12 mph tops by sanford university.
The equation for the gradient of a linear function mapped in a two dimensional, Cartesian coordinate space is as follows.
The easiest way is to either derive the function you use the gradient formula
(y2 - y1) / (x2 - x1)
were one co-ordinate is (x1, y1) and a second co-ordinate is (x2, y2)
This, however, is almost always referred to as the slope of the function and is a very specific example of a gradient. When one talks about the gradient of a scalar function, they are almost always referring to the vector field that results from taking the spacial partial derivatives of a scalar function, as shown below.
The equation for the gradient of a function, symbolized ∇f, depends on the coordinate system being used.
For the Cartesian coordinate system:
∇f(x,y,z) = ∂f/∂x i + ∂f/∂y j + ∂f/∂z k where ∂f/(∂x, ∂y, ∂z) is the partial derivative of f with respect to (x, y, z) and i, j, and k are the unit vectors in the x, y, and z directions, respectively.
For the cylindrical coordinate system:
∇f(ρ,θ,z) = ∂f/∂ρ iρ + (1/ρ)∂f/∂θ jθ + ∂f/∂z kz where ∂f/(∂ρ, ∂θ, ∂z) is the partial derivative of f with respect to (ρ, θ, z) and iρ, jθ, and kz are the unit vectors in the ρ, θ, and z directions, respectively.
For the spherical coordinate system:
∇f(r,θ,φ) = ∂f/∂r ir + (1/r)∂f/∂θ jθ + [1/(r sin(θ))]∂f/∂φ kφ where ∂f/(∂r, ∂θ, ∂φ) is the partial derivative of f with respect to (r, θ, φ) and ir, jθ, and kφ are the unit vectors in the r, θ, and φ directions, respectively.
Of course, the equation for ∇f can be generalized to any coordinate system in any n-dimensional space, but that is beyond the scope of this answer.
1/4 X 1/4 = 1/16
its around $15-$20. There is no such thing as an M8A1 bayonet. The M8A1 is the designation of the scabbard that holds the M-4, M-5 (series), M-6, and M-7 bayonet.