What are the vertices of triangle A'B'C' produced by (x, y) -> (x+5, y-2)?

The vertices of triangle A'B'C' produced by (x, y)  →  (x+5, y-2)  will be:

Step-by-step explanation:

Given the vertices of an original triangle ΔABC such that

As the rule of translation is given by

Translation: (x, y)  →  (x+5, y-2)

So, after the translation the vertices of the original triangle ΔABC will be translated as:

A = (3, 0)  → (x+5, y-2)  ⇒ (3+5, 0-2) = A'(8, -2)

B = (3, 2)  →  (x+5, y-2)  ⇒ (3+5, 2-2) = B'(8, 0)

C = (0, -3) → (x+5, y-2)  ⇒ (0+5, -3-2) = C'(5, -5)

Therefore, the vertices of triangle A'B'C' produced by (x, y)  →  (x+5, y-2)  will be:

i = 2 / - 5

step-by-step explanation:

evaluation of ( 2 - 4 ) ( p + q ) when p = 2 and q = 5 i, by putting values of p and q,                                                                    

the 8 in 8,420 is in the thousands place so it's value is 8 thousand.

the 8 in 6,870 is in the hundreds place so it's value is 8 hundred.

divide 8000 by 800 to find the answer:

8000 / 800 = 10

the 8 in 8,420 is 10 times greater than the 8 in 6,870