x^0 = 1
1/1 = 1
x0 = 1 because any number raised to the power of 0 is always equal to 1
Any number to the exponent of 0 is equal to 1. EXAMPLE x0=1
Any number to the power zero is equal to one. That can be derived from the following index law: xa*xb = xa+b (x not zero) Now let b = 0 so that the above becomes xa*x0 = xa+0 so xa*x0 = xa (since a+0 = a) That is, any number multiplied by x0 is the number itself. That can be true only if x0 is the multiplicative identity, that is, only if x0 = 1.
5
x0 = x(n -n), which is equal to xn/xn by the law of powers. This obvoiusly = 1
In fact, a non-zero number with an exponent of 0 is always equal to 1. This can be explained with a simple example. Let x = 2. x2=4 x2=4 Thus it follows: x2 / x2 = x0 And thus: x0 = 4 / 4 4 / 4 = 1 Therefore x0=1.
This derives from one of the laws of indices which states that, for any x (not = 0), xa * xb = xa+b Put b = 0 Then xa * x0 = xa+0 = xa (because a + 0 = a) But that means that x0 is the multiplicative identity. And since that is unique, and equal to 1, x0 = 1. This is true for all x. Put
It is a consequence of the definition of the index laws. xa * xb = xa+b If you put b = 0 in the above equation, then you get xa * x0 = xa+0 But a+0 = a so that the right hand side becomes xa Thus the equation now reads xa * x0 = xa For that to be true for all x, x0 must be the identity element for multiplication. That is x0 = 1 for all x.
The answer is -13 1/3ohere is the detailed calculation for the problem:Let x0 be the angle, then;(180 - x0) - 2[180 - (90 - x0)] =40(180 -x0) - 2[90+x0]=40180 -x0 - 180 - 2x0=40-3x0=40hencex0= -13 1/3oAny comments are welcome
In order to get the results of 0x1*2-1*x0 you will have to do a little math. The answer to this math problem is X equals one.
Long story short- x2 / x = x , x3 / x = x2 , so x1 /x = x0 , and any whole number besides zero divided by itself is equal to one, therefore x0 is equal to one.
The general equation for a linear approximation is f(x) ≈ f(x0) + f'(x0)(x-x0) where f(x0) is the value of the function at x0 and f'(x0) is the derivative at x0. This describes a tangent line used to approximate the function. In higher order functions, the same concept can be applied. f(x,y) ≈ f(x0,y0) + fx(x0,y0)(x-x0) + fy(x0,y0)(y-y0) where f(x0,y0) is the value of the function at (x0,y0), fx(x0,y0) is the partial derivative with respect to x at (x0,y0), and fy(x0,y0) is the partial derivative with respect to y at (x0,y0). This describes a tangent plane used to approximate a surface.