The position (resp. velocity, resp. acceleration) at time t of a particle of mass m is c(t) (resp. v(t), resp. a(t)). Let F(x, y, z) = (f(x, y, z), g(x, y, z), h(x, y, z)) be a C 1 mapping representing a force at (x, y, z). Then Newton’s second law states: F(c(t)) = ma(t). Show that d dt[mc(t) × v(t)] = c(t) × F(c(t))

(f(x, y, z), g(x, y, z), h(x, y, z))

Step-by-step explanation:

i think it is amswer

because 1.2 isnt half of 6 duh

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.