Given rst equals npq. rt = 7x-5 nq = 5x+11 find the length of rt and nq

Given :

As it is not given whether RST and NPQ are collinear points or vertices of Triangle.

Whatever , if these two shapes are congruent , then their corresponding parts whether side length or Part or whole of segment will be congruent with each other.

RT = N Q  [Corresponding part of congruent shapes]

→7 x -5 = 5 x + 11

Taking variable on one side and constant on another side

7 x - 5 x = 11 + 5

2 x = 16

Dividing both sides by 2, we get

x=8

RT = 7 × 8 -5= 56 -5 =51 Units

N Q= 5 × 8 + 11= 40 + 11= 51 Units

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3

definitely a. the others have almost all data points on one side of the line.

step-by-step explanation: