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Q: When x0 how many solutions is this?

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The equation of a sphere with radius r, centered at (x0 ,y0 ,z0 ) is (x-x0 )+(y-y0 )+(z-z0 )=r2

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.

There are two distinct real solutions.

If a system of equations is inconsistent, there are no solutions.

A quadratic equation can have either two real solutions or no real solutions.

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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 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.

0! You said x0! anything x0=0!

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

On a transformer connection H1 and H2 are the primary connections. X1 and X2 are the secondary connections. If your transformer has a split secondary that is grounded, that terminal is X0. The sequence is X1 - X0 - X2. The X0 usually indicates that there is a connection to a neutral wire along with the ground wire.

y2 there will me many solutions

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.

#include#include#include#includefloat eq(float);void falsi(float,float,int);void main(){float x0,x1;int iter;clrscr();cout

let f be a function and f' the first derivative.If f'>0 the function is genuinely ascending.If f'

The integral of e-2x is -1/2*e-2x + c but I am not sure what "for x0" in the question means.

Function f(x) x2=x*x f=x*x2-3*x2+2*x-1 return end function f1(x) f1=3*x*x-6*x+2 return end 5 write(*,*)'enter initial guess' read(*,*) x0 write(*,*)'enter tolerrence' read(*,*) eps 10 x1=x0-f(x0)/f1(x0) if(abs((x1-x0)/x0).lt.eps) then write(*,1) x1 1 format('solution=',f10.4) stop else x0=x1 go to 10 endif end

That would besqrt[ (x80 - x0)2 + (y80 - y0)2 ) at an angle of tan-1 (y80 - y0) / (x80 - x0)or(x80 - x0) i + (y80 - y0) j

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