Draw the image of quadrilateral ABCD under a translation by 5 units to the right

The draw in the attached figure

Step-by-step explanation:

we know that

The rule of the translation is equal to

(x,y) ----> (x+5,y)

we have the coordinates of the pre-image of quadrilateral ABCD

A(-1,2),B(-6,1),C(-7,-2),D(0,-5)

applying the rule of the translation to the coordinates ABCD to obtain the image A'B'C'D'

so

1) A(-1,2) -----> A'(-1+5,2)

A(-1,2) -----> A'(4,2)

2) B(-6,1) ----> B'(-6+5,1)

B(-6,1) ----> B'(-1,1)

3) C(-7,-2) ----> C'(-7+5,-2)

C(-7,-2) ----> C'(-2,-2)

4) D(0,-5) ----> D'(0+5,-5)

D(0,-5) ----> D'(5,-5)

The draw in the attached figure

= 5

step-by-step explanation:

to find the 15th term substitute n = 15 into the nth term formula

= (- 3 × 15 ) + 50 = - 45 + 50 = 5