A radius of a circle has endpoints (-4,3) and (1,-2). What is the equation that defines the circle if its center is at the second quadrant?​

false

step-by-step explanation:

rulers and compasses arent required in order to bisect an angle because when bisecting angles, you're making a "half circle" and considering in order to bisect an angle it wouldn't be straight, you wouldn't need either of the two

step-by-step explanation:

a'(12.5,11.25)

slope=(y1-y2)/(x1-x2)

p(x)= 8 x³ - 36 x² + 54 x - 27

q(x) = 2 x -3

you can check your mistake and then you can choose the correct option.

dividend =   8 x³ - 36 x² + 54 x - 27

divisor = 2 x -3

if you want to find quotient, proceed as follows

q(x)= 0

2 x - 3=0

x =

now , i have used division method , to find quotient and remainder

let's first notice, the center of the hyperbola is at the origin, since the focus points are equidistantly 3 units away from the origin, that also means that the distance "c" from the center to either focus is 3.

it should also be notable that, the traverse axis for this hyperbola is vertical, or namely over the y-axis and therefore, the positive fraction will be the one with y-variable.

the traverse "a" component or distance from the center to either vertex is not 2, is more than 2, so we can safely put it at √5 ≈ 2.24.