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

5 units

A point 4 up and 3 over from its original location has moved a distance that can be found using the Pythagorean theorem.

... d² = 4² +3² = 16 +9 = 25

... d = √25 = 5 . . . . units

5 units

Step-by-step explanation:

Assuming that any point on the initial triangle XYZ was the origin (0,0).

The distance of between the two points can be calculated by using the Pythagoras Theorem to solve the problem:

where is the hypotenuse and and are the legs of the triangle.

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

see attached

step-by-step explanation:

given the function as;

write the equation replacing f(x) with y

form   and fill a table containing values of y with corresponding values of x

you can take a range for   values of x as -3≤x≤3

solving for x=-3

for x=-2

repeat the same procedure to get the values of y for the remaining ,x=-1 to x=3

plot the values of y against values   of x and obtain the following graph