Riangle xyz is translated 4 units up and 3 units left to yield δx'y'z'. what is the distance between any two corresponding points on δxyz and δx'y'z′?

5 units

Step-by-step explanation:

According to the given statement  Δ XYZ is translated 4 units up and 3 units left to yield ΔX'Y'Z' which means that each point in ΔXYZ is moved 4 units up and moved 3 units left.

To find the distance of each corresponding point we will use the Pythagorean theorem which states that the square of the length of the Pythagorean of a right triangle is equal to the sum of the squares of the length of other legs

The square of the required distance = 4^2+3^2 = 16+9 =25

By taking root of 25 we get:

√25 = 5

Thus, we can conclude that the the distance between any two corresponding points on ΔXYZ and ΔX′Y′Z′  is 5 units. ..

it's equivalent, so we can say that its equal.

let's call the denominator x and the numerador is 18 less then denominator, x - 18.

multiply

2.x = 5.(x - 18)

2x = 5x - 90

90 = 5x - 2x

90 = 3x

x = 90/3

x = 30

so: