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The like terms in this expression are the terms that can be grouped together by having exactly the same set of unknown factors. Here, the sets of unknown factors are x 2 y , {\displaystyle x^{2}y,} y 2 , {\displaystyle y^{2},} and y . {\displaystyle y.} .
If there are like terms in an expression, one can simplify the expression by combining the like terms. One adds the coefficients and keeps the same variable. 4 x + 7 x + 2 x = 15 x {\displaystyle 4x+7x+2x=15x}
[c] For example, x 3 y 2 + 7x 2 y 3 − 3x 5 is homogeneous of degree 5. For more details, see Homogeneous polynomial. The commutative law of addition can be used to rearrange terms into any preferred order.
Therefore, let f(x) = g(x) = 2x + 1. Then, f(x)g(x) = 4x 2 + 4x + 1 = 1. Thus deg(f⋅g) = 0 which is not greater than the degrees of f and g (which each had degree 1). Since the norm function is not defined for the zero element of the ring, we consider the degree of the polynomial f(x) = 0 to also be undefined so that it follows the rules of a ...
A non-vertical line can be defined by its slope m, and its y-intercept y 0 (the y coordinate of its intersection with the y-axis). In this case, its linear equation can be written = +. If, moreover, the line is not horizontal, it can be defined by its slope and its x-intercept x 0. In this case, its equation can be written
The derivative of a constant term is 0, so when a term containing a constant term is differentiated, the constant term vanishes, regardless of its value. Therefore the antiderivative is only determined up to an unknown constant term, which is called "the constant of integration" and added in symbolic form (usually denoted as ).
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It is also possible to take the variable y to be the unknown, and then the equation is solved by y = x – 1. Or x and y can both be treated as unknowns, and then there are many solutions to the equation; a symbolic solution is (x, y) = (a + 1, a), where the variable a may take any value. Instantiating a symbolic solution with specific numbers ...