Need help please:
(3,0) (7,6)
(x-7)2 + (y-6)2 = 7(2)
(x-7)2 + (y-6)2 = 49
Of course! I'll be happy to help you. It seems like you have provided two sets of information. The first set consists of two points, (3,0) and (7,6), while the second set is an equation in the form of (x-7)² + (y-6)² = 7(2) and the third set is an equation in the form of (x-7)² + (y-6)² = 49.
From the first set of points, it appears that you have two coordinate pairs on a Cartesian plane. These points represent the x and y coordinates of two distinct points: (3,0) and (7,6).
The second and third sets of equations are given in the form of a circle equation, which is (x-h)² + (y-k)² = r². Here, (h,k) represents the coordinates of the center of the circle, and r represents the radius of the circle.
For the second equation, (x-7)² + (y-6)² = 7(2), we can observe that the center of the circle is at the point (7,6), and the radius is the square root of 7 times 2 (i.e., √(7*2)). To find the equation's solution, you can simplify the equation further or utilize the center-radius form of the equation to identify additional information about the circle.
For the third equation, (x-7)² + (y-6)² = 49, the center remains the same at (7,6), but the radius is different. Here, it is 7, which is the square root of 49.
If you have any specific questions or require further assistance, please let me know!