Answer: which one
Step-by-step explanation:
Plug it into the distance formula.
d= (sqrt) (x2-x1)^2 +(y2-y1)^2
d= (sqrt) (17-17)^2 + (33-(-21)^2
d=(sqrt) (0)^2 + (33 +21)^2
d= (sqrt) (54)^2
d= (sqrt) 2916
d=54
The distance is 54 units.
I hope this helps!
~kaikers
For a relation to be a function it must be


relation is not a function.
In the above options, C is

relation therefore cannot be a function.
The reason is that 3 alone maps onto -2 and 5, in the ordered pairs (3,-2) and (3,5). This disqualifies it from being a function.
Hence the graph of this relation will not pass the vertical line test
I think you have to first separate the integral:1/(1+v^2) + v/(1+v^2),
so the integral of the first term is ArcTan (v) and for the integral of the second term i recommend you to do a change of variable:
y= 1+v^2
so
dy= 2v
and
v= dy/2and then you substitute:v/(1+v^2) = (1/2)(dy/y)
and the integral is
(1/2) (In y)finally you plug in the initial variables:
(1/2)(In [1+v^2])
so the total integral is:
ArcTan (y) + (1/2)(In [1+v^2])