Z^2+ y^2 + 8x – 16y + 31 = 0
What is the center of this circle ?
What is the radius of this circle ?

(- 4, - 8 ), radius = 7

Step-by-step explanation:

The equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k) are the coordinates of the centre and r is the radius

Given

x² + y² + 8x - 16y + 31 = 0

Collect the x- terms, collect the y- terms together and subtract 31 from both sides.

x² + 8x + y² - 16y = - 31

Use the method of completing the square on the x and y terms

add ( half the coefficient of the x/ y term )² to both sides

x² + 2(4)x + 16 + y² + 2(- 8)y + 64 = - 31 + 16 + 64

(x + 4)² + (y - 8)² = 49 ← in standard form

with centre = (- 4, 8 ) and r = = 7

10a^3+277a^2-132a-792

step-by-step explanation:

(2a+11)(5a-12)(foil)

=10a^2-24a+55a-132 =10a^2+31a-132

(10a^2+31a-132)(a+6)

=10a^3+60a^2+31a^2+186a^2-132a-792 =10a^3+277a^2-132a-792

i hope this is what you are looking for : )