Given below are descriptions of Mya's workout runs for 3 days of the week. Each graph shows her distance traveled, in miles, as
a function of time,
t
, in minutes.
For each day, drag the graph that corresponds to the description of her run into the space below it.
t
, in minutes.
For each day, drag the graph that corresponds to the description of her run into the space below it.
1 answer:
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<u>Answer:</u>
37th term of given arithmetic sequence = 2.3; d = -2.3 is -80.5.
<u>Solution:</u>
Given that
Need to determine 37th term, when = 2.3, d = -2.3
Means first term of arithmetic sequence = = 2.3 and common difference d = -2.3
<u><em>Formula for nth term of arithmetic sequence is </em></u>
--- equation 1
We need to determine 37th term so n = 37.
On substituting given values in equation (1) we get
Hence 37th term of given arithmetic sequence is -80.5
Answer:
Joe's wife must drive at a rate of 45km/hour.
Step-by-step explanation:
We are given that Joe leaves home and bikes at a speed of 30km/hour. Joe's wife leaves home five minutes later by car, and we want to determine her speed in order for her to catch up to Joe in 10 minutes.
Since Joe bikes at a speed of 30km/hour, he bikes at the equivalent rate of 0.5km/min.
Then after five minutes, when his wife leaves, Joe is 5(0.5) or 2.5 km from the house. He will still be traveling at a rate of 0.5km/min, so his distance from the house can be given by:
Where <em>t</em> represents the time in minutes after his wife left the house.
And since we want to catch up in 10 minutes, Joe's distance from the house 10 minutes after his wife left will be:
Let <em>s</em> represent the wife's speed in km/min. So, her speed times 10 minutes must total 7.5 km:
Solve for <em>s: </em>
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Thus, Joe's wife must drive at a rate of 0.75km/min, or 45km/hour.