The larger of two intergers is 7 times the smaller if the sum of two intergers is greater than 40 find the smallest possible value of x

Hello,

Let's assume x the smaller integer,
 7x the larger.

x+7x>40==>8x>40==>x>5
The smallest value of x is 6 and the second integer is 7*6=42.

X^2 - 4x +(4/2^2) + y^2 + 8y + (8/2^2) = -4 + (4/2^2) + (8/2^2)(x + 2)^2 + (y + 4)^2 = -4 + 4 + 16your center is (-2, -4) and the radius is the sqrt of -4 + 4 + 16 which is 4

Letter c i’d say