If it took Carol 1/2 hour to cycle from his house to the library yesterday, was the distance that he cycled greater than 6 miles? (1mile = 5,280feet)
Question isd>6 --> asrt=d (wherer is the rate in miles per hour) then question becomes: isrt=d>6 --> or isr*\frac{1}{2}>6 , ast=\frac{1}{2} hours --> isr>12 miles/hour? -->12 \ miles/hour = \frac{12*5280}{60*60} \ feet/second = 17.6 \ feet/sec . Isr>17.6 feet/sec?
(1) The average speed at which Carlos cycled from his house to the library
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Question isd>6 --> asrt=d (wherer is the rate in miles per hour) then question becomes: isrt=d>6 --> or isr*\frac{1}{2}>6 , ast=\frac{1}{2} hours --> isr>12 miles/hour? -->12 \ miles/hour = \frac{12*5280}{60*60} \ feet/second = 17.6 \ feet/sec . Isr>17.6 feet/sec?
(1) The average speed at which Carlos cycled from his house to the library
...