Is a convex cone a convex polyhedron? Say that I have a convex cone

$C=\{t|Ax = t, x\geq 0\}$.

where $x\in R^n$, and $t\in R^m$, $A\in R^m\times R^n$. 

Can I say that this is a convex polyhedron? and why?

EDIT:
Just in order to avoid confusion, the definition of convex polyhedron I am thinking of is: the set of solutions to a finite number of linear equalities and inequalites.
Yes, or more precisely (as polyhedron may imply three dimensions) a convex polytope. You may readily verify that $C$ is convex, and also that its boundary is piecewise a part of a hyperplane.
No.  A cone is not a "polyhedron".  A polyhedron, by definition has planar (straight) sides, not curved sides.