Remember to Simplify Videos - Free Educational Videos for Students in K - 12

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Rational Exponents | MathHelp.com


By MathHelp.com

In this example, we’re asked to write “a” to the negative 3rd squared in simplest form without negative or zero exponents. Remember that the power rule tells us that when we have a power taken to another power, such as a to the negative 3rd squared, we multiply the exponents. So we have a to the -3 times 2, or a to the negative 6th. Finally, remember from our study of negative exponents that a to the negative 6th can be written as 1 over a to the positive 6th. So a to the negative 3rd squared simplifies to 1 over a to the 6th.

Negative Exponents | MathHelp.com


By MathHelp.com

In this example, we’re given the functions f(x) = 3x – 2 (read as “f of x equals…”) and g(x) = root x, and we’re asked to find the composite functions f(g(9)) (read as “f of g of 9”) and g(f(9). To find f(g(9)), we first find g(9). Since g(x) = root x, we can find g(9) by substituting a 9 in for the x in the function, to get g(9) = root 9, and the square root of 9 is 3, so g(9) = 3. Now, since g(9) = 3, f(g(9)) is the same thing as f(3), so our next step is to find f(3). And remember that f(x) = 3x – 2, so to find f(3), we substitute a 3 in for the x in the function, and we have f(3) = 3 times 3 minus 2. Notice that I always use parentheses when substituting a value into a function, in this case 3. Finally, 3 times 3 minus 2 simplifies to 9 minus 2, or 7, so f(3) = 7. Therefore, f(g(9)) = 7. Next, to find g(f(9), we first find f(9). Since f(x) = 3x - 2, we find f(9) by substituting a 9 in for the x in the function, to get f(9) = 3 times 9 minus 2, which simplifies to 27 – 2, or 25, so f(9) = 25. Now, since f(9) = 25, g(f(9)) is the same thing as g(25), so our next step is to find g(25). And remember that g(x) = root x, so to find g(25), we substitute a 25 in for the x in the function, to get g(25) = root 25. Finally, the square root of 25 is 5, so g(25) = 5. Therefore, g(f(9)) = 5. It’s important to recognize that

Parallel Lines | MathHelp.com


By MathHelp.com

This lesson covers imaginary numbers. Students learn that the imaginary number "i" is equal to the square root of -1, which means that i^2 is equal to (the square root of -1) squared, which equals -1. Students also learn to simplify imaginary numbers. For example, to simplify the square root of -81, think of it as the square root of -1 times the square root of 81, which simplifies to i times 9, or 9i. To simplify 11/8i, the first step is to get rid of the "i" in the denominator by multiplying both the numerator and the denominator of the fraction by i, to get 11i/8i^2, and remember that i^2 = -1, so we have 11i/8(-1), or 11i/-8, or -11i/8.

Logarithm Rules: Expanding Logarithms | MathHelp.com


By MathHelp.com

To simplify 81 to the ½, remember from our study of rational exponents that an exponent of ½ means that we take the square root of the base. In other words, 81 to the ½ means the same thing as the square root of 81. And the square root of 81 is 9. So 81 to the ½ is 9

Imaginary Numbers | MathHelp.com


By MathHelp.com

This lesson covers estimating quotients. Students learn to estimate a product or quotient by first rounding each number to one non-zero digit. For example, to estimate 8,291 x 27, first round 8,291 down to 8,000, and round 27 up to 30, then multiply 8,000 x 30 to get 240,000.

Quadratic Word Problems | MathHelp.com


By MathHelp.com

A number is 56 less than its square. Find the number. To solve this problem, let’s translate the first sentence into an equation. A number, that’s x, is, =, 56 less than it’s square, that’s x squared – 56. Remember that “less than” switches the order around. In other words, “56 less than its square” is not 56 minus x squared, it’s x squared minus 56. Next, since we have an x squared term in our equation, we set it equal to 0 by subtracting x from both sides, and we have 0 = x squared – x – 56. Next, we factor the right side as the product of two binomials. In the first position of each binomial, we have the factors of x squared, x and x. In the second position of each binomial, we’re looking for the factors of -56 that add to -1, which are -8 and positive 7. So we have 0 = x - 8 times x + 7, which means that either 0 = x – 8 or 0 = x + 7. Finally, in the first equation, we add 8 to both sides, to get 8 = x. And in the second equation, we subtract 7 from both sides, to get -7 = x. So 8 = x or -7 = x. It’s important to understand that both of these answers work. Plugging an 8 back into the original problem, we have 8 is 56 less than 8 squared, or 8 = 8 squared – 56, which simplifies to 8 = 64 – 56, or 8 = 8, which is a true statement. And plugging a -7 back into the original problem, we have -7 is 56 less than -7 squared, or -7 = -7 squared – 56, which simplifies to -7 = 49 – 56, or -7 = -7, which is also a true statement.

Complex Numbers | MathHelp.com


By MathHelp.com

MathHelp.com offers custom 10th grade math courses, including Algebra 1, Geometry, and Algebra 2. Perfect for homeschoolers or any student who needs lots of extra help getting back up to speed in school.

16 - Simplify Logarithms - Part 1 (Log Bases, Calculate Logarithms & More)


By Math and Science

Quality Math And Science Videos that feature step-by-step example problems!

Systems of Inequalities | MathHelp.com


By MathHelp.com

Here we’re asked to write the equation of a circle with center (-2, 7) and radius 5. Remember from the previous example that the formula for the equation of a circle is (x – h)^2 + (y – k)^2 = r^2, where (h, k) is the center of the circle, and r is the radius. And in this problem, since we’re given that the center is (-2, 7), which represents (h, k), and radius is 5, which represents r, we simply plug these values into the formula. And we have x – parentheses -2 squared….+ y – parentheses -7 squared…= parentheses 5 squared. Notice that I changed the parentheses that were in the original formula to brackets so that we wouldn’t have too many sets of parentheses when plugging our values into the formula. Finally, we simplify inside the brackets. x minus a negative 2 simplifies to x + 2, so we have parentheses x plus 2 squared + parentheses y – 7 squared = 5 squared, or 25. Notice that I changed the brackets back to parentheses in the final answer. So the equation of a circle with center (-2, 7) and radius 5 is (x plus 2)^2 + (y minus 7)^2 = 25. Notice that the negative 2, positive 7 in our given center becomes a positive 2, negative 7 in our formula, and don’t forget to square the radius of 5 to get 25.

Multiplying Scientific Notation | MathHelp.com


By MathHelp.com

In this example, which involves natural logarithms, we’re asked to solve each of the following equations for x, and leave our answers in terms of e. To solve for x in the first equation, ln x = 3, we simply switch the equation from logarithmic to exponential form. Remember that ln x means the natural logarithm of x, and a natural log has a base of e. So, to convert the given 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 e…to the 3rd…= x, and we’ve solved for x. Notice that our answer, e cubed, is written in terms of e, which is what the problem asks us to do. Now, let’s take a look at the second equation, ln x squared = 8. Again, we solve for x by switching the equation from logarithmic to exponential form. Ln x squared means the natural logarithm of x squared, and a natural log has a base of e. So, converting the equation to exponential form, we have e…to the 8th…= x squared. Next, since x is squared, we take the square root of both sides. On the right, the square root of x squared is x. On the left, however, there are a couple of things to watch out for. First, remember that the square root of e to the 8th is the same thing as e to the 8th to the ½, which simplifies to e to the 8 times ½, or e to the 4th. Also, remember that when we take the square root of both sides of an equation, we use plus or minus, so our final answer is plus or minus e to the 4th = x.