The two velocities shown in the diagram.

The first vector has a magnitude of $3 \text{ m/s}$ at $40^\circ$ above the horizontal, pointing to the left.

$$v_{1x} = -3 \cos(40^\circ)$$ $$v_{1y} = 3 \sin(40^\circ)$$

Using approximate values:

$$v_{1x} = -3 \times 0.7660 \approx -2.298 \text{ m/s}$$ $$v_{1y} = 3 \times 0.6428 \approx 1.928 \text{ m/s}$$

The second vector has a magnitude of $6 \text{ m/s}$ at $20^\circ$ above the horizontal, pointing to the right.

$$v_{2x} = 6 \cos(20^\circ)$$ $$v_{2y} = 6 \sin(20^\circ)$$

Using approximate values:

$$v_{2x} = 6 \times 0.9397 \approx 5.638 \text{ m/s}$$ $$v_{2y} = 6 \times 0.3420 \approx 2.052 \text{ m/s}$$

$$R_x = v_{1x} + v_{2x} = -2.298 + 5.638 = 3.340 \text{ m/s}$$ $$R_y = v_{1y} + v_{2y} = 1.928 + 2.052 = 3.980 \text{ m/s}$$

$$|\vec{R}| = \sqrt{R_x^2 + R_y^2}$$ $$|\vec{R}| = \sqrt{(3.340)^2 + (3.980)^2}$$ $$|\vec{R}| = \sqrt{11.1556 + 15.8404}$$ $$|\vec{R}| = \sqrt{26.996}$$ $$|\vec{R}| \approx 5.196 \text{ m/s}$$

Since both $R_x$ and $R_y$ are positive, the resultant vector is in the first quadrant.

$$\tan\phi = \frac{R_y}{R_x} = \frac{3.980}{3.340}$$ $$\tan\phi \approx 1.1916$$ $$\phi = \arctan(1.1916)$$ $$\phi \approx 50.0^\circ$$

The resultant velocity has a magnitude of $\boxed{5.20 \text{ m/s}}$ and a direction of $\boxed{50.0^\circ \text{ above the horizontal}}$.