The table shows three values of $x$ and their corresponding values of $y$ for the linear relationship between $x$ and $y$. The $x$-intercept occurs at $x = -4$. The number $d$ is a negative constant, and $k$ is a constant. We are tasked with finding the value of $dk$.

Solution:

The equation for the slope of the line is given by:

$$\text{slope} = \frac{y_2 - y_1}{x_2 - x_1}$$

Using the points $(-7, d+3)$ and $(-4, 4k+6d)$:

$$\frac{(4k+6d) - (d+3)}{-4 - (-7)} = \frac{8 - (4k+6d)}{(1-3d) - (-4)}$$

Equating the two slopes:

$$\frac{-d - 3}{3} = \frac{8}{5}$$

Cross-multiply and simplify:

$$(-d - 3)(-3d + 5) = 24$$ $$3d^2 + 4d - 39 = 0$$

Solving for $d$ using the quadratic formula:

$$d = \frac{-4 \pm \sqrt{4^2 - 4(3)(-39)}}{2(3)} = \frac{-4 \pm \sqrt{484}}{6}$$ $$d = \frac{-4 \pm 22}{6}$$

Since $d$ is negative:

$$d = \frac{-13}{3}$$

Substituting $d$ into the equation $4k + 6d = 0$:

$$4k + 6\left(-\frac{13}{3}\right) = 0$$ $$4k - 26 = 0$$ $$k = \frac{13}{2}$$

Finally, calculate $dk$:

$$dk = \left(-\frac{13}{3}\right) \cdot \frac{13}{2} = -\frac{169}{6}$$

Answer: B) $-\frac{169}{6}$