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To solve the problem, let's analyze the given quadratic equation and the conditions provided.

The quadratic equation is:

\[
-52x^2 + (13s - 4)t x + st = 0
\]

where \( s \) and \( t \) are positive constants. We are given that the sum of the solutions to this equation is:

\[
\frac{13s - 4}{d}
\]

where \( d \) is a constant. We are asked to find the value of \( d \).

### Step-by-Step Solution

1. **Identify the sum of the solutions of the quadratic equation:**

   The general form of a quadratic equation is \( ax^2 + bx + c = 0 \). The sum of the solutions (roots) for a quadratic equation \( ax^2 + bx + c = 0 \) is given by:

   \[
   -\frac{b}{a}
   \]

   In our equation, \( a = -52 \), \( b = (13s - 4)t \), and \( c = st \).

   Therefore, the sum of the solutions is:

   \[
   -\frac{(13s - 4)t}{-52} = \frac{(13s - 4)t}{52}
   \]

2. **Set up the equation with the given sum:**

   According to the problem, the sum of the solutions is also given by \( \frac{13s - 4}{d} \). So, we set the two expressions for the sum equal to each other:

   \[
   \frac{(13s - 4)t}{52} = \frac{13s - 4}{d}
   \]

3. **Simplify and solve for \( d \):**

   To solve for \( d \), cross-multiply:

   \[
   (13s - 4)t \cdot d = 52(13s - 4)
   \]

   Now, divide both sides by \( (13s - 4) \) (assuming \( 13s - 4 \neq 0 \)):

   \[
   td = 52
   \]

   Solving for \( d \):

   \[
   d = \frac{52}{t}
   \]

4. **Determine the correct answer choice:**

   The problem asks for the value of \( d \) in terms of the constant values given. The correct answer choice must match \( \frac{52}{t} \). Looking at the answer choices:

   -

Explore the solution to the quadratic equation -52x^2 + (13s - 4)t x + st = 0. Understand how to find the value of constant d using step-by-step methods and quadratic equation formulas. Suitable for high school students.

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