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Simplify rational expressions by canceling monomial factors
By Khan Academy
Sal simplifies (28b^6)/(7b) as 4b^5.
Simplify rational expressions by canceling monomial factors
By Khan Academy
Sal simplifies (28b^6)/(7b) as 4b^5.
Simplify rational expressions by canceling binomial factors
By Khan Academy
Sal solves the following problem:������������Given a rectangle with length a^2+6a+27 and width a^2-9, write the ratio of the width to the length as a simplified rational expression.
Simplify rational expressions by canceling binomial factors
By Khan Academy
Sal������������simplifies and states the domain of (x^2-36)/(6-x).
Simplify rational expressions by canceling monomial factors
By Khan Academy
Sal explains what it means to simplify a rational expression and why we would want to do that. Just don't forget the excluded values!
Simplify rational expressions using advanced factorization methods
By Khan Academy
Sal������������simplifies and states the domain of (2x^2+13x+20)/(2x^2+17x+30).
Simplify rational expressions using advanced factorization methods
By Khan Academy
Sal������������simplifies and states the domain of (2x^2+13x+20)/(2x^2+17x+30).
Simplify rational expressions by canceling binomial factors
By Khan Academy
Sal solves the following problem:������������Given a rectangle with length a^2+6a+27 and width a^2-9, write the ratio of the width to the length as a simplified rational expression.
Simplify rational expressions by canceling binomial factors
By Khan Academy
Sal������������simplifies and states the domain of (x^2-36)/(6-x).
Simplify rational expressions by canceling monomial factors
By Khan Academy
Sal explains what it means to simplify a rational expression and why we would want to do that. Just don't forget the excluded values!
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.
Increasing/Decreasing , Local Maximums/Minimums
By PatrickJMT
This course on First Semester Calculus includes Basic idea of Limits, Finding limits from a graph, Calculating a limit by factoring and canceling, Calculating a limit by Expanding and Simplifying, Calculating a limit by getting a common denominator, Limits involving Absolute Value, The Squeeze Theorem for Limits, Infinite Limits and more.