If the distance between the points (3, 0) and (0, y) is 5 units and y is positive, then what is the
value of y?
Answers
Answered by
2
Step-by-step explanation:
(3-0)^2+(0-y)^2=5^2
9+y^2=25
y^2=25-9
y^2=16
y=4
Similar questions