For ellipse x^2/16 + y^2/9 = 1, the foci are at (±√7, 0).

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Multiple Choice

For ellipse x^2/16 + y^2/9 = 1, the foci are at (±√7, 0).

Explanation:
In an ellipse, the foci lie along the major axis at a distance c from the center, where c^2 = a^2 − b^2. Here the equation is x^2/16 + y^2/9 = 1, so a^2 = 16 and b^2 = 9. The major axis is horizontal, and the center is at (0,0). Compute c: c^2 = 16 − 9 = 7, so c = √7. Therefore the foci are at (±√7, 0) along the x-axis. The other options correspond to vertices or co-vertices (±4, 0 are the endpoints of the major axis, ±3, 0 aren’t foci, and (0, ±√7) would be foci only if the major axis were vertical).

In an ellipse, the foci lie along the major axis at a distance c from the center, where c^2 = a^2 − b^2. Here the equation is x^2/16 + y^2/9 = 1, so a^2 = 16 and b^2 = 9. The major axis is horizontal, and the center is at (0,0). Compute c: c^2 = 16 − 9 = 7, so c = √7. Therefore the foci are at (±√7, 0) along the x-axis.

The other options correspond to vertices or co-vertices (±4, 0 are the endpoints of the major axis, ±3, 0 aren’t foci, and (0, ±√7) would be foci only if the major axis were vertical).

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