Point r is the midpoint of line segment xy. if xr=3n+1 and yr=16-2n, what is the length of line segment xy?

explain your work.

1) 3n+1=16-2n - add 2n to both sides - you should now be left with 5n+1=16 (remember the 2n gets canceled out) - subtract 1 from both sides - you should now be left with 5n=15 - now you divide by 5 to get n by itself - 15 divided by 5 is 3 - n=3 2) substitute n with 3 in to the original equation - so 3n+1=16-2n now becomes 3(5)+1=16-2(3) - you should get 10 for both so 10=10 3) according to the segment addition postualte , XR + YR = XY. - add 10+10 which stands for the lengths of XR and YR Answer - XY = 20

Sin(a)+cos(a)+cot(a)-1=2(csc(x)+1)