Which equations represent circles that have a diameter of 12 units and a center that lies on the y-axis? check all that apply. x2 + (y – 3)2 = 36 x2 + (y – 5)2 = 6 (x – 4)² + y² = 36 (x + 6)² + y² = 144 x2 + (y + 8)2 = 36

we know that

The equation of the circle in the center-radius form is equal to

where

(h,k) is the center of the circle

r is the radius of the circle

In this problem we have

so

The center lies on the y-axis

that means

the coordinate x of the center is equal to zero

we are going to verify the coordinates of the center and the value of the radius of each equation to determine the solution of the problem

case A)

In this case the center of the circle is

the point > the center lies on the y-axis

the radius is

therefore

The equation represent a circle that have a diameter of units and a center that lies on the y-axis

case B)

In this case the center of the circle is

the point > the center lies on the y-axis

the radius is

therefore

The equation   does not represent a circle having a diameter of units

case C)

In this case the center of the circle is

the point > the center lies on the x-axis

the radius is

therefore

The equation   does not represent a circle that has a center on the y-axis

case D)

In this case the center of the circle is

the point > the center lies on the x-axis

the radius is

therefore

The equation   does not represent a circle having a diameter of and a center on the y-axis.

case E)

In this case the center of the circle is

the point > the center lies on the y-axis

the radius is

therefore

The equation  represent a circle that have a diameter of units and a center that lies on the y-axis

therefore

the answer is

it is A and E

Step-by-step explanation:

r(0,90)(x,y) . t(-5,0)

last option

step-by-step explanation:

∆def to ∆d"e"f"

- rotate 90 counterclockwise degrees

- then translated 5 to the left (x --> x - 5)

answer is the last option

answer: integers are just whole numbers: positive, negative, and zero.

i = {-∞, -2, -1, 0, 1, 2, 3, ∞}

let x = 1st integer

let y = 2nd integer

x = 3y + 17   {equation 1}

xy = -24         {equation 2}

substitute 3y+17 in place of x n equation 2

and solve for y.

(3y+17)y = -24

3y2 + 17y = -24

3y2 + 17y + 24 = 0

from quadratic equation:

y = [-17±√(172-4(3)(24))]/(2(3))

y = (-17 ±√1)/6

y = (-17+1)/6   or     y = (-17-1)/6

y = -16/6           or     y = -18/6

y = -2 2/3         or   y = -3

since we are dealing with

y = -3

x = 3y+17 = -9+17 = 8

the integers are -3, 8

step-by-step explanation: