What is the center of a circle represented by the equation (x+9Mrs. Culland is finding the center of a circle whose equation is x2 + y2 + 6x + 4y – 3 = 0 by completing the square. Her work is shown. x2 + y2 + 6x + 4y – 3 = 0

x2 + 6x + y2 + 4y – 3 = 0

(x2 + 6x) + (y2 + 4y) = 3

(x2 + 6x + 9) + (y2 + 4y + 4) = 3 + 9 + 4)2+(y−6)2=102?

The center of the circle is at  the point (-3, -2) on the plane

Step-by-step explanation:

The first four steps in the completing of the squares to find the standard equation of a circle are correct. Let's start from there and actually complete the squares correctly:

Therefore, this is a circle centered at x = -3 [based on the horizontal translation defined by (x+3)], and at y = -2 [based on the vertical translation defined by (y+2)], and the circle has radius 4 based on the numerical constant squared on the right side of the equal sign.

the answer is the option b

the range is all real numbers

step-by-step explanation:

we know that

the absolute value parent function is equal to

see the attached figure to better understand the problem

so

a. it is v-shaped > is true

b. the range is all real numbers > is not true

because the range is the > [0,∞)

all real numbers greater than or equal to zero

c. it goes through the origin > is true

d. the domain is all real numbers > is true

step-by-step explanation:

answer: 1 + 1 = 11

step-by-step explanation: