Math problem for algebra, please help!?
2016-09-27 7:40 pm
Mary starts jogging at five miles per hour. one-half an hour later Gene starts jogging on the same route at six miles per hour. How long will it take Gene to catch Mary?
回答 (3)
2016-09-27 7:55 pm
Mary 5Mi/H @ 30(60) = X
Gene 6 Mi/H @ T = X
6Mi/ H x Time = 5 MI/H (30x60+Time)
Time = 5mi/h / 6mi/h ( 30x60+time)
Time = 5/6(1800+Time)
Time = 5/6(1800) + 5/6(Time)
Time - 5/6(Time) = 5/6( 1800)
(6T-5T)/6 = 1T/6
T/6 = 5/6(1800)
T= 6(5/6)(1800)
Hopefully this is the correct solution.
Gene 6 Mi/H @ T = X
6Mi/ H x Time = 5 MI/H (30x60+Time)
Time = 5mi/h / 6mi/h ( 30x60+time)
Time = 5/6(1800+Time)
Time = 5/6(1800) + 5/6(Time)
Time - 5/6(Time) = 5/6( 1800)
(6T-5T)/6 = 1T/6
T/6 = 5/6(1800)
T= 6(5/6)(1800)
Hopefully this is the correct solution.
2016-09-27 7:42 pm
Mary has been jogging at 5 mph for ½ hour.
D = RT
D = (5) * 1/2
D = 2½ miles
So Mary has a 2.5 mile head start.
Now Gene starts running. Since he is running 1 mph faster than Mary, he will be closing this 2.5 mile gap at the rate of 1 mph.
T = D/R
T = 2.5 / 1
T = 2.5 hours
Answer:
2½ hours
D = RT
D = (5) * 1/2
D = 2½ miles
So Mary has a 2.5 mile head start.
Now Gene starts running. Since he is running 1 mph faster than Mary, he will be closing this 2.5 mile gap at the rate of 1 mph.
T = D/R
T = 2.5 / 1
T = 2.5 hours
Answer:
2½ hours
2016-09-27 7:41 pm
After half an hour, Mary is (1/2)*5 = 2.5 miles ahead.
The difference in their speeds is 6 - 5 = 1 mile per hour, so it will take Gene 2.5 / 1 = 2.5 hours to catch up.
The difference in their speeds is 6 - 5 = 1 mile per hour, so it will take Gene 2.5 / 1 = 2.5 hours to catch up.
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