x0 = 1 because any number raised to the power of 0 is always equal to 1
Zero. x0 = 1.
x0 = x(n -n), which is equal to xn/xn by the law of powers. This obvoiusly = 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.
Any number to the exponent of 0 is equal to 1. EXAMPLE x0=1
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.
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.
3 over 12 and 1 over 12 are not equal.
8 over -7 equal = 1
The integral of e-2x is -1/2*e-2x + c but I am not sure what "for x0" in the question means.
x3 = x times x times x x0 = x divided by x ---- (another explanation) x0 = 1 provided x in not 0 because this definition is consistent with all other definition of exponents. The easiest is the rule for multiplying powers of a like base by adding the exponents: (xp)(xq) = xp+q Suppose q = 0 and you apply the rule to get (xp)(x0) = xp+0 = xp (1) . Cancel xp from both sides and get x0 = 1. Another rule says to divide powers, subtract the exponents: (xp)/(xq) = x p-q Suppose you apply the rule when p = q: You get (xp)/(xp) = x p-p = x0. But xp/xp = 1 so x0 must be 1. A more complicated reason is that the limit as x goes to infinity of x(1/n) is 1. so it makes sense to define 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.
1 over 6
Assuming you want the equation of the straight line between the two points (x0, y0) and (x1, y1), the equation is: y - y0 = m(x - x0) where m is the gradient between the two points: m = (y1 - y0) ÷ (x1 - x0) Note: if the two x coordinates are equal, that is x0 = x1, then the equation of the line is x = x0.
It's a method used in Numerical Analysis to find increasingly more accurate solutions to the roots of an equation. x1 = x0 - f(x0)/f'(x0) where f'(x0) is the derivative of f(x0)
33 over 33 is 1 and 999 over 999 is 1, so yes, they are equal.
1 trillion is equal to 1 trillion over 1, smartypants
No, 1 over 2.
0! You said x0! anything x0=0!
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.
2 over 6 or 1 over 3