Answer the questions by drawing on the coordinate plane below. a. draw the image of δpqr after a rotation of 90° clockwise about the origin. label the image δp’q’r’. b. draw the image of δpqr after a reflection across the x-axis. label the image δp”q”r”.

Coordinates of the given triangle ΔPQR P(1,-1), Q(3,-2), R(3,-4).

a) Let us discuss the rule for rotation of 90° clockwise about the origin.

If given coordinate is (h,k) is rotated 90° clockwise about the origin the position of rotated coordinate will be (k, -h).

x and y coordinates would switch their values and after switching x any values, the y value of the coordinates would reverse in sign.

So, the coordinates of rotated ΔPQR P(1,-1), Q(3,-2), R(3,-4) would become

P(1,-1) --> P'(-1,-1)

Q(3,-2) --> Q'( -2,-3)

R(3,-4) --> R'(-4,-3)

b) If given coordinate is (h,k), the coordinates after reflection across the x-axis would besome (h,-k). (Just change in the sign of y-coordinate).

So, the coordinates of rotated ΔPQR P(1,-1), Q(3,-2), R(3,-4) would become

P(1,-1) --> P"(1,1)

Q(3,-2) --> Q"( 3,2)

R(3,-4) --> R"(3,4)

set each factor equal to 0

4x-3 = 0

2x-1 = 0

solve for x.

4x-3=0

4x=3

x= 3/4

solve for x. 

2x-1= 0

2x= 1

x= 1/2

your answer should be 1/2 is less than or equal to x and 3/4 is greater than or equal to x

the solution is x=1

step-by-step explanation:

we are given our   equation as

let's assume

left side function is

right side function is

now, we can draw graph

and then we can get intersection point

the intersection point will be solution

so, the solution is

x=1