Question

which of the following graphs represents the solution(s) of the following system?
$x^{2}+y = 7$
$x^{2}+y^{2}=49$
done

Explanation:

Step1: Analyze the first equation

The first equation \(x^{2}+y = 7\) can be rewritten as \(y=-x^{2}+7\), which is a parabola opening downwards with vertex at \((0,7)\).

Step2: Analyze the second equation

The second equation \(x^{2}+y^{2}=49\) is a circle with center \((0,0)\) and radius \(r = 7\) (since \(r^{2}=49\)).

Step3: Check the intersection points

Substitute \(x^{2}=7 - y\) from the first equation into the second equation:

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

The parabola \(y=-x^{2}+7\) and the circle \(x^{2}+y^{2}=49\) intersect at \((0,7)\), \((\sqrt{13},-6)\) and \((-\sqrt{13},-6)\). The parabola \(y=-x^{2}+7\) opens downwards and the circle is centered at the origin with radius \(7\).

Answer:

The graph where the parabola \(y=-x^{2}+7\) (opening downwards) and the circle \(x^{2}+y^{2}=49\) (radius \(7\) centered at origin) intersect at the appropriate points (as calculated above) is the correct one. (Since the actual options are not labeled with letters in a standard way in the problem statement, but based on the analysis of the equations: parabola \(y=-x^{2}+7\) and circle \(x^{2}+y^{2}=49\), the graph with a downward - opening parabola (vertex at \((0,7)\)) and a circle of radius \(7\) centered at the origin is the answer)