Which shows the image of quadrilateral abcd after the transformation r0, 90°?

The answer is A. on edgen

Step-by-step explanation:

A. the first graph

The second picture shows the transformation.

Step-by-step explanation:

To find the image lets apply the rotation rule counterclockwise about the origin:

R0,90° = R90°(x,y) = (-y,x)

We switch the coordinates of the point and change the sign of the y-coordinate

The coordinates of our quadrilateral ABCD are:

A = (-1, 0), B = (0, -1), C = (-2, -3), and D = (-3 , -2)

Now solve the coordinates by applying rotation rule:

A=(-1,0)

A' = (-y,x) = (-0,-1)=(0,-1)

B=(0,-1)

B'= (-y,x) = (-(-1),0)=(1,0)

C = (-2, -3)

C' = (3,-2)

D=(-3,-2)

D'=(2,-3)

Now the coordinates after the rotation of 90 degree are:

A'=(0,-1) , B'=(1,0) , C'=(3,-2) , D'=(2,-3)

The second picture shows the transformation after 90 degree

(x3 + 6x2 + 2x - 6) • (x - 1)

step-by-step explanation:

polynomial long division :

3.2     polynomial long division

dividing :   x4+5x3-4x2-8x+6

                              ("dividend")

by         :     x-1     ("divisor")

dividend     x4   +   5x3   -   4x2   -   8x   +   6

- divisor   * x3     x4   -   x3            

remainder         6x3   -   4x2   -   8x   +   6

- divisor   * 6x2         6x3   -   6x2        

remainder             2x2   -   8x   +   6

- divisor   * 2x1             2x2   -   2x    

remainder               -   6x   +   6

- divisor   * -6x0               -   6x   +   6

remainder                     0

quotient :   x3+6x2+2x-6   remainder:   0

246

step-by-step explanation: