Answer:
(x-3)^2 + (y+2)^2 = 9^2
Step-by-step explanation:
x^2 -6x+y^2+4y-68 = 0
Complete the square
x^2 -6x+y^2+4y-68+68 = 0+68
x^2 -6x+y^2+4y = 68
Find the term to add for x
-6 /2 = -3 -3^2 = 9
Find the term to add for y
4/2 =2 2^2 = 4
Add 9 and 4
x^2 -6x+9+y^2+4y+4 = 68+9+4
(x-3)^2 + (y+2)^2 = 81
(x-3)^2 + (y+2)^2 = 9^2
The standard form is
(x-h)^2 + (y-k)^2 = r^2 where (h,k) is the center and r is the radius
Answer:
i. 15.4
ii.45.2
Step-by-step explanation:
correct me if im wrong
tnx
Hi!! What do you need help with? Solving the problems? Or understanding them? Also 1-9 is a bit blurry, so if you could retake it and repost it, I could solve for them. :). But if you don’t want to, don’t worry.
If the circumference is half of the perimeter than the area for both circles would be half of them find one circles area and half it
To find the third side of a triangle (there are two answers you could get), you need to use the Pythagorean formula: a^2+b^2=c^2. First answer, 20^2+30^2=c^2. Solve for c. c=36.06. The second answer, 20^2+b^2=30^2. Solve for b. b=26.36.
