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damaskus [11]
3 years ago
9

emily leaves her house at exactly 8:25 am to bike to her school, which is 3.42 miles away. when she passes the post office, whic

h is 3/4 mile away from her home, she looks at her watch and sees that it is 30 seconds past 8:29 am. If emily's school starts at 8:50 am can emily make it on time without increasing her rate of speed? show and/or explain the work necessary to support your answer.
Mathematics
1 answer:
NNADVOKAT [17]3 years ago
6 0
We first calculate Emily's rate by dividing the distance from her home to the post office by the time it took her to travel that distance. From 8:25 to 8:29 (and 30 seconds), there are exactly 4.5 minutes. Thus, the rate is equal to,
                                   r = (0.75 miles) / (4.5 minutes) = 1/6 miles/min
Then, we divide the given distance from her home to school by the calculate rate to determine the total number of minutes it will take her to travel to school.
                               t = (3.42 miles)/(1/6 miles/min) = 20.52 minutes
Adding this to 8:25 will give us an answer of 8:45 and 31.2 seconds. She will not be late. 
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either q = 90 + 45 = 135° which is equivalent to \frac{3 \pi }{4}
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The y intercept is 3 because of the point (0,3)

We go from y = mx+b to y = 4x+3

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