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See It, Say It, Sign It | Letter Sounds| ASL Alphabet |Jack Hartmann
By Jack Hartmann Kids Music Channel
See It, Say It, Sign It by Jack Hartmann teaches sign language for each letter and the letter sounds for each letter of the alphabet with American sign language in this Jack Hartmann video. Jack shows the sign for each letter and the letter sounds for each letter as he sings the letter name and
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
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.
Definition of Perpendicular Lines | MathHelp.com
By MathHelp.com
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Solving Algebra Problems | MathHelp.com
By MathHelp.com
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Subtraction | Subtracting Whole Numbers | MathHelp.com
By MathHelp.com
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