1. The problem is to analyze the ellipse given by the equation $$\frac{(x+5)^2}{100} + \frac{(y-5)^2}{49} = 1.$$\n\n2. This is the standard form of an ellipse centered at $(-5, 5)$, where the denominators represent the squares of the semi-axes lengths.\n\n3. The formula for an ellipse centered at $(h, k)$ is $$\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1,$$ where $a$ is the semi-major axis and $b$ is the semi-minor axis.\n\n4. Here, $h = -5$, $k = 5$, $a^2 = 100$, so $a = 10$, and $b^2 = 49$, so $b = 7$.\n\n5. Since $a > b$, the major axis is horizontal with length $2a = 20$, and the minor axis is vertical with length $2b = 14$.\n\n6. The ellipse is centered at $(-5, 5)$, with horizontal radius 10 and vertical radius 7, confirming the shape and position described.