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This page provides a list of educational videos related to Parts of a Set. You can also use this page to find sample questions, apps, worksheets, lessons , infographics and presentations related to Parts of a Set.
Elements of a Story - Setting
By David Hunter
This video explains the two essential parts of a setting in a narrative: time and place. It illustrates both with photographs and movie stills.
Of Mice and Men - In Context (Part 2 of 2)
By TheRoborovsky
This video is the second of two parts that give geographical, social, and historical context of the novel, Of Mice and Men by John Steinbeck. Like the first part, it features films of the Salinas Valley where the novel is set, historic footage of the Dust Bowl, clips from movie versions of the book,and interviews. The audio includes a recording of Woody Guthrie singing Do Re Mi as well as sections of the book read aloud
Intermediate Algebra | MathHelp.com
By MathHelp.com
This lesson covers motion problems. Students learn to solve advanced "motion" word problems -- for example, riding a bike to pick up a car and driving back, or biking part of a trip and taking a boat the rest of the trip. Students should first draw a diagram to represent the relationship between the distances involved in the problem, then set up a chart based on the formula rate times time = distance. The chart is then used to set up the equation.
Of Mice and Men - In Context (Part 1 of 2)
By TheRoborovsky
This video, the first of two parts, gives the geographic, social, and historical context of the novel, Of Mice and Men by John Steinbeck. It features films of the novel's setting, the Salinas Valley in California as well as historic footage and photographs. The audio includes passages from the book, interviews, and even a recording of Woody Guthrie describing his travels across the country during the depression.
How To Integrate Using U-Substitution
By The Organic Chemistry Tutor
This calculus video tutorial provides a basic introduction into u-substitution. It explains how to integrate using u-substitution. You need to determine which part of the function to set equal to the u variable and you to find the derivative of u to get du and solve for dx. After replacing all x variables with u variables, find the antiderivative of f(u) and substitute u in the new function with x variables. This video contains plenty of examples and practice problems of finding the indefinite integral using u-substitution. Examples include polynomial functions, trigonometric functions, exponential functions, square root functions, and rational functions.
Evaluating Logarithms | MathHelp.com
By MathHelp.com
In this example, notice that we have a polynomial divided by a binomial, and our binomial is in the form of an x term minus a constant term, or x – c. In this situation, instead of having to use long division, like we did in the previous lesson, we can divide the polynomials using synthetic division, which is a much more efficient method. Here’s how it works. We start by finding the value of c. Since –c = -3, we know that c = 3. Next, we put the value of c inside a box, so we put the 3 inside a box. It’s very important to understand that the number that goes inside the box always uses the opposite sign as the constant term in the binomial. In other words, since the constant term in the binomial is -3, the number that goes inside the box, is positive 3. Next, we write the coefficients of the dividend, which are 2, -7, 4, and 5. Be very careful with your signs. Now, we’re ready to start our synthetic division. First, we bring down the 2. Next, we multiply the 3 in the box times 2 to get 6, and we put the 6 under the -7. Next, we add -7 + 6 to get -1. Next, we multiply the 3 in the box times -1 to get -3, and we put the -3 under the 4. Next, we add 4 + -3 to get 1. Next, we multiply the 3 in the box times 1 to get 3, and we put the 3 under the 5. Finally, we add 5 + 3 to get 8. Now, notice that we have a 2, -1, 1, and 8 in the bottom row of our synthetic division. These values will give us our answer: the first 3 numbers represent the coefficients of the quotient, and the last number is the remainder. And it’s important to understand that our answer will be one degree less than the dividend. In other words, since our dividend starts with x cubed, and we’re dividing by x, our answer will start with x squared. So our answer is 2x squared – 1x + 1 + 8 over x – 3. Notice that we always use descending order of powers in our quotient. In this case x squared, x, and the constant. Finally, remember that we add the remainder over the divisor, just like we did in the previous lesson on long division, and we have our answer. It’s important to understand that we’ll get the same answer whether we use synthetic division or long division. However, synthetic division is much faster.
