#### Answer

The x-component of the vector is 12 m/s

#### Work Step by Step

Since the $40^{\circ}$ angle is below the positive x-axis, the x-component of the vector is positive.
We can find the x-component. Let $y$ be the magnitude of the y-component;
$tan(\theta) = \frac{y}{x}$
$x = \frac{y}{tan(\theta)}$
$x = \frac{10~m/s}{tan(40^{\circ})}$
$x = 12~m/s$
The x-component of the vector is thus 12 m/s.