Person A leaves his home to visit his cousin, person B, who lives 162miles away. He travels at an average rate of 60miles per hour. One half-hour later, person B leaves her house to visit person A, traveling at an average rate of 72miles per hour. How long after person B leaves will it be before they meet?

0.727 hours

Step-by-step explanation:

Call A’s distance from A’s house “a(t)” when t hours have passed. Then a(t) = 60t, because distance = rate x time.

Call B’s distance from A’s house “b(t)”. Then b(t) = 162 - 72t, because they start 162 miles away and subtract miles as time elapses.

The time at which they meet will be when a(t) = b(t). (Think about it). So now we just solve for t:

60t = 162 - 72t

132t = 162

t = 1.227 hours.

HOWEVER this isn’t quite what they asked for. They want the time they meet since B left (not A), and since B left .5 hours after, we must subtract .5. Thus the answer is 0.727 hours.

the cat is blue because the probability of extension towards the prefix of the diameter of the circle could only mathematically equal to the sum of the roots of the carrots from   your garden.

step-by-step explanation: