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.

step-by-step explanation:

the original amount doesn't ''re just looking for the time it takes for 1/2 of the substance to we have

(1/2) =)     take the ln   of both sides

ln (1/2)   = )   and we can write

ln(1/2) = ) lne     and lne   = 1     so we have

ln(1/2) = -.1092t     divide both sides by -.1092

t = ln(1/2)/-.1092 = about 6.347 days = about 6.3 days (rounded)

how am i supposed to answer

step-by-step explanation: