On a coordinate plane, triangles P Q R and S T U are shown. Triangle P Q R has points (4, 4), (negative 2, 0), (negative 2, 4). Triangle S T U has points (2, negative 4), (negative 1, negative 2), (negative 1, negative 4). Complete the statements to verify that the triangles are similar.

StartFraction Q R Over T U EndFraction =

StartFraction P R Over S U EndFraction = ?

StartFraction P Q Over S T EndFraction = ?
StartFraction StartRoot 52 EndRoot Over StartRoot 13 EndRoot EndFraction = ?

Therefore, △PQR ~ △STU by the
theorem.

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3

a - base

b - exponent

so,

δabc ≅ δdcb by sas because

db ≅ ac . . given (side)

bc ≅ bc . . reflexive property (side)

∠dbc ≅ ∠ acb . . alternate interior angles where transversal bc crosses parallel lines bd and ac (included angle)