1. answer the questions by drawing on the coordinate plane below. (2 points possible) (a) draw the image of ∆jkl after a rotation of 90°counterclockwise about the origin. label the image ∆j′k′ l′. (b) draw the image of ∆jkl after a reflection across the y-axis. label the image ∆j′′k′′l′′..

90 degrees counterclockwise

Invert the x and y points  and the y becomes negative (x, y) > (-y, x).

J( -4, 1) becomes (-1,-4)

K( -4,-2) becomes (2,-4)

L (-3,-1) becomes (1,-3)

reflection across the y axis

the x becomes negative (x, y) > (-x, y).

J( -4, 1) becomes (4,-1)

K( -4,-2) becomes (4,-2)

L (-3,-1) becomes (3,1)

(a) The vertices of image are J'(-1,-4), K'(2,-4) and L'(1,-3).

(b) The vertices of image are J''(4,1), K''(4,-2) and L''(3,-1).

Step-by-step explanation:

From the given figure it is clear that J(-4,1), K(-4,-2) and L(-3,-1).

(a)

If a figure rotated 90°counterclockwise about the origin, then

The vertices of image of ∆JKL after a rotation of 90°counterclockwise about the origin are

Therefore the vertices of image are J'(-1,-4), K'(2,-4) and L'(1,-3).

(b)

If a figure reflected across the y-axis, then

The vertices of image of ∆JKL after a reflection across the y-axis are

Therefore the vertices of image are J''(4,1), K''(4,-2) and L''(3,-1).

x= -3 and

step-by-step explanation:

to find the roots of a cubic, use division or factoring to find the factors then set equal to 0. as this is a difference of cubes, it factors into

where a is the cube root of .

here   this is 27 and the cube root of 27 is 3. so we write using the form:

.

set each factor to 0 and solve for x.

x-3=0

x=3

the quadratic expression does not factor and must be solved using the quadratic formula.

where a = 1, b=3, and c=9

this cubic has two complex roots.