This is a related rates problem involving a ladder leaning against a wall. Let's break it down step by step:

Given:

• The ladder length is constant at 14 ft (hypotenuse of the right triangle).
• The top of the ladder is sliding down the wall at a rate of $-dy/dt$ ft/sec.
• We need to find how fast the base of the ladder is moving away from the wall, denoted as $dx/dt$, when the base is a certain distance ($x$) from the wall.

The relationship between the variables is based on the Pythagorean Theorem: $x^2 + y^2 = 14^2$

Differentiate with respect to time $t$:

$2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$

Simplify: $x \frac{dx}{dt} + y \frac{dy}{dt} = 0$

Solve for $dx/dt$:

$\frac{dx}{dt} = -\frac{y \frac{dy}{dt}}{x}$

To find $dx/dt$, we need the values of $x$, $y$, and $dy/dt$.

Problem Completion:

1. Input Values: You need to fill in the missing values for:

   • $x$: the base distance of the ladder from the wall.
   • $dy/dt$: the rate at which the ladder slides down the wall.

2. Determine $y$: Using the Pythagorean Theorem, calculate $y = \sqrt{14^2 - x^2}$.

3. Substitute all values into the formula to calculate $dx/dt$: $\frac{dx}{dt} = -\frac{y \cdot \frac{dy}{dt}}{x}$

Would you like me to solve for $dx/dt$ once you provide the missing rate $dy/dt$ and distance $x$?