Solving Logarithmic Equations | MathHelp.com
By MathHelp.com
Here we’re asked to evaluate each of the following logarithms. In part a, we have log base 7 of 49. To evaluate this logarithm, we set it equal to x. In other words, log base 7 of 49 = what? Notice that we now have an equation written in logarithmic form, so let’s see if we can solve the equation by converting it to exponential form. Remember that the base of the log represents the base of the power, the right side of the equation represents the exponent, and the number inside the log represents the result, so we have 7…to the x…= 49. Next, we solve for x. Notice that 7 and 49 have a like base of 7, so we rewrite 49 as 7 squared, and we have 7 to the x = 7 squared, so x must equal 2. In part b, we have log base 3 of 1/27. Again, to evaluate this logarithm, we set it equal to x, and convert the logarithmic equation to exponential form. Remember that the base of the log represents the base of the power, the right side of the equation represents the exponent, and the number inside the log represents the result, so we have 3…to the x…= 1/27. Next, we solve for x. Notice that 3 and 1/27 have a like base of 3, so we rewrite 1/27 as 1 over 3 cubed, and we have 3 to the x = 1 over 3 cubed. Next, 1 over 3 cubed is the same thing as 3 to the negative 3, so we have 3 to the x = 3 to the negative 3, which means that x must equal -3. Therefore, log base 3 of 1/27 = -3. So remember the following rule. To evaluate a logarithm, set it equal to x, convert to exponential form, and solve the equation using like bases.
Reading Comprehension Strategies: Compare and Contrast
By Anovellife
"Compare and Contrast. Compare and Contrast.If they are the same, they go in the middle, and if they are different, they go on the sides."
OST 2020-21 Blueprint and Best Practices | 10 Min. webinar organized by EdShorts
By Lumos Learning
This Free 10 Min power-packed webinar organized by EdShorts on Jan 27th provides all the information available about 2021 OST Assessments - such as testing guidelines, blueprint changes, testing windows, and more! Join the EdShorts Facebook community today: https://www.facebook.com/groups/60370... for more bite-sized power-packed webinars every week!
Adding and Subtracting Polynomials | MathHelp.com
By MathHelp.com
In this example, notice that each of our variables, x, y, and z, appears in all three equations. To solve this system, we use the addition method. In other words, let’s start with our first two equations, x + y + z = 4, and x – y + z = 2. Notice that if we add these equations together, the +y and –y will cancel out, and we have 2x + 2z = 6. So, in our new equation, 2x + 2z = 6, we’ve eliminated the variable y. Unfortunately, we still haven’t solved for any of our variables. However, if we can create another equation with just x and z in it, then we’ll have a system of equations in two variables, which we can use to solve for x and z. To create another equation with just x and z in it, we need to eliminate y. We can’t add the first and second equations together, because we’ve already done that. However, notice that if we add the first and third equations together, the first equation has a +y and the third equation has a –y, so we’ll be able to eliminate the y. So we have our first equation, x + y + z = 4, and our third equation, x – y – z = 0, and adding them together, notice that the +y – y cancels out, and, as a bonus, the +z – z also cancels out, so we have 2x = 4, and dividing both sides by 2, x = 2. Now, since we know that x = 2, notice that if we plug a 2 in for x in the equation that we created earlier, we can solve for z. And we have 2(2) + 2z = 6, or 4 + 2z = 6, and subtracting 4 from both sides, we have 2z = 2, and dividing both sides by 2, z = 1. So x = 2 and z = 1, and to find the value of y, we simply plug our values of x and z into any of the equations in the original system. Let’s use the first equation, x + y + z = 4. Since x = 2 and z = 1, we plug a 2 in for x and a 1 in for z, and we have 2 + y + 1 = 4, or 3 + y = 4, and subtracting 3 from both sides, y = 1. So x = 2, y = 1, and z = 1, and finally, we write our answer as the ordered triple, x, y, z, or (2, 1, 1).
