The side that is equivalent to side SR is side AB.
<h3>How to carry out Rotational Transformation?</h3>
When rotating a point 180 degrees counterclockwise about the origin, our point A(x, y) becomes A'(-x, -y). Thus, all we do is make both x and y negative.
The coordinates of the original square are;
P(2, 4); Q(5, 5); R(5, 1); S(2, 1)
Now, applying the transformation rule above gives us;
P(-2, -4); Q(-5, -5); R(-5, -1); S(-2, -1)
The side that is therefore equivalent to side SR is side AB.
Read more about Rotational Transformation at; brainly.com/question/26249005
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Answer:
![(x-7)^2+(y-2)^2=130](https://tex.z-dn.net/?f=%28x-7%29%5E2%2B%28y-2%29%5E2%3D130)
Step-by-step explanation:
The standard form of a circle is
![(x-h)^2+(y-k)^2=r^2](https://tex.z-dn.net/?f=%28x-h%29%5E2%2B%28y-k%29%5E2%3Dr%5E2)
where x and y are points on the circle, h and k are the center, and r is the radius. We have what we need to fill in everything but the radius, but we will solve for that like this:
![(-2-7)^2+(-5-2)^2=r^2](https://tex.z-dn.net/?f=%28-2-7%29%5E2%2B%28-5-2%29%5E2%3Dr%5E2)
Simplifying gives you
![(-9)^2+(-7)^2=r^2](https://tex.z-dn.net/?f=%28-9%29%5E2%2B%28-7%29%5E2%3Dr%5E2)
and
![81+49=r^2](https://tex.z-dn.net/?f=81%2B49%3Dr%5E2)
so ![r^2=130](https://tex.z-dn.net/?f=r%5E2%3D130)
That means the equation for this circle is
![(x-7)^2+(y-2)^2=130](https://tex.z-dn.net/?f=%28x-7%29%5E2%2B%28y-2%29%5E2%3D130)
The answer I think at least would be C
The answer is actually 15