A line has slope $\frac{2}{5}.$  Which of the following vectors are possible direction vectors for the line?

[asy]
usepackage("amsmath");

unitsize(1 cm);

pair x = (3,0), y = (0,2);

label("(A) $\begin{pmatrix} 2 \\ 5 \end{pmatrix}$", y);
label("(B) $\begin{pmatrix} 5 \\ 2 \end{pmatrix}$", x + y);
label("(C) $\begin{pmatrix} 0 \\ 0 \end{pmatrix}$", 2*x + y);
label("(D) $\begin{pmatrix} -2 \\ -5 \end{pmatrix}$", 3*x + y);
label("(E) $\begin{pmatrix} -5 \\ -2 \end{pmatrix}$", (0,0));
label("(F) $\begin{pmatrix} 2/5 \\ 1 \end{pmatrix}$", x);
label("(G) $\begin{pmatrix} 40 \\ 16 \end{pmatrix}$", 2*x);
label("(H) $\begin{pmatrix} 4 \\ 10 \end{pmatrix}$", 3*x);
[/asy]

Enter the letters of the correct options, separated by commas.
Since the slope of the line is $\frac{2}{5},$ the line rises 2 units vertically for every 5 horizontal units.  Thus, a possible direction vector is $\begin{pmatrix} 5 \\ 2 \end{pmatrix}.$

[asy]
unitsize(1 cm);

pair A, B, C;

A = (0,0);
B = (5,0);
C = (5,2);

draw(A--B--C);
draw(A--C,red,Arrow(6));

label("$5$", (A + B)/2, S);
label("$2$", (B + C)/2, E);
[/asy]

This means any nonzero scalar multiple of $\begin{pmatrix} 5 \\ 2 \end{pmatrix}$ is a possible direction vector.  The possible options are then $\boxed{\text{B, E, G}}.$