Part A
Let d be the distance from the hive to the flowerbed. The units are in feet.
The bee flies at a speed of 9 ft/sec when going from the hive to the flowerbed. This means,
distance = rate*time
d = r*t
d = 9t
t = d/9
The time the bee spends flying at that speed is d/9 seconds. We'll use this expression later.
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When the bee flies back to the hive, it does so at a speed of 6 ft/sec and the time in flight is....
d = r*t
d = 6t
t = d/6
The expression d/6 is the number of seconds the bee is flying back (from the flowerbed to the hive).
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Combining those two time expressions will lead us to the total amount of time the bee was in the air.
(d/9) + (d/6)
(2d/18) + (3d/18)
(2d+3d)/18
5d/18
If we knew what d was, then computing 5d/18 will get us the total flight time. This time duration is in seconds.
Next, we convert the values 19 minutes and 21 minutes to seconds. To do so, we multiply each value by 60
- 19 min = 19*60 = 1140 sec
- 21 min = 21*60 = 1260 sec
We do this because the 5d/18 is the flight time in seconds. Everything must be the same time unit if we want to add up the times and have them make sense. In other words, we can't say something like 5 seconds + 2 minutes = 7 seconds because it doesn't make sense.
So we can form the equation 5d/18 + 1140 = 1260
We simply add the total flight time (5d/18 sec) with the time spent at the flower bed (1140 sec) to get the total time the bee was away from the hive (1260 sec).
<h3>Answer: 5d/18 + 1140 = 1260</h3>
The 1140 is not part of the denominator. Other equations are possible.
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Part B
We solve the equation found in part A to get the distance d.
5d/18 + 1140 = 1260
5d/18 = 1260-1140
5d/18 = 120
5d = 18*120
5d = 2160
d = 2160/5
d = 432
<h3>Answer: 432 feet </h3>
