Joe runs one lap around a track in 80 seconds. Joe and Jim meet after 30 seconds.We know how long does it take for Joe to run one lap, so we can calculate what distance he runs in 30 seconds:
He runs half lap for 40 seconds. He runs quarter lap for 20 minutes and he runs 1/8 lap for ten minutes. So for 30 seconds he runs 3 * 1/8 lap=3/8 lap.
Because Jim runs in opposite direction this means that he will meat with Joe at 1-3/8=5/8 lap. So, Jim runs 5/8 lap for 30 seconds.
1/8 lap Jim runs for 30/5=6 seconds, so Jim needs 8*6=48 seconds to run the lap.
After 30 seconds Joe has run e runs
Hope this helps
3,-13
-1,3
0,-1
3,-13
By the Intersecting Secants Theorem we have
2x(2x+7)=x(x+23)
4x^2+14x=x^2+23x
3x^2=9x
3x=9
x=3
Since DY is length 2x,
the length of DY is 6.
(x+20)-(x+1)=18
x+20-x-1=18
19=18
It has no solution
