. We find out that y is equal to 2. We are going to take this answer we got for y and plug it back into either one of the original equations to solve for x.

 

I chose to plug this back into the first equation: x = 2y – 3. So we are going to take our y is equal to 2, put it where we see the y and go ahead and solve for x. 2 times 2 is 4. Subtract 3 and x is equal to 1. And now we know both solutions. X is equal to 1 and y is equal to 2.

 

Sometimes a variable will not already by isolated, or by itself, so you have to work with an equation to get one of the variables by itself. In this example the second equation could easily be turned into one where y is by itself if we just subtract 3x from both sides. So it becomes y = -3x + 5. And we can take that, just like we did before, and plug that into the other equation where-ever we see a y.

 

4x – 2; plug in the y; -3x +5 is equal to 10. Now all we have are x’s and we can solve for x. Distribute the -2; So we get plus 6x minus 10 is equal to 10. And we’ll continue this on another page. We’ll combine our like terms the 4x and the 6x leaves us with 10x and at the same time I’ll add 10 to both sides. 10 plus 10 is 20. We’ll divide both sides by 10 to get x by itself and have our answer for x is 2. Now that we know what x is we can go back to one of the original equations, plug in 2 where we see an x and solve for y.

 

Here I plugged in 2 into the first equation and we will solve for y. 4 times 2 is 8 minus 2y is equal to 10. We’ll subtract 8 from both sides and be left with a -2y on the left, a +2 on the right. Divide both sides by a -2 and y is equal to -1.

 

That’s all there is for solving systems of equations by substitution. They get a little more complex especially when the variables are not already by themselves. But follow this process and you’ll be able to solve any of them.

 

I hope you found this lesson useful. If you have any questions leave a comment on the blog at www.YourTutorOnline.com If you have any lesson suggestions send an email to podcast@yourtutoronline.com Thanks for watching, class dismissed.

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