Question

7
What is the distance from 33, 5) to (3 7.-1

Answer 1

7.211

Explanation:

I'm going to assume that's supposed to say (33, 5) to (37, -1)

Use the pythagorean theorem: a^2 + b^2 = c^2

"a" is going to be your difference in x values (37 - 33) = 4

"b" is going to be your difference in y values (5 - (-1)) = 6

4^2 + 6^2 = c^2

16 + 36 = c^2

52 = c^2

sqrt(52) = c

7.211 = c