## Precalculus (6th Edition)

$45$ minutes ( or 0.75 hours)
Distance traveled by either runner = (average speed)(time of travel). If the speed is given in mph, the time is expressed in hours. After t hours, Mary covers $7t$ miles. Janet covers $5t$ miles. We want the time t at which Mary has covered 1.5 mi more than Janet (Mary's distance) = (Janet's distance) +1.5 $7t=5t+1.5$ $2t=1.5$ $t=0.75$ hours In minutes, $0.75(60)=45$ minutes