We are given the equation relating the pressure $p$ and the depth $d$ in seawater:

$33p - 18d = 495$

We need to find the pressure at a depth of $d = 12,460$ feet.

Step 1: Substitute $d = 12,460$ into the equation.

$33p - 18(12,460) = 495$

Step 2: Calculate $18 \times 12,460$.

$18 \times 12,460 = 224,280$

So, the equation becomes:

$33p - 224,280 = 495$

Step 3: Add $224,280$ to both sides to isolate the term with $p$.

$33p = 495 + 224,280$

$33p = 224,775$

Step 4: Solve for $p$ by dividing both sides by 33.

$p = \frac{224,775}{33}$

$p = 6,810.61 \, \text{psi}$

Final Answer:

The pressure at a depth of 12,460 feet is approximately $6,810.61 \, \text{psi}$.

Do you want further details or have any questions?

Here are 5 related questions for practice:

1. If the depth is doubled, what will be the pressure?
2. At what depth would the pressure reach 10,000 psi?
3. How does changing the coefficient of $d$ affect the pressure-depth relationship?
4. What would be the pressure at 5,000 feet using the same equation?
5. How is the relationship between depth and pressure affected by different liquids?

Tip: When working with equations, isolating variables step by step helps ensure accuracy and prevents mistakes.