which of the following graphs represents the solution(s) of the following system?\n$x^{2}+y =…

Explanation:

Step1: Rewrite the first - equation

From $x^{2}+y = 7$, we can get $x^{2}=7 - y$.

Step2: Substitute into the second - equation

Substitute $x^{2}=7 - y$ into $x^{2}+y^{2}=49$, we have $(7 - y)+y^{2}=49$. Rearrange it to the standard quadratic form: $y^{2}-y - 42 = 0$.

Step3: Solve the quadratic equation

Factor the quadratic equation $y^{2}-y - 42=(y - 7)(y + 6)=0$. So $y=7$ or $y=-6$.

Step4: Find the corresponding x - values

When $y = 7$, from $x^{2}=7 - y$, we have $x^{2}=7 - 7=0$, so $x = 0$. When $y=-6$, from $x^{2}=7 - y$, we have $x^{2}=7-(-6)=13$, so $x=\pm\sqrt{13}$.

The first equation $x^{2}+y = 7$ can be rewritten as $y=-x^{2}+7$, which is a parabola opening downwards with vertex at $(0,7)$. The second equation $x^{2}+y^{2}=49$ is a circle centered at the origin $(0,0)$ with radius $r = 7$. The parabola $y=-x^{2}+7$ and the circle $x^{2}+y^{2}=49$ intersect at $(0,7),(\sqrt{13},-6),(-\sqrt{13},-6)$.

The graph that shows a parabola opening downwards with vertex at $(0,7)$ intersecting a circle centered at the origin with radius 7 at these three points is the correct one.

Answer:

The graph that shows a parabola opening downwards with vertex at $(0,7)$ intersecting a circle centered at the origin with radius 7 at $(0,7),(\sqrt{13},-6),(-\sqrt{13},-6)$ (you need to visually identify this among the given options).