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.
shut up and do your hw
1 anything to the power of 0 equals 1 100=1 x0=1
It is a consequence of the isomorphism between powers of numbers under multiplication and their indices under addition. This leads to the definition of x-a as the [multiplicative] inverse of xa. Then xa * x-a = xa-a = x0 But since x-a is the inverse of xa, their product is 1. That is to say, x0 = 1.
x0 = 1 because any number raised to the power of 0 is always equal to 1
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.
so 10 x0 equals 0 people its times and x is diffrent to plus.
f'(x)= 0x^-1=0anything multiplied by zero equals zero
shut up and do your hw
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.
1 anything to the power of 0 equals 1 100=1 x0=1
Any number to the power of zero equals one (x0 = 1).
The exponent rule for multiplication is xa * xb = xa+b Now, if you put b = 0, then a+b = a so that the above reads: xa * x0 = xa which only works if x0 = 1.
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)
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
Pythagoras can be used to find the distance between any two points (x0, y0) and (x1, y1): Distance = √((x1 - x0)² + (y1 - y0)²) → distance = √((5 - 5)² + (13 - 9)²) = √(0 + 4²) = √(4²) = 4.
0! You said x0! anything x0=0!