A rational expression is not defined whenever the denominator of the expression equals zero. These will be the roots or zeros of the denominator.
No. 0/3 is well defined.
5/5 is 1. Whenever the numerator and denominator is the exact same number, it is equal to 1.
I think you are asking "what is the derivative of [sin(x)]^0=sin^0(x)?" and I shall answer this accordingly. Recall that x^0 = 1 whenever x is not 0. On the other hand, also notice that 0^0 is generally left undefined. Thus, sin^0(x) is the function f(x) such that f(x) is undefined when x = n(pi) and 1 everywhere else. As a result, on every open interval not containing a multiple of pi, i.e. on (n(pi), (n+1)(pi)) the derivative will be zero, since f is just a constant function on these intervals, and whenever x is a multiple of pi, the derivative at x will be undefined. Thus, [d/dx]sin^0(x) is undefined whenever x = n(pi) and 0 everywhere else. In some cases, mathematicians define 0^0 to be 1, and if we were to use this convention, sin^0(x) = 1 for all x, and its derivative would just be 0.
x2 - 4 is a special expression that is referred to as a difference of squares. Whenever an expression is in the format: a2 - b2 it can be factored out as: (a + b)(a - b) In the case of x2 + 4, that would be: (x + 2)(x - 2)
A rational expression is not defined whenever the denominator of the expression equals zero. These will be the roots or zeros of the denominator.
False
Yes, it is true.
A rational number is any number that can be expressed as a fraction. It becomes meaningless or undefined when the lower number, the denominator, its 0 (zero)
false
No. 0/3 is well defined.
Whenever we are dealing with rational fractions.
Whenever you solve a rational equation you must make sure the result obtained for an answer does not allow the denominator of one of the rational expressions to assume a value of ZERO, as division by zero is undefined and therefore prohibited. For example if we have 2x/(x-3) =(x2 -9x)/ x when we multiply out by x(x-3) we get 2x(x) = (x2 -9x)(x-3) so 2x2 = x(x-9)(x-3) 2x2 = x(x2 - 12x + 27) 2x2 = x3 - 12x2 + 27x so 0 = x3 - 14x2 + 27x 0 = x(x2 - 14x + 27) so solutions are 0 and 7 + √22 and 7 -√22 but 0 makes right hand side expression have zero in denominator so it is not a solution. We actually have to look at all obtained solutions to be sure they ae not extraneous. Suppose we had obtained a 3 for a solution. That would make the left side denominator equal zero and we would have to dismiss that, if 3 was obtained. The two irrational solutions we have obtained are genuine solutions as neither introduces a zero to a denominator
Yes it does: whenever you find an equivalent fraction.
cause whenever he sees little boys he makes that facial expression
-Pi is irrational, because it does not terminate or repeat. Whenever you multiply an irrational number by a rational number (-1), the result is an irrational number.
Since any number divided by zero is undefined, a computer cannot base a calculation on this. Whenever a program attempts to divide by zero, this error is generated.