Thank you for the opportunity to help you with your question!
First 8 min
At the speed of 70 mph = (70 mph/60 minutes)=1.16 miles per min total distance =8 min*1.16 miles per min =9.33 miles
Second 8 min
At the speed of 40 mph = (40 mph/60 minutes)=0.66 miles per min total distance =8 min*0.66 miles per min =5.33 miles
Difference 9.33-5.33 = 3.99
Distance traveled in first 8 min =9.33 = Distance in second 8 min *1.75 (1.75* 5.33)
Yes distance traveled in second 8 min is 1.75 times grater than first 8 min
I have mentioned the answers in bold. I am pleased to answer your questions.
Did you mean, distance traveled in the 1st 8 min. is 1.75 times greater than the second 8 min.?
Thank you for all your help by the way!!
Thank you. I see that to get 1.75 times greater, I would divide 9.33 by 5.33 and get 1.75
If the cheetah made a round-trip and took half the amount of time on the return trip as on the front end of the trip, what would be the relationship between the average rates on each leg of the trip? Using a complete sentence, explain how you arrived at this conclusion.
Front end of trip r = d/t The return trip only took half the time which means
the rate was doubled coming back.
Return trip 2(r) = d/tAverage Rate = 2d / d / r1 + d / r2
I was not online when you accepted the bid.I thought to answer your question.Sorry for the late reply.
Average rate is not what you have indicated. It's mph of kmph.
Sentence: The cheetah went the same distance in the front end and in the return trip but the time spent for the return trip is half than the front end, for this to be true the average rate of the return trip should be twice as the average rate of the front end.
I withdrew because I needed the answer sooner. I had to hand my answer in. I figured it out. It wasn't as nicely put as your answer. I hope my teacher finds it satisfactory.
I will definitely try your services again on another equation that I might get stuck on. Thank you for your time.
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