Ans is 0, because when anything multiply with zero the answer is zero.
10x-2y=30
It is simply y0/y.
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.
To solve the equation ( xy^0 + 5x + 2y - 3 = 0 ), note that ( y^0 = 1 ), simplifying the equation to ( x + 5x + 2y - 3 = 0 ) or ( 6x + 2y - 3 = 0 ). Rearranging gives ( 2y = 3 - 6x ), leading to ( y = \frac{3 - 6x}{2} ). This represents a linear relationship between ( x ) and ( y ).
Use Pythagoras: For a line joining two points (x0, y0) to (x1, y1) there is a right angle triangle: One side (leg) joins (x0, y0) to (x1, y0) with length (x1 - x0) One side (leg) joins (x1, y0) to (x1, y1) with length (y1 - y0) The hypotenuse joins (x0, y0) to (x1, y1) → length hypotenuse = √((x1 - x0)² + (y1 - y0)²) → length between (0, 0) and (-15, 8) is given by: distance = √((-15 - 0)² + (8 - 0)²) = √((-15)² + (8)²) = √(225 + 64) = √289 = 17 units
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
y0
in ohio y0
linear (A+)
The equation of a sphere with radius r, centered at (x0 ,y0 ,z0 ) is (x-x0 )+(y-y0 )+(z-z0 )=r2
You get bad boy spankings from y0 mama
5x-5x = 0