(SAT Prep) Rays OM
,
ON
, and
OP
coincide at time t=0. At the same instant in time,
OM
and
OP
begin to rotate in the plane about point O, but in opposite directions. Ray
ON
remains fixed. Ray
OM
rotates at a constant rate of 3° per second with respect to
ON
, while
OP
rotates at a constant rate of 6° per second. What is the smallest positive value of t, in seconds, in which rays
OM
and
OP
will coincide again? When t=30 min, how many more revolutions has ray
OP
completed than ray
OM
?

T=40 sec. OP does 15 revolutions more than OM.

Step-by-step explanation:

h

know the answer!  This is my first answer!!! I hope its correct, i trple checked it! :)

HOPE YOU ENJO!!

answer: are you

step-by-step explanation:

step-by-step explanation:

given that ab is divided into equal parts as

ac, cd, de, ef, fg, gh, hi, ib.

from this we find that f is the midpoint of ab.

so using mid point formula we find

f = (3,7)

similarly considering fb, we get

h = mid point of fb = ((6,6) and g, mid point of fh is

g = (9/2, 13/2)

i is mid point of hb

so i =(15/2,11/2)