Write the equation of a circle with center (0,−6) and radius 4 in general form.

x^2 + (y + 6)^2 = 4^2

Step-by-step explanation:

The general form we need here is (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center and r is the radius.

If the center is at (0, -6) and the radius is 4, then the desired equation is

(x - 0)^2 + (y + 6)^2 = 4^2, or just

x^2 + (y + 6)^2 = 4^2

step-by-step explanation:

∆abc has the points a(1, 7), b(-2, 2), and c(4, 2) as its vertices. formed with the point d(1, 2) as its third vertex, then ∆abd is triangle.

step-by-step explanation:

use pothagarean therom

a^2+b^2=c^2 so first ac

acm is a right triangle because c= 90°

and am (the hypotenuse or c) =16+9 =25

but there is not enough info we need another angle or side length