A cube has eight vertices (corners) and twelve edges.  A segment, such as $x$, which joins two vertices not joined by an edge is called a diagonal.  Segment $y$ is also a diagonal.  How many diagonals does a cube have?

$\text{(A)}\ 6 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 16$
On each face, there are $2$ diagonals like $x$.  There are $6$ faces on a cube.  Thus, there are $2\times 6 = 12$ diagonals that are "x-like".
Every "y-like" diagonal must connect the bottom of the cube to the top of the cube.  Thus, for each of the $4$ bottom vertices of the cube, there is a different "y-like" diagonal.  So there are $4$ "y-like" diagonals.
This gives a total of $12 + 4 = 16$ diagonals on the cube, which is answer $\boxed{E}$.