Answer:
a. [tex] \frac{- 42,500}{(v + 60)(v - 40)} [/tex]
Step-by-step Explanation:
Given:
Distance Ottawa to Québec = 425 km
Initial flight rate = v + 60
Return flight rate = v - 40
[tex] t = \frac{d}{r} [/tex]
Required:
Flight times difference of the initial and return flights
Solution:
=>Flight time of the initial flight:
[tex] t = \frac{d}{r} [/tex]
[tex] t = \frac{425}{v + 60} [/tex]
=>Flight time of the return flight:
[tex] t = \frac{425}{v - 40} [/tex]
=>Difference in flight times:
[tex] \frac{425}{v + 60} - \frac{425}{v - 40} [/tex]
[tex] \frac{425(v - 40) -425(v + 60)}{(v + 60)(v - 40)} [/tex]
[tex] \frac{425(v) - 425(40) -425(v) -425(+60)}{(v + 60)(v - 40)} [/tex]
[tex] \frac{425v - 17000 -425v - 25500}{(v + 60)(v - 40)} [/tex]
[tex] \frac{425v - 425v - 17000 - 25500}{(v + 60)(v - 40)} [/tex]
[tex] \frac{- 42,500}{(v + 60)(v - 40)} [/tex]
